Chapter 6 Application Of Derivatives

Given:

The equation of the curve is $y = 5x - 2x^3$.

The rate at which $x$ is changing with respect to time $t$ is $\frac{dx}{dt} = 2$ units/sec.

To Find:

The rate at which the slope of the curve is changing when $x = 3$. This is $\frac{d}{dt}\left(\frac{dy}{dx}\right)\Bigr|_{x=3}$.

Solution:

The slope of the curve at any point $(x, y)$ is given by the first derivative of $y$ with respect to $x$. Let $m$ denote the slope.

$m = \frac{dy}{dx}$

Differentiating the equation of the curve with respect to $x$:

$m = \frac{d}{dx}(5x - 2x^3)$

$m = 5 - 2(3x^{3-1})$

$m = 5 - 6x^2$

We are asked to find the rate of change of the slope with respect to time $t$, which is $\frac{dm}{dt}$.

Using the chain rule, we can relate $\frac{dm}{dt}$ to $\frac{dm}{dx}$ and $\frac{dx}{dt}$.

$\frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt}$

First, we find $\frac{dm}{dx}$ by differentiating the expression for $m$ with respect to $x$:

$\frac{dm}{dx} = \frac{d}{dx}(5 - 6x^2)$

$\frac{dm}{dx} = 0 - 6(2x^{2-1})$

$\frac{dm}{dx} = -12x$

Now, substitute the values of $\frac{dm}{dx}$ and the given $\frac{dx}{dt}$ into the chain rule formula:

$\frac{dm}{dt} = (-12x) \cdot (2)$

$\frac{dm}{dt} = -24x$

We need to find the rate of change of the slope when $x = 3$. Substitute $x=3$ into the expression for $\frac{dm}{dt}$:

$\frac{dm}{dt}\Bigr|_{x=3} = -24(3)$

$\frac{dm}{dt}\Bigr|_{x=3} = -72$

The rate at which the slope of the curve is changing when $x = 3$ is $-72$ units/sec.

The negative sign indicates that the slope is decreasing at this rate.

Final Answer: The slope of the curve is changing at the rate of $-72$ units/sec when $x = 3$.

Given:

The semi-vertical angle of the conical funnel is $\alpha = \frac{\pi}{4}$.

Water is dripping out at the rate of 2 cm$^2$/sec in surface area. This means the rate of decrease of the surface area of the water is 2 cm$^2$/sec.

Let $A$ be the surface area of the water in the funnel and $t$ be time in seconds. Then, $\frac{dA}{dt} = -2$ cm$^2$/sec.

We are given that the slant height of the water cone is $l = 4$ cm at a particular moment.

To Find:

The rate of decrease of the slant height of the water, $\frac{dl}{dt}$, when $l = 4$ cm.

Solution:

Let $r$ be the radius of the water surface and $l$ be the slant height of the water cone at time $t$.

The semi-vertical angle is $\alpha = \frac{\pi}{4}$. In a right cone, the radius $r$, height $h$, and slant height $l$ are related by $r = l \sin(\alpha)$ and $h = l \cos(\alpha)$.

The exposed surface area of the water is the lateral surface area of the cone formed by the water, given by $A = \pi r l$.

Substitute $r = l \sin(\alpha)$ into the area formula:

$A = \pi (l \sin(\alpha)) l = \pi l^2 \sin(\alpha)$

Given $\alpha = \frac{\pi}{4}$, we have $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

So, the surface area of the water is $A = \pi l^2 \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{\sqrt{2}} l^2$.

We need to find the rate of change of the slant height, $\frac{dl}{dt}$, given the rate of change of the surface area, $\frac{dA}{dt}$. We differentiate the equation for $A$ with respect to time $t$ using the chain rule:

$\frac{dA}{dt} = \frac{d}{dt}\left(\frac{\pi}{\sqrt{2}} l^2\right)$

$\frac{dA}{dt} = \frac{\pi}{\sqrt{2}} \cdot \frac{d}{dt}(l^2)$

$\frac{dA}{dt} = \frac{\pi}{\sqrt{2}} \cdot (2l) \cdot \frac{dl}{dt}$

$\frac{dA}{dt} = \frac{2\pi l}{\sqrt{2}} \frac{dl}{dt} = \sqrt{2}\pi l \frac{dl}{dt}$

We are given $\frac{dA}{dt} = -2$ cm$^2$/sec (since the water is dripping out, the area is decreasing) and we want to find $\frac{dl}{dt}$ when $l = 4$ cm.

Substitute these values into the related rates equation:

$-2 = \sqrt{2}\pi (4) \frac{dl}{dt}$

Now, solve for $\frac{dl}{dt}$:

$\frac{dl}{dt} = \frac{-2}{4\sqrt{2}\pi}$

$\frac{dl}{dt} = \frac{-1}{2\sqrt{2}\pi}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$\frac{dl}{dt} = \frac{-1}{2\sqrt{2}\pi} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{-\sqrt{2}}{2 \cdot 2 \cdot \pi} = \frac{-\sqrt{2}}{4\pi}$

The rate of change of the slant height is $\frac{-\sqrt{2}}{4\pi}$ cm/sec.

The question asks for the rate of decrease of the slant height. The negative sign indicates that the slant height is decreasing. Therefore, the rate of decrease is the absolute value of this rate.

Final Answer:

The rate of decrease of the slant height of water is $\frac{\sqrt{2}}{4\pi}$ cm/sec when the slant height is 4 cm.

Given:

The equations of the curves are $y^2 = x$ and $x^2 = y$.

To Find:

The angle of intersection of the two curves.

Solution:

To find the angle of intersection between two curves, we first need to find the points where the curves intersect. Then, we find the slopes of the tangents to each curve at these intersection points. The angle between the curves at a point is the angle between their tangents at that point.

Finding the Points of Intersection:

We have the equations of the two curves:

Substitute the expression for $x$ from equation (1) into equation (2):

$(y^2)^2 = y$

$y^4 = y$

Move all terms to one side:

$y^4 - y = 0$

Factor out $y$:

$y(y^3 - 1) = 0$

This equation holds true if $y = 0$ or $y^3 - 1 = 0$.

If $y = 0$, substitute into equation (1) to find $x$:

$x = (0)^2 = 0$.

This gives the intersection point $(0, 0)$.

If $y^3 - 1 = 0$, then $y^3 = 1$. The real solution for $y$ is $y = 1$.

Substitute $y = 1$ into equation (1) to find $x$:

$x = (1)^2 = 1$.

This gives the intersection point $(1, 1)$.

The curves intersect at two points: $(0, 0)$ and $(1, 1)$.

Finding the Slopes of the Tangents:

To find the slope of the tangent line to each curve, we differentiate the equation of the curve with respect to $x$.

For the first curve, $y^2 = x$, differentiate implicitly with respect to $x$:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(x)$

$2y \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{2y}$

For the second curve, $x^2 = y$, differentiate directly with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x^2)$

$\frac{dy}{dx} = 2x$

Angle of Intersection at $(0, 0)$:

At the point $(0, 0)$:

For the first curve $y^2 = x$, the slope is $m_1 = \frac{dy}{dx}\Bigr|_{(0,0)} = \frac{1}{2(0)}$. This is undefined, which means the tangent line at $(0, 0)$ is vertical. The vertical line passing through $(0, 0)$ is the y-axis ($x=0$).

For the second curve $x^2 = y$, the slope is $m_2 = \frac{dy}{dx}\Bigr|_{(0,0)} = 2(0) = 0$. This means the tangent line at $(0, 0)$ is horizontal. The horizontal line passing through $(0, 0)$ is the x-axis ($y=0$).

The angle between the y-axis and the x-axis is $90^\circ$ or $\frac{\pi}{2}$ radians.

Angle of Intersection at $(1, 1)$:

At the point $(1, 1)$:

For the first curve $y^2 = x$, the slope is $m_1 = \frac{dy}{dx}\Bigr|_{(1,1)} = \frac{1}{2(1)} = \frac{1}{2}$.

For the second curve $x^2 = y$, the slope is $m_2 = \frac{dy}{dx}\Bigr|_{(1,1)} = 2(1) = 2$.

Let $\theta$ be the acute angle between the tangents with slopes $m_1 = \frac{1}{2}$ and $m_2 = 2$. The formula for the angle between two lines with slopes $m_1$ and $m_2$ is given by:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Substitute the values of $m_1$ and $m_2$:

$\tan \theta = \left| \frac{\frac{1}{2} - 2}{1 + (\frac{1}{2})(2)} \right|$

$\tan \theta = \left| \frac{\frac{1 - 4}{2}}{1 + 1} \right|$

$\tan \theta = \left| \frac{-\frac{3}{2}}{2} \right|$

$\tan \theta = \left| -\frac{3}{2} \times \frac{1}{2} \right|$

$\tan \theta = \left| -\frac{3}{4} \right|$

$\tan \theta = \frac{3}{4}$

The angle $\theta$ is the arctangent of $\frac{3}{4}$.

$\theta = \tan^{-1}\left(\frac{3}{4}\right)$

Final Answer:

The curves $y^2 = x$ and $x^2 = y$ intersect at two points: $(0, 0)$ and $(1, 1)$.

At the point $(0, 0)$, the angle of intersection is $\frac{\pi}{2}$ or $90^\circ$.

At the point $(1, 1)$, the angle of intersection is $\tan^{-1}\left(\frac{3}{4}\right)$.

Given:

The function is $f(x) = \tan x - 4x$.

The interval is $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

To Prove:

The function $f(x)$ is strictly decreasing on the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

Proof:

A function $f(x)$ is strictly decreasing on an interval if its derivative $f'(x)$ is strictly negative for all $x$ in that interval.

First, we find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\tan x - 4x)$

Using the standard differentiation rules, $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(4x) = 4$.

$f'(x) = \sec^2 x - 4$

Now, we need to determine the sign of $f'(x)$ on the given interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

Consider the interval $-\frac{\pi}{3} < x < \frac{\pi}{3}$.

For any $x$ in this interval, the value of $\cos x$ is positive and ranges from $\cos(\frac{\pi}{3}) = \frac{1}{2}$ to $\cos(0) = 1$. Since the interval is open, $\cos x$ is strictly between $\frac{1}{2}$ and $1$, except at $x=0$ where $\cos(0)=1$. More precisely, for $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$, we have $\frac{1}{2} < \cos x \leq 1$.

Squaring the inequality (all values are positive):

$\left(\frac{1}{2}\right)^2 < \cos^2 x \leq (1)^2$

$\frac{1}{4} < \cos^2 x \leq 1$

Now consider $\sec^2 x = \frac{1}{\cos^2 x}$. Taking the reciprocal of the inequality and reversing the signs:

$\frac{1}{1} \leq \frac{1}{\cos^2 x} < \frac{1}{\frac{1}{4}}$

$1 \leq \sec^2 x < 4$

This inequality $1 \leq \sec^2 x < 4$ holds for all $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

Now substitute this into the expression for $f'(x)$:

$f'(x) = \sec^2 x - 4$

Since $\sec^2 x < 4$ for all $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$, it follows that $\sec^2 x - 4 < 0$ for all $x$ in this interval.

Thus, $f'(x) < 0$ for all $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

By the definition of a strictly decreasing function, if $f'(x) < 0$ on an interval, then the function $f(x)$ is strictly decreasing on that interval.

Therefore, the function $f(x) = \tan x - 4x$ is strictly decreasing on the interval $\left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$.

Conclusion:

The derivative $f'(x) = \sec^2 x - 4$ is negative for all $x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$, which proves that $f(x)$ is strictly decreasing on the given interval.

Given:

The function is $y = f(x) = x^4 - \frac{4x^3}{3}$.

To Determine:

The intervals of $x$ for which the function $f(x)$ is increasing and the intervals for which it is decreasing.

Solution:

A function $f(x)$ is increasing on an interval if $f'(x) > 0$ for all $x$ in that interval (except possibly at discrete points where $f'(x) = 0$), and strictly increasing if $f'(x) > 0$. A function $f(x)$ is decreasing on an interval if $f'(x) < 0$ for all $x$ in that interval (except possibly at discrete points where $f'(x) = 0$), and strictly decreasing if $f'(x) < 0$. For intervals of monotonicity, we often include endpoints where the derivative is zero, provided the function is continuous there.

First, we find the derivative of the function $y = x^4 - \frac{4x^3}{3}$ with respect to $x$:

$f'(x) = \frac{d}{dx}\left(x^4 - \frac{4}{3}x^3\right)$

$f'(x) = \frac{d}{dx}(x^4) - \frac{4}{3}\frac{d}{dx}(x^3)$

$f'(x) = 4x^{4-1} - \frac{4}{3}(3x^{3-1})$

$f'(x) = 4x^3 - \frac{4}{\cancel{3}}\cancel{3}x^2$

$f'(x) = 4x^3 - 4x^2$

To find the critical points, we set the derivative equal to zero and solve for $x$:

$f'(x) = 0$

$4x^3 - 4x^2 = 0$

Factor out $4x^2$:

$4x^2(x - 1) = 0$

This equation is satisfied if $4x^2 = 0$ or $x - 1 = 0$.

If $4x^2 = 0$, then $x^2 = 0$, which means $x = 0$.

If $x - 1 = 0$, then $x = 1$.

The critical points are $x = 0$ and $x = 1$. These points divide the number line into three intervals:

$(-\infty, 0)$, $(0, 1)$, and $(1, \infty)$.

Now, we examine the sign of $f'(x) = 4x^2(x - 1)$ in each interval:

Interval $(-\infty, 0)$: Choose a test value, say $x = -1$.

$f'(-1) = 4(-1)^2(-1 - 1) = 4(1)(-2) = -8$.

Since $f'(-1) < 0$, the function is decreasing on $(-\infty, 0)$.

Interval $(0, 1)$: Choose a test value, say $x = 0.5$.

$f'(0.5) = 4(0.5)^2(0.5 - 1) = 4(0.25)(-0.5) = 1(-0.5) = -0.5$.

Since $f'(0.5) < 0$, the function is decreasing on $(0, 1)$.

Interval $(1, \infty)$: Choose a test value, say $x = 2$.

$f'(2) = 4(2)^2(2 - 1) = 4(4)(1) = 16$.

Since $f'(2) > 0$, the function is increasing on $(1, \infty)$.

The function is decreasing on both $(-\infty, 0)$ and $(0, 1)$. Since $f'(0) = 0$ and the function continues to decrease across $x=0$, we can say the function is decreasing on the union of these intervals including the endpoint $x=0$, which is $(-\infty, 1]$.

Final Answer:

The function $f(x) = x^4 - \frac{4x^3}{3}$ is decreasing on the interval $(-\infty, 1]$.

The function $f(x) = x^4 - \frac{4x^3}{3}$ is increasing on the interval $[1, \infty)$.

Given:

The function is $f(x) = 4x^3 - 18x^2 + 27x - 7$.

To Prove:

The function $f(x)$ has neither a local maximum nor a local minimum.

Proof:

To find local maxima or minima, we first need to find the critical points of the function. Critical points are the points where the first derivative is either zero or undefined. For a polynomial function like this, the derivative is always defined for all real numbers.

First, find the first derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(4x^3 - 18x^2 + 27x - 7)$

$f'(x) = \frac{d}{dx}(4x^3) - \frac{d}{dx}(18x^2) + \frac{d}{dx}(27x) - \frac{d}{dx}(7)$

$f'(x) = 4(3x^2) - 18(2x) + 27(1) - 0$

$f'(x) = 12x^2 - 36x + 27$

To find the critical points, set $f'(x) = 0$:

$12x^2 - 36x + 27 = 0$

We can factor out a common factor of 3 from the equation:

$3(4x^2 - 12x + 9) = 0$

Divide by 3:

$4x^2 - 12x + 9 = 0$

This is a quadratic equation. We can check if it is a perfect square or use the quadratic formula. Notice that $4x^2 = (2x)^2$, $9 = 3^2$, and $12x = 2 \times (2x) \times 3$. This fits the form $(a-b)^2 = a^2 - 2ab + b^2$, with $a = 2x$ and $b = 3$.

So, the equation can be written as:

$(2x - 3)^2 = 0$

Taking the square root of both sides:

$2x - 3 = 0$

Solving for $x$:

$2x = 3$

$x = \frac{3}{2}$

The only critical point for the function is $x = \frac{3}{2}$.

For a function to have a local maximum or minimum at a critical point, the sign of the first derivative must change as $x$ passes through that point. Let's examine the sign of $f'(x) = 3(2x - 3)^2$ around $x = \frac{3}{2}$.

The term $(2x - 3)^2$ is always non-negative for any real value of $x$ (since it is a square).

$(2x - 3)^2 \geq 0$ for all $x \in \mathbb{R}$.

Multiplying by 3 (a positive number), we get:

$f'(x) = 3(2x - 3)^2 \geq 0$ for all $x \in \mathbb{R}$.

Specifically:

For $x < \frac{3}{2}$, $2x - 3 < 0$, so $(2x - 3)^2 > 0$, and $f'(x) = 3(2x - 3)^2 > 0$.

For $x > \frac{3}{2}$, $2x - 3 > 0$, so $(2x - 3)^2 > 0$, and $f'(x) = 3(2x - 3)^2 > 0$.

At $x = \frac{3}{2}$, $f'\left(\frac{3}{2}\right) = 3\left(2\left(\frac{3}{2}\right) - 3\right)^2 = 3(3 - 3)^2 = 3(0)^2 = 0$.

The sign of $f'(x)$ does not change as $x$ passes through the critical point $x = \frac{3}{2}$ (it is positive before $\frac{3}{2}$ and positive after $\frac{3}{2}$). Since the derivative is positive on either side of the critical point, the function is increasing before and after $x = \frac{3}{2}$. The point $x = \frac{3}{2}$ is an inflection point where the tangent is horizontal, not a local extremum.

A function has local maxima or minima only at critical points where the sign of the first derivative changes (from positive to negative for a maximum, or negative to positive for a minimum). Since the sign of $f'(x)$ does not change at $x = \frac{3}{2}$, the function has neither a local maximum nor a local minimum at this point or any other point (as there are no other critical points).

Alternatively, we can use the second derivative test at the critical point.

Find the second derivative of $f(x)$:

$f''(x) = \frac{d}{dx}(12x^2 - 36x + 27)$

$f''(x) = 12(2x) - 36(1) + 0$

$f''(x) = 24x - 36$

Evaluate the second derivative at the critical point $x = \frac{3}{2}$:

$f''\left(\frac{3}{2}\right) = 24\left(\frac{3}{2}\right) - 36$

$f''\left(\frac{3}{2}\right) = \frac{\cancel{24}^{12} \times 3}{\cancel{2}_{1}} - 36$

$f''\left(\frac{3}{2}\right) = 12 \times 3 - 36$

$f''\left(\frac{3}{2}\right) = 36 - 36 = 0$

Since the second derivative $f''\left(\frac{3}{2}\right) = 0$, the second derivative test is inconclusive at this critical point. We must rely on the first derivative test, which we have already performed and shown that the sign of $f'(x)$ does not change at $x = \frac{3}{2}$.

Therefore, the function $f(x) = 4x^3 - 18x^2 + 27x - 7$ has neither a local maximum nor a local minimum.

Conclusion:

Since the first derivative $f'(x)$ is non-negative for all $x$ and its sign does not change at the only critical point, the function is always increasing (or non-decreasing) and thus has neither local maxima nor local minima.

Given:

We need to find the approximate value of $\sqrt{0.082}$ using differentials.

To Find:

Approximate value of $\sqrt{0.082}$ using differentials.

Solution:

To use differentials, we define a function $f(x)$ such that the value we want to approximate is of the form $f(x + \Delta x)$.

Let $f(x) = \sqrt{x}$.

We need to choose a value of $x$ close to $0.082$ for which the square root is easy to calculate. A suitable value is $x = 0.09$, because $\sqrt{0.09} = 0.3$.

Let $x = 0.09$. Then $x + \Delta x = 0.082$.

So, $\Delta x = 0.082 - x = 0.082 - 0.09 = -0.008$.

The differential approximation formula is:

$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$

First, find the value of $f(x)$ at $x = 0.09$:

$f(0.09) = \sqrt{0.09} = 0.3$

Next, find the derivative of $f(x) = \sqrt{x}$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2})$

$f'(x) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

Evaluate $f'(x)$ at $x = 0.09$:

$f'(0.09) = \frac{1}{2\sqrt{0.09}} = \frac{1}{2(0.3)} = \frac{1}{0.6}$

$f'(0.09) = \frac{10}{6} = \frac{5}{3}$

Now substitute the values of $f(x)$, $f'(x)$, and $\Delta x$ into the differential approximation formula:

$\sqrt{0.082} = f(0.09 + (-0.008)) \approx f(0.09) + f'(0.09) \times (-0.008)$

$\sqrt{0.082} \approx 0.3 + \left(\frac{5}{3}\right) \times (-0.008)$

$\sqrt{0.082} \approx 0.3 - \frac{5 \times 0.008}{3}$

$\sqrt{0.082} \approx 0.3 - \frac{0.040}{3}$

Now, calculate the value of $\frac{0.040}{3}$.

$\frac{0.040}{3} \approx 0.01333...$

$\sqrt{0.082} \approx 0.3 - 0.01333$

$\sqrt{0.082} \approx 0.28667$

Rounding to a few decimal places, we get approximately 0.287.

Final Answer:

The approximate value of $\sqrt{0.082}$ using differentials is $0.287$ (rounded to three decimal places).

Given:

The equations of the two curves are:

Curve 1: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Curve 2: $xy = c^2$

Assume $a \neq 0$ and $b \neq 0$ for the first equation to represent a hyperbola.

To Find:

The condition on $a$, $b$, and $c$ such that the curves intersect orthogonally.

Solution:

Two curves intersect orthogonally at a point if their tangent lines at that point are perpendicular. The slopes of perpendicular lines (that are not horizontal or vertical) multiply to -1. A more general approach is to use the dot product of the gradients of the functions defining the curves. If the dot product of the gradients is zero at an intersection point, the curves are orthogonal at that point.

Let's rewrite the equations of the curves in the form $F(x, y) = 0$ and $G(x, y) = 0$.

Curve 1: $F(x, y) = \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1 = 0$

Curve 2: $G(x, y) = xy - c^2 = 0$

The gradient of $F$ is $\nabla F = \left\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y} \right\rangle$.

$\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x^2}{a^2} - \frac{y^2}{b^2} - 1\right) = \frac{2x}{a^2}$

$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}\left(\frac{x^2}{a^2} - \frac{y^2}{b^2} - 1\right) = -\frac{2y}{b^2}$

So, $\nabla F = \left\langle \frac{2x}{a^2}, -\frac{2y}{b^2} \right\rangle$.

The gradient of $G$ is $\nabla G = \left\langle \frac{\partial G}{\partial x}, \frac{\partial G}{\partial y} \right\rangle$.

$\frac{\partial G}{\partial x} = \frac{\partial}{\partial x}(xy - c^2) = y$

$\frac{\partial G}{\partial y} = \frac{\partial}{\partial y}(xy - c^2) = x$

So, $\nabla G = \langle y, x \rangle$.

The curves intersect orthogonally at a point $(x, y)$ if the dot product of their gradients at that point is zero, provided the gradients are non-zero (which is true at any intersection point for these curves, except possibly at the origin which is not on the hyperbola).

$\nabla F \cdot \nabla G = \left\langle \frac{2x}{a^2}, -\frac{2y}{b^2} \right\rangle \cdot \langle y, x \rangle = 0$

$\left(\frac{2x}{a^2}\right)(y) + \left(-\frac{2y}{b^2}\right)(x) = 0$

$\frac{2xy}{a^2} - \frac{2xy}{b^2} = 0$

$2xy \left(\frac{1}{a^2} - \frac{1}{b^2}\right) = 0$

This equation must hold at every point $(x, y)$ where the two curves intersect. A point $(x, y)$ of intersection must satisfy both curve equations.

Since $(x, y)$ is a point of intersection, it must satisfy $xy = c^2$. Substitute this into the orthogonality condition:

$2(c^2) \left(\frac{1}{a^2} - \frac{1}{b^2}\right) = 0$

$2c^2 \left(\frac{b^2 - a^2}{a^2 b^2}\right) = 0$

Assuming $a \neq 0$ and $b \neq 0$, for this equation to be true, we must have either $c^2 = 0$ or $b^2 - a^2 = 0$.

$c^2 = 0 \implies c = 0$

$b^2 - a^2 = 0 \implies a^2 = b^2$ (since $a, b$ are usually real and non-zero in this context)

So, the condition for the curves to intersect orthogonally is that for every intersection point, $c=0$ or $a^2=b^2$. This means that either $c=0$ (which makes $xy=0$, and intersection points occur on the axes, leading to orthogonal tangents regardless of $a, b$) or $a^2=b^2$ (which makes the product of slopes $-1$ at any non-axial intersection point, and also handles the axial case when $c=0$).

Thus, the condition for the curves to intersect orthogonally at all points of intersection is that $c = 0$ or $a^2 = b^2$. This can be written as $c^2(a^2 - b^2) = 0$.

Final Answer:

The condition for the curves $\frac{x^2}{a^2} − \frac{y^2}{b^2} = 1$ and $xy = c^2$ to intersect orthogonally is $a^2 = b^2$ or $c = 0$. This can be expressed as $c^2(a^2 - b^2) = 0$.

Given:

The function is $f(x) = - \frac{3}{4} x^4 - 8x^3 - \frac{45}{2} x^2 + 105$.

To Find:

All points of local maxima and local minima of the function $f(x)$.

Solution:

To find the points of local maxima and local minima, we first need to find the critical points of the function. Critical points are the points where the first derivative $f'(x)$ is either zero or undefined. For a polynomial function, the derivative is always defined.

Find the first derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}\left(- \frac{3}{4} x^4 - 8x^3 - \frac{45}{2} x^2 + 105\right)$

$f'(x) = - \frac{3}{4} (4x^3) - 8(3x^2) - \frac{45}{2} (2x) + 0$

$f'(x) = -3x^3 - 24x^2 - 45x$

To find the critical points, set $f'(x) = 0$:

$-3x^3 - 24x^2 - 45x = 0$

Factor out $-3x$:

$-3x(x^2 + 8x + 15) = 0$

Factor the quadratic expression $x^2 + 8x + 15$. We look for two numbers that multiply to 15 and add up to 8. These numbers are 3 and 5.

$x^2 + 8x + 15 = (x + 3)(x + 5)$

So the factored form of the derivative is:

$f'(x) = -3x(x + 3)(x + 5) = 0$

The critical points are the values of $x$ for which the factors are zero:

$-3x = 0 \implies x = 0$

$x + 3 = 0 \implies x = -3$

$x + 5 = 0 \implies x = -5$

The critical points are $x = -5$, $x = -3$, and $x = 0$. These points divide the number line into four intervals: $(-\infty, -5)$, $(-5, -3)$, $(-3, 0)$, and $(0, \infty)$.

Now, we use the first derivative test to determine the sign of $f'(x)$ in each interval and classify the critical points.

Interval $(-\infty, -5)$: Choose a test value, say $x = -6$.

$f'(-6) = -3(-6)(-6 + 3)(-6 + 5) = 18(-3)(-1) = 54$

Since $f'(-6) > 0$, the function is increasing on $(-\infty, -5)$.

Interval $(-5, -3)$: Choose a test value, say $x = -4.

$f'(-4) = -3(-4)(-4 + 3)(-4 + 5) = 12(-1)(1) = -12$

Since $f'(-4) < 0$, the function is decreasing on $(-5, -3)$.

At $x = -5$, the function changes from increasing to decreasing. Thus, there is a local maximum at $x = -5$.

Interval $(-3, 0)$: Choose a test value, say $x = -1.

$f'(-1) = -3(-1)(-1 + 3)(-1 + 5) = 3(2)(4) = 24$

Since $f'(-1) > 0$, the function is increasing on $(-3, 0)$.

At $x = -3$, the function changes from decreasing to increasing. Thus, there is a local minimum at $x = -3$.

Interval $(0, \infty)$: Choose a test value, say $x = 1.

$f'(1) = -3(1)(1 + 3)(1 + 5) = -3(4)(6) = -72$

Since $f'(1) < 0$, the function is decreasing on $(0, \infty)$.

At $x = 0$, the function changes from increasing to decreasing. Thus, there is a local maximum at $x = 0$.

Now, we find the value of the function at these critical points to get the local maximum and minimum values.

Local maximum at $x = -5$:

$f(-5) = - \frac{3}{4}(-5)^4 - 8(-5)^3 - \frac{45}{2}(-5)^2 + 105$

$f(-5) = - \frac{3}{4}(625) - 8(-125) - \frac{45}{2}(25) + 105$

$f(-5) = -\frac{1875}{4} + 1000 - \frac{1125}{2} + 105$

$f(-5) = -\frac{1875}{4} + \frac{4000}{4} - \frac{2250}{4} + \frac{420}{4}$

$f(-5) = \frac{-1875 + 4000 - 2250 + 420}{4} = \frac{2125 - 2250 + 420}{4} = \frac{-125 + 420}{4} = \frac{295}{4}$

Local minimum at $x = -3$:

$f(-3) = - \frac{3}{4}(-3)^4 - 8(-3)^3 - \frac{45}{2}(-3)^2 + 105$

$f(-3) = - \frac{3}{4}(81) - 8(-27) - \frac{45}{2}(9) + 105$

$f(-3) = -\frac{243}{4} + 216 - \frac{405}{2} + 105$

$f(-3) = -\frac{243}{4} + \frac{864}{4} - \frac{810}{4} + \frac{420}{4}$

$f(-3) = \frac{-243 + 864 - 810 + 420}{4} = \frac{621 - 810 + 420}{4} = \frac{-189 + 420}{4} = \frac{231}{4}$

Local maximum at $x = 0$:

$f(0) = - \frac{3}{4}(0)^4 - 8(0)^3 - \frac{45}{2}(0)^2 + 105$

$f(0) = 0 - 0 - 0 + 105 = 105$

Final Answer:

The points of local maxima are at $x = -5$ and $x = 0$. The corresponding local maximum values are $f(-5) = \frac{295}{4}$ and $f(0) = 105$.

The point of local minimum is at $x = -3$. The corresponding local minimum value is $f(-3) = \frac{231}{4}$.

Given:

The function is $f(x) = x + \frac{1}{x}$.

To Show:

The local maximum value of $f(x)$ is less than its local minimum value.

Solution:

To find the local maximum and minimum values, we first find the critical points of the function. The critical points are where the first derivative $f'(x)$ is zero or undefined. The function $f(x)$ is defined for all real numbers except $x=0$.

Find the first derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}\left(x + \frac{1}{x}\right) = \frac{d}{dx}(x + x^{-1})$

$f'(x) = 1 + (-1)x^{-2} = 1 - x^{-2} = 1 - \frac{1}{x^2}$

Set the first derivative equal to zero to find critical points:

$f'(x) = 0$

$1 - \frac{1}{x^2} = 0$

$1 = \frac{1}{x^2}$

$x^2 = 1$

$x = \pm 1$

The critical points are $x = -1$ and $x = 1$. The derivative is undefined at $x=0$, but this point is not in the domain of the function, so it's not considered for local extrema.

We can use the first derivative test or the second derivative test to classify these critical points.

Using the First Derivative Test:

Examine the sign of $f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$ in the intervals determined by the critical points and where the function is undefined: $(-\infty, -1)$, $(-1, 0)$, $(0, 1)$, and $(1, \infty)$.

In the intervals $(-\infty, -1)$ and $(1, \infty)$, $x^2 > 1$, so $x^2 - 1 > 0$. Since $x^2 > 0$ for $x \neq 0$, $f'(x) = \frac{x^2 - 1}{x^2} > 0$. The function is increasing on $(-\infty, -1)$ and $(1, \infty)$.

In the intervals $(-1, 0)$ and $(0, 1)$, $x^2 < 1$, so $x^2 - 1 < 0$. Since $x^2 > 0$ for $x \neq 0$, $f'(x) = \frac{x^2 - 1}{x^2} < 0$. The function is decreasing on $(-1, 0)$ and $(0, 1)$.

At $x = -1$, the function changes from increasing to decreasing. Thus, there is a local maximum at $x = -1$.

At $x = 1$, the function changes from decreasing to increasing. Thus, there is a local minimum at $x = 1$.

Using the Second Derivative Test:

Find the second derivative of $f(x)$:

$f''(x) = \frac{d}{dx}\left(1 - x^{-2}\right) = 0 - (-2)x^{-3} = 2x^{-3} = \frac{2}{x^3}$

Evaluate $f''(x)$ at the critical points:

At $x = -1$: $f''(-1) = \frac{2}{(-1)^3} = \frac{2}{-1} = -2$. Since $f''(-1) < 0$, there is a local maximum at $x = -1$.

At $x = 1$: $f''(1) = \frac{2}{(1)^3} = \frac{2}{1} = 2$. Since $f''(1) > 0$, there is a local minimum at $x = 1$.

Both tests confirm the nature of the critical points.

Calculating Local Maximum and Minimum Values:

The local maximum value occurs at $x = -1$.

$f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$

The local minimum value occurs at $x = 1$.

$f(1) = 1 + \frac{1}{1} = 1 + 1 = 2$

Comparing the local maximum value and the local minimum value:

Local maximum value = $-2$

Local minimum value = $2$

We observe that $-2 < 2$.

Thus, the local maximum value of the function $f(x) = x + \frac{1}{x}$ is less than its local minimum value.

Conclusion:

The local maximum value of $f(x)$ is $-2$, which is less than the local minimum value of $2$.

Given:

The rate of decrease of the volume of water is $\frac{dV}{dt} = -1$ cm$^3$/sec (negative because volume is decreasing).

The slant height of the water at a specific moment is $l = 4$ cm.

The semi-vertical angle of the conical vessel is $\alpha = \frac{\pi}{6}$.

To Find:

The rate of decrease of the slant height of the water, $\frac{dl}{dt}$, when $l = 4$ cm.

Solution:

Let $V$ be the volume of water in the cone at time $t$, $r$ be the radius of the water surface, and $h$ be the height of the water.

The volume of a cone is given by $V = \frac{1}{3}\pi r^2 h$.

In a right cone with semi-vertical angle $\alpha$, we have the relationships $r = l \sin(\alpha)$ and $h = l \cos(\alpha)$, where $l$ is the slant height.

Substitute these relationships into the volume formula:

$V = \frac{1}{3}\pi (l \sin(\alpha))^2 (l \cos(\alpha))$

$V = \frac{1}{3}\pi l^2 \sin^2(\alpha) l \cos(\alpha)$

$V = \frac{1}{3}\pi (\sin^2(\alpha) \cos(\alpha)) l^3$

Given the semi-vertical angle $\alpha = \frac{\pi}{6}$, we have:

$\sin(\alpha) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

$\cos(\alpha) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

$\sin^2(\alpha) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Now substitute these values into the volume equation:

$V = \frac{1}{3}\pi \left(\frac{1}{4} \cdot \frac{\sqrt{3}}{2}\right) l^3$

$V = \frac{1}{3}\pi \left(\frac{\sqrt{3}}{8}\right) l^3$

$V = \frac{\pi\sqrt{3}}{24} l^3$

We need to find the rate of change of the slant height $\frac{dl}{dt}$, given $\frac{dV}{dt}$. We differentiate the equation for $V$ with respect to time $t$ using the chain rule:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi\sqrt{3}}{24} l^3\right)$

$\frac{dV}{dt} = \frac{\pi\sqrt{3}}{24} \frac{d}{dt}(l^3)$

$\frac{dV}{dt} = \frac{\pi\sqrt{3}}{24} (3l^2) \frac{dl}{dt}$

$\frac{dV}{dt} = \frac{3\pi\sqrt{3}}{24} l^2 \frac{dl}{dt}$

$\frac{dV}{dt} = \frac{\pi\sqrt{3}}{8} l^2 \frac{dl}{dt}$

We are given $\frac{dV}{dt} = -1$ cm$^3$/sec and $l = 4$ cm. Substitute these values into the equation:

$-1 = \frac{\pi\sqrt{3}}{8} (4)^2 \frac{dl}{dt}$

$-1 = \frac{\pi\sqrt{3}}{8} (16) \frac{dl}{dt}$

$-1 = 2\pi\sqrt{3} \frac{dl}{dt}$

Solve for $\frac{dl}{dt}$:

$\frac{dl}{dt} = \frac{-1}{2\pi\sqrt{3}}$

To rationalize the denominator:

$\frac{dl}{dt} = \frac{-1}{2\pi\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{-\sqrt{3}}{2\pi(3)} = \frac{-\sqrt{3}}{6\pi}$

The rate of change of the slant height is $\frac{-\sqrt{3}}{6\pi}$ cm/sec. The negative sign indicates that the slant height is decreasing.

The question asks for the rate of decrease of the slant height, which is the positive value of this rate.

Final Answer:

The rate of decrease of the slant height of water is $\frac{\sqrt{3}}{6\pi}$ cm/sec when the slant height is 4 cm.

Given:

The equation of the curve is $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$.

The tangents are parallel to the line $x + 2y = 0$.

To Find:

The equations of all such tangent lines.

Solution:

First, find the slope of the given line $x + 2y = 0$. Rewrite the equation in slope-intercept form $y = mx + c$:

$2y = -x$

$y = -\frac{1}{2}x$

The slope of this line is $m_{line} = -\frac{1}{2}$.

Since the tangent lines are parallel to this line, the slope of the tangent lines is also $m_{tangent} = -\frac{1}{2}$.

Next, find the general expression for the slope of the tangent to the curve $y = \cos(x + y)$ by differentiating implicitly with respect to $x$.

$\frac{dy}{dx} = \frac{d}{dx}(\cos(x + y))$

Using the chain rule:

$\frac{dy}{dx} = -\sin(x + y) \cdot \frac{d}{dx}(x + y)$

$\frac{dy}{dx} = -\sin(x + y) \cdot \left(1 + \frac{dy}{dx}\right)$

Distribute $-\sin(x + y)$:

$\frac{dy}{dx} = -\sin(x + y) - \sin(x + y)\frac{dy}{dx}$

Move the $\frac{dy}{dx}$ term to the left side:

$\frac{dy}{dx} + \sin(x + y)\frac{dy}{dx} = -\sin(x + y)$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx}(1 + \sin(x + y)) = -\sin(x + y)$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-\sin(x + y)}{1 + \sin(x + y)}$

We set the slope of the tangent equal to the required slope, which is $-\frac{1}{2}$:

Assuming $1 + \sin(x + y) \neq 0$ (i.e., $\sin(x + y) \neq -1$), we can multiply both sides by $-(1 + \sin(x + y))$:

$\sin(x + y) = \frac{1}{2}(1 + \sin(x + y))$

$2\sin(x + y) = 1 + \sin(x + y)$

$2\sin(x + y) - \sin(x + y) = 1$

$\sin(x + y) = 1$

The general solution for $\sin \theta = 1$ is $\theta = 2n\pi + \frac{\pi}{2}$, where $n$ is an integer.

So, $x + y = 2n\pi + \frac{\pi}{2}$.

We also need the points $(x, y)$ to lie on the original curve $y = \cos(x + y)$.

Substitute the condition $x + y = 2n\pi + \frac{\pi}{2}$ into the curve equation:

$y = \cos\left(2n\pi + \frac{\pi}{2}\right)$

Using the property $\cos(\theta + 2n\pi) = \cos \theta$:

$y = \cos\left(\frac{\pi}{2}\right)$

$y = 0$

Now substitute $y = 0$ back into the condition $x + y = 2n\pi + \frac{\pi}{2}$:

$x + 0 = 2n\pi + \frac{\pi}{2}$

$x = 2n\pi + \frac{\pi}{2}$

The points of tangency are of the form $\left(2n\pi + \frac{\pi}{2}, 0\right)$ for integer values of $n$.

We are given the restriction $-2\pi \leq x \leq 2\pi$. We find the integer values of $n$ that satisfy this condition for $x$:

$-2\pi \leq 2n\pi + \frac{\pi}{2} \leq 2\pi$

Divide the inequality by $\pi$:

$-2 \leq 2n + \frac{1}{2} \leq 2$

Subtract $\frac{1}{2}$ from all parts:

$-2 - \frac{1}{2} \leq 2n \leq 2 - \frac{1}{2}$

$-\frac{5}{2} \leq 2n \leq \frac{3}{2}$

Divide by 2:

$-\frac{5}{4} \leq n \leq \frac{3}{4}$

$-1.25 \leq n \leq 0.75$

Since $n$ must be an integer, the possible values are $n = -1$ and $n = 0$.

For $n = -1$:

$x = 2(-1)\pi + \frac{\pi}{2} = -2\pi + \frac{\pi}{2} = -\frac{3\pi}{2}$

$y = 0$

Point of tangency: $\left(-\frac{3\pi}{2}, 0\right)$. The x-coordinate $-\frac{3\pi}{2}$ is in the range $[-2\pi, 2\pi]$.

For $n = 0$:

$x = 2(0)\pi + \frac{\pi}{2} = \frac{\pi}{2}$

$y = 0$

Point of tangency: $\left(\frac{\pi}{2}, 0\right)$. The x-coordinate $\frac{\pi}{2}$ is in the range $[-2\pi, 2\pi]$.

Now, we write the equation of the tangent line for each point using the point-slope form $y - y_0 = m(x - x_0)$, where $m = -\frac{1}{2}$.

Tangent at $\left(-\frac{3\pi}{2}, 0\right)$:

$y - 0 = -\frac{1}{2}\left(x - \left(-\frac{3\pi}{2}\right)\right)$

$y = -\frac{1}{2}\left(x + \frac{3\pi}{2}\right)$

$y = -\frac{1}{2}x - \frac{3\pi}{4}$

Multiply by 4 to clear fractions:

$4y = -2x - 3\pi$

Rearrange into the form $Ax + By + C = 0$:

$2x + 4y + 3\pi = 0$

Tangent at $\left(\frac{\pi}{2}, 0\right)$:

$y - 0 = -\frac{1}{2}\left(x - \frac{\pi}{2}\right)$

$y = -\frac{1}{2}x + \frac{\pi}{4}$

Multiply by 4 to clear fractions:

$4y = -2x + \pi$

Rearrange into the form $Ax + By + C = 0$:

$2x + 4y - \pi = 0$

We should check if the case where $1 + \sin(x + y) = 0$ (i.e., $\sin(x+y) = -1$) could yield tangents with slope $-1/2$. If $\sin(x+y) = -1$, then $x+y = 2n\pi - \frac{\pi}{2}$. Substituting into the curve equation $y = \cos(x+y)$, we get $y = \cos(2n\pi - \frac{\pi}{2}) = \cos(-\frac{\pi}{2}) = 0$. If $y=0$, then $x = 2n\pi - \frac{\pi}{2}$. At these points $(2n\pi - \frac{\pi}{2}, 0)$, the derivative $\frac{dy}{dx} = \frac{-\sin(x + y)}{1 + \sin(x + y)} = \frac{-(-1)}{1 + (-1)} = \frac{1}{0}$, which corresponds to a vertical tangent. Vertical tangents have infinite slope, not $-1/2$. So, these points do not yield the desired tangents.

Final Answer:

The equations of the tangents to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$ that are parallel to the line $x + 2y = 0$ are $2x + 4y + 3\pi = 0$ and $2x + 4y - \pi = 0$.

Given:

The equations of the two curves are:

To Find:

The angle(s) of intersection of the two curves.

Solution:

To find the angle of intersection between two curves, we first find their intersection points. Then, we find the slopes of the tangent lines to each curve at each intersection point. The angle between the curves at an intersection point is the angle between their tangents at that point.

Finding the Points of Intersection:

From equation (1), $x = \frac{y^2}{4a}$ (assuming $a \neq 0$). Substitute this expression for $x$ into equation (2):

$\left(\frac{y^2}{4a}\right)^2 = 4by$

$\frac{y^4}{16a^2} = 4by$

$y^4 = 64a^2by$

$y^4 - 64a^2by = 0$

Factor out $y$:

$y(y^3 - 64a^2b) = 0$

This gives two possibilities:

Case 1: $y = 0$

Substitute $y = 0$ into equation (1): $(0)^2 = 4ax \implies 0 = 4ax$. If $a \neq 0$, then $x = 0$. The point is $(0, 0)$.

Case 2: $y^3 - 64a^2b = 0$

$y^3 = 64a^2b$

$y = \sqrt[3]{64a^2b} = 4a^{2/3}b^{1/3}$

Substitute this value of $y$ back into equation (1) to find the corresponding $x$ value:

$x = \frac{y^2}{4a} = \frac{(4a^{2/3}b^{1/3})^2}{4a} = \frac{16a^{4/3}b^{2/3}}{4a} = 4a^{4/3 - 1}b^{2/3} = 4a^{1/3}b^{2/3}$

The second intersection point is $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$. Note that this assumes $a$ and $b$ are non-zero and have the same sign for real intersection points other than the origin.

The two curves intersect at $(0, 0)$ and $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$.

Finding the Slopes of the Tangents:

Differentiate equation (1), $y^2 = 4ax$, implicitly with respect to $x$:

$2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$ (Let $m_1 = \frac{2a}{y}$)

Differentiate equation (2), $x^2 = 4by$, implicitly with respect to $x$:

$2x = 4b \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{2x}{4b} = \frac{x}{2b}$ (Let $m_2 = \frac{x}{2b}$)

Angle of Intersection at $(0, 0)$:

At the point $(0, 0)$:

For the curve $y^2 = 4ax$ (assuming $a \neq 0$), the tangent slope is $m_1 = \frac{2a}{y}$. As $(x, y) \to (0, 0)$ along the curve (e.g., $y = \sqrt{4ax}$), the slope $\to \infty$. The tangent line is vertical (the y-axis, $x=0$).

For the curve $x^2 = 4by$ (assuming $b \neq 0$), the tangent slope is $m_2 = \frac{x}{2b}$. As $(x, y) \to (0, 0)$ along the curve (e.g., $x = \sqrt{4by}$), the slope $\to 0$. The tangent line is horizontal (the x-axis, $y=0$).

If $a \neq 0$ and $b \neq 0$, the tangent lines at $(0, 0)$ are the x-axis and the y-axis, which are perpendicular. The angle of intersection is $90^\circ$ or $\frac{\pi}{2}$.

Angle of Intersection at $(x_0, y_0) = (4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$:

At this point, the slope of the tangent to the first curve is:

$m_1 = \frac{2a}{y_0} = \frac{2a}{4a^{2/3}b^{1/3}} = \frac{1}{2} a^{1/3} b^{-1/3} = \frac{1}{2} \left(\frac{a}{b}\right)^{1/3}$

The slope of the tangent to the second curve is:

$m_2 = \frac{x_0}{2b} = \frac{4a^{1/3}b^{2/3}}{2b} = 2 a^{1/3} b^{-1/3} = 2 \left(\frac{a}{b}\right)^{1/3}$

Let $\theta$ be the acute angle between the tangents. The formula for the angle between two lines with slopes $m_1$ and $m_2$ is:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Substitute the values of $m_1$ and $m_2$:

$\tan \theta = \left| \frac{\frac{1}{2} (\frac{a}{b})^{1/3} - 2 (\frac{a}{b})^{1/3}}{1 + (\frac{1}{2} (\frac{a}{b})^{1/3}) (2 (\frac{a}{b})^{1/3})} \right|$

$\tan \theta = \left| \frac{(\frac{1}{2} - 2) (\frac{a}{b})^{1/3}}{1 + (\frac{a}{b})^{1/3 + 1/3}} \right|$

$\tan \theta = \left| \frac{-\frac{3}{2} (\frac{a}{b})^{1/3}}{1 + (\frac{a}{b})^{2/3}} \right|$

Assuming $a$ and $b$ have the same sign (so $(a/b)^{1/3}$ is real and positive), the expression inside the absolute value is negative. We take the positive value for the acute angle.

$\tan \theta = \frac{\frac{3}{2} (a/b)^{1/3}}{1 + (a/b)^{2/3}} = \frac{3 (a/b)^{1/3}}{2(1 + (a/b)^{2/3})}$

The angle of intersection is $\theta = \tan^{-1} \left( \frac{3 (a/b)^{1/3}}{2(1 + (a/b)^{2/3})} \right)$.

Final Answer:

The curves intersect at two points:

1. At the origin $(0, 0)$, the angle of intersection is $\frac{\pi}{2}$ or $90^\circ$ (assuming $a \neq 0$ and $b \neq 0$).

2. At the point $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$, the angle of intersection is $\tan^{-1} \left( \frac{3 (a/b)^{1/3}}{2(1 + (a/b)^{2/3})} \right)$.

Given:

The parametric equations of the curve are $x = 3\cos \theta - \cos^3 \theta$ and $y = 3\sin \theta - \sin^3 \theta$.

To Show:

The equation of the normal at any point on the curve is $4 (y \cos^3 \theta - x \sin^3 \theta) = 3 \sin 4\theta$.

Solution:

To find the equation of the normal line at a point on a parametric curve, we first need to find the slope of the tangent line $\frac{dy}{dx}$ at that point. The slope of the normal line is the negative reciprocal of the slope of the tangent line.

We have $x = 3\cos \theta - \cos^3 \theta$ and $y = 3\sin \theta - \sin^3 \theta$. We need to find $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.

Find $\frac{dx}{d\theta}$:

$\frac{dx}{d\theta} = \frac{d}{d\theta}(3\cos \theta - \cos^3 \theta)$

$\frac{dx}{d\theta} = 3(-\sin \theta) - 3\cos^2 \theta (-\sin \theta)$

$\frac{dx}{d\theta} = -3\sin \theta + 3\cos^2 \theta \sin \theta$

Factor out $3\sin \theta$:

$\frac{dx}{d\theta} = 3\sin \theta (\cos^2 \theta - 1)$

Using the identity $\cos^2 \theta - 1 = -\sin^2 \theta$:

$\frac{dx}{d\theta} = 3\sin \theta (-\sin^2 \theta) = -3\sin^3 \theta$

Find $\frac{dy}{d\theta}$:

$\frac{dy}{d\theta} = \frac{d}{d\theta}(3\sin \theta - \sin^3 \theta)$

$\frac{dy}{d\theta} = 3(\cos \theta) - 3\sin^2 \theta (\cos \theta)$

$\frac{dy}{d\theta} = 3\cos \theta (1 - \sin^2 \theta)$

Using the identity $1 - \sin^2 \theta = \cos^2 \theta$:

$\frac{dy}{d\theta} = 3\cos \theta (\cos^2 \theta) = 3\cos^3 \theta$

Now find the slope of the tangent $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3\cos^3 \theta}{-3\sin^3 \theta} = -\frac{\cos^3 \theta}{\sin^3 \theta} = -\cot^3 \theta$

The slope of the normal line at the point $(x, y)$ corresponding to the parameter $\theta$ is the negative reciprocal of the slope of the tangent, provided the slope is not zero or undefined.

Slope of the normal ($m_{normal}$) $= -\frac{1}{m_{tangent}} = -\frac{1}{-\cot^3 \theta} = \frac{1}{\cot^3 \theta} = \tan^3 \theta$

The equation of the normal line at the point $(x_0, y_0)$ with slope $m$ is $y - y_0 = m(x - x_0)$. Here, the point is $(3\cos \theta - \cos^3 \theta, 3\sin \theta - \sin^3 \theta)$ and the slope is $\tan^3 \theta = \frac{\sin^3 \theta}{\cos^3 \theta}$.

$y - (3\sin \theta - \sin^3 \theta) = \frac{\sin^3 \theta}{\cos^3 \theta} (x - (3\cos \theta - \cos^3 \theta))$

Multiply both sides by $\cos^3 \theta$:

$(y - 3\sin \theta + \sin^3 \theta)\cos^3 \theta = \sin^3 \theta (x - 3\cos \theta + \cos^3 \theta)$

$y\cos^3 \theta - 3\sin \theta \cos^3 \theta + \sin^3 \theta \cos^3 \theta = x\sin^3 \theta - 3\cos \theta \sin^3 \theta + \cos^3 \theta \sin^3 \theta$

Subtract $\sin^3 \theta \cos^3 \theta$ from both sides:

$y\cos^3 \theta - 3\sin \theta \cos^3 \theta = x\sin^3 \theta - 3\cos \theta \sin^3 \theta$

Rearrange the terms to group $x$ and $y$ terms:

$y\cos^3 \theta - x\sin^3 \theta = 3\sin \theta \cos^3 \theta - 3\cos \theta \sin^3 \theta$

Factor out $3\sin \theta \cos \theta$ from the right side:

$y\cos^3 \theta - x\sin^3 \theta = 3\sin \theta \cos \theta (\cos^2 \theta - \sin^2 \theta)$

Using the double angle identities: $\sin(2\theta) = 2\sin \theta \cos \theta$ and $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$:

The right side is $3 \left(\frac{1}{2} \sin(2\theta)\right) (\cos(2\theta)) = \frac{3}{2} \sin(2\theta) \cos(2\theta)$

Using the identity $\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$, so $\sin(2\theta)\cos(2\theta) = \frac{1}{2}\sin(4\theta)$:

The right side is $\frac{3}{2} \left(\frac{1}{2} \sin(4\theta)\right) = \frac{3}{4}\sin(4\theta)$

So, the equation is:

$y\cos^3 \theta - x\sin^3 \theta = \frac{3}{4}\sin(4\theta)$

Multiply both sides by 4:

$4(y\cos^3 \theta - x\sin^3 \theta) = 3\sin(4\theta)$

This is the required equation of the normal line.

Conclusion:

We have derived the equation of the normal line at any point on the given parametric curve and shown that it matches the required form $4 (y \cos^3 \theta – x \sin^3 \theta) = 3 \sin 4\theta$.

Given:

The function is $f(x) = \sec x + \log \cos^2 x$.

The interval is $0 < x < 2\pi$.

To Find:

The local maximum and local minimum values of the function $f(x)$ on the given interval.

Solution:

The function is $f(x) = \sec x + \log \cos^2 x$.

We can rewrite $\log \cos^2 x$ as $\log (\cos x)^2 = 2 \log |\cos x|$.

The function $f(x)$ is defined when $\cos x \neq 0$, which means $x \neq \frac{\pi}{2}$ and $x \neq \frac{3\pi}{2}$ in the interval $(0, 2\pi)$. Also, $\log |\cos x|$ is defined when $|\cos x| > 0$, which is the same condition $\cos x \neq 0$. Thus, the domain of the function on $(0, 2\pi)$ is $(0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$.

To find local extrema, we find the critical points by setting the first derivative $f'(x)$ to zero.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sec x + \log \cos^2 x)$

Using the chain rule and the derivative of $\log u$, we have $\frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx}$. Let $u = \cos^2 x$.

$f'(x) = \frac{d}{dx}(\sec x) + \frac{1}{\cos^2 x} \frac{d}{dx}(\cos^2 x)$

$f'(x) = \sec x \tan x + \frac{1}{\cos^2 x} (2\cos x (-\sin x))$

$f'(x) = \sec x \tan x - \frac{2\cos x \sin x}{\cos^2 x}$

$f'(x) = \sec x \tan x - 2 \frac{\sin x}{\cos x}$

$f'(x) = \sec x \tan x - 2 \tan x$

Factor out $\tan x$:

$f'(x) = \tan x (\sec x - 2)$

Set $f'(x) = 0$ to find critical points:

$\tan x (\sec x - 2) = 0$

This implies $\tan x = 0$ or $\sec x - 2 = 0$.

Case 1: $\tan x = 0$

In the interval $(0, 2\pi)$, $\tan x = 0$ when $x = \pi$.

Case 2: $\sec x - 2 = 0$

$\sec x = 2 \implies \frac{1}{\cos x} = 2 \implies \cos x = \frac{1}{2}$.

In the interval $(0, 2\pi)$, $\cos x = \frac{1}{2}$ when $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.

The critical points in the domain of $f(x)$ are $x = \frac{\pi}{3}$, $x = \pi$, and $x = \frac{5\pi}{3}$.

We use the first derivative test to determine the nature of these critical points by examining the sign of $f'(x) = \tan x (\sec x - 2)$ in the intervals defined by the critical points and the points where the function is undefined ($x=\pi/2, 3\pi/2$). The relevant intervals are $(0, \pi/3)$, $(\pi/3, \pi/2)$, $(\pi/2, \pi)$, $(\pi, 3\pi/2)$, $(3\pi/2, 5\pi/3)$, $(5\pi/3, 2\pi)$.

$f'(x) = \tan x (\sec x - 2)$. The sign depends on $\tan x$ and $(\sec x - 2)$.

Let's analyze the sign in the relevant intervals:

Interval $(0, \pi/3)$: $\tan x > 0$. $\cos x \in (\frac{1}{2}, 1)$, so $\sec x \in (1, 2)$. $\sec x - 2 < 0$. $f'(x) = (+)(-) = -$. Function is decreasing.

Interval $(\pi/3, \pi/2)$: $\tan x > 0$. $\cos x \in (0, \frac{1}{2})$, so $\sec x \in (2, \infty)$. $\sec x - 2 > 0$. $f'(x) = (+)(+) = +$. Function is increasing.

At $x = \pi/3$, $f'(x)$ changes from negative to positive. Thus, there is a local minimum at $x = \pi/3$.

Interval $(\pi/2, \pi)$: $\tan x < 0$. $\cos x \in (-1, 0)$, so $\sec x \in (-\infty, -1)$. $\sec x - 2 < 0$. $f'(x) = (-)(-) = +$. Function is increasing.

Interval $(\pi, 3\pi/2)$: $\tan x > 0$. $\cos x \in (-1, 0)$, so $\sec x \in (-\infty, -1)$. $\sec x - 2 < 0$. $f'(x) = (+)(-) = -$. Function is decreasing.

At $x = \pi$, $f'(x)$ changes from positive to negative. Thus, there is a local maximum at $x = \pi$.

Interval $(3\pi/2, 5\pi/3)$: $\tan x < 0$. $\cos x \in (0, \frac{1}{2})$, so $\sec x \in (2, \infty)$. $\sec x - 2 > 0$. $f'(x) = (-)(+) = -$. Function is decreasing.

Interval $(5\pi/3, 2\pi)$: $\tan x < 0$. $\cos x \in (\frac{1}{2}, 1)$, so $\sec x \in (1, 2)$. $\sec x - 2 < 0$. $f'(x) = (-)(-) = +$. Function is increasing.

At $x = 5\pi/3$, $f'(x)$ changes from negative to positive. Thus, there is a local minimum at $x = 5\pi/3$.

Now, we calculate the values of the function at these local extrema.

Local minimum values at $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$:

$f\left(\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) + \log \cos^2\left(\frac{\pi}{3}\right) = 2 + \log \left(\frac{1}{2}\right)^2 = 2 + \log \frac{1}{4} = 2 + \log (2^{-2}) = 2 - 2\log 2$

$f\left(\frac{5\pi}{3}\right) = \sec\left(\frac{5\pi}{3}\right) + \log \cos^2\left(\frac{5\pi}{3}\right) = 2 + \log \left(\frac{1}{2}\right)^2 = 2 + \log \frac{1}{4} = 2 - 2\log 2$

The local minimum value is $2 - 2\log 2$.

Local maximum value at $x = \pi$:

$f(\pi) = \sec(\pi) + \log \cos^2(\pi) = -1 + \log (-1)^2 = -1 + \log 1 = -1 + 0 = -1$

The local maximum value is $-1$.

Comparing the local maximum and minimum values:

Local maximum value = $-1$

Local minimum value = $2 - 2\log 2$

Since $\log 2 \approx 0.693$, $2\log 2 \approx 1.386$. Thus, $2 - 2\log 2 \approx 2 - 1.386 = 0.614$.

We see that $-1 < 0.614$.

So, the local maximum value ($-1$) is less than the local minimum value ($2 - 2\log 2$).

Note: The function approaches $\pm\infty$ as $x$ approaches $\frac{\pi}{2}$ or $\frac{3\pi}{2}$. Therefore, there are no global maximum or minimum values on the open interval $(0, 2\pi)$. The question asks for "the maximum and minimum values", which in this context refers to the local extrema.

Final Answer:

The local maximum value of the function is $-1$, occurring at $x = \pi$.

The local minimum value of the function is $2 - 2\log 2$, occurring at $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

To Find:

The area of the greatest rectangle that can be inscribed in the ellipse.

Solution:

Consider a rectangle inscribed in the ellipse with sides parallel to the coordinate axes. Let the vertices of the rectangle be $(\pm x, \pm y)$. Since these points lie on the ellipse, they must satisfy the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The lengths of the sides of the rectangle are $2x$ and $2y$. The area of the rectangle, denoted by $A$, is given by:

$A = (2x)(2y) = 4xy$

We need to maximize this area subject to the constraint that the point $(x, y)$ is on the ellipse in the first quadrant (where $x \ge 0, y \ge 0$). From the ellipse equation, we can express $y$ in terms of $x$:

$\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2}$

$y^2 = \frac{b^2}{a^2} (a^2 - x^2)$

$y = \frac{b}{a} \sqrt{a^2 - x^2}$ (Since we are in the first quadrant, $y \ge 0$)

Substitute this expression for $y$ into the area formula:

$A(x) = 4x \left(\frac{b}{a} \sqrt{a^2 - x^2}\right) = \frac{4b}{a} x \sqrt{a^2 - x^2}$

The possible values for $x$ range from $0$ to $a$ (when $x=a$, $y=0$, the rectangle degenerates to a line). We want to maximize $A(x)$ for $x \in [0, a]$.

To maximize $A(x)$, it is equivalent to maximize $A(x)^2$, as the area is non-negative. This avoids dealing with the square root during differentiation.

Let $Z(x) = A(x)^2 = \left(\frac{4b}{a}\right)^2 x^2 (a^2 - x^2) = \frac{16b^2}{a^2} (a^2x^2 - x^4)$.

Now, find the derivative of $Z(x)$ with respect to $x$ and set it to zero to find critical points:

$Z'(x) = \frac{d}{dx}\left(\frac{16b^2}{a^2} (a^2x^2 - x^4)\right)$

$Z'(x) = \frac{16b^2}{a^2} (2a^2x - 4x^3)$

Set $Z'(x) = 0$:

$\frac{16b^2}{a^2} (2a^2x - 4x^3) = 0$

Since $\frac{16b^2}{a^2} \neq 0$ (assuming $a, b \neq 0$), we must have:

$2a^2x - 4x^3 = 0$

Factor out $2x$:

$2x(a^2 - 2x^2) = 0$

This gives possible solutions $x = 0$ or $a^2 - 2x^2 = 0$.

$x = 0$ corresponds to a degenerate rectangle with zero area.

$a^2 - 2x^2 = 0 \implies 2x^2 = a^2 \implies x^2 = \frac{a^2}{2} \implies x = \pm \frac{a}{\sqrt{2}}$

Since we are considering $x \ge 0$ in the first quadrant, the relevant critical point is $x = \frac{a}{\sqrt{2}}$. This value is in the interval $(0, a)$.

We can use the first derivative test for $Z(x)$ around $x = \frac{a}{\sqrt{2}}$.

$Z'(x) = \frac{16b^2}{a^2} (2x(a^2 - 2x^2)) = \frac{16b^2}{a^2} (2x(\sqrt{a^2} - \sqrt{2}x)(\sqrt{a^2} + \sqrt{2}x))$

$Z'(x) = \frac{16b^2}{a^2} (2x(a - \sqrt{2}x)(a + \sqrt{2}x))$

For $x \in (0, a)$, $2x > 0$ and $a + \sqrt{2}x > 0$. The sign of $Z'(x)$ depends on the sign of $(a - \sqrt{2}x)$.

If $0 < x < \frac{a}{\sqrt{2}}$, then $\sqrt{2}x < a$, so $a - \sqrt{2}x > 0$. $Z'(x) > 0$. $Z(x)$ is increasing.

If $\frac{a}{\sqrt{2}} < x < a$, then $\sqrt{2}x > a$, so $a - \sqrt{2}x < 0$. $Z'(x) < 0$. $Z(x)$ is decreasing.

Since $Z(x)$ changes from increasing to decreasing at $x = \frac{a}{\sqrt{2}}$, $Z(x)$ has a local maximum at this point. This corresponds to the maximum area of the rectangle.

Find the corresponding value of $y$ when $x = \frac{a}{\sqrt{2}}$:

$y = \frac{b}{a} \sqrt{a^2 - x^2} = \frac{b}{a} \sqrt{a^2 - \left(\frac{a}{\sqrt{2}}\right)^2} = \frac{b}{a} \sqrt{a^2 - \frac{a^2}{2}} = \frac{b}{a} \sqrt{\frac{a^2}{2}} = \frac{b}{a} \frac{a}{\sqrt{2}} = \frac{b}{\sqrt{2}}$

The dimensions of the rectangle with the greatest area are $2x = 2\left(\frac{a}{\sqrt{2}}\right) = \sqrt{2}a$ and $2y = 2\left(\frac{b}{\sqrt{2}}\right) = \sqrt{2}b$.

The greatest area is $A = 4xy = 4\left(\frac{a}{\sqrt{2}}\right)\left(\frac{b}{\sqrt{2}}\right) = 4 \frac{ab}{2} = 2ab$.

Final Answer:

The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $2ab$.

Given:

The function is $f(x) = \sin 2x - x$.

The closed interval is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

To Find:

The difference between the greatest (maximum) and least (minimum) values of the function on the given interval.

Solution:

To find the absolute maximum and minimum values of a continuous function on a closed interval, we need to evaluate the function at the critical points within the interval and at the endpoints of the interval. The critical points are where the first derivative is zero or undefined. The function $f(x)$ is a sum of trigonometric and polynomial functions, so its derivative is defined everywhere.

Find the first derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sin 2x - x)$

$f'(x) = \cos(2x) \cdot \frac{d}{dx}(2x) - 1$

$f'(x) = 2\cos(2x) - 1$

Set $f'(x) = 0$ to find critical points:

$2\cos(2x) - 1 = 0$

$2\cos(2x) = 1$

$\cos(2x) = \frac{1}{2}$

We need to find the values of $x$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\cos(2x) = \frac{1}{2}$.

Let $u = 2x$. The interval for $x$ is $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$. Multiplying by 2, the interval for $u$ is $2(-\frac{\pi}{2}) \leq 2x \leq 2(\frac{\pi}{2})$, which is $-\pi \leq u \leq \pi$.

In the interval $[-\pi, \pi]$, the solutions for $\cos u = \frac{1}{2}$ are $u = -\frac{\pi}{3}$ and $u = \frac{\pi}{3}$.

Substitute back $u = 2x$:

$2x = -\frac{\pi}{3} \implies x = -\frac{\pi}{6}$

$2x = \frac{\pi}{3} \implies x = \frac{\pi}{6}$

Both critical points $x = -\frac{\pi}{6}$ and $x = \frac{\pi}{6}$ lie within the given interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Now, evaluate the function $f(x)$ at the critical points and the endpoints of the interval:

Endpoints: $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$.

Critical points: $x = -\frac{\pi}{6}$ and $x = \frac{\pi}{6}$.

Evaluate $f(x) = \sin 2x - x$ at these points:

$f\left(-\frac{\pi}{2}\right) = \sin\left(2 \cdot -\frac{\pi}{2}\right) - \left(-\frac{\pi}{2}\right) = \sin(-\pi) + \frac{\pi}{2} = 0 + \frac{\pi}{2} = \frac{\pi}{2}$

$f\left(\frac{\pi}{2}\right) = \sin\left(2 \cdot \frac{\pi}{2}\right) - \frac{\pi}{2} = \sin(\pi) - \frac{\pi}{2} = 0 - \frac{\pi}{2} = -\frac{\pi}{2}$

$f\left(-\frac{\pi}{6}\right) = \sin\left(2 \cdot -\frac{\pi}{6}\right) - \left(-\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{3}\right) + \frac{\pi}{6} = -\sin\left(\frac{\pi}{3}\right) + \frac{\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{\pi}{6}$

$f\left(\frac{\pi}{6}\right) = \sin\left(2 \cdot \frac{\pi}{6}\right) - \frac{\pi}{6} = \sin\left(\frac{\pi}{3}\right) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6}$

Now, we compare these values to find the greatest and least values.

We have the values: $\frac{\pi}{2}$, $-\frac{\pi}{2}$, $-\frac{\sqrt{3}}{2} + \frac{\pi}{6}$, and $\frac{\sqrt{3}}{2} - \frac{\pi}{6}$.

Approximate values:

$\frac{\pi}{2} \approx \frac{3.14159}{2} \approx 1.5708$

$-\frac{\pi}{2} \approx -1.5708$

$-\frac{\sqrt{3}}{2} + \frac{\pi}{6} \approx -\frac{1.732}{2} + \frac{3.14159}{6} \approx -0.866 + 0.5236 \approx -0.3424$

$\frac{\sqrt{3}}{2} - \frac{\pi}{6} \approx \frac{1.732}{2} - \frac{3.14159}{6} \approx 0.866 - 0.5236 \approx 0.3424$

Comparing the values: $-1.5708$, $-0.3424$, $0.3424$, $1.5708$.

The greatest value is $\frac{\pi}{2}$.

The least value is $-\frac{\pi}{2}$.

The difference between the greatest and least values is:

Difference = Greatest Value - Least Value

Difference = $\frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$

Final Answer:

The greatest value of the function is $\frac{\pi}{2}$, and the least value is $-\frac{\pi}{2}$.

The difference between the greatest and least values is $\pi$.

Given:

An isosceles triangle is inscribed in a circle of radius $a$.

The vertical angle of the triangle is $2\theta$.

To Show:

The area of the triangle is maximum when $\theta = \frac{\pi}{6}$.

Solution:

Let the isosceles triangle be ABC, with the vertical angle at A being $2\theta$. Since it is an isosceles triangle, the base angles at B and C are equal. The sum of angles in a triangle is $\pi$ (or $180^\circ$), so each base angle is $\frac{\pi - 2\theta}{2} = \frac{\pi}{2} - \theta$. Thus, $\angle B = \angle C = \frac{\pi}{2} - \theta$.

For a non-degenerate triangle inscribed in a circle, the angles must be positive and less than $\pi$. So, $2\theta > 0$, $\frac{\pi}{2} - \theta > 0$. This implies $\theta > 0$ and $\theta < \frac{\pi}{2}$. Thus, the domain for $\theta$ is $(0, \frac{\pi}{2})$. We will consider the closed interval $[0, \frac{\pi}{2}]$ and check the endpoints.

The area of a triangle inscribed in a circle with radius $R$ is given by the formula $K = 2R^2 \sin A \sin B \sin C$, where A, B, C are the angles of the triangle.

In this case, $R = a$, $A = 2\theta$, $B = \frac{\pi}{2} - \theta$, and $C = \frac{\pi}{2} - \theta$.

The area of the triangle is:

$K(\theta) = 2a^2 \sin(2\theta) \sin\left(\frac{\pi}{2} - \theta\right) \sin\left(\frac{\pi}{2} - \theta\right)$

Using the identity $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$:

$K(\theta) = 2a^2 \sin(2\theta) \cos\theta \cos\theta$

$K(\theta) = 2a^2 \sin(2\theta) \cos^2\theta$

Using the double angle identity $\sin(2\theta) = 2\sin\theta \cos\theta$:

$K(\theta) = 2a^2 (2\sin\theta \cos\theta) \cos^2\theta$

$K(\theta) = 4a^2 \sin\theta \cos^3\theta$

To find the maximum area, we find the critical points by taking the derivative of $K(\theta)$ with respect to $\theta$ and setting it to zero.

Find $K'(\theta)$:

$K'(\theta) = \frac{d}{d\theta}(4a^2 \sin\theta \cos^3\theta)$

$K'(\theta) = 4a^2 \frac{d}{d\theta}(\sin\theta \cos^3\theta)$

Using the product rule:

$K'(\theta) = 4a^2 \left[ (\cos\theta)(\cos^3\theta) + (\sin\theta)(3\cos^2\theta (-\sin\theta)) \right]$

$K'(\theta) = 4a^2 \left[ \cos^4\theta - 3\sin^2\theta \cos^2\theta \right]$

Set $K'(\theta) = 0$:

$4a^2 (\cos^4\theta - 3\sin^2\theta \cos^2\theta) = 0$

Since $a \neq 0$, we can divide by $4a^2$:

$\cos^4\theta - 3\sin^2\theta \cos^2\theta = 0$

Factor out $\cos^2\theta$:

$\cos^2\theta (\cos^2\theta - 3\sin^2\theta) = 0$

This equation is satisfied if $\cos^2\theta = 0$ or $\cos^2\theta - 3\sin^2\theta = 0$.

If $\cos^2\theta = 0$, then $\cos\theta = 0$. In the interval $[0, \frac{\pi}{2}]$, this occurs at $\theta = \frac{\pi}{2}$. At $\theta = \frac{\pi}{2}$, the vertical angle is $\pi$, and the base angles are 0, which means the triangle degenerates into a line segment (a diameter), and the area is $K(\pi/2) = 4a^2 \sin(\pi/2) \cos^3(\pi/2) = 4a^2(1)(0)^3 = 0$. This is a minimum area.

If $\cos^2\theta - 3\sin^2\theta = 0$:

$\cos^2\theta = 3\sin^2\theta$

Assuming $\cos\theta \neq 0$ (i.e., $\theta \neq \pi/2$), we can divide by $\cos^2\theta$:

$1 = 3 \frac{\sin^2\theta}{\cos^2\theta}$

$1 = 3 \tan^2\theta$

$\tan^2\theta = \frac{1}{3}$

$\tan\theta = \pm \frac{1}{\sqrt{3}}$

Since we are considering $0 < \theta < \frac{\pi}{2}$, $\tan\theta$ must be positive.

$\tan\theta = \frac{1}{\sqrt{3}}$

This occurs when $\theta = \frac{\pi}{6}$.

So, the critical point in the interval $(0, \frac{\pi}{2})$ is $\theta = \frac{\pi}{6}$.

To determine if this critical point corresponds to a maximum, we can use the first derivative test. The sign of $K'(\theta) = 4a^2 \cos^2\theta (\cos^2\theta - 3\sin^2\theta)$ is the same as the sign of $(\cos^2\theta - 3\sin^2\theta)$ for $0 < \theta < \pi/2$ (where $\cos^2\theta > 0$).

$\cos^2\theta - 3\sin^2\theta = \cos^2\theta - 3(1 - \cos^2\theta) = \cos^2\theta - 3 + 3\cos^2\theta = 4\cos^2\theta - 3$.

The sign of $K'(\theta)$ depends on $4\cos^2\theta - 3$.

If $0 < \theta < \frac{\pi}{6}$, then $\cos\theta > \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. So $\cos^2\theta > \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$. Then $4\cos^2\theta > 3$, so $4\cos^2\theta - 3 > 0$. $K'(\theta) > 0$. The area is increasing.

If $\frac{\pi}{6} < \theta < \frac{\pi}{2}$, then $0 < \cos\theta < \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. So $0 < \cos^2\theta < \frac{3}{4}$. Then $4\cos^2\theta < 3$, so $4\cos^2\theta - 3 < 0$. $K'(\theta) < 0$. The area is decreasing.

Since $K'(\theta)$ changes sign from positive to negative as $\theta$ passes through $\frac{\pi}{6}$, there is a local maximum at $\theta = \frac{\pi}{6}$.

We check the values at the endpoints: $K(0) = 0$ and $K(\pi/2) = 0$. The area at the critical point $\theta = \pi/6$ is $K(\pi/6) = 4a^2 \sin(\pi/6) \cos^3(\pi/6) = 4a^2 (\frac{1}{2}) (\frac{\sqrt{3}}{2})^3 = 4a^2 \cdot \frac{1}{2} \cdot \frac{3\sqrt{3}}{8} = a^2 \frac{3\sqrt{3}}{4}$. This is a positive value, greater than the endpoint values.

Therefore, the maximum area of the triangle occurs at $\theta = \frac{\pi}{6}$.

Conclusion:

The derivative of the area function $K(\theta)$ is positive for $0 < \theta < \frac{\pi}{6}$ and negative for $\frac{\pi}{6} < \theta < \frac{\pi}{2}$. This indicates that the function has a maximum at $\theta = \frac{\pi}{6}$.

Given:

The equation of the curve is $3y = 6x - 5x^3$.

The normal at a point $(x_0, y_0)$ on the curve passes through the origin $(0, 0)$.

To Find:

The abscissa ($x$-coordinate) of the point on the curve.

Solution:

Let the point on the curve be $(x_0, y_0)$. Since the point is on the curve, it satisfies the equation:

$3y_0 = 6x_0 - 5x_0^3$

Find the slope of the tangent to the curve at the point $(x_0, y_0)$ by differentiating the curve equation with respect to $x$:

$\frac{d}{dx}(3y) = \frac{d}{dx}(6x - 5x^3)$

$3 \frac{dy}{dx} = 6 - 15x^2$

$\frac{dy}{dx} = \frac{6 - 15x^2}{3} = 2 - 5x^2$

The slope of the tangent at $(x_0, y_0)$ is $m_{tangent} = 2 - 5x_0^2$.

The slope of the normal at $(x_0, y_0)$ is the negative reciprocal of the tangent slope (assuming the tangent is not horizontal or vertical):

$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{2 - 5x_0^2}$

We are given that the normal at $(x_0, y_0)$ passes through the origin $(0, 0)$. The slope of the line passing through $(x_0, y_0)$ and $(0, 0)$ is given by the formula $\frac{y_2 - y_1}{x_2 - x_1}$.

Slope of the line passing through $(x_0, y_0)$ and $(0, 0)$ is $\frac{y_0 - 0}{x_0 - 0} = \frac{y_0}{x_0}$.

This slope must be equal to the slope of the normal:

From the curve equation, we have $y_0 = \frac{1}{3}(6x_0 - 5x_0^3)$. Substitute this into equation (1):

$\frac{\frac{1}{3}(6x_0 - 5x_0^3)}{x_0} = -\frac{1}{2 - 5x_0^2}$

Assuming $x_0 \neq 0$ (if $x_0 = 0$, then $y_0 = 0$, the point is $(0,0)$. The tangent at $(0,0)$ is $3y = 6x$, slope is 2. Normal slope is $-1/2$. The line through $(0,0)$ with slope $-1/2$ is $y = -x/2$. $(0,0)$ is on this line. So $(0,0)$ is a point where the normal passes through origin. The abscissa is 0, which is not among the options. Thus $x_0 \neq 0$).

Simplify the left side:

$\frac{6x_0 - 5x_0^3}{3x_0} = \frac{x_0(6 - 5x_0^2)}{3x_0} = \frac{6 - 5x_0^2}{3}$

So, the equation becomes:

$\frac{6 - 5x_0^2}{3} = -\frac{1}{2 - 5x_0^2}$

Cross-multiply:

$(6 - 5x_0^2)(2 - 5x_0^2) = -3$

Let $u = 2 - 5x_0^2$. Then $6 - 5x_0^2 = (2 - 5x_0^2) + 4 = u + 4$.

$(u + 4)u = -3$

$u^2 + 4u = -3$

$u^2 + 4u + 3 = 0$

Factor the quadratic equation:

$(u + 1)(u + 3) = 0$

So, $u + 1 = 0$ or $u + 3 = 0$.

Case A: $u + 1 = 0 \implies u = -1$.

Substitute back $u = 2 - 5x_0^2$:

$2 - 5x_0^2 = -1$

$5x_0^2 = 2 + 1 = 3$

$x_0^2 = \frac{3}{5}$

$x_0 = \pm \sqrt{\frac{3}{5}}$

Case B: $u + 3 = 0 \implies u = -3$.

Substitute back $u = 2 - 5x_0^2$:

$2 - 5x_0^2 = -3$

$5x_0^2 = 2 + 3 = 5$

$x_0^2 = \frac{5}{5} = 1$

$x_0 = \pm \sqrt{1} = \pm 1$

The possible values for the abscissa $x_0$ are $\pm \sqrt{\frac{3}{5}}$, $+1$, and $-1$. We check which of these values are among the given options.

The options are (A) 1, (B) $\frac{1}{3}$, (C) 2, (D) $\frac{1}{2}$.

Comparing our results with the options, we see that $x_0 = 1$ is one of the possible values.

We should verify that the denominator $2 - 5x_0^2$ is not zero for $x_0 = \pm 1$. For $x_0 = \pm 1$, $2 - 5(\pm 1)^2 = 2 - 5(1) = -3 \neq 0$.

We should also verify that the point $(x_0, y_0)$ is not the origin $(0,0)$, as we assumed $x_0 \neq 0$. For $x_0 = 1$, $3y_0 = 6(1) - 5(1)^3 = 6 - 5 = 1$, so $y_0 = 1/3$. The point is $(1, 1/3)$. This is not the origin. For $x_0 = -1$, $3y_0 = 6(-1) - 5(-1)^3 = -6 - 5(-1) = -6 + 5 = -1$, so $y_0 = -1/3$. The point is $(-1, -1/3)$. This is not the origin. For $x_0 = \pm \sqrt{3/5}$, $x_0 \neq 0$, so the point is not the origin.

Since 1 is one of the obtained abscissa values and is among the options, it is the correct answer.

Final Answer:

The abscissa of the point on the curve is 1.

The correct option is (A) 1.

Given:

The equations of the two curves are:

Curve 1: $x^3 - 3xy^2 + 2 = 0$

Curve 2: $3x^2y - y^3 = 2$

To Determine:

The angle of intersection between the two curves.

Solution:

To find the angle of intersection, we first need to find the point(s) of intersection and then the slopes of the tangent lines at those points.

Finding the Points of Intersection:

We have the system of equations:

At a point of intersection $(x, y)$, both equations must be satisfied. This means:

$x^3 - 3xy^2 + 2 = 3x^2y - y^3$

$x^3 - 3xy^2 - 3x^2y + y^3 + 2 = 0$

Notice that $x^3 - 3xy^2$ is the real part of $(x + iy)^3$ if $y$ is imaginary, and $3x^2y - y^3$ is the imaginary part. Let's consider polar coordinates. Let $x = r\cos\phi$ and $y = r\sin\phi$.

Equation (1): $(r\cos\phi)^3 - 3(r\cos\phi)(r\sin\phi)^2 + 2 = 0$

$r^3\cos^3\phi - 3r^3\cos\phi \sin^2\phi + 2 = 0$

$r^3(\cos^3\phi - 3\cos\phi \sin^2\phi) + 2 = 0$

$r^3\cos(3\phi) + 2 = 0$

$r^3\cos(3\phi) = -2$

Equation (2): $3(r\cos\phi)^2(r\sin\phi) - (r\sin\phi)^3 = 2$

$3r^3\cos^2\phi \sin\phi - r^3\sin^3\phi = 2$

$r^3(3\cos^2\phi \sin\phi - \sin^3\phi) = 2$

$r^3\sin(3\phi) = 2$

So we have: $r^3\cos(3\phi) = -2$ and $r^3\sin(3\phi) = 2$.

Divide the second equation by the first:

$\frac{r^3\sin(3\phi)}{r^3\cos(3\phi)} = \frac{2}{-2}$

$\tan(3\phi) = -1$

This implies $3\phi = n\pi - \frac{\pi}{4}$ for some integer $n$.

For the simplest intersection point, let's look for a real solution $(x, y)$. Notice that if we add the two equations: $(x^3 - 3xy^2 + 2) + (3x^2y - y^3 - 2) = 0$, we get $x^3 - 3xy^2 + 3x^2y - y^3 = 0$. This is not helpful for finding intersection points.

Let's try to find at least one intersection point by inspection or by recognizing special forms. Consider the point $(1, 1)$.

For Curve 1: $(1)^3 - 3(1)(1)^2 + 2 = 1 - 3 + 2 = 0$. The point $(1, 1)$ is on Curve 1.

For Curve 2: $3(1)^2(1) - (1)^3 = 3 - 1 = 2$. The point $(1, 1)$ is on Curve 2.

So, the point $(1, 1)$ is an intersection point.

Finding the Slopes of the Tangents at $(1, 1)$:

Differentiate Curve 1 implicitly with respect to $x$:

$\frac{d}{dx}(x^3 - 3xy^2 + 2) = \frac{d}{dx}(0)$

$3x^2 - (3y^2 + 3x(2y \frac{dy}{dx})) + 0 = 0$ (using product rule for $3xy^2$)

$3x^2 - 3y^2 - 6xy \frac{dy}{dx} = 0$

$-6xy \frac{dy}{dx} = 3y^2 - 3x^2$

$\frac{dy}{dx} = \frac{3y^2 - 3x^2}{-6xy} = \frac{y^2 - x^2}{-2xy} = \frac{x^2 - y^2}{2xy}$

Slope of the tangent to Curve 1 at $(1, 1)$, $m_1$:

$m_1 = \frac{(1)^2 - (1)^2}{2(1)(1)} = \frac{1 - 1}{2} = \frac{0}{2} = 0$

The tangent to Curve 1 at $(1, 1)$ is a horizontal line with slope 0.

Differentiate Curve 2 implicitly with respect to $x$:

$\frac{d}{dx}(3x^2y - y^3) = \frac{d}{dx}(2)$

$(6xy + 3x^2 \frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0$ (using product rule for $3x^2y$)

$6xy + (3x^2 - 3y^2) \frac{dy}{dx} = 0$

$(3x^2 - 3y^2) \frac{dy}{dx} = -6xy$

$\frac{dy}{dx} = \frac{-6xy}{3x^2 - 3y^2} = \frac{-2xy}{x^2 - y^2}$

Slope of the tangent to Curve 2 at $(1, 1)$, $m_2$:

$m_2 = \frac{-2(1)(1)}{(1)^2 - (1)^2} = \frac{-2}{1 - 1} = \frac{-2}{0}$

The slope is undefined, which means the tangent to Curve 2 at $(1, 1)$ is a vertical line.

At the point $(1, 1)$, the tangent to Curve 1 is horizontal, and the tangent to Curve 2 is vertical. A horizontal line and a vertical line are perpendicular.

Therefore, the curves cut at a right angle at the point $(1, 1)$.

Let's check if there are other intersection points using the polar results: $r^3\cos(3\phi) = -2$ and $r^3\sin(3\phi) = 2$.

Square both equations and add them: $(r^3\cos(3\phi))^2 + (r^3\sin(3\phi))^2 = (-2)^2 + (2)^2$

$r^6(\cos^2(3\phi) + \sin^2(3\phi)) = 4 + 4$

$r^6(1) = 8$

$r^6 = 8$

$r = \sqrt[6]{8} = \sqrt{2}$ (taking the real positive root for the magnitude).

Substitute $r^3 = (\sqrt{2})^3 = 2\sqrt{2}$ back into the equations:

$2\sqrt{2}\cos(3\phi) = -2 \implies \cos(3\phi) = -\frac{1}{\sqrt{2}}$

$2\sqrt{2}\sin(3\phi) = 2 \implies \sin(3\phi) = \frac{1}{\sqrt{2}}$

The angles $3\phi$ that satisfy $\cos(3\phi) = -\frac{1}{\sqrt{2}}$ and $\sin(3\phi) = \frac{1}{\sqrt{2}}$ are in the second quadrant.

$3\phi = \frac{3\pi}{4} + 2k\pi$, where $k$ is an integer.

$\phi = \frac{\pi}{4} + \frac{2k\pi}{3}$

For $k = 0$, $\phi_0 = \frac{\pi}{4}$.

The intersection point $(x_0, y_0) = (r\cos\phi_0, r\sin\phi_0) = (\sqrt{2}\cos(\frac{\pi}{4}), \sqrt{2}\sin(\frac{\pi}{4})) = (\sqrt{2} \cdot \frac{1}{\sqrt{2}}, \sqrt{2} \cdot \frac{1}{\sqrt{2}}) = (1, 1)$. This confirms the point we found earlier.

For $k = 1$, $\phi_1 = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{3\pi + 8\pi}{12} = \frac{11\pi}{12}$.

The intersection point $(x_1, y_1) = (r\cos\phi_1, r\sin\phi_1) = (\sqrt{2}\cos(\frac{11\pi}{12}), \sqrt{2}\sin(\frac{11\pi}{12}))$. Since the equations have real coefficients, intersection points must come in conjugate pairs unless they are real. The point $(1, 1)$ is a real intersection point. The other intersection points derived from complex roots of $z^3 = -2 + 2i$ (where $z = x + iy$ would satisfy some combined equation) might not lie on the original real curves.

Let's confirm the derivative calculations in terms of $x$ and $y$.

For Curve 1: $\frac{dy}{dx} = \frac{x^2 - y^2}{2xy}$. At $(1, 1)$, $\frac{1^2 - 1^2}{2(1)(1)} = 0$. Correct.

For Curve 2: $\frac{dy}{dx} = \frac{-2xy}{x^2 - y^2}$. At $(1, 1)$, $\frac{-2(1)(1)}{1^2 - 1^2} = \frac{-2}{0}$, undefined. Correct.

Since the tangents at $(1, 1)$ are perpendicular, the curves intersect at a right angle at this point.

The problem asks for the angle of intersection of "the two curves", implying we need to check at all intersection points. However, often such problems imply the angle at a specific point or that the angle is the same at all intersection points. Given the options are specific angles, it's likely they intersect at a right angle at every real intersection point.

Final Answer:

The curves intersect at the point $(1, 1)$, where the slope of the tangent to the first curve is 0 and the slope of the tangent to the second curve is undefined. This means the tangents are horizontal and vertical, respectively, and are therefore perpendicular.

The curves cut at a right angle.

The correct option is (B) cut at right angle.

Given:

The parametric equations of the curve are $x = e^t \cos t$ and $y = e^t \sin t$.

We need to find the angle of the tangent at $t = \frac{\pi}{4}$ with the x-axis.

To Find:

The angle $\alpha$ that the tangent line makes with the positive x-axis at $t = \frac{\pi}{4}$.

Solution:

The slope of the tangent to a parametric curve at a point is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. The angle $\alpha$ that the tangent makes with the x-axis satisfies $\tan \alpha = \frac{dy}{dx}$.

First, find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Find $\frac{dx}{dt}$ using the product rule:

$\frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = \frac{d}{dt}(e^t)\cos t + e^t \frac{d}{dt}(\cos t)$

$\frac{dx}{dt} = e^t \cos t + e^t (-\sin t) = e^t (\cos t - \sin t)$

Find $\frac{dy}{dt}$ using the product rule:

$\frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = \frac{d}{dt}(e^t)\sin t + e^t \frac{d}{dt}(\sin t)$

$\frac{dy}{dt} = e^t \sin t + e^t (\cos t) = e^t (\sin t + \cos t)$

Now, find the slope of the tangent $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{e^t (\sin t + \cos t)}{e^t (\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t}$

We need to find the slope at $t = \frac{\pi}{4}$. Substitute $t = \frac{\pi}{4}$ into the expression for $\frac{dy}{dx}$:

Slope at $t = \frac{\pi}{4}$ $= \frac{\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})}{\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4})}$

We know that $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

Slope at $t = \frac{\pi}{4}$ $= \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}} = \frac{\frac{2}{\sqrt{2}}}{0}$

The denominator is 0, and the numerator is non-zero ($\frac{2}{\sqrt{2}} = \sqrt{2}$). This means the slope is undefined.

When the slope of the tangent is undefined, the tangent line is vertical, i.e., it is parallel to the y-axis.

A vertical line makes an angle of $\frac{\pi}{2}$ or $90^\circ$ with the positive x-axis.

The angle $\alpha$ such that $\tan \alpha$ is undefined is $\frac{\pi}{2}$.

Let's check the values of $x$ and $y$ at $t = \frac{\pi}{4}$:

$x\left(\frac{\pi}{4}\right) = e^{\pi/4} \cos(\frac{\pi}{4}) = e^{\pi/4} \cdot \frac{1}{\sqrt{2}}$

$y\left(\frac{\pi}{4}\right) = e^{\pi/4} \sin(\frac{\pi}{4}) = e^{\pi/4} \cdot \frac{1}{\sqrt{2}}$

The point of tangency is $\left(\frac{e^{\pi/4}}{\sqrt{2}}, \frac{e^{\pi/4}}{\sqrt{2}}\right)$. Since $e^{\pi/4} > 0$, the point is in the first quadrant. The tangent is a vertical line passing through this point.

Final Answer:

The slope of the tangent at $t = \frac{\pi}{4}$ is undefined, which means the tangent line is vertical.

A vertical line makes an angle of $\frac{\pi}{2}$ with the x-axis.

The correct option is (D) $\frac{π}{2}$.

Given:

The equation of the curve is $y = \sin x$.

The point is $(0, 0)$. We verify that this point is on the curve: $0 = \sin(0)$, which is true. So the point $(0, 0)$ is on the curve.

To Find:

The equation of the normal to the curve at $(0, 0)$.

Solution:

To find the equation of the normal line, we first need to find the slope of the tangent line at the given point. The slope of the normal is the negative reciprocal of the slope of the tangent.

Find the derivative of the curve equation $y = \sin x$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Evaluate the slope of the tangent at the point $(0, 0)$ by substituting $x = 0$ into the derivative:

Slope of the tangent at $(0, 0)$ $= \cos(0) = 1$.

The slope of the normal at $(0, 0)$ is the negative reciprocal of the tangent slope:

Slope of the normal ($m_{normal}$) $= -\frac{1}{\text{slope of tangent}} = -\frac{1}{1} = -1$.

Now, use the point-slope form of the equation of a line: $y - y_0 = m(x - x_0)$, where $(x_0, y_0) = (0, 0)$ and $m = -1$.

$y - 0 = -1(x - 0)$

$y = -x$

Rearrange the equation into the form $Ax + By + C = 0$ or compare with the given options:

$x + y = 0$

Final Answer:

The equation of the normal to the curve $y = \sin x$ at $(0, 0)$ is $x + y = 0$.

The correct option is (C) x + y = 0.

Given:

The equation of the curve is $y^2 = x$.

The tangent to the curve makes an angle of $\frac{\pi}{4}$ with the x-axis.

To Find:

The point $(x, y)$ on the curve where the tangent has this property.

Solution:

The slope of the tangent line to a curve at a point $(x, y)$ is given by the derivative $\frac{dy}{dx}$ evaluated at that point. The angle $\alpha$ that the tangent makes with the positive x-axis is related to the slope by $m = \tan \alpha$.

We are given that the angle is $\alpha = \frac{\pi}{4}$.

The slope of the tangent is $m = \tan\left(\frac{\pi}{4}\right) = 1$.

Now, find the derivative $\frac{dy}{dx}$ for the curve $y^2 = x$ by differentiating implicitly with respect to $x$:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(x)$

$2y \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{2y}$

Set the slope of the tangent equal to the value we found, $m = 1$:

$\frac{1}{2y} = 1$

Solve for $y$:

$1 = 2y$

$y = \frac{1}{2}$

Now that we have the $y$-coordinate of the point, we can find the corresponding $x$-coordinate by substituting this value back into the equation of the curve $y^2 = x$:

$x = y^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

The point on the curve where the tangent makes an angle of $\frac{\pi}{4}$ with the x-axis is $\left(\frac{1}{4}, \frac{1}{2}\right)$.

Let's check if this point is among the given options.

The options are (A) $\left( \frac{1}{2}, \frac{1}{4} \right)$, (B) $\left( \frac{1}{4}, \frac{1}{2} \right)$, (C) (4, 2), (D) (1, 1).

Our calculated point $\left(\frac{1}{4}, \frac{1}{2}\right)$ matches option (B).

Final Answer:

The point on the curve $y^2 = x$ where the tangent makes an angle of $\frac{\pi}{4}$ with the x-axis is $\left(\frac{1}{4}, \frac{1}{2}\right)$.

The correct option is (B) $\left( \frac{1}{4}, \frac{1}{2} \right)$.

Given:

The curve is given by the equation $y = x^2 + ax + 25$.

The curve touches the x-axis.

To Find:

The values of the constant $a$ for which this occurs.

Solution:

A curve touches the x-axis if it intersects the x-axis at exactly one point. The x-axis is the line $y = 0$. So, we need to find the values of $a$ for which the equation $x^2 + ax + 25 = 0$ has exactly one solution for $x$.

The equation $x^2 + ax + 25 = 0$ is a quadratic equation in the form $Ax^2 + Bx + C = 0$, where $A = 1$, $B = a$, and $C = 25$.

A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

The discriminant $\Delta$ (or $D$) is given by $\Delta = B^2 - 4AC$.

In this case, $\Delta = a^2 - 4(1)(25)$.

$\Delta = a^2 - 100$

For the curve to touch the x-axis, the discriminant must be zero:

$a^2 - 100 = 0$

$a^2 = 100$

Take the square root of both sides:

$a = \pm \sqrt{100}$

$a = \pm 10$

The values of $a$ for which the curve $y = x^2 + ax + 25$ touches the axis of x are $10$ and $-10$.

Final Answer:

The values of a for which y = x² + ax + 25 touches the axis of x are $\underline{10, -10}$.

Given:

The function is $f(x) = \frac{1}{4x^2 + 2x + 1}$.

To Find:

The maximum value of the function $f(x)$.

Solution:

To find the maximum value of $f(x)$, we need to find the minimum value of the denominator $g(x) = 4x^2 + 2x + 1$. This is because for a fraction $\frac{1}{D}$, the value of the fraction is maximum when the denominator $D$ is minimum (assuming $D > 0$).

The denominator $g(x) = 4x^2 + 2x + 1$ is a quadratic expression in the form $ax^2 + bx + c$, where $a = 4$, $b = 2$, and $c = 1$. Since the coefficient of $x^2$ ($a=4$) is positive, the parabola opens upwards, and the quadratic function has a minimum value.

The minimum value of a quadratic function $ax^2 + bx + c$ with $a > 0$ occurs at the vertex, where $x = -\frac{b}{2a}$.

For $g(x) = 4x^2 + 2x + 1$, the x-coordinate of the vertex is:

$x = -\frac{2}{2(4)} = -\frac{2}{8} = -\frac{1}{4}$

The minimum value of $g(x)$ occurs at $x = -\frac{1}{4}$. Substitute this value into $g(x)$ to find the minimum value of the denominator:

$g\left(-\frac{1}{4}\right) = 4\left(-\frac{1}{4}\right)^2 + 2\left(-\frac{1}{4}\right) + 1$

$g\left(-\frac{1}{4}\right) = 4\left(\frac{1}{16}\right) - \frac{2}{4} + 1$

$g\left(-\frac{1}{4}\right) = \frac{4}{16} - \frac{1}{2} + 1$

$g\left(-\frac{1}{4}\right) = \frac{1}{4} - \frac{2}{4} + \frac{4}{4}$

$g\left(-\frac{1}{4}\right) = \frac{1 - 2 + 4}{4} = \frac{3}{4}$

The minimum value of the denominator is $\frac{3}{4}$. Since the minimum value is positive, the denominator $4x^2 + 2x + 1$ is always positive for all real $x$. This can also be seen from the discriminant of the denominator: $\Delta = 2^2 - 4(4)(1) = 4 - 16 = -12 < 0$. Since the discriminant is negative and the leading coefficient is positive, the quadratic is always positive.

The maximum value of $f(x) = \frac{1}{g(x)}$ occurs when $g(x)$ is minimum.

Maximum value of $f(x) = \frac{1}{\text{minimum value of } g(x)}$

Maximum value of $f(x) = \frac{1}{\frac{3}{4}} = 1 \times \frac{4}{3} = \frac{4}{3}$

Final Answer:

If $f (x) = \frac{1}{4x^2 + 2x + 1}$ , then its maximum value is $\underline{\frac{4}{3}}$.

Given:

A function $f$ has a second derivative at a point $c$.

At this point, the first derivative is zero, $f'(c) = 0$.

The second derivative is positive, $f''(c) > 0$.

To Determine:

What type of point $c$ is, based on the given conditions.

Solution:

The conditions $f'(c) = 0$ means that $c$ is a critical point of the function $f$. A critical point is a candidate for a local maximum, a local minimum, or a point of inflection.

The second derivative test helps classify critical points based on the sign of the second derivative at that point.

The second derivative test states:

If $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at $c$.

If $f'(c) = 0$ and $f''(c) < 0$, then $f$ has a local maximum at $c$.

If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive (the point could be a local maximum, local minimum, or inflection point).

In the given problem, we have $f'(c) = 0$ and $f''(c) > 0$. According to the second derivative test, these conditions indicate that the function $f$ has a local minimum at the point $x = c$.

Final Answer:

Let f have second deriative at c such that f ‘(c) = 0 and f'' (c) > 0, then c is a point of local minimum.

Given:

The function is $f(x) = \sin x$.

The closed interval is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

To Find:

The minimum value of the function on the given interval.

Solution:

To find the absolute minimum value of a continuous function on a closed interval, we evaluate the function at the critical points within the interval and at the endpoints of the interval. The critical points are where the first derivative is zero or undefined. The derivative of $\sin x$ is $\cos x$, which is defined for all real $x$.

Find the first derivative of $f(x) = \sin x$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sin x) = \cos x$

Set $f'(x) = 0$ to find critical points:

$\cos x = 0$

In the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, the value of $x$ for which $\cos x = 0$ is $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$. These are the endpoints of the interval.

Since the critical points are only at the endpoints, we only need to evaluate the function at the endpoints to find the minimum and maximum values on the closed interval.

Evaluate $f(x) = \sin x$ at the endpoints $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$:

$f\left(-\frac{\pi}{2}\right) = \sin\left(-\frac{\pi}{2}\right) = -1$

$f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1$

Comparing the values $f(-\frac{\pi}{2}) = -1$ and $f(\frac{\pi}{2}) = 1$, the minimum value is $-1$ and the maximum value is $1$. This is expected, as the sine function increases monotonically from $-1$ to $1$ on the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Final Answer:

Minimum value of f if $f (x) = \sin x$ in $\left[ −\frac{π}{2}, \frac{π}{2} \right]$ is $\underline{-1}$.

Given:

The function is $f(x) = \sin x + \cos x$.

To Find:

The maximum value of the function $f(x)$.

Solution:

We can find the maximum value using calculus or by trigonometric manipulation.

Method 1: Using Calculus

To find the maximum value, we find the critical points by setting the first derivative $f'(x)$ to zero.

$f'(x) = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$

Set $f'(x) = 0$:

$\cos x - \sin x = 0$

$\cos x = \sin x$

Assuming $\cos x \neq 0$, divide by $\cos x$:

$\frac{\sin x}{\cos x} = 1$

$\tan x = 1$

The general solution for $\tan x = 1$ is $x = n\pi + \frac{\pi}{4}$, where $n$ is an integer.

To determine if these critical points correspond to a maximum or minimum, we use the second derivative test.

$f''(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x = -(\sin x + \cos x)$

At the critical points where $\tan x = 1$, we have $\sin x = \cos x$.

$f''(x) = -(\sin x + \cos x)$

Substitute $\sin x = \cos x$:

$f''(x) = -(\cos x + \cos x) = -2\cos x$

At $x = n\pi + \frac{\pi}{4}$:

If $n$ is even, say $n=2k$, then $x = 2k\pi + \frac{\pi}{4}$. $\cos x = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

$f''(x) = -2\left(\frac{1}{\sqrt{2}}\right) = -\sqrt{2} < 0$. This indicates a local maximum.

If $n$ is odd, say $n=2k+1$, then $x = (2k+1)\pi + \frac{\pi}{4} = 2k\pi + \pi + \frac{\pi}{4}$. $\cos x = \cos(\pi + \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.

$f''(x) = -2\left(-\frac{1}{\sqrt{2}}\right) = \sqrt{2} > 0$. This indicates a local minimum.

The maximum values occur when $x = 2k\pi + \frac{\pi}{4}$. The maximum value is $f\left(2k\pi + \frac{\pi}{4}\right) = \sin\left(2k\pi + \frac{\pi}{4}\right) + \cos\left(2k\pi + \frac{\pi}{4}\right)$.

Using periodicity, $\sin(2k\pi + \theta) = \sin \theta$ and $\cos(2k\pi + \theta) = \cos \theta$:

Maximum value $= \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

Method 2: Using Trigonometric Identity

The expression $\sin x + \cos x$ can be rewritten in the form $R\sin(x + \alpha)$ or $R\cos(x + \alpha)$.

Let $\sin x + \cos x = R(\sin x \cos \alpha + \cos x \sin \alpha) = (R\cos \alpha)\sin x + (R\sin \alpha)\cos x$.

Comparing coefficients, $R\cos \alpha = 1$ and $R\sin \alpha = 1$.

Squaring and adding: $(R\cos \alpha)^2 + (R\sin \alpha)^2 = 1^2 + 1^2$

$R^2(\cos^2 \alpha + \sin^2 \alpha) = 1 + 1$

$R^2(1) = 2 \implies R = \sqrt{2}$ (taking the positive value for amplitude).

Dividing: $\frac{R\sin \alpha}{R\cos \alpha} = \frac{1}{1} \implies \tan \alpha = 1$. The simplest value for $\alpha$ is $\frac{\pi}{4}$.

So, $\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$.

The sine function $\sin \theta$ has a maximum value of 1. Therefore, the maximum value of $\sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$ is $\sqrt{2} \times 1 = \sqrt{2}$.

Final Answer:

The maximum value of sin x + cos x is $\underline{\sqrt{2}}$.

Given:

The radius of the sphere is $r = 2$ cm.

To Find:

The rate of change of the volume of a sphere with respect to its surface area, i.e., $\frac{dV}{dS}$, when $r = 2$ cm.

Solution:

The volume of a sphere with radius $r$ is given by $V = \frac{4}{3}\pi r^3$.

The surface area of a sphere with radius $r$ is given by $S = 4\pi r^2$.

We need to find $\frac{dV}{dS}$. We can do this by finding the derivatives of $V$ and $S$ with respect to the radius $r$ and then using the chain rule relationship $\frac{dV}{dS} = \frac{dV/dr}{dS/dr}$.

Find the derivative of the volume with respect to $r$:

$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi (3r^{3-1}) = 4\pi r^2$

Find the derivative of the surface area with respect to $r$:

$\frac{dS}{dr} = \frac{d}{dr}(4\pi r^2) = 4\pi (2r^{2-1}) = 8\pi r$

Now, use the chain rule to find $\frac{dV}{dS}$:

$\frac{dV}{dS} = \frac{dV/dr}{dS/dr} = \frac{4\pi r^2}{8\pi r}$

Simplify the expression:

$\frac{dV}{dS} = \frac{\cancel{4\pi} r^{\cancel{2}}}{\cancel{8\pi} \cancel{r}} = \frac{r}{2}$

We need to evaluate this rate of change when the radius is $r = 2$ cm.

$\frac{dV}{dS}\Bigr|_{r=2} = \frac{2}{2} = 1$

The units of $\frac{dV}{dS}$ are the units of volume divided by the units of surface area, which are cm$^3$ / cm$^2$ = cm.

Final Answer:

The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is $\underline{1 \text{ cm}}$.

Let $V$ be the volume, $A$ the surface area, and $r$ the radius of the spherical ball of salt at any time $t$.

The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$.

The formula for the surface area of a sphere is $A = 4\pi r^2$.

According to the problem, the rate of decrease of the volume is proportional to the surface area. This can be written as:

$-\frac{dV}{dt} \propto A$

Introducing a constant of proportionality, let $k$ be a positive constant such that:

$-\frac{dV}{dt} = kA$

Now, we differentiate the volume $V$ with respect to time $t$ using the chain rule:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt}$

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Substitute this expression for $\frac{dV}{dt}$ and the formula for $A$ into the proportionality equation $-\frac{dV}{dt} = kA$:

$-(4\pi r^2 \frac{dr}{dt}) = k(4\pi r^2)$

$-4\pi r^2 \frac{dr}{dt} = 4\pi k r^2$

Assuming $r > 0$ (as long as the ball exists and has a volume and surface area), we can divide both sides of the equation by $4\pi r^2$:

$-\frac{dr}{dt} = k$

Multiplying by $-1$, we get the rate of change of the radius:

$\frac{dr}{dt} = -k$

Since $k$ is a positive constant, $-k$ is a constant value. This shows that the rate of change of the radius, $\frac{dr}{dt}$, is a negative constant. A negative rate of change indicates that the radius is decreasing.

Since $\frac{dr}{dt}$ is a constant value, the radius is decreasing at a constant rate.

Hence, it is proved that the radius is decreasing at a constant rate.

Let $A$ be the area, $P$ the perimeter, and $r$ the radius of the circle at any time $t$.

The formula for the area of a circle is $A = \pi r^2$.

The formula for the perimeter of a circle is $P = 2\pi r$.

We are given that the area of the circle increases at a uniform rate. Let this constant rate be $k$, where $k$ is a constant ($k > 0$). Thus,

Differentiating the area $A$ with respect to time $t$, we get:

$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2)$

Using the chain rule, $\frac{dA}{dt} = \pi \cdot 2r \frac{dr}{dt} = 2\pi r \frac{dr}{dt}$.

Substituting this into the given condition:

From equation (i), we can find the rate of change of the radius $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{k}{2\pi r}$

Now, we want to examine how the perimeter $P$ varies. Let's find the rate of change of the perimeter with respect to time $t$:

$\frac{dP}{dt} = \frac{d}{dt}(2\pi r)$

Using the chain rule, $\frac{dP}{dt} = 2\pi \frac{dr}{dt}$.

Substitute the expression for $\frac{dr}{dt}$ we found:

$\frac{dP}{dt} = 2\pi \left(\frac{k}{2\pi r}\right)$

$\frac{dP}{dt} = \frac{k}{r}$

Since $k$ is a constant, the rate of change of the perimeter ($\frac{dP}{dt}$) is proportional to $\frac{1}{r}$.

Therefore, the rate at which the perimeter increases varies inversely as the radius.

This proves the statement under the standard interpretation that "perimeter varies inversely as the radius" refers to the rate of change of the perimeter.

Given:

Height of the kite from the ground = $151.5$ m

Height of the boy = $1.5$ m

Horizontal speed of the kite, $\frac{dx}{dt} = 10$ m/s

Distance of the kite from the boy (length of string), $s = 250$ m

Let $h$ be the constant vertical height of the kite above the boy's hand level.

$h = \text{Height of kite from ground} - \text{Height of boy}$

$h = 151.5 - 1.5 = 150$ m

Let $x$ be the horizontal distance of the kite from the boy at any time $t$, and $s$ be the length of the string at the same time $t$. We can consider a right-angled triangle formed by the horizontal distance ($x$), the vertical height ($h$), and the length of the string ($s$).

By the Pythagorean theorem, we have:

$x^2 + h^2 = s^2$

Since $h = 150$ m is constant, the relationship is:

$x^2 + (150)^2 = s^2$

$x^2 + 22500 = s^2$

To Find:

The rate at which the string is being let out, $\frac{ds}{dt}$, when $s = 250$ m.

Solution:

Differentiate the relationship $x^2 + 22500 = s^2$ with respect to time $t$:

$\frac{d}{dt}(x^2 + 22500) = \frac{d}{dt}(s^2)$

$2x \frac{dx}{dt} + 0 = 2s \frac{ds}{dt}$

$2x \frac{dx}{dt} = 2s \frac{ds}{dt}$

Divide by 2:

We need to find the value of $x$ at the instant when $s = 250$ m.

Using the Pythagorean theorem again:

$x^2 + (150)^2 = (250)^2$

$x^2 + 22500 = 62500$

$x^2 = 62500 - 22500$

$x^2 = 40000$

$x = \sqrt{40000}$

$x = 200$ m (Since $x$ represents a distance, it must be positive)

Now, substitute the values of $x$, $s$, and $\frac{dx}{dt}$ into equation (i):

$x = 200$ m

$s = 250$ m

$\frac{dx}{dt} = 10$ m/s

$2000 = 250 \frac{ds}{dt}$

To find $\frac{ds}{dt}$, divide both sides by 250:

$\frac{ds}{dt} = \frac{2000}{250}$

$\frac{ds}{dt} = \frac{200}{25}$

$\frac{ds}{dt} = 8$ m/s

Thus, the string is being let out at a rate of $8$ m/s when the kite is $250$ m away from the boy.

Given:

Two men A and B start from a junction (O) simultaneously.

Velocity of man A = $v$

Velocity of man B = $v$

Angle between the two roads = $45^\circ$

To Find:

The rate at which the distance between the two men is increasing.

Solution:

Let O be the junction from which the two men start.

Let the roads be along the lines OA and OB, such that the angle between them, $\angle AOB = 45^\circ$.

Let $r_A(t)$ be the distance of man A from O at time $t$, and $r_B(t)$ be the distance of man B from O at time $t$.

Since they start at the same time from O with velocity $v$, at any time $t$ from the start:

$r_A(t) = vt$

$r_B(t) = vt$

Let $s(t)$ be the distance between the two men at time $t$, i.e., $s(t) = AB$.

Consider the triangle OAB. We have sides OA = $vt$, OB = $vt$, and the included angle $\angle AOB = 45^\circ$.

We can use the Law of Cosines in $\triangle AOB$ to find the relationship between $s$, $r_A$, $r_B$, and the angle.

According to the Law of Cosines:

$s^2 = r_A^2 + r_B^2 - 2 r_A r_B \cos(\angle AOB)$

Substitute $r_A = vt$, $r_B = vt$, and $\angle AOB = 45^\circ$ into the equation:

$s^2 = (vt)^2 + (vt)^2 - 2(vt)(vt) \cos(45^\circ)$

$s^2 = v^2 t^2 + v^2 t^2 - 2v^2 t^2 \left(\frac{1}{\sqrt{2}}\right)$

$s^2 = 2v^2 t^2 - \sqrt{2} v^2 t^2$

$s^2 = v^2 t^2 (2 - \sqrt{2})$

We want to find the rate at which they are being separated, which is $\frac{ds}{dt}$. We differentiate the equation $s^2 = v^2 t^2 (2 - \sqrt{2})$ with respect to time $t$. Note that $v$ and the term $(2 - \sqrt{2})$ are constants.

$\frac{d}{dt}(s^2) = \frac{d}{dt}(v^2 t^2 (2 - \sqrt{2}))$

$2s \frac{ds}{dt} = v^2 (2 - \sqrt{2}) \frac{d}{dt}(t^2)$

From the equation $s^2 = v^2 t^2 (2 - \sqrt{2})$, we can find an expression for $s$. Assuming $s \ge 0$, $v \ge 0$, $t \ge 0$, we have:

$s = \sqrt{v^2 t^2 (2 - \sqrt{2})} = \sqrt{v^2 t^2} \sqrt{2 - \sqrt{2}} = vt \sqrt{2 - \sqrt{2}}$

Now, substitute this expression for $s$ into equation (i):

$2(vt \sqrt{2 - \sqrt{2}}) \frac{ds}{dt} = 2 v^2 t (2 - \sqrt{2})$

Assuming the men are moving away from the junction ($v \neq 0$) and $t > 0$, we can divide both sides of the equation by $2vt$:

$\sqrt{2 - \sqrt{2}} \frac{ds}{dt} = v (2 - \sqrt{2})$

Now, solve for $\frac{ds}{dt}$:

$\frac{ds}{dt} = \frac{v (2 - \sqrt{2})}{\sqrt{2 - \sqrt{2}}}$

Since $2 > \sqrt{2}$, the term $2 - \sqrt{2}$ is positive. We can simplify the expression by recognizing that $X = (\sqrt{X})^2$ for $X>0$. So, $2 - \sqrt{2} = (\sqrt{2 - \sqrt{2}})^2$ is incorrect logic here. The correct simplification is $\frac{Y}{\sqrt{Y}} = \sqrt{Y}$ for $Y>0$. Let $Y = 2-\sqrt{2}$.

$\frac{ds}{dt} = v \sqrt{2 - \sqrt{2}}$

Since $v$ is a constant velocity and $\sqrt{2 - \sqrt{2}}$ is a constant value, the rate of separation $\frac{ds}{dt}$ is a constant.

The rate at which they are being separated is $v \sqrt{2 - \sqrt{2}}$.

Let $\theta$ be the angle at time $t$. We are given that the rate of increase of $\theta$ is twice the rate of increase of $\sin\theta$.

In terms of derivatives with respect to time $t$, this can be written as:

Now, we differentiate $\sin\theta$ with respect to $t$ using the chain rule:

$\frac{d}{dt}(\sin\theta) = \frac{d(\sin\theta)}{d\theta} \cdot \frac{d\theta}{dt} = \cos\theta \frac{d\theta}{dt}$

Substitute this expression back into equation (i):

$\frac{d\theta}{dt} = 2 \left(\cos\theta \frac{d\theta}{dt}\right)$

Rearrange the equation to solve for $\theta$:

$\frac{d\theta}{dt} - 2 \cos\theta \frac{d\theta}{dt} = 0$

Factor out $\frac{d\theta}{dt}$:

$\frac{d\theta}{dt} (1 - 2 \cos\theta) = 0$

For this equation to hold, either $\frac{d\theta}{dt} = 0$ or $1 - 2 \cos\theta = 0$.

If $\frac{d\theta}{dt} = 0$, the angle is not changing, which contradicts the condition that the angle "increases". Therefore, we consider the other possibility:

$1 - 2 \cos\theta = 0$

$1 = 2 \cos\theta$

$\cos\theta = \frac{1}{2}$

We are looking for an angle $\theta$ in the interval $0 < \theta < \frac{\pi}{2}$ (the first quadrant) such that $\cos\theta = \frac{1}{2}$.

The angle in this interval whose cosine is $\frac{1}{2}$ is $\theta = \frac{\pi}{3}$.

Check the interval: $0 < \frac{\pi}{3} < \frac{\pi}{2}$. This condition is satisfied.

Thus, the angle $\theta$ which increases twice as fast as its sine in the interval $0 < \theta < \frac{\pi}{2}$ is $\frac{\pi}{3}$ radians or $60^\circ$.

We want to find the approximate value of $(1.999)^5$. This can be done using differentials or linear approximation.

Let $y = f(x) = x^5$. We want to approximate $f(1.999)$.

We can choose a nearby value for $x$ where the calculation is easy. Let $x = 2$.

The value of the function at $x=2$ is:

$f(2) = 2^5 = 32$

The change in $x$ from the chosen value is $\Delta x = 1.999 - 2 = -0.001$.

The derivative of the function $f(x)$ is:

$f'(x) = \frac{d}{dx}(x^5) = 5x^4$

Now, evaluate the derivative at $x=2$:

$f'(2) = 5(2)^4 = 5(16) = 80$

Using the linear approximation formula, $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$, we can approximate $f(1.999)$.

Here, $x = 2$ and $\Delta x = -0.001$.

$f(1.999) \approx f(2) + f'(2)(-0.001)$

$f(1.999) \approx 32 + (80)(-0.001)$

$f(1.999) \approx 32 - 0.080$

$f(1.999) \approx 31.920$

Alternatively, using differentials, we have $dy = f'(x) dx$. For a small change $\Delta x$, $\Delta y \approx dy$.

$\Delta y \approx f'(2) \times (-0.001) = 80 \times (-0.001) = -0.080$.

The approximate value of $f(1.999)$ is $f(2) + \Delta y \approx 32 + (-0.080) = 31.920$.

The approximate value of $(1.999)^5$ is $31.920$.

Given:

Internal radius of the spherical shell, $r = 3$ cm.

External radius of the spherical shell = $3.0005$ cm.

Let $V(r)$ be the volume of a solid sphere with radius $r$. The formula for the volume of a sphere is:

$V(r) = \frac{4}{3}\pi r^3$

The thickness of the shell is the difference between the external and internal radii.

Let $\Delta r$ be the thickness. $\Delta r = 3.0005 - 3 = 0.0005$ cm.

The volume of the metal in the hollow spherical shell is the difference between the volume of the outer sphere and the volume of the inner sphere.

The outer sphere has radius $r + \Delta r = 3 + 0.0005 = 3.0005$ cm.

The volume of the metal is $V(r + \Delta r) - V(r)$.

We can approximate the change in volume, $\Delta V = V(r + \Delta r) - V(r)$, using differentials.

The approximate change in volume is given by $dV = V'(r) dr$. In this context, $dr = \Delta r$.

First, find the derivative of the volume function $V(r)$ with respect to $r$:

$V'(r) = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi (3r^2) = 4\pi r^2$

Now, substitute the values of $r$ and $\Delta r$ into the approximation formula for the volume of the shell:

Approximate Volume $\approx V'(r) \Delta r$

Approximate Volume $\approx (4\pi r^2) \Delta r$

Substitute $r = 3$ cm and $\Delta r = 0.0005$ cm:

Approximate Volume $\approx 4\pi (3)^2 (0.0005)$

Approximate Volume $\approx 4\pi (9) (0.0005)$

Approximate Volume $\approx 36\pi (0.0005)$

Calculate the numerical value:

$36 \times 0.0005 = 0.0180$

Approximate Volume $\approx 0.018\pi$ cubic cm.

The approximate volume of metal in the hollow spherical shell is $0.018\pi$ cubic cm.

Given:

Height of the man, $h = 2$ m.

Height of the street light, $H = 5 \frac{1}{3}$ m = $\frac{16}{3}$ m.

Speed of the man towards the light = $1 \frac{2}{3}$ m/s = $\frac{5}{3}$ m/s.

Let O be the base of the street light, L be the top of the street light, M be the position of the man's feet, M' be the top of the man's head, and S' be the tip of the man's shadow.

Let $x$ be the distance of the man from the base of the light at time $t$ (i.e., OM = $x$). Since the man is walking towards the light, $x$ is decreasing, so $\frac{dx}{dt} = -\frac{5}{3}$ m/s.

Let $s$ be the length of the man's shadow at time $t$ (i.e., MS' = $s$).

Let $y$ be the distance of the tip of the shadow from the base of the light at time $t$ (i.e., OS' = $y$).

From the diagram, the tip of the shadow S' is on the ground. The man is between the light and the tip of his shadow. Thus, $y = x + s$.

Consider the triangles $\triangle LOS'$ (formed by the light, the ground, and the line from the top of the light to the shadow tip) and $\triangle MM'S'$ (formed by the man, the ground, and the line from the top of the light to the shadow tip).

These two triangles are similar right-angled triangles because the man and the light are vertical, and the angle at S' is common.

By the property of similar triangles, the ratio of corresponding sides is equal:

$\frac{OL}{MM'} = \frac{OS'}{MS'}$

Substituting the variables and given heights:

$\frac{H}{h} = \frac{y}{s}$

$\frac{16/3}{2} = \frac{y}{s}$

$\frac{16}{6} = \frac{y}{s}$

From equation (i), we have $8s = 3y$.

To Find:

1. Rate at which the tip of his shadow is moving ($\left|\frac{dy}{dt}\right|$).

2. Rate at which the length of the shadow is changing ($\frac{ds}{dt}$) when $x = 3 \frac{1}{3}$ m.

Solution:

We have the relationship $y = x + s$. Substitute this into equation (i):

$\frac{8}{3} = \frac{x+s}{s}$

$8s = 3(x+s)$

$8s = 3x + 3s$

Differentiate equation (ii) with respect to time $t$ to find the rate of change of the shadow's length:

$\frac{d}{dt}(5s) = \frac{d}{dt}(3x)$

$5 \frac{ds}{dt} = 3 \frac{dx}{dt}$

Substitute the given value $\frac{dx}{dt} = -\frac{5}{3}$ m/s:

$5 \frac{ds}{dt} = 3 \left(-\frac{5}{3}\right)$

$5 \frac{ds}{dt} = -5$

$\frac{ds}{dt} = \frac{-5}{5}$

$\frac{ds}{dt} = -1$ m/s

The rate at which the length of the shadow is changing is $-1$ m/s. The negative sign indicates that the length of the shadow is decreasing.

Note that this rate is constant and does not depend on the specific distance $x$ of the man from the light. So, the rate is $-1$ m/s when he is $3 \frac{1}{3}$ m from the base of the light, and indeed at any other distance (until he reaches the light).

Now, we find the rate at which the tip of the shadow is moving. This is the rate of change of the distance $y$ from the base of the light.

From equation (ii), $3x = 5s$. Substitute $s = y - x$:

$3x = 5(y - x)$

$3x = 5y - 5x$

Differentiate equation (iii) with respect to time $t$:

$\frac{d}{dt}(8x) = \frac{d}{dt}(5y)$

$8 \frac{dx}{dt} = 5 \frac{dy}{dt}$

Substitute the given value $\frac{dx}{dt} = -\frac{5}{3}$ m/s:

$8 \left(-\frac{5}{3}\right) = 5 \frac{dy}{dt}$

$-\frac{40}{3} = 5 \frac{dy}{dt}$

$\frac{dy}{dt} = \frac{-40/3}{5} = -\frac{40}{3} \times \frac{1}{5}$

$\frac{dy}{dt} = -\frac{8}{3}$ m/s

The rate of change of the distance of the shadow tip from the light base is $-\frac{8}{3}$ m/s. The negative sign indicates that the tip of the shadow is moving towards the base of the light.

The question asks for the "rate at which the tip of his shadow is moving". This is the speed of the shadow tip relative to the ground, which is the magnitude of its velocity:

Rate of tip moving = $\left|\frac{dy}{dt}\right| = \left|-\frac{8}{3}\right| = \frac{8}{3}$ m/s.

Final Answer:

The rate at which the tip of his shadow is moving is $\frac{8}{3}$ m/s.

The rate at which the length of the shadow is changing is $-1$ m/s when he is $3 \frac{1}{3}$ m from the base of the light.

Given:

The volume of water in the pool at time $t$ seconds is given by $L(t) = 200(10 - t)^2$ litres.

The rate at which the water is running out is the negative of the rate of change of the volume in the pool, i.e., $-\frac{dL}{dt}$.

First, we find the derivative of $L(t)$ with respect to $t$:

$\frac{dL}{dt} = \frac{d}{dt} [200(10 - t)^2]$

Using the chain rule, $\frac{d}{dt} [c \cdot u^n] = c \cdot n \cdot u^{n-1} \cdot \frac{du}{dt}$, where $c=200$, $n=2$, and $u = 10 - t$.

$\frac{dL}{dt} = 200 \cdot 2 (10 - t)^{2-1} \cdot \frac{d}{dt}(10 - t)$

$\frac{dL}{dt} = 400 (10 - t) \cdot (-1)$

The rate at which the water is running out is $-\frac{dL}{dt}$.

$-\frac{dL}{dt} = -[-400(10 - t)] = 400(10 - t)$ litres/s.

Part 1: How fast is the water running out at the end of 5 seconds?

We need to evaluate the rate of water running out at $t = 5$ seconds.

Rate out at $t=5 = 400(10 - 5)$

Rate out at $t=5 = 400(5)$

Rate out at $t=5 = 2000$ litres/s.

Part 2: What is the average rate at which the water flows out during the first 5 seconds?

The first 5 seconds correspond to the time interval from $t=0$ to $t=5$.

The average rate of flow is the total volume of water drained during this interval divided by the time taken.

Volume of water at $t=0$: $L(0) = 200(10 - 0)^2 = 200(10^2) = 200(100) = 20000$ litres.

Volume of water at $t=5$: $L(5) = 200(10 - 5)^2 = 200(5^2) = 200(25) = 5000$ litres.

Total volume of water drained during the first 5 seconds = $L(0) - L(5) = 20000 - 5000 = 15000$ litres.

Time taken = $5 - 0 = 5$ seconds.

Average rate of flow = $\frac{\text{Total Volume Drained}}{\text{Time Taken}}$

Average rate of flow = $\frac{15000 \text{ litres}}{5 \text{ seconds}}$

Average rate of flow = $3000$ litres/s.

Final Answer:

The water is running out at a rate of $2000$ litres/s at the end of 5 seconds.

The average rate at which the water flows out during the first 5 seconds is $3000$ litres/s.

Given:

The volume of a cube increases at a constant rate.

To Prove:

The increase in its surface area varies inversely as the length of the side.

Solution:

Let $x$ be the length of the side of the cube at time $t$.

Let $V$ be the volume of the cube at time $t$.

Let $A$ be the surface area of the cube at time $t$.

The formula for the volume of a cube is $V = x^3$.

The formula for the surface area of a cube is $A = 6x^2$.

We are given that the volume of the cube increases at a constant rate. Let this constant rate be $k$, where $k$ is a constant and $k > 0$ (since the volume is increasing).

Differentiate the volume $V$ with respect to time $t$:

$\frac{dV}{dt} = \frac{d}{dt}(x^3)$

Using the chain rule, $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$.

Substitute this into the given condition $\frac{dV}{dt} = k$:

From equation (i), we can find the rate of change of the side length $\frac{dx}{dt}$:

$\frac{dx}{dt} = \frac{k}{3x^2}$

Now, we want to find the rate of increase of the surface area, which is $\frac{dA}{dt}$. Differentiate the surface area $A$ with respect to time $t$:

$\frac{dA}{dt} = \frac{d}{dt}(6x^2)$

Using the chain rule, $\frac{dA}{dt} = 6 \cdot 2x \frac{dx}{dt} = 12x \frac{dx}{dt}$.

Substitute the expression for $\frac{dx}{dt}$ from the previous step into the equation for $\frac{dA}{dt}$:

$\frac{dA}{dt} = 12x \left(\frac{k}{3x^2}\right)$

Simplify the expression:

$\frac{dA}{dt} = \frac{12kx}{3x^2}$

Assuming the side length $x > 0$ (since it's a physical cube), we can cancel an $x$ term from the numerator and denominator:

$\frac{dA}{dt} = \frac{4k}{x}$

The rate of increase in the surface area is $\frac{dA}{dt} = \frac{4k}{x}$.

Since $k$ is a constant, $4k$ is also a constant. Let $C = 4k$. Then $\frac{dA}{dt} = \frac{C}{x}$.

This shows that the rate of increase in the surface area is proportional to $\frac{1}{x}$.

Therefore, the increase in its surface area varies inversely as the length of the side.

Hence, it is proved.

Given:

Side of the first square = $x$

Side of the second square = $y$

Relationship between sides: $y = x - x^2$

To Find:

The rate of change of the area of the second square with respect to the area of the first square, which is $\frac{dA_2}{dA_1}$.

Solution:

Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.

The area of the first square is $A_1 = x^2$.

The area of the second square is $A_2 = y^2$.

Substitute the given relationship $y = x - x^2$ into the expression for $A_2$:

$A_2 = (x - x^2)^2$

Expand the expression for $A_2$:

$A_2 = x^2 (1 - x)^2 = x^2 (1 - 2x + x^2)$

$A_2 = x^2 - 2x^3 + x^4$

We need to find the rate of change of $A_2$ with respect to $A_1$, which is $\frac{dA_2}{dA_1}$. We can use the chain rule, considering both $A_1$ and $A_2$ as functions of $x$:

$\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx}$

First, find the derivative of $A_1$ with respect to $x$:

$\frac{dA_1}{dx} = \frac{d}{dx}(x^2) = 2x$

Next, find the derivative of $A_2$ with respect to $x$:

$\frac{dA_2}{dx} = \frac{d}{dx}(x^4 - 2x^3 + x^2)$

$\frac{dA_2}{dx} = 4x^{4-1} - 2 \cdot 3x^{3-1} + 2x^{2-1}$

$\frac{dA_2}{dx} = 4x^3 - 6x^2 + 2x$

Now, substitute these derivatives into the chain rule formula $\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx}$:

$\frac{dA_2}{dA_1} = \frac{4x^3 - 6x^2 + 2x}{2x}$

Assuming $x \ne 0$ (otherwise, the squares have no area and the rates are trivial or undefined), we can factor out $2x$ from the numerator and simplify:

$\frac{dA_2}{dA_1} = \frac{2x(2x^2 - 3x + 1)}{2x}$

$\frac{dA_2}{dA_1} = 2x^2 - 3x + 1$

The rate of change of the area of the second square with respect to the area of the first square is $2x^2 - 3x + 1$.

Given:

The equations of the two curves are:

Curve 1: $2x = y^2$

Curve 2: $2xy = k$

To Find:

The condition on $k$ such that the two curves intersect orthogonally.

Solution:

Two curves intersect orthogonally at a point of intersection if the product of the slopes of their tangent lines at that point is $-1$.

First, we find the points of intersection of the two curves. From Curve 1, $x = \frac{y^2}{2}$. Substitute this into Curve 2:

$2 \left(\frac{y^2}{2}\right) y = k$

This equation gives the y-coordinates of the intersection points. From this, $y = k^{1/3}$. Substituting this back into $x = \frac{y^2}{2}$, we get $x = \frac{(k^{1/3})^2}{2} = \frac{k^{2/3}}{2}$.

So, if $k \ne 0$, there is one intersection point $\left(\frac{k^{2/3}}{2}, k^{1/3}\right)$.

If $k = 0$, equation (i) becomes $y^3 = 0$, so $y = 0$. Substituting $y=0$ into $2x = y^2$ gives $2x = 0$, so $x = 0$. Thus, if $k=0$, the only intersection point is $(0,0)$.

Next, we find the slopes of the tangent lines to each curve by implicit differentiation with respect to $x$.

For Curve 1: $2x = y^2$

Differentiating with respect to $x$:

$\frac{d}{dx}(2x) = \frac{d}{dx}(y^2)$

$2 = 2y \frac{dy}{dx}$

The slope of the tangent to Curve 1 is $m_1 = \frac{dy}{dx} = \frac{2}{2y} = \frac{1}{y}$ (provided $y \ne 0$).

For Curve 2: $2xy = k$

Differentiating with respect to $x$ using the product rule:

$\frac{d}{dx}(2xy) = \frac{d}{dx}(k)$

$2 \left(1 \cdot y + x \frac{dy}{dx}\right) = 0$

$y + x \frac{dy}{dx} = 0$

$x \frac{dy}{dx} = -y$

The slope of the tangent to Curve 2 is $m_2 = \frac{dy}{dx} = -\frac{y}{x}$ (provided $x \ne 0$).

For orthogonal intersection, the product of the slopes at the point of intersection $(x_0, y_0)$ must be $-1$, i.e., $m_1 m_2 = -1$. This applies when neither tangent is vertical or horizontal, i.e., $x_0 \ne 0$ and $y_0 \ne 0$. This corresponds to the case when $k \ne 0$.

At the intersection point $(x_0, y_0)$: $\left(\frac{1}{y_0}\right) \left(-\frac{y_0}{x_0}\right) = -1$

$-\frac{1}{x_0} = -1$

$\frac{1}{x_0} = 1$

$\implies x_0 = 1$

Now we use the fact that $(x_0, y_0)$ is on both curves and $x_0=1$ to find the condition on $k$.

Substitute $x_0 = 1$ into the equation for Curve 1: $2(1) = y_0^2 \implies y_0^2 = 2 \implies y_0 = \pm \sqrt{2}$.

Substitute $x_0 = 1$ and $y_0$ into the equation for Curve 2: $2(1)y_0 = k \implies k = 2y_0$.

If $y_0 = \sqrt{2}$, then $k = 2\sqrt{2}$. The intersection point is $(1, \sqrt{2})$.

If $y_0 = -\sqrt{2}$, then $k = -2\sqrt{2}$. The intersection point is $(1, -\sqrt{2})$.

In both these cases ($k = \pm 2\sqrt{2}$), the intersection occurs at $x_0 = 1 \ne 0$ and $y_0 = \pm \sqrt{2} \ne 0$, and the condition $m_1 m_2 = -1$ is satisfied. The condition $k^2 = (2\sqrt{2})^2 = (-2\sqrt{2})^2 = 8$ covers these cases where $k \ne 0$.

Now, consider the special case when $k = 0$. The only intersection point is $(0,0)$.

For Curve 1 ($2x = y^2$) at $(0,0)$: Differentiating $2x=y^2$ with respect to $y$, $2 \frac{dx}{dy} = 2y$, so $\frac{dx}{dy} = y$. At $(0,0)$, $\frac{dx}{dy} = 0$. This means the tangent is a vertical line (slope is undefined), which is the line $x=0$ (the y-axis).

For Curve 2 ($2xy = 0$) at $(0,0)$: The equation $2xy=0$ represents the union of two lines: $x=0$ (the y-axis) and $y=0$ (the x-axis). Both of these lines pass through the origin $(0,0)$. The tangents to $2xy=0$ at $(0,0)$ are the lines $x=0$ and $y=0$.

The tangent to Curve 1 at $(0,0)$ is the line $x=0$ (vertical). One of the tangents to Curve 2 at $(0,0)$ is the line $y=0$ (horizontal). Since a vertical line and a horizontal line are orthogonal, the curves intersect orthogonally at $(0,0)$ when $k=0$.

So, the curves intersect orthogonally when $k=0$ or when $k^2=8$. These conditions can be combined into a single equation.

If $k=0$, then $k^2 = 0$. If $k^2=8$, then $k \ne 0$.

The set of values for $k$ is $\{0\} \cup \{\pm 2\sqrt{2}\}$.

Consider the equation $k^2(k^2 - 8) = 0$.

This equation is satisfied if and only if $k^2 = 0$ or $k^2 - 8 = 0$.

$k^2 = 0 \implies k = 0$

$k^2 - 8 = 0 \implies k^2 = 8$

Thus, the condition $k^2(k^2-8)=0$ covers both cases where the curves intersect orthogonally.

The condition for the curves to intersect orthogonally is $k^2(k^2 - 8) = 0$.

Given:

Curve 1: $xy = 4$

Curve 2: $x^2 + y^2 = 8$

To Prove:

The two curves touch each other.

Solution:

Two curves touch each other at a point if they intersect at that point and have the same tangent (and thus the same slope) at that point.

First, find the points of intersection of the two curves.

From Curve 1, $y = \frac{4}{x}$ (assuming $x \ne 0$). Substitute this into Curve 2:

$x^2 + \left(\frac{4}{x}\right)^2 = 8$

$x^2 + \frac{16}{x^2} = 8$

Multiply by $x^2$ (assuming $x \ne 0$):

$x^4 + 16 = 8x^2$

Rearrange into a quadratic equation in $x^2$:

$x^4 - 8x^2 + 16 = 0$

This is a perfect square trinomial: $(x^2)^2 - 2(x^2)(4) + 4^2 = 0$

$(x^2 - 4)^2 = 0$

So, $x^2 - 4 = 0 \implies x^2 = 4 \implies x = \pm 2$.

Now, find the corresponding y-values using $y = \frac{4}{x}$:

If $x = 2$, $y = \frac{4}{2} = 2$. Intersection point is $(2, 2)$.

If $x = -2$, $y = \frac{4}{-2} = -2$. Intersection point is $(-2, -2)$.

The two curves intersect at two distinct points: $(2, 2)$ and $(-2, -2)$.

Next, find the slopes of the tangents to each curve at these intersection points.

For Curve 1: $xy = 4$

Differentiate implicitly with respect to $x$ using the product rule:

$1 \cdot y + x \frac{dy}{dx} = 0$

$x \frac{dy}{dx} = -y$

Slope $m_1 = \frac{dy}{dx} = -\frac{y}{x}$ (for $x \ne 0$).

At $(2, 2)$: $m_1 = -\frac{2}{2} = -1$.

At $(-2, -2)$: $m_1 = -\frac{-2}{-2} = -1$.

For Curve 2: $x^2 + y^2 = 8$

Differentiate implicitly with respect to $x$:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(8)$

$2x + 2y \frac{dy}{dx} = 0$

$2y \frac{dy}{dx} = -2x$

Slope $m_2 = \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$ (for $y \ne 0$).

At $(2, 2)$: $m_2 = -\frac{2}{2} = -1$.

At $(-2, -2)$: $m_2 = -\frac{-2}{-2} = -1$.

At the intersection point $(2, 2)$, the slope of the tangent to Curve 1 is $-1$, and the slope of the tangent to Curve 2 is $-1$. Since the slopes are equal ($m_1 = m_2 = -1$), the curves have the same tangent line at $(2, 2)$.

At the intersection point $(-2, -2)$, the slope of the tangent to Curve 1 is $-1$, and the slope of the tangent to Curve 2 is $-1$. Since the slopes are equal ($m_1 = m_2 = -1$), the curves have the same tangent line at $(-2, -2)$.

Since the curves intersect at these points and have the same tangent slope at each point of intersection, they touch each other at these points.

Hence, it is proved that the curves $xy = 4$ and $x^2 + y^2 = 8$ touch each other.

Given:

The equation of the curve is $\sqrt{x} + \sqrt{y} = 4$.

To Find:

The coordinates of the point(s) on the curve where the tangent is equally inclined to the axes.

Solution:

A line is equally inclined to the coordinate axes if its angle with the positive x-axis is $\theta$ such that $\tan\theta = \pm 1$. The slope of the tangent line is given by $\frac{dy}{dx}$.

So, we need to find the points on the curve where $\frac{dy}{dx} = 1$ or $\frac{dy}{dx} = -1$.

First, differentiate the equation of the curve implicitly with respect to $x$:

$\frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sqrt{y}) = \frac{d}{dx}(4)$

$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$

$\frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}$

$\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \cdot 2\sqrt{y}$

Case 1: The tangent is equally inclined to the axes, and its slope is $1$.

If $\frac{dy}{dx} = 1$, then from equation (i):

$1 = -\sqrt{\frac{y}{x}}$

Squaring both sides gives $1 = \frac{y}{x}$, so $y = x$.

Now, substitute $y = x$ into the equation of the curve $\sqrt{x} + \sqrt{y} = 4$:

$\sqrt{x} + \sqrt{x} = 4$

$2\sqrt{x} = 4$

$\sqrt{x} = 2$

Squaring both sides gives $x = 4$.

Since $y = x$, we have $y = 4$.

The point is $(4, 4)$. Check if this point is on the curve: $\sqrt{4} + \sqrt{4} = 2 + 2 = 4$. The point is on the curve.

Also check the slope at this point: $\frac{dy}{dx} = -\sqrt{\frac{4}{4}} = -\sqrt{1} = -1$.

This means the slope is $-1$, not $1$. So, the tangent is equally inclined, but its slope is $-1$. This point falls under Case 2.

Case 2: The tangent is equally inclined to the axes, and its slope is $-1$.

If $\frac{dy}{dx} = -1$, then from equation (i):

$-1 = -\sqrt{\frac{y}{x}}$

$1 = \sqrt{\frac{y}{x}}$

Squaring both sides gives $1 = \frac{y}{x}$, so $y = x$.

Substitute $y = x$ into the equation of the curve $\sqrt{x} + \sqrt{y} = 4$:

$\sqrt{x} + \sqrt{x} = 4$

$2\sqrt{x} = 4$

$\sqrt{x} = 2$

$x = 4$

Since $y = x$, we have $y = 4$.

The point is $(4, 4)$. The slope at this point is indeed $-1$, as calculated in Case 1.

The point on the curve $\sqrt{x} + \sqrt{y} = 4$ at which the tangent is equally inclined to the axes is $(4, 4)$.

Given:

The equations of the two curves are:

Curve 1: $y = 4 - x^2$

Curve 2: $y = x^2$

To Find:

The angle of intersection of the two curves.

Solution:

First, we find the points of intersection of the two curves by setting their y-values equal:

$4 - x^2 = x^2$

$4 = 2x^2$

$x^2 = 2$

$x = \pm \sqrt{2}$

For $x = \sqrt{2}$, the corresponding y-value from $y = x^2$ is $y = (\sqrt{2})^2 = 2$. Point 1: $(\sqrt{2}, 2)$.

For $x = -\sqrt{2}$, the corresponding y-value from $y = x^2$ is $y = (-\sqrt{2})^2 = 2$. Point 2: $(-\sqrt{2}, 2)$.

The two curves intersect at the points $(\sqrt{2}, 2)$ and $(-\sqrt{2}, 2)$.

Next, we find the slopes of the tangent lines to each curve at the points of intersection by differentiating each equation with respect to $x$.

For Curve 1: $y = 4 - x^2$

$\frac{dy}{dx} = \frac{d}{dx}(4 - x^2) = -2x$

Let $m_1$ be the slope of the tangent to Curve 1. $m_1 = -2x$.

For Curve 2: $y = x^2$

$\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x$

Let $m_2$ be the slope of the tangent to Curve 2. $m_2 = 2x$.

Now, evaluate the slopes at each intersection point:

At the point $(\sqrt{2}, 2)$:

$m_1 = -2(\sqrt{2}) = -2\sqrt{2}$

$m_2 = 2(\sqrt{2}) = 2\sqrt{2}$

At the point $(-\sqrt{2}, 2)$:

$m_1 = -2(-\sqrt{2}) = 2\sqrt{2}$

$m_2 = 2(-\sqrt{2}) = -2\sqrt{2}$

The angle of intersection $\theta$ between two curves at a point of intersection with tangent slopes $m_1$ and $m_2$ is given by the formula:

$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

We can use the slopes calculated at either intersection point, as the pair of slopes $\{m_1, m_2\}$ is the same up to sign swapping, which does not affect the absolute value in the formula.

Let's use the slopes at $(\sqrt{2}, 2)$: $m_1 = -2\sqrt{2}$ and $m_2 = 2\sqrt{2}$.

$m_1 m_2 = (-2\sqrt{2})(2\sqrt{2}) = -4 \times 2 = -8$

$1 + m_1 m_2 = 1 + (-8) = -7$

$m_1 - m_2 = -2\sqrt{2} - 2\sqrt{2} = -4\sqrt{2}$

$\tan\theta = \left|\frac{-4\sqrt{2}}{-7}\right| = \left|\frac{4\sqrt{2}}{7}\right| = \frac{4\sqrt{2}}{7}$

The angle of intersection $\theta$ is $\arctan\left(\frac{4\sqrt{2}}{7}\right)$.

Final Answer:

The angle of intersection of the curves is $\arctan\left(\frac{4\sqrt{2}}{7}\right)$.

Given:

Curve 1: $y^2 = 4x$

Curve 2: $x^2 + y^2 – 6x + 1 = 0$

Point of potential touching: $(1, 2)$.

To Prove:

The curves touch each other at the point $(1, 2)$.

Solution:

Two curves touch each other at a point if they both pass through that point and have the same slope of the tangent line at that point.

Step 1: Verify that the point (1, 2) lies on both curves.

For Curve 1, $y^2 = 4x$:

Substitute $x=1$ and $y=2$ into the equation:

$(2)^2 = 4(1)$

$4 = 4$

The equation holds true. So, the point $(1, 2)$ lies on Curve 1.

For Curve 2, $x^2 + y^2 – 6x + 1 = 0$:

Substitute $x=1$ and $y=2$ into the equation:

$(1)^2 + (2)^2 - 6(1) + 1 = 0$

$1 + 4 - 6 + 1 = 0$

$6 - 6 = 0$

$0 = 0$

The equation holds true. So, the point $(1, 2)$ lies on Curve 2.

Since the point $(1, 2)$ satisfies the equations of both curves, it is a point of intersection.

Step 2: Find the slope of the tangent to Curve 1 at (1, 2).

The equation of Curve 1 is $y^2 = 4x$.

Differentiate both sides with respect to $x$ using implicit differentiation:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(4x)$

$2y \frac{dy}{dx} = 4$

$\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$

Let $m_1$ be the slope of the tangent to Curve 1. Evaluate $m_1$ at the point $(1, 2)$ by substituting $y=2$:

$m_1 = \frac{2}{2} = 1$

Step 3: Find the slope of the tangent to Curve 2 at (1, 2).

The equation of Curve 2 is $x^2 + y^2 – 6x + 1 = 0$.

Differentiate both sides with respect to $x$ using implicit differentiation:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(1) = \frac{d}{dx}(0)$

$2x + 2y \frac{dy}{dx} - 6 + 0 = 0$

$2y \frac{dy}{dx} = 6 - 2x$

$\frac{dy}{dx} = \frac{6 - 2x}{2y} = \frac{2(3 - x)}{2y} = \frac{3 - x}{y}$

Let $m_2$ be the slope of the tangent to Curve 2. Evaluate $m_2$ at the point $(1, 2)$ by substituting $x=1$ and $y=2$:

$m_2 = \frac{3 - 1}{2} = \frac{2}{2} = 1$

Step 4: Compare the slopes.

At the point $(1, 2)$, the slope of the tangent to Curve 1 is $m_1 = 1$, and the slope of the tangent to Curve 2 is $m_2 = 1$.

Since $m_1 = m_2$, the slopes of the tangent lines to both curves are equal at the point $(1, 2)$.

Because the curves intersect at $(1, 2)$ and have the same tangent slope at this point, they touch each other at $(1, 2)$.

Hence, it is proved that the curves $y^2 = 4x$ and $x^2 + y^2 – 6x + 1 = 0$ touch each other at the point $(1, 2)$.

Given:

Equation of the curve: $3x^2 - y^2 = 8$

Equation of the line parallel to the normal lines: $x + 3y = 4$

To Find:

The equations of the normal lines to the curve that are parallel to $x + 3y = 4$.

Solution:

The slope of the given line $x + 3y = 4$ can be found by rewriting it in slope-intercept form $y = mx + c$:

$3y = -x + 4$

$y = -\frac{1}{3}x + \frac{4}{3}$

The slope of the given line is $m_{\text{line}} = -\frac{1}{3}$.

Since the normal lines are parallel to this line, the slope of the normal lines must be equal to the slope of this line.

Slope of the normal lines, $m_{\text{normal}} = -\frac{1}{3}$.

Now, we find the slope of the tangent line to the curve $3x^2 - y^2 = 8$ at any point $(x, y)$ by differentiating implicitly with respect to $x$:

$\frac{d}{dx}(3x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx}(8)$

$6x - 2y \frac{dy}{dx} = 0$

$6x = 2y \frac{dy}{dx}$

The slope of the tangent line at $(x, y)$ is $m_{\text{tangent}} = \frac{dy}{dx} = \frac{6x}{2y} = \frac{3x}{y}$ (provided $y \ne 0$).

The slope of the normal line is the negative reciprocal of the slope of the tangent line (provided the tangent is not horizontal or vertical):

$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{3x/y} = -\frac{y}{3x}$ (provided $x \ne 0$).

We set the slope of the normal line equal to the required slope:

$-\frac{y}{3x} = -\frac{1}{3}$

Multiply both sides by $-3$:

$\frac{y}{x} = 1$

$y = x$

This means that the points on the curve where the normal line has a slope of $-\frac{1}{3}$ must satisfy the condition $y = x$. Now we find these points by substituting $y = x$ into the equation of the curve $3x^2 - y^2 = 8$:

$3x^2 - (x)^2 = 8$

$3x^2 - x^2 = 8$

$2x^2 = 8$

$x^2 = 4$

$x = \pm 2$

Since $y = x$, the corresponding y-coordinates are $y = \pm 2$.

The points on the curve where the normal line is parallel to $x + 3y = 4$ are $(2, 2)$ and $(-2, -2)$.

Now, we write the equation of the normal line for each point using the point-slope form $Y - y_0 = m_{\text{normal}}(X - x_0)$, where $(x_0, y_0)$ is a point on the line and $m_{\text{normal}} = -\frac{1}{3}$.

For the point $(2, 2)$:

$Y - 2 = -\frac{1}{3}(X - 2)$

Multiply by 3:

$3(Y - 2) = -(X - 2)$

$3Y - 6 = -X + 2$

Rearrange to the form $AX + BY + C = 0$:

$X + 3Y - 6 - 2 = 0$

$X + 3Y - 8 = 0$

For the point $(-2, -2)$:

$Y - (-2) = -\frac{1}{3}(X - (-2))$

$Y + 2 = -\frac{1}{3}(X + 2)$

Multiply by 3:

$3(Y + 2) = -(X + 2)$

$3Y + 6 = -X - 2$

Rearrange to the form $AX + BY + C = 0$:

$X + 3Y + 6 + 2 = 0$

$X + 3Y + 8 = 0$

Thus, there are two normal lines to the curve $3x^2 - y^2 = 8$ which are parallel to the line $x + 3y = 4$.

The equations of these normal lines are $X + 3Y - 8 = 0$ and $X + 3Y + 8 = 0$.

Given:

The equation of the curve is $x^2 + y^2 - 2x - 4y + 1 = 0$.

To Find:

The points on the curve where the tangents are parallel to the y-axis.

Solution:

A tangent line is parallel to the y-axis if its slope is undefined. The slope of the tangent is given by $\frac{dy}{dx}$.

For $\frac{dy}{dx}$ to be undefined, the denominator of the derivative must be zero. Let's find $\frac{dy}{dx}$ by implicit differentiation of the curve equation with respect to $x$.

Differentiate the equation $x^2 + y^2 - 2x - 4y + 1 = 0$ with respect to $x$:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2x) - \frac{d}{dx}(4y) + \frac{d}{dx}(1) = \frac{d}{dx}(0)$

$2x + 2y \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} + 0 = 0$

$2y \frac{dy}{dx} - 4 \frac{dy}{dx} = 2 - 2x$

Factor out $\frac{dy}{dx}$ from the terms on the left side:

$\frac{dy}{dx} (2y - 4) = 2 - 2x$

$\frac{dy}{dx} = \frac{2 - 2x}{2y - 4}$

Simplify the expression:

$\frac{dy}{dx} = \frac{2(1 - x)}{2(y - 2)}$

The tangent is parallel to the y-axis when the slope $\frac{dy}{dx}$ is undefined. This occurs when the denominator of the expression for $\frac{dy}{dx}$ is zero, provided the numerator is non-zero at that point (if both are zero, the slope is indeterminate).

Set the denominator equal to zero:

$y - 2 = 0$

$y = 2$

Now, substitute $y = 2$ into the equation of the curve to find the corresponding x-coordinates:

$x^2 + (2)^2 - 2x - 4(2) + 1 = 0$

$x^2 + 4 - 2x - 8 + 1 = 0$

$x^2 - 2x - 3 = 0$

Solve this quadratic equation for $x$. We can factor it or use the quadratic formula.

Factoring: We look for two numbers that multiply to $-3$ and add to $-2$. These numbers are $-3$ and $1$.

$(x - 3)(x + 1) = 0$

This gives two possible values for $x$: $x - 3 = 0 \implies x = 3$ or $x + 1 = 0 \implies x = -1$.

The points on the curve where $y=2$ are $(3, 2)$ and $(-1, 2)$.

Check the numerator of $\frac{dy}{dx}$ at these points:

At $(3, 2)$: Numerator is $1 - x = 1 - 3 = -2 \ne 0$. Denominator is $y - 2 = 2 - 2 = 0$. Slope is $\frac{-2}{0}$, which is undefined. The tangent is vertical.

At $(-1, 2)$: Numerator is $1 - x = 1 - (-1) = 1 + 1 = 2 \ne 0$. Denominator is $y - 2 = 2 - 2 = 0$. Slope is $\frac{2}{0}$, which is undefined. The tangent is vertical.

In both cases, the slope is undefined, confirming that the tangent is parallel to the y-axis.

The points on the curve where the tangents are parallel to the y-axis are $(3, 2)$ and $(-1, 2)$.

Given:

Equation of the line: $\frac{x}{a} + \frac{y}{b} = 1$

Equation of the curve: $y = b e^{\frac{-x}{a}}$

To Show:

The line touches the curve at the point where the curve intersects the y-axis.

Solution:

The axis of y is the line $x = 0$.

Step 1: Find the point where the curve intersects the y-axis.

Substitute $x = 0$ into the equation of the curve $y = b e^{\frac{-x}{a}}$:

$y = b e^{\frac{-0}{a}} = b e^0 = b(1) = b$

The curve intersects the y-axis at the point $(0, b)$.

Step 2: Verify that the point $(0, b)$ lies on the given line.

Substitute $x = 0$ and $y = b$ into the equation of the line $\frac{x}{a} + \frac{y}{b} = 1$:

$\frac{0}{a} + \frac{b}{b} = 1$

$0 + 1 = 1$

$1 = 1$

The equation holds true. So, the point $(0, b)$ lies on the line.

Since the point $(0, b)$ lies on both the curve and the line, it is an intersection point.

Step 3: Find the slope of the tangent to the curve at the point $(0, b)$.

The equation of the curve is $y = b e^{\frac{-x}{a}}$.

Differentiate $y$ with respect to $x$ using the chain rule ($\frac{d}{dx}(e^{f(x)}) = e^{f(x)} f'(x)$):

$\frac{dy}{dx} = \frac{d}{dx}\left(b e^{\frac{-x}{a}}\right)$

$\frac{dy}{dx} = b \cdot e^{\frac{-x}{a}} \cdot \frac{d}{dx}\left(\frac{-x}{a}\right)$

$\frac{dy}{dx} = b e^{\frac{-x}{a}} \cdot \left(-\frac{1}{a}\right)$

Evaluate the slope of the tangent at the point $(0, b)$ by substituting $x = 0$ into equation (i):

Slope of tangent $m_{\text{curve}} = -\frac{b}{a} e^{\frac{-0}{a}} = -\frac{b}{a} e^0 = -\frac{b}{a}(1) = -\frac{b}{a}$.

Step 4: Find the slope of the given line.

The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$. We can rewrite it in slope-intercept form $y = mx + c$ to find the slope.

$\frac{y}{b} = 1 - \frac{x}{a}$

$y = b\left(1 - \frac{x}{a}\right) = b - \frac{b}{a}x$

$y = -\frac{b}{a}x + b$

The slope of the line is $m_{\text{line}} = -\frac{b}{a}$.

Step 5: Compare the slopes.

At the intersection point $(0, b)$, the slope of the tangent to the curve is $m_{\text{curve}} = -\frac{b}{a}$.

The slope of the given line is $m_{\text{line}} = -\frac{b}{a}$.

Since $m_{\text{curve}} = m_{\text{line}} = -\frac{b}{a}$, the tangent to the curve at $(0, b)$ has the same slope as the given line.

Because the line passes through the point $(0, b)$ and has the same slope as the tangent to the curve at that point, the line is the tangent line to the curve at $(0, b)$.

Therefore, the line touches the curve at the point where the curve intersects the y-axis.

Hence, it is shown.

Given:

The function $f(x) = 2x + \cot^{-1}x + \log\left(\sqrt{1+x^2} - x\right)$.

To Show:

$f(x)$ is increasing in R (the set of all real numbers).

Solution:

A function $f(x)$ is increasing on an interval if $f'(x) \ge 0$ for all $x$ in that interval. We need to find the derivative of $f(x)$ and determine its sign.

The domain of $f(x)$:

The term $2x$ is defined for all $x \in \mathbb{R}$.

The term $\cot^{-1}x$ is defined for all $x \in \mathbb{R}$.

The term $\log\left(\sqrt{1+x^2} - x\right)$ requires $\sqrt{1+x^2} - x > 0$.

$\sqrt{1+x^2} > x$

If $x \le 0$, the inequality $\sqrt{1+x^2} > x$ is always true since $\sqrt{1+x^2} \ge \sqrt{1+0} = 1 > x$.

If $x > 0$, we can square both sides:

$(\sqrt{1+x^2})^2 > x^2$

$1+x^2 > x^2$

$1 > 0$

This is always true. So, $\sqrt{1+x^2} - x > 0$ for all $x \in \mathbb{R}$.

Thus, the domain of $f(x)$ is $\mathbb{R}$.

Now, we find the derivative $f'(x)$. We differentiate each term separately:

$\frac{d}{dx}(2x) = 2$

$\frac{d}{dx}(\cot^{-1}x) = -\frac{1}{1 + x^2}$

For the third term, $\log\left(\sqrt{1+x^2} - x\right)$, we use the chain rule. $\frac{d}{dx}(\log(u)) = \frac{1}{u} \frac{du}{dx}$, where $u = \sqrt{1+x^2} - x$.

$\frac{du}{dx} = \frac{d}{dx}(\sqrt{1+x^2} - x) = \frac{d}{dx}((1+x^2)^{1/2}) - \frac{d}{dx}(x)$

Using the chain rule for the first term: $\frac{d}{dx}((1+x^2)^{1/2}) = \frac{1}{2}(1+x^2)^{-1/2} \cdot \frac{d}{dx}(1+x^2) = \frac{1}{2\sqrt{1+x^2}} \cdot (2x) = \frac{x}{\sqrt{1+x^2}}$.

$\frac{d}{dx}(x) = 1$.

So, $\frac{du}{dx} = \frac{x}{\sqrt{1+x^2}} - 1 = \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}}$.

Now, substitute this into the derivative of the log term:

$\frac{d}{dx}\left(\log\left(\sqrt{1+x^2} - x\right)\right) = \frac{1}{\sqrt{1+x^2} - x} \cdot \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}}$

$\frac{d}{dx}\left(\log\left(\sqrt{1+x^2} - x\right)\right) = \frac{-( \sqrt{1+x^2} - x)}{\sqrt{1+x^2} - x} \cdot \frac{1}{\sqrt{1+x^2}}$

Assuming $\sqrt{1+x^2} - x \ne 0$ (which we established is true for all real $x$), we can cancel the term $(\sqrt{1+x^2} - x)$:

$\frac{d}{dx}\left(\log\left(\sqrt{1+x^2} - x\right)\right) = -1 \cdot \frac{1}{\sqrt{1+x^2}} = -\frac{1}{\sqrt{1+x^2}}$

Now, combine the derivatives of all terms to get $f'(x)$:

$f'(x) = 2 + \left(-\frac{1}{1 + x^2}\right) + \left(-\frac{1}{\sqrt{1+x^2}}\right)$

$f'(x) = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{1+x^2}}$

We need to determine if $f'(x) \ge 0$ for all $x \in \mathbb{R}$.

$f'(x) = 2 - \left(\frac{1}{1 + x^2} + \frac{1}{\sqrt{1+x^2}}\right)$

Consider the terms $\frac{1}{1+x^2}$ and $\frac{1}{\sqrt{1+x^2}}$.

For any real number $x$, $x^2 \ge 0$.

$1 + x^2 \ge 1$

$\sqrt{1 + x^2} \ge \sqrt{1} = 1$

So, $1 + x^2 \ge 1$ and $\sqrt{1+x^2} \ge 1$.

This implies:

$0 < \frac{1}{1 + x^2} \le 1$ (Equality holds only if $x=0$)

$0 < \frac{1}{\sqrt{1+x^2}} \le 1$ (Equality holds only if $x=0$)

Let's examine the sum of these two terms:

$\frac{1}{1 + x^2} + \frac{1}{\sqrt{1+x^2}}$

Since $x^2 \ge 0$, we have $1+x^2 \ge \sqrt{1+x^2} \ge 1$.

Therefore, $\frac{1}{1+x^2} \le \frac{1}{\sqrt{1+x^2}}$.

Let's analyze the sum $\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}}$ compared to 2.

Consider the case when $x=0$.

$f'(0) = 2 - \frac{1}{1+0^2} - \frac{1}{\sqrt{1+0^2}} = 2 - \frac{1}{1} - \frac{1}{1} = 2 - 1 - 1 = 0$.

So, $f'(0) = 0$.

Consider the case when $x \ne 0$.

If $x \ne 0$, then $x^2 > 0$, so $1+x^2 > 1$ and $\sqrt{1+x^2} > 1$.

Thus, if $x \ne 0$:

$0 < \frac{1}{1 + x^2} < 1$

$0 < \frac{1}{\sqrt{1+x^2}} < 1$

We need to show that $2 - \left(\frac{1}{1 + x^2} + \frac{1}{\sqrt{1+x^2}}\right) \ge 0$, or equivalently, $\frac{1}{1 + x^2} + \frac{1}{\sqrt{1+x^2}} \le 2$.

Since $x^2 \ge 0$, we know $1+x^2 \ge 1$. Also, since $x^2 \ge 0$, $\sqrt{1+x^2} \ge 1$.

For any real $x$, $1+x^2 \ge 1$ and $\sqrt{1+x^2} \ge 1$.

This means $\frac{1}{1+x^2} \le 1$ and $\frac{1}{\sqrt{1+x^2}} \le 1$.

Therefore, $\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}} \le 1 + 1 = 2$.

The equality $\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}} = 2$ occurs only when $\frac{1}{1+x^2} = 1$ AND $\frac{1}{\sqrt{1+x^2}} = 1$.

$\frac{1}{1+x^2} = 1 \implies 1 = 1+x^2 \implies x^2 = 0 \implies x=0$.

$\frac{1}{\sqrt{1+x^2}} = 1 \implies 1 = \sqrt{1+x^2} \implies 1 = 1+x^2 \implies x^2 = 0 \implies x=0$.

So, $\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}} = 2$ only when $x = 0$.

For all $x \ne 0$, we have strict inequalities $0 < \frac{1}{1 + x^2} < 1$ and $0 < \frac{1}{\sqrt{1+x^2}} < 1$.

Thus, for $x \ne 0$, $\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}} < 1 + 1 = 2$.

So, $\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}} \le 2$ for all $x \in \mathbb{R}$, with equality only at $x=0$.

Therefore, $f'(x) = 2 - \left(\frac{1}{1 + x^2} + \frac{1}{\sqrt{1+x^2}}\right) \ge 2 - 2 = 0$ for all $x \in \mathbb{R}$.

$f'(x) \ge 0$ for all $x \in \mathbb{R}$.

This means that the function $f(x)$ is increasing on $\mathbb{R}$.

Since $f'(x) = 0$ only at a single point ($x=0$) and $f'(x) > 0$ elsewhere, the function is strictly increasing on $\mathbb{R}$ except at $x=0$ where it is neither strictly increasing nor decreasing. Overall, it is considered an increasing function on $\mathbb{R}$.

Hence, it is shown that $f(x)$ is increasing in R.

Given:

The function $f(x) = \sqrt{3} \sin x - \cos x - 2ax + b$, where $a \ge 1$ and $b$ is a constant.

To Show:

$f(x)$ is decreasing in R (the set of all real numbers).

Solution:

A function $f(x)$ is decreasing on an interval if $f'(x) \le 0$ for all $x$ in that interval. We need to find the derivative of $f(x)$ and determine its sign.

Find the derivative $f'(x)$ by differentiating each term with respect to $x$:

$f'(x) = \frac{d}{dx}(\sqrt{3} \sin x) - \frac{d}{dx}(\cos x) - \frac{d}{dx}(2ax) + \frac{d}{dx}(b)$

$f'(x) = \sqrt{3} \frac{d}{dx}(\sin x) - \frac{d}{dx}(\cos x) - 2a \frac{d}{dx}(x) + 0$

$f'(x) = \sqrt{3} (\cos x) - (-\sin x) - 2a (1)$

We need to determine the sign of $f'(x)$. Consider the term $\sqrt{3} \cos x + \sin x$. This expression can be written in the form $R \cos(x - \alpha)$ or $R \sin(x + \alpha)$.

Let's write it in the form $R \cos(x - \alpha) = R (\cos x \cos\alpha + \sin x \sin\alpha)$.

Comparing $\sqrt{3} \cos x + \sin x$ with $R \cos x \cos\alpha + R \sin x \sin\alpha$, we get:

$R \cos\alpha = \sqrt{3}$

$R \sin\alpha = 1$

Square and add the equations: $(R \cos\alpha)^2 + (R \sin\alpha)^2 = (\sqrt{3})^2 + 1^2$

$R^2 (\cos^2\alpha + \sin^2\alpha) = 3 + 1$

$R^2 (1) = 4 \implies R = 2$ (since R is usually taken as positive)

Divide the equations: $\frac{R \sin\alpha}{R \cos\alpha} = \frac{1}{\sqrt{3}} \implies \tan\alpha = \frac{1}{\sqrt{3}}$.

Since $\cos\alpha = \frac{\sqrt{3}}{2} > 0$ and $\sin\alpha = \frac{1}{2} > 0$, $\alpha$ is in the first quadrant. So $\alpha = \frac{\pi}{6}$.

Therefore, $\sqrt{3} \cos x + \sin x = 2 \cos\left(x - \frac{\pi}{6}\right)$.

Substitute this back into the expression for $f'(x)$ (equation i):

$f'(x) = 2 \cos\left(x - \frac{\pi}{6}\right) - 2a$

The range of the cosine function is $[-1, 1]$.

So, $-1 \le \cos\left(x - \frac{\pi}{6}\right) \le 1$ for all $x \in \mathbb{R}$.

Multiply the inequality by 2:

$-2 \le 2 \cos\left(x - \frac{\pi}{6}\right) \le 2$

Now, consider the expression for $f'(x)$: $f'(x) = 2 \cos\left(x - \frac{\pi}{6}\right) - 2a$.

We know that $2 \cos\left(x - \frac{\pi}{6}\right) \le 2$.

So, $f'(x) \le 2 - 2a$.

We are given the condition $a \ge 1$.

Multiply the inequality $a \ge 1$ by $-2$: $-2a \le -2$.

Now, add 2 to both sides:

$2 - 2a \le 2 - 2 = 0$

$2 - 2a \le 0$

Combining the inequalities, we have $f'(x) \le 2 - 2a$ and $2 - 2a \le 0$.

Therefore, $f'(x) \le 0$ for all $x \in \mathbb{R}$, given that $a \ge 1$.

Since $f'(x) \le 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is decreasing in R.

Hence, it is shown.

Given:

The function $f(x) = \tan^{-1}(\sin x + \cos x)$.

To Show:

$f(x)$ is an increasing function in the interval $\left(0, \frac{\pi}{4}\right)$.

Solution:

A function $f(x)$ is increasing on an interval if $f'(x) \ge 0$ for all $x$ in that interval. For strict increasing, we need $f'(x) > 0$. We need to find the derivative of $f(x)$ and determine its sign in the given interval.

Find the derivative $f'(x)$. We use the chain rule for $\tan^{-1}(u)$, where $u = \sin x + \cos x$. The derivative of $\tan^{-1}(u)$ is $\frac{1}{1 + u^2} \frac{du}{dx}$.

First, find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$

Now, substitute $u = \sin x + \cos x$ and $\frac{du}{dx} = \cos x - \sin x$ into the chain rule formula:

$f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)$

$f'(x) = \frac{\cos x - \sin x}{1 + (\sin^2 x + \cos^2 x + 2 \sin x \cos x)}$

Using the identity $\sin^2 x + \cos^2 x = 1$ and $2 \sin x \cos x = \sin(2x)$:

$f'(x) = \frac{\cos x - \sin x}{1 + (1 + \sin(2x))}$

Now, determine the sign of $f'(x)$ for $x \in \left(0, \frac{\pi}{4}\right)$.

Consider the denominator $2 + \sin(2x)$. The range of $\sin(2x)$ is $[-1, 1]$.

So, $2 + \sin(2x)$ will always be in the range $[2 - 1, 2 + 1] = [1, 3]$.

Thus, the denominator $2 + \sin(2x)$ is always positive for all $x \in \mathbb{R}$.

Consider the numerator $\cos x - \sin x$ in the interval $\left(0, \frac{\pi}{4}\right)$.

In the interval $0 < x < \frac{\pi}{4}$ (which is $0^\circ < x < 45^\circ$), the value of $\cos x$ is greater than the value of $\sin x$. For example, at $x = \frac{\pi}{6}$ ($30^\circ$), $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Since $\sqrt{3} > 1$, $\frac{\sqrt{3}}{2} > \frac{1}{2}$.

In fact, for $0 < x < \frac{\pi}{4}$, the graph of $\cos x$ is above the graph of $\sin x$.

Thus, for $x \in \left(0, \frac{\pi}{4}\right)$, $\cos x > \sin x$.

This implies $\cos x - \sin x > 0$ for $x \in \left(0, \frac{\pi}{4}\right)$.

Since the numerator ($\cos x - \sin x$) is positive and the denominator ($2 + \sin(2x)$) is positive in the interval $\left(0, \frac{\pi}{4}\right)$, their ratio $f'(x)$ must be positive.

$f'(x) = \frac{\text{positive}}{\text{positive}} > 0$ for all $x \in \left(0, \frac{\pi}{4}\right)$.

Since $f'(x) > 0$ for all $x \in \left(0, \frac{\pi}{4}\right)$, the function $f(x)$ is strictly increasing in this interval.

Hence, it is shown that $f (x) = \tan^{–1} (\sin x + \cos x)$ is an increasing function in $\left( 0, \frac{π}{4} \right)$.

Given:

The equation of the curve is $y = -x^3 + 3x^2 + 9x - 27$.

To Find:

1. The point on the curve where the slope is maximum.

2. The maximum slope.

Solution:

The slope of the curve at any point $(x, y)$ is given by the first derivative $\frac{dy}{dx}$.

Let $m(x)$ denote the slope of the curve at $x$.

$m(x) = \frac{dy}{dx} = \frac{d}{dx}(-x^3 + 3x^2 + 9x - 27)$

$m(x) = -3x^2 + 6x + 9$

We want to find the value of $x$ for which the slope $m(x)$ is maximum. To find the maximum (or minimum) value of a function, we find its critical points by setting its derivative equal to zero.

Let $m'(x) = \frac{d}{dx}(m(x))$ be the derivative of the slope with respect to $x$.

$m'(x) = \frac{d}{dx}(-3x^2 + 6x + 9)$

$m'(x) = -6x + 6$

Set $m'(x) = 0$ to find the critical points:

$-6x + 6 = 0$

$-6x = -6$

$x = 1$

To determine if this value of $x$ corresponds to a maximum or minimum slope, we can use the second derivative test on $m(x)$. Find the second derivative of the slope, $m''(x)$.

$m''(x) = \frac{d}{dx}(m'(x)) = \frac{d}{dx}(-6x + 6) = -6$

Evaluate $m''(x)$ at $x = 1$. $m''(1) = -6$.

Since $m''(1) = -6 < 0$, the slope $m(x)$ has a local maximum at $x = 1$. Since the second derivative is negative for all $x$, this local maximum is indeed a global maximum.

The x-coordinate of the point where the slope is maximum is $x = 1$. To find the full coordinates of the point on the curve, substitute $x = 1$ back into the equation of the curve $y = -x^3 + 3x^2 + 9x - 27$:

$y = -(1)^3 + 3(1)^2 + 9(1) - 27$

$y = -1 + 3(1) + 9 - 27$

$y = -1 + 3 + 9 - 27$

$y = 2 + 9 - 27$

$y = 11 - 27$

$y = -16$

The point on the curve where the slope is maximum is $(1, -16)$.

Now, calculate the maximum slope by substituting $x = 1$ into the expression for the slope $m(x) = -3x^2 + 6x + 9$:

Maximum slope = $m(1) = -3(1)^2 + 6(1) + 9$

Maximum slope = $-3(1) + 6 + 9$

Maximum slope = $-3 + 6 + 9$

Maximum slope = $3 + 9$

Maximum slope = $12$

Final Answer:

The point on the curve where the slope is maximum is $(1, -16)$.

The maximum slope is $12$.

Given:

The function $f(x) = \sin x + \sqrt{3} \cos x$.

To Prove:

$f(x)$ has a maximum value at $x = \frac{\pi}{6}$.

Solution:

To find the maximum value of a function, we find its critical points by setting the first derivative equal to zero and then using the second derivative test or analyzing the sign of the first derivative.

Find the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(\sin x + \sqrt{3} \cos x)$

$f'(x) = \cos x + \sqrt{3}(-\sin x)$

$f'(x) = \cos x - \sqrt{3} \sin x$

Set $f'(x) = 0$ to find the critical points:

$\cos x - \sqrt{3} \sin x = 0$

$\cos x = \sqrt{3} \sin x$

Assuming $\cos x \ne 0$, divide both sides by $\cos x$:

$1 = \sqrt{3} \frac{\sin x}{\cos x}$

$1 = \sqrt{3} \tan x$

$\tan x = \frac{1}{\sqrt{3}}$

The general solution for $\tan x = \frac{1}{\sqrt{3}}$ is $x = n\pi + \frac{\pi}{6}$, where $n$ is an integer.

Now, we need to determine if these critical points correspond to a maximum value. We can use the second derivative test.

Find the second derivative $f''(x)$:

$f''(x) = \frac{d}{dx}(\cos x - \sqrt{3} \sin x)$

$f''(x) = -\sin x - \sqrt{3} \cos x$

Evaluate $f''(x)$ at the critical points $x = n\pi + \frac{\pi}{6}$.

Let's evaluate at $x = \frac{\pi}{6}$ (which corresponds to $n=0$):

$f''\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) - \sqrt{3} \cos\left(\frac{\pi}{6}\right)$

$f''\left(\frac{\pi}{6}\right) = -\left(\frac{1}{2}\right) - \sqrt{3} \left(\frac{\sqrt{3}}{2}\right)$

$f''\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{3}{2} = -\frac{4}{2} = -2$

Since $f''\left(\frac{\pi}{6}\right) = -2 < 0$, the function $f(x)$ has a local maximum at $x = \frac{\pi}{6}$.

To confirm it's a global maximum (or a repeated maximum), we can express $f(x)$ in the form $R \sin(x + \alpha)$ or $R \cos(x + \alpha)$.

$f(x) = \sin x + \sqrt{3} \cos x$. Compare with $R \sin(x + \alpha) = R(\sin x \cos\alpha + \cos x \sin\alpha)$.

$R \cos\alpha = 1$

$R \sin\alpha = \sqrt{3}$

$R^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4 \implies R = 2$ (taking the positive value).

$\tan\alpha = \frac{\sqrt{3}}{1} = \sqrt{3}$. Since $\cos\alpha = \frac{1}{2} > 0$ and $\sin\alpha = \frac{\sqrt{3}}{2} > 0$, $\alpha$ is in the first quadrant. So $\alpha = \frac{\pi}{3}$.

Thus, $f(x) = 2 \sin\left(x + \frac{\pi}{3}\right)$.

The maximum value of $\sin(\theta)$ is $1$, which occurs when $\theta = \frac{\pi}{2} + 2n\pi$.

So, the maximum value of $f(x)$ is $2 \times 1 = 2$. This maximum occurs when $x + \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi$.

$x = \frac{\pi}{2} - \frac{\pi}{3} + 2n\pi = \frac{3\pi - 2\pi}{6} + 2n\pi = \frac{\pi}{6} + 2n\pi$.

For $n=0$, we get $x = \frac{\pi}{6}$.

The function $f(x)$ is periodic with period $2\pi$. The absolute maximum value of $f(x)$ is $2$, and this occurs at points $x = \frac{\pi}{6} + 2n\pi$ for any integer $n$.

Specifically, at $x = \frac{\pi}{6}$, $f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) + \sqrt{3} \cos\left(\frac{\pi}{6}\right) = \frac{1}{2} + \sqrt{3} \left(\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$.

Since the maximum value of $f(x)$ is 2, and this value is achieved at $x = \frac{\pi}{6}$, $f(x)$ has a maximum value at $x = \frac{\pi}{6}$.

Hence, it is proved.

Given:

In a right-angled triangle, the sum of the length of the hypotenuse and a side is given as a constant.

To Show:

The area of the triangle is maximum when the angle between the hypotenuse and that side is $\frac{\pi}{3}$.

Solution:

Let the right-angled triangle be ABC, with the right angle at B. Let the hypotenuse be AC, and let the sides be AB and BC.

Let $h$ be the length of the hypotenuse AC, and $x$ be the length of the side AB. Let $y$ be the length of the other side BC.

We are given that the sum of the lengths of the hypotenuse and one side is a constant. Let this constant be $S$. There are two possibilities for the given side:

Case 1: The given side is one of the legs (e.g., AB or BC).

Case 2: The given side is adjacent to the angle in question, and it's a leg.

The problem states "the angle between them", referring to the angle between the hypotenuse and the given side. This implies the given side is a leg adjacent to one of the non-right angles. Let the given side be AB, and the angle between the hypotenuse AC and the side AB is $\theta$ (i.e., $\angle BAC = \theta$).

In this case, the given constant sum is $S = h + x$.

In the right-angled triangle ABC:

$x = h \cos\theta$

$y = h \sin\theta$

The given condition is $h + x = S$, where $S$ is a constant.

Substitute $x = h \cos\theta$ into the condition:

$h + h \cos\theta = S$

$h(1 + \cos\theta) = S$

The area of the triangle is $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} xy$.

Substitute $x = h \cos\theta$ and $y = h \sin\theta$ into the area formula:

$A = \frac{1}{2} (h \cos\theta) (h \sin\theta) = \frac{1}{2} h^2 \sin\theta \cos\theta$

Using the identity $\sin(2\theta) = 2 \sin\theta \cos\theta$, we have $\sin\theta \cos\theta = \frac{1}{2} \sin(2\theta)$.

$A = \frac{1}{2} h^2 \left(\frac{1}{2} \sin(2\theta)\right) = \frac{1}{4} h^2 \sin(2\theta)$

Substitute the expression for $h$ from equation (i) into the area formula:

$A(\theta) = \frac{1}{4} \left(\frac{S}{1 + \cos\theta}\right)^2 \sin(2\theta)$

$A(\theta) = \frac{1}{4} \frac{S^2}{(1 + \cos\theta)^2} (2 \sin\theta \cos\theta)$

$A(\theta) = \frac{S^2}{2} \frac{\sin\theta \cos\theta}{(1 + \cos\theta)^2}$

Using half-angle identities: $\sin\theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$ and $1 + \cos\theta = 2 \cos^2(\frac{\theta}{2})$.

$A(\theta) = \frac{S^2}{2} \frac{(2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})) (\cos\theta)}{(2 \cos^2(\frac{\theta}{2}))^2}$

$A(\theta) = \frac{S^2}{2} \frac{2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2}) \cos\theta}{4 \cos^4(\frac{\theta}{2})}$

$A(\theta) = \frac{S^2}{4} \frac{\sin(\frac{\theta}{2}) \cos\theta}{\cos^3(\frac{\theta}{2})}$

This expression looks complicated to differentiate. Let's use an alternative approach by expressing the area in terms of $x$ and $S$.

We have $h = S - x$.

In the right triangle, $h^2 = x^2 + y^2$.

$(S - x)^2 = x^2 + y^2$

$S^2 - 2Sx + x^2 = x^2 + y^2$

$y^2 = S^2 - 2Sx$

$y = \sqrt{S^2 - 2Sx} = \sqrt{S(S - 2x)}$

For $y$ to be real, $S(S - 2x) \ge 0$. Since $S = h+x > 0$ (lengths are positive), we must have $S - 2x \ge 0$, which means $x \le \frac{S}{2}$. Also, $x$ must be a side length, so $x > 0$. From $x = h \cos\theta$, we have $x < h$, so $x < S-x$, which means $2x < S$, or $x < S/2$. So the range for $x$ is $0 < x < \frac{S}{2}$.

The area is $A = \frac{1}{2} xy = \frac{1}{2} x \sqrt{S^2 - 2Sx} = \frac{1}{2} x \sqrt{S(S - 2x)}$.

To maximize $A$, we can maximize $A^2$, which is often simpler as it removes the square root.

$A^2 = \left(\frac{1}{2} x \sqrt{S(S - 2x)}\right)^2 = \frac{1}{4} x^2 S(S - 2x) = \frac{S}{4} (Sx^2 - 2x^3)$.

Let $f(x) = Sx^2 - 2x^3$. We want to maximize $f(x)$ for $0 < x < \frac{S}{2}$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(Sx^2 - 2x^3) = 2Sx - 6x^2$

Set $f'(x) = 0$ to find critical points:

$2Sx - 6x^2 = 0$

$2x(S - 3x) = 0$

This gives $x = 0$ or $S - 3x = 0$.

$S - 3x = 0 \implies 3x = S \implies x = \frac{S}{3}$.

The critical point in the interval $0 < x < \frac{S}{2}$ is $x = \frac{S}{3}$. Note that $0 < \frac{S}{3} < \frac{S}{2}$ since $\frac{1}{3} < \frac{1}{2}$.

Use the second derivative test to check if $x = \frac{S}{3}$ corresponds to a maximum.

Find the second derivative of $f(x)$:

$f''(x) = \frac{d}{dx}(2Sx - 6x^2) = 2S - 12x$

Evaluate $f''(x)$ at $x = \frac{S}{3}$:

$f''\left(\frac{S}{3}\right) = 2S - 12\left(\frac{S}{3}\right) = 2S - 4S = -2S$

Since $S > 0$, $f''\left(\frac{S}{3}\right) = -2S < 0$. This indicates a local maximum at $x = \frac{S}{3}$. Since it's the only critical point in the interval, it's a global maximum.

The area is maximum when $x = \frac{S}{3}$.

Now, we relate this value of $x$ back to the angle $\theta$.

We have $x = h \cos\theta$ and $h = S - x$.

Substitute $x = \frac{S}{3}$ into $h = S - x$:

$h = S - \frac{S}{3} = \frac{3S - S}{3} = \frac{2S}{3}$

Now use $x = h \cos\theta$:

$\frac{S}{3} = \left(\frac{2S}{3}\right) \cos\theta$

Assuming $S \ne 0$ (a triangle with zero sum of sides is not interesting), divide both sides by $\frac{2S}{3}$:

$\frac{S/3}{2S/3} = \cos\theta$

$\frac{1}{2} = \cos\theta$

The angle $\theta$ in a right-angled triangle must be between $0$ and $\frac{\pi}{2}$ (exclusive of the right angle). The angle $\theta$ between the hypotenuse and a leg must satisfy $0 < \theta < \frac{\pi}{2}$.

The value of $\theta$ in $(0, \frac{\pi}{2})$ for which $\cos\theta = \frac{1}{2}$ is $\theta = \frac{\pi}{3}$.

Thus, the area of the triangle is maximum when the angle between the hypotenuse and the side is $\frac{\pi}{3}$.

Hence, it is shown.

Given:

The function is $f(x) = x^5 - 5x^4 + 5x^3 - 1$.

To Find:

Points of local maxima, local minima, and points of inflection, and the corresponding local maximum and minimum values.

Solution:

To find local maxima and minima, we first find the critical points by setting the first derivative $f'(x)$ equal to zero.

Find the first derivative $f'(x)$:

$f'(x) = \frac{d}{dx}(x^5 - 5x^4 + 5x^3 - 1)$

$f'(x) = 5x^{5-1} - 5 \cdot 4x^{4-1} + 5 \cdot 3x^{3-1} - 0$

Set $f'(x) = 0$ to find the critical numbers:

$5x^4 - 20x^3 + 15x^2 = 0$

Factor out the common term $5x^2$:

$5x^2 (x^2 - 4x + 3) = 0$

Factor the quadratic expression $x^2 - 4x + 3$:

$5x^2 (x - 1)(x - 3) = 0$

The critical numbers are the values of $x$ for which $f'(x) = 0$: $x = 0$, $x - 1 = 0 \implies x = 1$, and $x - 3 = 0 \implies x = 3$.

The critical numbers are $0, 1, 3$.

To classify these critical points (as local maximum, minimum, or neither), we use the second derivative test. Find the second derivative $f''(x)$:

$f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}(5x^4 - 20x^3 + 15x^2)$

$f''(x) = 5 \cdot 4x^{4-1} - 20 \cdot 3x^{3-1} + 15 \cdot 2x^{2-1}$

Evaluate $f''(x)$ at each critical number:

• At $x = 0$:

$f''(0) = 20(0)^3 - 60(0)^2 + 30(0) = 0 - 0 + 0 = 0$.

Since $f''(0) = 0$, the second derivative test is inconclusive at $x=0$. We can use the first derivative test or analyze the sign of $f'(x)$ around $x=0$. $f'(x) = 5x^2(x-1)(x-3)$. For $x$ near $0$, $5x^2$ is positive (unless $x=0$), $(x-1)$ is negative, and $(x-3)$ is negative. So $f'(x)$ is positive for $x<0$ (but near 0) and positive for $x>0$ (but near 0). Since $f'(x)$ does not change sign at $x=0$, there is neither a local maximum nor a local minimum at $x=0$.

---

• At $x = 1$:

$f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10$.

Since $f''(1) = -10 < 0$, there is a local maximum at $x = 1$.

---

• At $x = 3$:

$f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 20(27) - 60(9) + 90 = 540 - 540 + 90 = 90$.

Since $f''(3) = 90 > 0$, there is a local minimum at $x = 3$.

Now, find the corresponding local maximum and local minimum values by evaluating $f(x)$ at $x=1$ and $x=3$.

Local maximum value at $x = 1$:

$f(1) = (1)^5 - 5(1)^4 + 5(1)^3 - 1 = 1 - 5(1) + 5(1) - 1 = 1 - 5 + 5 - 1 = 0$.

The local maximum point is $(1, 0)$. The local maximum value is $0$.

Local minimum value at $x = 3$:

$f(3) = (3)^5 - 5(3)^4 + 5(3)^3 - 1 = 243 - 5(81) + 5(27) - 1 = 243 - 405 + 135 - 1 = 378 - 406 = -28$.

The local minimum point is $(3, -28)$. The local minimum value is $-28$.

To find the points of inflection, we set the second derivative $f''(x)$ equal to zero and check for a sign change in $f''(x)$ around these points.

Set $f''(x) = 0$:

$20x^3 - 60x^2 + 30x = 0$

Factor out the common term $10x$:

$10x (2x^2 - 6x + 3) = 0$

This gives $x = 0$ or $2x^2 - 6x + 3 = 0$.

Use the quadratic formula to solve $2x^2 - 6x + 3 = 0$:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(3)}}{2(2)}$

$x = \frac{6 \pm \sqrt{36 - 24}}{4}$

$x = \frac{6 \pm \sqrt{12}}{4}$

$x = \frac{6 \pm 2\sqrt{3}}{4}$

$x = \frac{3 \pm \sqrt{3}}{2}$

The potential points of inflection are $x = 0$, $x = \frac{3 - \sqrt{3}}{2}$, and $x = \frac{3 + \sqrt{3}}{2}$.

We need to check the sign of $f''(x) = 20x(x - \frac{3 - \sqrt{3}}{2})(x - \frac{3 + \sqrt{3}}{2})$ around these values. Let $p_1 = \frac{3 - \sqrt{3}}{2}$ and $p_2 = \frac{3 + \sqrt{3}}{2}$. Approximately, $p_1 \approx 0.634$ and $p_2 \approx 2.366$.

The potential inflection points are $0, p_1, p_2$. The factors of $f''(x)$ are $x$, $(x-p_1)$, and $(x-p_2)$.

Since the concavity changes at $x=0$, $x=p_1$, and $x=p_2$, all three are points of inflection.

Find the y-coordinates of the points of inflection.

• At $x = 0$:

$f(0) = (0)^5 - 5(0)^4 + 5(0)^3 - 1 = -1$.

Point of inflection is $(0, -1)$.

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• At $x = \frac{3 - \sqrt{3}}{2}$:

Let $x_1 = \frac{3 - \sqrt{3}}{2}$. This value is a root of $2x^2 - 6x + 3 = 0$, so $2x_1^2 - 6x_1 + 3 = 0$. This implies $2x_1^2 = 6x_1 - 3$, or $x_1^2 = 3x_1 - \frac{3}{2}$.

We found earlier by polynomial division that $f(x) = -\frac{39}{4}x + \frac{23}{4}$ at the roots of $2x^2 - 6x + 3 = 0$.

So, $f\left(\frac{3 - \sqrt{3}}{2}\right) = -\frac{39}{4}\left(\frac{3 - \sqrt{3}}{2}\right) + \frac{23}{4} = \frac{-39(3 - \sqrt{3}) + 2(23)}{8} = \frac{-117 + 39\sqrt{3} + 46}{8} = \frac{-71 + 39\sqrt{3}}{8}$.

Point of inflection is $\left(\frac{3 - \sqrt{3}}{2}, \frac{-71 + 39\sqrt{3}}{8}\right)$.

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• At $x = \frac{3 + \sqrt{3}}{2}$:

Let $x_2 = \frac{3 + \sqrt{3}}{2}$. This value is also a root of $2x^2 - 6x + 3 = 0$.

$f\left(\frac{3 + \sqrt{3}}{2}\right) = -\frac{39}{4}\left(\frac{3 + \sqrt{3}}{2}\right) + \frac{23}{4} = \frac{-39(3 + \sqrt{3}) + 2(23)}{8} = \frac{-117 - 39\sqrt{3} + 46}{8} = \frac{-71 - 39\sqrt{3}}{8}$.

Point of inflection is $\left(\frac{3 + \sqrt{3}}{2}, \frac{-71 - 39\sqrt{3}}{8}\right)$.

Summary:

• Local maximum at point $(1, 0)$ with local maximum value $0$.
• Local minimum at point $(3, -28)$ with local minimum value $-28$.
• Points of inflection are $(0, -1)$, $\left(\frac{3 - \sqrt{3}}{2}, \frac{-71 + 39\sqrt{3}}{8}\right)$, and $\left(\frac{3 + \sqrt{3}}{2}, \frac{-71 - 39\sqrt{3}}{8}\right)$.

Given:

Initial number of subscribers = 500

Initial annual charge per subscriber = $\textsf{₹}300$

For every $\textsf{₹}1$ increase in charge, 1 subscriber discontinues the service.

To Find:

The increase in annual subscription that will bring maximum profit.

Solution:

Let $x$ be the increase in the annual subscription fee in Rupees.

If the increase is $x$ Rupees, the new annual charge per subscriber will be $(300 + x)$ Rupees.

For every $\textsf{₹}1$ increase, the number of subscribers decreases by 1. So, for an increase of $x$ Rupees, the number of subscribers will decrease by $x$.

The new number of subscribers will be $(500 - x)$.

The total annual profit (or revenue, assuming no costs or fixed costs are included in this model, which is typical for this type of problem) is the product of the number of subscribers and the charge per subscriber.

Let $P(x)$ be the total annual profit as a function of the increase $x$.

$P(x) = (\text{Number of subscribers}) \times (\text{Charge per subscriber})$

$P(x) = (500 - x)(300 + x)$

Expand the expression for $P(x)$:

$P(x) = 500 \times 300 + 500x - 300x - x^2$

$P(x) = 150000 + 200x - x^2$

We want to find the value of $x$ that maximizes the profit $P(x)$. To find the maximum, we find the critical points by setting the first derivative $P'(x)$ equal to zero.

Find the first derivative $P'(x)$:

$P'(x) = \frac{d}{dx}(150000 + 200x - x^2)$

$P'(x) = 0 + 200(1) - 2x$

$P'(x) = 200 - 2x$

Set $P'(x) = 0$ to find the critical number(s):

$200 - 2x = 0$

$200 = 2x$

$x = 100$

To determine if this value of $x$ corresponds to a maximum, we use the second derivative test. Find the second derivative $P''(x)$:

$P''(x) = \frac{d}{dx}(200 - 2x) = -2$

Evaluate $P''(x)$ at $x = 100$. $P''(100) = -2$.

Since $P''(100) = -2 < 0$, the profit function $P(x)$ has a local maximum at $x = 100$. Since the second derivative is constant and negative, this local maximum is indeed a global maximum.

The increase in annual subscription that maximizes the profit is $x = 100$ Rupees.

We should also consider the practical domain for $x$. The number of subscribers $500 - x$ must be non-negative, so $500 - x \ge 0 \implies x \le 500$. The increase $x$ is also typically non-negative, $x \ge 0$. So the relevant interval is $0 \le x \le 500$. Our critical point $x=100$ is within this interval. The endpoints $x=0$ and $x=500$ give profits $P(0) = 150000$ and $P(500) = (500-500)(300+500) = 0$. The profit at $x=100$ is $P(100) = (500-100)(300+100) = (400)(400) = 160000$. Since $160000 > 150000$ and $160000 > 0$, the maximum profit occurs at $x=100$.

The increase in annual subscription that will bring maximum profit is $\textsf{₹}100$.

Given:

Equation of the line: $x \cos \alpha + y \sin \alpha = p$

Equation of the curve: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (This is an ellipse)

The line touches the curve.

To Prove:

$a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2$

Solution:

If the line touches the curve, it is the tangent line to the curve at the point of contact. The slope of the tangent to the curve at the point of contact must be equal to the slope of the given line.

First, find the slope of the given line $x \cos \alpha + y \sin \alpha = p$. Rewrite the equation in slope-intercept form $y = mx + c$ (assuming $\sin \alpha \ne 0$):

$y \sin \alpha = -x \cos \alpha + p$

$y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{p}{\sin \alpha}$

$y = -(\cot \alpha) x + \frac{p}{\sin \alpha}$

The slope of the line is $m_{\text{line}} = -\cot \alpha = -\frac{\cos \alpha}{\sin \alpha}$ (for $\sin \alpha \ne 0$).

Next, find the slope of the tangent to the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ by implicit differentiation with respect to $x$:

$\frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = \frac{d}{dx}(1)$

$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$

$\frac{2y}{b^2} \frac{dy}{dx} = -\frac{2x}{a^2}$

$\frac{dy}{dx} = -\frac{2x}{a^2} \cdot \frac{b^2}{2y} = -\frac{b^2 x}{a^2 y}$ (provided $y \ne 0$).

The slope of the tangent to the curve at a point $(x, y)$ is $m_{\text{curve}} = -\frac{b^2 x}{a^2 y}$.

If the line touches the curve at a point $(x_0, y_0)$, the slope of the tangent at this point must equal the slope of the line:

$m_{\text{curve}} |_{(x_0, y_0)} = m_{\text{line}}$

$-\frac{b^2 x_0}{a^2 y_0} = -\frac{\cos \alpha}{\sin \alpha}$

$\frac{b^2 x_0}{a^2 y_0} = \frac{\cos \alpha}{\sin \alpha}$

Since the point $(x_0, y_0)$ is the point of contact, it lies on both the curve and the line.

Substitute $(x_0, y_0)$ into the equation of the line:

$x_0 \cos \alpha + y_0 \sin \alpha = p$

Substitute $(x_0, y_0)$ into the equation of the curve:

From equation (i), we can express the ratio $\frac{x_0}{y_0}$. We can use this to relate $x_0$ and $y_0$ to $\alpha$.

From (i), $\frac{x_0}{a^2 \cos \alpha} = \frac{y_0}{b^2 \sin \alpha}$. Let this ratio be $\lambda$.

$\frac{x_0}{a^2 \cos \alpha} = \lambda \implies x_0 = \lambda a^2 \cos \alpha$

$\frac{y_0}{b^2 \sin \alpha} = \lambda \implies y_0 = \lambda b^2 \sin \alpha$

Substitute these expressions for $x_0$ and $y_0$ into equation (ii):

$(\lambda a^2 \cos \alpha) \cos \alpha + (\lambda b^2 \sin \alpha) \sin \alpha = p$

$\lambda a^2 \cos^2 \alpha + \lambda b^2 \sin^2 \alpha = p$

$\lambda (a^2 \cos^2 \alpha + b^2 \sin^2 \alpha) = p$

Substitute the expressions for $x_0$ and $y_0$ into equation (iii):

$\frac{(\lambda a^2 \cos \alpha)^2}{a^2} + \frac{(\lambda b^2 \sin \alpha)^2}{b^2} = 1$

$\frac{\lambda^2 a^4 \cos^2 \alpha}{a^2} + \frac{\lambda^2 b^4 \sin^2 \alpha}{b^2} = 1$

$\lambda^2 a^2 \cos^2 \alpha + \lambda^2 b^2 \sin^2 \alpha = 1$

$\lambda^2 (a^2 \cos^2 \alpha + b^2 \sin^2 \alpha) = 1$

Now we have two expressions for $\lambda$. From equation (iv), $\lambda = \frac{p}{a^2 \cos^2 \alpha + b^2 \sin^2 \alpha}$. From equation (v), $\lambda^2 = \frac{1}{a^2 \cos^2 \alpha + b^2 \sin^2 \alpha}$.

Square the expression for $\lambda$ from equation (iv):

$\lambda^2 = \left(\frac{p}{a^2 \cos^2 \alpha + b^2 \sin^2 \alpha}\right)^2 = \frac{p^2}{(a^2 \cos^2 \alpha + b^2 \sin^2 \alpha)^2}$

Equate the two expressions for $\lambda^2$ (from the squared version of (iv) and from (v)):

$\frac{p^2}{(a^2 \cos^2 \alpha + b^2 \sin^2 \alpha)^2} = \frac{1}{a^2 \cos^2 \alpha + b^2 \sin^2 \alpha}$

Assume $a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \ne 0$ (otherwise the line equation $x \cos \alpha + y \sin \alpha = p$ would imply $p=0$ and the curve would pass through $(0,0)$, which is not the case for an ellipse unless $a=b=0$). Multiply both sides by $(a^2 \cos^2 \alpha + b^2 \sin^2 \alpha)^2$:

$p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$

This is the required condition. We should also consider the cases where $\sin \alpha = 0$ or $\cos \alpha = 0$.

If $\sin \alpha = 0$, then $\alpha = n\pi$ for some integer $n$. $\cos \alpha = \pm 1$.

The line is $x(\pm 1) + y(0) = p$, so $\pm x = p$, i.e., $x = \pm p$. These are vertical lines. For a vertical line to touch the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, it must be the tangent at the endpoints of the major or minor axis if it's horizontal/vertical. The vertical tangents are at $x = \pm a$. So, $p = \pm a$, which means $p^2 = a^2$. The condition $a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2$ becomes $a^2 (\pm 1)^2 + b^2 (0)^2 = p^2 \implies a^2 = p^2$, which holds. This case is consistent.

If $\cos \alpha = 0$, then $\alpha = \frac{\pi}{2} + n\pi$ for some integer $n$. $\sin \alpha = \pm 1$.

The line is $x(0) + y(\pm 1) = p$, so $\pm y = p$, i.e., $y = \pm p$. These are horizontal lines. The horizontal tangents to the ellipse are at $y = \pm b$. So, $p = \pm b$, which means $p^2 = b^2$. The condition $a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2$ becomes $a^2 (0)^2 + b^2 (\pm 1)^2 = p^2 \implies b^2 = p^2$, which holds. This case is also consistent.

The derived condition $a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2$ holds in all cases where the line $x \cos \alpha + y \sin \alpha = p$ touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Hence, it is proved.

Given:

An open box with a square base is to be made.

The total surface area of the cardboard used is $c^2$.

To Show:

The maximum volume of the box is $\frac{c^3}{6\sqrt{3}}$ cubic units.

Solution:

Let $x$ be the length of the side of the square base of the box.

Let $h$ be the height of the box.

Since the box is open, it has a square base and four rectangular sides.

Area of the base = $x^2$.

Area of each side = $x \cdot h$.

Total surface area of the open box = Area of base + Area of 4 sides

Given that the total surface area is $c^2$, we have:

$x^2 + 4xh = c^2$

We need to express the height $h$ in terms of $x$ and $c^2$ from this equation:

$4xh = c^2 - x^2$

Assuming $x \ne 0$ (otherwise, there is no box), we have:

For the height to be positive, $c^2 - x^2 > 0$, which means $x^2 < c^2$. Since $x$ is a length, $x > 0$. So, $0 < x < c$.

The volume of the box is $V = (\text{Area of base}) \times \text{height} = x^2 h$.

Substitute the expression for $h$ from equation (i) into the volume formula:

$V(x) = x^2 \left(\frac{c^2 - x^2}{4x}\right)$

Assuming $x \ne 0$, we can simplify:

$V(x) = \frac{x(c^2 - x^2)}{4} = \frac{c^2 x - x^3}{4}$

We want to find the value of $x$ that maximizes the volume $V(x)$ in the interval $(0, c)$. To find the maximum, we find the critical points by setting the first derivative $V'(x)$ equal to zero.

Find the first derivative $V'(x)$ with respect to $x$:

$V'(x) = \frac{d}{dx}\left(\frac{c^2 x - x^3}{4}\right) = \frac{1}{4} \frac{d}{dx}(c^2 x - x^3)$

$V'(x) = \frac{1}{4} (c^2 - 3x^2)$

Set $V'(x) = 0$ to find the critical number(s):

$\frac{1}{4} (c^2 - 3x^2) = 0$

$c^2 - 3x^2 = 0$

$3x^2 = c^2$

$x^2 = \frac{c^2}{3}$

Since $x$ is a length, $x > 0$. So, $x = \sqrt{\frac{c^2}{3}} = \frac{c}{\sqrt{3}}$.

Check if this critical point is in the interval $(0, c)$. Since $\frac{1}{\sqrt{3}} < 1$, $0 < \frac{c}{\sqrt{3}} < c$. So the critical point is valid.

To determine if this value of $x$ corresponds to a maximum, we can use the second derivative test. Find the second derivative $V''(x)$:

$V''(x) = \frac{d}{dx}\left(\frac{1}{4} (c^2 - 3x^2)\right) = \frac{1}{4} (0 - 3 \cdot 2x) = -\frac{6x}{4} = -\frac{3x}{2}$

Evaluate $V''(x)$ at $x = \frac{c}{\sqrt{3}}$:

$V''\left(\frac{c}{\sqrt{3}}\right) = -\frac{3}{2} \left(\frac{c}{\sqrt{3}}\right) = -\frac{3c}{2\sqrt{3}} = -\frac{\sqrt{3}c}{2}$

Since $c$ represents an area, $c > 0$. Thus, $V''\left(\frac{c}{\sqrt{3}}\right) = -\frac{\sqrt{3}c}{2} < 0$.

By the second derivative test, the volume $V(x)$ has a local maximum at $x = \frac{c}{\sqrt{3}}$. Since it's the only critical point in the domain $(0, c)$, it is the global maximum.

Now, find the maximum volume by substituting $x = \frac{c}{\sqrt{3}}$ into the volume formula $V(x) = \frac{c^2 x - x^3}{4}$:

Maximum Volume $= V\left(\frac{c}{\sqrt{3}}\right) = \frac{1}{4} \left(c^2 \left(\frac{c}{\sqrt{3}}\right) - \left(\frac{c}{\sqrt{3}}\right)^3\right)$

Maximum Volume $= \frac{1}{4} \left(\frac{c^3}{\sqrt{3}} - \frac{c^3}{(\sqrt{3})^3}\right)$

Maximum Volume $= \frac{1}{4} \left(\frac{c^3}{\sqrt{3}} - \frac{c^3}{3\sqrt{3}}\right)$

Combine the terms inside the parenthesis by finding a common denominator:

Maximum Volume $= \frac{1}{4} \left(\frac{3c^3 - c^3}{3\sqrt{3}}\right)$

Maximum Volume $= \frac{1}{4} \left(\frac{2c^3}{3\sqrt{3}}\right)$

Maximum Volume $= \frac{2c^3}{12\sqrt{3}} = \frac{c^3}{6\sqrt{3}}$

The maximum volume of the box is $\frac{c^3}{6\sqrt{3}}$ cubic units.

Hence, it is shown.

Given:

Perimeter of the rectangle = 36 cm.

To Find:

1. The dimensions of the rectangle that maximize the volume swept out when revolved about one of its sides.

2. The maximum volume.

Solution:

Let the dimensions of the rectangle be length $l$ and width $w$.

The perimeter of the rectangle is $2(l + w) = 36$.

$l + w = 18$

So, $w = 18 - l$.

Since $l$ and $w$ are lengths, they must be positive. $l > 0$ and $w > 0$.

$18 - l > 0 \implies l < 18$.

Thus, the possible range for the length $l$ is $0 < l < 18$. The width $w$ will also be in the range $0 < w < 18$.

When the rectangle is revolved about one of its sides, it sweeps out a cylinder. There are two possibilities for the axis of revolution:

Case 1: Revolved about the side with length $l$.

In this case, the height of the cylinder is $l$, and the radius of the base is $w$.

The volume of the cylinder is $V_1 = \pi (\text{radius})^2 (\text{height}) = \pi w^2 l$.

Substitute $w = 18 - l$ into the volume formula:

$V_1(l) = \pi (18 - l)^2 l = \pi (324 - 36l + l^2) l = \pi (324l - 36l^2 + l^3)$

Case 2: Revolved about the side with length $w$.

In this case, the height of the cylinder is $w$, and the radius of the base is $l$.

The volume of the cylinder is $V_2 = \pi (\text{radius})^2 (\text{height}) = \pi l^2 w$.

Substitute $w = 18 - l$ into the volume formula:

$V_2(l) = \pi l^2 (18 - l) = \pi (18l^2 - l^3)$

We need to find the dimensions that maximize the volume swept out, which means finding the maximum value of $V_1(l)$ and $V_2(l)$ and comparing them.

Let's analyze $V_1(l) = \pi (l^3 - 36l^2 + 324l)$ for $0 < l < 18$.

Find the first derivative $V_1'(l)$:

$V_1'(l) = \pi (3l^2 - 72l + 324)$

Set $V_1'(l) = 0$ to find critical points:

$3l^2 - 72l + 324 = 0$

Divide by 3:

$l^2 - 24l + 108 = 0$

Factor the quadratic equation: We look for two numbers that multiply to 108 and add to -24. These are -6 and -18.

$(l - 6)(l - 18) = 0$

This gives critical points $l = 6$ and $l = 18$.

The critical point $l = 18$ is at the boundary of the interval $(0, 18)$, where $w=0$, which is not a rectangle. The critical point in the interval is $l = 6$.

Use the second derivative test for $V_1(l)$. Find $V_1''(l)$:

$V_1''(l) = \pi (6l - 72)$

Evaluate $V_1''(6)$: $V_1''(6) = \pi (6(6) - 72) = \pi (36 - 72) = -36\pi$.

Since $V_1''(6) = -36\pi < 0$, there is a local maximum at $l = 6$.

When $l = 6$, $w = 18 - 6 = 12$. Dimensions are $6 \times 12$.

The volume swept out is $V_1(6) = \pi (6)^3 - 36(6)^2 + 324(6) = \pi(216 - 36(36) + 1944) = \pi(216 - 1296 + 1944) = \pi(864)$. This calculation is wrong. $V_1(6) = \pi (18-6)^2 (6) = \pi (12)^2 (6) = \pi (144)(6) = 864\pi$.

Now analyze $V_2(l) = \pi (18l^2 - l^3)$ for $0 < l < 18$.

Find the first derivative $V_2'(l)$:

$V_2'(l) = \pi (36l - 3l^2)$

Set $V_2'(l) = 0$ to find critical points:

$36l - 3l^2 = 0$

$3l(12 - l) = 0$

This gives critical points $l = 0$ and $l = 12$.

The critical point $l = 0$ is at the boundary, where $w=18$ but $l=0$, not a rectangle. The critical point in the interval is $l = 12$.

Use the second derivative test for $V_2(l)$. Find $V_2''(l)$:

$V_2''(l) = \pi (36 - 6l)$

Evaluate $V_2''(12)$: $V_2''(12) = \pi (36 - 6(12)) = \pi (36 - 72) = -36\pi$.

Since $V_2''(12) = -36\pi < 0$, there is a local maximum at $l = 12$.

When $l = 12$, $w = 18 - 12 = 6$. Dimensions are $12 \times 6$.

The volume swept out is $V_2(12) = \pi (12)^2 (18 - 12) = \pi (144)(6) = 864\pi$.

Comparing the maximum volumes from the two cases:

Case 1 (revolved about side of length $l=6$, width $w=12$): Volume $V_1 = 864\pi$. The dimensions are $6 \times 12$. Revolved about the side of length 6.

Case 2 (revolved about side of length $w=6$, length $l=12$): Volume $V_2 = 864\pi$. The dimensions are $12 \times 6$. Revolved about the side of length 6.

Both cases yield the same maximum volume $864\pi$ when the side of length 6 is the axis of revolution.

The dimensions of the rectangle are $12$ cm and $6$ cm.

The maximum volume is obtained when the side of length 6 cm is used as the axis of revolution.

The dimensions of the rectangle are 6 cm and 12 cm.

The maximum volume is $864\pi$ cubic units.

Given:

The sum of the surface areas of a cube and a sphere is constant.

To Find:

The ratio of an edge of the cube to the diameter of the sphere when the sum of their volumes is minimum.

Solution:

Let $x$ be the length of an edge of the cube.

Let $r$ be the radius of the sphere.

Surface area of the cube, $A_{\text{cube}} = 6x^2$.

Surface area of the sphere, $A_{\text{sphere}} = 4\pi r^2$.

The sum of the surface areas is constant. Let this constant be $K$.

Volume of the cube, $V_{\text{cube}} = x^3$.

Volume of the sphere, $V_{\text{sphere}} = \frac{4}{3}\pi r^3$.

The sum of their volumes is $V = V_{\text{cube}} + V_{\text{sphere}} = x^3 + \frac{4}{3}\pi r^3$.

We want to minimize the sum of the volumes $V$. We need to express $V$ as a function of a single variable, either $x$ or $r$, using the constraint from equation (i).

From equation (i), we can express $r^2$ in terms of $x$ and $K$:

$4\pi r^2 = K - 6x^2$

$r^2 = \frac{K - 6x^2}{4\pi}$

For $r$ to be a real radius, $r^2 \ge 0$, so $K - 6x^2 \ge 0$. Also, $x$ must be a length, so $x > 0$. Since $r^2 = \frac{K - 6x^2}{4\pi}$, we must have $x^2 \le \frac{K}{6}$. So $0 < x \le \sqrt{\frac{K}{6}}$.

We need $r$ for the volume formula, not $r^2$. $r = \sqrt{\frac{K - 6x^2}{4\pi}} = \frac{\sqrt{K - 6x^2}}{2\sqrt{\pi}}$.

Substituting this into the volume sum formula seems complicated due to the cube of the square root.

From equation (i):

$6x^2 = K - 4\pi r^2$

$x^2 = \frac{K - 4\pi r^2}{6}$

$x = \sqrt{\frac{K - 4\pi r^2}{6}}$ (since $x > 0$)

For $x$ to be real, $K - 4\pi r^2 \ge 0$, so $4\pi r^2 \le K$, which means $r^2 \le \frac{K}{4\pi}$. So $0 \le r \le \sqrt{\frac{K}{4\pi}}$. Note that $r=0$ implies $x^2=K/6$, and $x=0$ implies $r^2=K/4\pi$. These are boundary cases.

Now, express the sum of volumes $V$ as a function of $r$:

$V(r) = x^3 + \frac{4}{3}\pi r^3 = \left(\sqrt{\frac{K - 4\pi r^2}{6}}\right)^3 + \frac{4}{3}\pi r^3$

$V(r) = \left(\frac{K - 4\pi r^2}{6}\right)^{3/2} + \frac{4}{3}\pi r^3$

We need to find the value of $r$ that minimizes $V(r)$ in the interval $\left(0, \sqrt{\frac{K}{4\pi}}\right)$. To find the minimum, we find the critical points by setting the first derivative $V'(r)$ equal to zero.

Find the first derivative $V'(r)$ with respect to $r$:

$V'(r) = \frac{d}{dr}\left[\left(\frac{K - 4\pi r^2}{6}\right)^{3/2}\right] + \frac{d}{dr}\left[\frac{4}{3}\pi r^3\right]$

Using the chain rule for the first term: $\frac{d}{dr}(u^{3/2}) = \frac{3}{2} u^{1/2} \frac{du}{dr}$, where $u = \frac{K - 4\pi r^2}{6}$.

$\frac{du}{dr} = \frac{1}{6} \frac{d}{dr}(K - 4\pi r^2) = \frac{1}{6} (0 - 4\pi \cdot 2r) = -\frac{8\pi r}{6} = -\frac{4\pi r}{3}$.

Derivative of the first term = $\frac{3}{2} \left(\frac{K - 4\pi r^2}{6}\right)^{1/2} \left(-\frac{4\pi r}{3}\right)$

= $\frac{3}{2} \sqrt{\frac{K - 4\pi r^2}{6}} \left(-\frac{4\pi r}{3}\right)$

= $\frac{3}{2} \frac{x}{\sqrt{6}} \left(-\frac{4\pi r}{3}\right)$ (using $x = \sqrt{\frac{K - 4\pi r^2}{6}}$)

= $-\frac{12\pi rx}{6\sqrt{6}} = -\frac{2\pi rx}{\sqrt{6}}$

Derivative of the second term = $\frac{4}{3}\pi \cdot 3r^2 = 4\pi r^2$.

So, $V'(r) = -\frac{2\pi rx}{\sqrt{6}} + 4\pi r^2$. Let's go back to using $r$ only, not $x$ here, as we need to set $V'(r) = 0$.

Derivative of the first term = $\frac{3}{2} \left(\frac{K - 4\pi r^2}{6}\right)^{1/2} \left(-\frac{4\pi r}{3}\right) = \frac{3}{2} \frac{\sqrt{K - 4\pi r^2}}{\sqrt{6}} \left(-\frac{4\pi r}{3}\right)$

= $-\frac{12\pi r \sqrt{K - 4\pi r^2}}{6 \sqrt{6}} = -\frac{2\pi r \sqrt{K - 4\pi r^2}}{\sqrt{6}}$. This still looks complicated.

Let's reconsider the derivatives in terms of $x$ and $r$ and then use the relationship between them.

We have $V = x^3 + \frac{4}{3}\pi r^3$. Differentiate with respect to one variable, say $x$, treating $r$ as a function of $x$. From $6x^2 + 4\pi r^2 = K$, differentiate with respect to $x$:

$\frac{d}{dx}(6x^2) + \frac{d}{dx}(4\pi r^2) = \frac{d}{dx}(K)$

$12x + 8\pi r \frac{dr}{dx} = 0$

Now, differentiate the volume sum $V$ with respect to $x$:

$\frac{dV}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dx} = 3x^2 + \frac{4}{3}\pi (3r^2) \frac{dr}{dx}$

$\frac{dV}{dx} = 3x^2 + 4\pi r^2 \frac{dr}{dx}$

Substitute the expression for $\frac{dr}{dx}$ from equation (ii) into the equation for $\frac{dV}{dx}$:

$\frac{dV}{dx} = 3x^2 + 4\pi r^2 \left(-\frac{3x}{2\pi r}\right)$

$\frac{dV}{dx} = 3x^2 - \frac{12\pi r^2 x}{2\pi r}$

$\frac{dV}{dx} = 3x^2 - 6rx$

Set $\frac{dV}{dx} = 0$ to find the critical points for the minimum volume:

$3x^2 - 6rx = 0$

$3x(x - 2r) = 0$

This gives $x = 0$ or $x - 2r = 0$.

$x = 0$ would mean the cube has no volume, so the total volume is just the volume of the sphere. This would be a boundary case. Let's consider $x - 2r = 0$, which means $x = 2r$.

$x = 2r$ implies the edge of the cube is equal to the diameter of the sphere.

To determine if $x = 2r$ corresponds to a minimum volume, we use the second derivative test on $V(x)$. This requires expressing $V$ solely in terms of $x$, which was the complicated expression $\left(\frac{K - 6x^2}{4\pi}\right)^{3/2} + \frac{4}{3}\pi \left(\frac{K - 6x^2}{4\pi}\right)^{3/2}$... this is not $V(x)$ in terms of $x$ only. $V = x^3 + \frac{4}{3}\pi r^3$, and $r$ is a function of $x$.

Let's use the critical condition $x = 2r$ and check the sign of the second derivative $\frac{d^2V}{dx^2}$.

$\frac{dV}{dx} = 3x^2 - 6rx$. Differentiate with respect to $x$, treating $r$ as a function of $x$:

$\frac{d^2V}{dx^2} = \frac{d}{dx}(3x^2) - \frac{d}{dx}(6rx)$

$\frac{d^2V}{dx^2} = 6x - 6\left(1 \cdot r + x \frac{dr}{dx}\right)$

$\frac{d^2V}{dx^2} = 6x - 6r - 6x \frac{dr}{dx}$

Substitute $\frac{dr}{dx} = -\frac{3x}{2\pi r}$ from equation (ii):

$\frac{d^2V}{dx^2} = 6x - 6r - 6x \left(-\frac{3x}{2\pi r}\right)$

$\frac{d^2V}{dx^2} = 6x - 6r + \frac{18x^2}{2\pi r} = 6x - 6r + \frac{9x^2}{\pi r}$

Now evaluate the second derivative at the critical condition $x = 2r$: Substitute $x = 2r$ into the expression for $\frac{d^2V}{dx^2}$.

$\frac{d^2V}{dx^2} |_{x=2r} = 6(2r) - 6r + \frac{9(2r)^2}{\pi r}$

= $12r - 6r + \frac{9(4r^2)}{\pi r}$

= $6r + \frac{36r^2}{\pi r}$

= $6r + \frac{36r}{\pi}$ (assuming $r \ne 0$. If $r=0$, then $x=0$ which is a boundary case.)

= $6r \left(1 + \frac{6}{\pi}\right)$

Since $r > 0$ (for a sphere to exist) and $\pi > 0$, the expression $6r \left(1 + \frac{6}{\pi}\right)$ is positive.

Since $\frac{d^2V}{dx^2} > 0$ at $x = 2r$, the sum of the volumes is minimum when $x = 2r$.

The condition for minimum volume is that the edge of the cube is equal to the diameter of the sphere.

The edge of the cube is $x$.

The diameter of the sphere is $2r$.

The ratio of an edge of the cube to the diameter of the sphere is $\frac{\text{edge of cube}}{\text{diameter of sphere}} = \frac{x}{2r}$.

When the volume sum is minimum, we have the condition $x = 2r$.

So, the ratio is $\frac{2r}{2r} = 1$.

The ratio of an edge of the cube to the diameter of the sphere is $1:1$.

Alternatively, consider the endpoint cases for the interval $0 \le r \le \sqrt{\frac{K}{4\pi}}$.

If $r = 0$, the object is just a cube with surface area $K = 6x^2$. Volume is $V = x^3 = (\sqrt{K/6})^3 = (K/6)^{3/2}$.

If $r = \sqrt{\frac{K}{4\pi}}$, the object is just a sphere with surface area $K = 4\pi r^2$. Volume is $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\sqrt{\frac{K}{4\pi}}\right)^3 = \frac{4}{3}\pi \frac{K^{3/2}}{(4\pi)^{3/2}} = \frac{4}{3}\pi \frac{K^{3/2}}{4\pi \sqrt{4\pi}} = \frac{K^{3/2}}{3\sqrt{4\pi}} = \frac{K^{3/2}}{6\sqrt{\pi}}$.

At the critical point $x=2r$, $6x^2 + 4\pi r^2 = K$ becomes $6(2r)^2 + 4\pi r^2 = K$, so $6(4r^2) + 4\pi r^2 = K$, $24r^2 + 4\pi r^2 = K$, $r^2(24 + 4\pi) = K$, $r^2 = \frac{K}{24 + 4\pi}$.

The volume at this point is $V = x^3 + \frac{4}{3}\pi r^3 = (2r)^3 + \frac{4}{3}\pi r^3 = 8r^3 + \frac{4}{3}\pi r^3 = r^3\left(8 + \frac{4\pi}{3}\right) = r^3\left(\frac{24 + 4\pi}{3}\right)$.

$V = \left(\frac{K}{24 + 4\pi}\right)^{3/2} \left(\frac{24 + 4\pi}{3}\right) = \frac{K^{3/2}}{(24 + 4\pi)^{3/2}} (24 + 4\pi) \frac{1}{3} = \frac{K^{3/2}}{\sqrt{24 + 4\pi}} \frac{1}{3}$.

$V = \frac{K^{3/2}}{3 \sqrt{4(6 + \pi)}} = \frac{K^{3/2}}{3 \cdot 2 \sqrt{6 + \pi}} = \frac{K^{3/2}}{6 \sqrt{6 + \pi}}$.

Compare $\frac{K^{3/2}}{(6)^{3/2}} = \frac{K^{3/2}}{6\sqrt{6}}$, $\frac{K^{3/2}}{6\sqrt{\pi}}$, and $\frac{K^{3/2}}{6\sqrt{6 + \pi}}$.

We compare $\sqrt{6}$, $\sqrt{\pi}$, and $\sqrt{6 + \pi}$.

$\pi \approx 3.14$. $6 > \pi$, so $\sqrt{6} > \sqrt{\pi}$.

$6 + \pi$ is greater than both 6 and $\pi$. So $\sqrt{6+\pi}$ is the largest denominator.

The smallest denominator is $\sqrt{\pi}$. This corresponds to the volume of a sphere only ($x=0$).

The largest denominator is $\sqrt{6+\pi}$, corresponding to the critical point $x=2r$. A larger denominator means a smaller fraction. So the minimum volume is at $x=2r$.

The ratio of an edge of the cube to the diameter of the sphere is $\frac{x}{2r}$. At minimum volume, $x = 2r$.

Ratio = $\frac{2r}{2r} = 1$.

The ratio of an edge of the cube to the diameter of the sphere is $1:1$ when the sum of their volumes is minimum.

Given:

AB is a diameter of a circle.

C is any point on the circle.

To Show:

The area of triangle ABC is maximum when the triangle is isosceles.

Solution:

Since AB is the diameter and C is a point on the circle, the angle subtended by the diameter at any point on the circumference is $90^\circ$. Therefore, $\angle ACB = 90^\circ$.

Thus, $\triangle ABC$ is a right-angled triangle with the right angle at C.

Let the radius of the circle be $R$. The length of the diameter AB is $2R$.

Let the coordinates of the center of the circle be $(0, 0)$. We can place A at $(-R, 0)$ and B at $(R, 0)$.

Let the coordinates of point C be $(x, y)$, where $x^2 + y^2 = R^2$ since C is on the circle.

In the right-angled triangle ABC, the base can be taken as the diameter AB, and the height is the perpendicular distance from C to the diameter AB. The height is simply the absolute value of the y-coordinate of point C, i.e., $|y|$.

The length of the base AB is $2R$.

The height corresponding to the base AB is $|y| = \sqrt{R^2 - x^2}$. Note that $y^2 = R^2 - x^2$, and since height must be positive, we take the positive square root.

The area of $\triangle ABC$, denoted by $A$, is given by:

$A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2R) \times |y|$

Since area must be positive, we can assume $y \ge 0$ without loss of generality, which means we consider the upper semicircle. In this case, $|y| = y = \sqrt{R^2 - x^2}$.

$A(x) = R \sqrt{R^2 - x^2}$

The domain of $x$ is $[-R, R]$ for point C to be on the circle. We are interested in $C \ne A$ and $C \ne B$, so $-R < x < R$.

To maximize the area $A(x)$, we can maximize $A(x)^2 = R^2 (R^2 - x^2)$ since $A(x) \ge 0$ and $R^2$ is a positive constant. Maximizing $R^2(R^2 - x^2)$ is equivalent to maximizing $R^2 - x^2$, or minimizing $x^2$.

The minimum value of $x^2$ in the interval $(-R, R)$ is $0^2 = 0$, which occurs at $x = 0$.

Alternatively, let's use calculus on $A(x)$.

$A'(x) = \frac{d}{dx}(R \sqrt{R^2 - x^2}) = R \cdot \frac{1}{2\sqrt{R^2 - x^2}} \cdot \frac{d}{dx}(R^2 - x^2)$

$A'(x) = R \cdot \frac{1}{2\sqrt{R^2 - x^2}} \cdot (-2x) = -\frac{Rx}{\sqrt{R^2 - x^2}}$

Set $A'(x) = 0$ to find the critical points:

$-\frac{Rx}{\sqrt{R^2 - x^2}} = 0$

This implies $Rx = 0$. Since $R > 0$, we must have $x = 0$.

To confirm that $x=0$ corresponds to a maximum, we can use the first derivative test or the second derivative test (though the second derivative might be complex). Let's use the first derivative test by checking the sign of $A'(x)$ around $x=0$.

If $x < 0$ (and near 0), $A'(x) = -\frac{R(\text{negative})}{\text{positive}} = \text{positive}$.

If $x > 0$ (and near 0), $A'(x) = -\frac{R(\text{positive})}{\text{positive}} = \text{negative}$.

Since $A'(x)$ changes sign from positive to negative at $x = 0$, there is a local maximum at $x = 0$. Since it's the only critical point in the interval $(-R, R)$, it's a global maximum.

The area is maximum when $x = 0$. When $x = 0$, the point C is $(0, y)$. Since $x^2 + y^2 = R^2$, we have $0^2 + y^2 = R^2$, so $y^2 = R^2$, which means $y = \pm R$.

If we chose the upper semicircle, $y = R$. The point C is $(0, R)$.

Let's check the triangle ABC when C is at $(0, R)$, A is at $(-R, 0)$, and B is at $(R, 0)$.

The lengths of the sides AC and BC are:

$AC = \sqrt{(0 - (-R))^2 + (R - 0)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$

$BC = \sqrt{(0 - R)^2 + (R - 0)^2} = \sqrt{(-R)^2 + R^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$

The length of AB is $2R$.

Since $AC = BC = R\sqrt{2}$, the triangle ABC is an isosceles triangle when the area is maximum.

The maximum area is $A(0) = R \sqrt{R^2 - 0^2} = R \sqrt{R^2} = R \cdot R = R^2$.

This also matches $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2R) \times R = R^2$.

Thus, the area of $\triangle ABC$ is maximum when the triangle is isosceles with $AC = BC$.

Hence, it is shown.

Given:

The box has a square base and vertical sides.

Volume of the box, $V = 1024$ cm$^3$.

Cost of material for top and bottom = $\textsf{₹}5$ per cm$^2$.

Cost of material for sides = $\textsf{₹}2.50$ per cm$^2$.

To Find:

The least cost of the box.

Solution:

Let $x$ be the length of the side of the square base in cm.

Let $h$ be the height of the box in cm.

The volume of the box is $V = (\text{Area of base}) \times \text{height} = x^2 h$.

We are given $V = 1024$ cm$^3$.

From equation (i), we can express the height $h$ in terms of $x$:

Assuming $x \ne 0$, $h = \frac{1024}{x^2}$.

Since $x$ and $h$ are physical lengths, we must have $x > 0$ and $h > 0$. The condition $x > 0$ ensures $h$ is positive.

Now, calculate the surface area of each part of the box and the associated cost.

Area of the base = $x^2$ cm$^2$. Cost of base = $x^2 \times 5 = 5x^2$ Rupees.

Area of the top = $x^2$ cm$^2$. Cost of top = $x^2 \times 5 = 5x^2$ Rupees.

Area of one side = $x \cdot h$ cm$^2$. There are 4 sides.

Total area of sides = $4xh$ cm$^2$. Cost of sides = $4xh \times 2.50 = 10xh$ Rupees.

The total cost of the box, $C(x)$, is the sum of the costs of the top, bottom, and sides:

$C(x) = \text{Cost of base} + \text{Cost of top} + \text{Cost of sides}$

$C(x) = 5x^2 + 5x^2 + 10xh = 10x^2 + 10xh$

Substitute the expression for $h = \frac{1024}{x^2}$ into the cost function to express $C$ in terms of $x$ only:

$C(x) = 10x^2 + 10x \left(\frac{1024}{x^2}\right)$

Assuming $x \ne 0$, simplify:

We want to find the value of $x > 0$ that minimizes $C(x)$. To find the minimum, we find the critical points by setting the first derivative $C'(x)$ equal to zero.

Find the first derivative $C'(x)$ with respect to $x$:

$C'(x) = \frac{d}{dx}\left(10x^2 + 10240x^{-1}\right)$

$C'(x) = 10 \cdot 2x^{2-1} + 10240 \cdot (-1)x^{-1-1}$

$C'(x) = 20x - 10240x^{-2} = 20x - \frac{10240}{x^2}$

Set $C'(x) = 0$ to find the critical number(s):

$20x - \frac{10240}{x^2} = 0$

$20x = \frac{10240}{x^2}$

Multiply by $x^2$ (assuming $x \ne 0$):

$20x^3 = 10240$

$x^3 = \frac{10240}{20} = \frac{1024}{2}$

$x^3 = 512$

Since $x$ is a length, $x = \sqrt[3]{512} = 8$.

To determine if $x = 8$ corresponds to a minimum cost, we use the second derivative test. Find the second derivative $C''(x)$:

$C''(x) = \frac{d}{dx}\left(20x - 10240x^{-2}\right)$

$C''(x) = 20(1) - 10240(-2)x^{-2-1}$

$C''(x) = 20 + 20480x^{-3} = 20 + \frac{20480}{x^3}$

Evaluate $C''(x)$ at $x = 8$:

$C''(8) = 20 + \frac{20480}{(8)^3} = 20 + \frac{20480}{512}$

$20480 \div 512$. $512 \times 10 = 5120$. $5120 \times 4 = 20480$. So $20480/512 = 40$.

$C''(8) = 20 + 40 = 60$.

Since $C''(8) = 60 > 0$, the cost function $C(x)$ has a local minimum at $x = 8$. Since it's the only critical point in the domain $x > 0$, it is the global minimum.

The dimensions that minimize the cost are when the side of the base is $x = 8$ cm. The corresponding height is $h = \frac{1024}{x^2} = \frac{1024}{8^2} = \frac{1024}{64}$.

$1024 \div 64$. $1024 = 32 \times 32$. $64 = 8 \times 8$. $1024/64 = (32/8) \times (32/8) = 4 \times 4 = 16$.

So, the height is $h = 16$ cm.

Now, find the least cost by evaluating the cost function $C(x)$ at $x = 8$:

$C(8) = 10(8)^2 + \frac{10240}{8}$

$C(8) = 10(64) + 1280$

$C(8) = 640 + 1280$

$C(8) = 1920$

The least cost of the box is $\textsf{₹}1920$.

Given:

A rectangular parallelopiped with sides $x$, $2x$, and $\frac{x}{3}$.

A sphere with radius $r$.

The sum of the surface areas of the parallelopiped and the sphere is a constant.

To Prove:

1. The sum of their volumes is minimum if $x = 3r$.

2. Find the minimum value of the sum of their volumes.

Solution:

Let the dimensions of the rectangular parallelopiped be $l = x$, $w = 2x$, and $h = \frac{x}{3}$.

The surface area of the parallelopiped, $A_p$, is $2(lw + lh + wh)$.

$A_p = 2\left(x(2x) + x\left(\frac{x}{3}\right) + 2x\left(\frac{x}{3}\right)\right)$

$A_p = 2\left(2x^2 + \frac{x^2}{3} + \frac{2x^2}{3}\right)$

$A_p = 2\left(2x^2 + \frac{3x^2}{3}\right) = 2(2x^2 + x^2) = 2(3x^2) = 6x^2$

The volume of the parallelopiped, $V_p$, is $lwh$.

$V_p = x(2x)\left(\frac{x}{3}\right) = \frac{2x^3}{3}$

The surface area of the sphere, $A_s$, with radius $r$, is $4\pi r^2$.

The volume of the sphere, $V_s$, is $\frac{4}{3}\pi r^3$.

The sum of the surface areas is given to be constant. Let this constant be $K$.

The sum of their volumes is $V = V_p + V_s$.

We want to minimize $V$. From the constraint (i), we can express $r$ as a function of $x$, or vice versa, and substitute into the volume equation. A simpler approach is to differentiate equation (ii) with respect to $x$, treating $r$ as a function of $x$, and use implicit differentiation on equation (i) to find $\frac{dr}{dx}$.

Differentiate equation (i) with respect to $x$:

$\frac{d}{dx}(6x^2 + 4\pi r^2) = \frac{d}{dx}(K)$

$12x + 8\pi r \frac{dr}{dx} = 0$

$8\pi r \frac{dr}{dx} = -12x$

Now, differentiate equation (ii) with respect to $x$:

$\frac{dV}{dx} = \frac{d}{dx}\left(\frac{2x^3}{3}\right) + \frac{d}{dx}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dx} = \frac{2}{3} \cdot 3x^2 + \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dx}$

Substitute the expression for $\frac{dr}{dx}$ from equation (iii) into equation (iv):

$\frac{dV}{dx} = 2x^2 + 4\pi r^2 \left(-\frac{3x}{2\pi r}\right)$

$\frac{dV}{dx} = 2x^2 - \frac{12\pi r^2 x}{2\pi r}$

$\frac{dV}{dx} = 2x^2 - 6rx$

To find the values of $x$ and $r$ that minimize the volume, we set $\frac{dV}{dx} = 0$:

$2x^2 - 6rx = 0$

Factor out $2x$:

$2x(x - 3r) = 0$

This gives two possibilities: $x = 0$ or $x - 3r = 0$.

If $x = 0$, the parallelopiped has zero volume. From the constraint $6(0)^2 + 4\pi r^2 = K$, we get $4\pi r^2 = K$, so $r = \sqrt{\frac{K}{4\pi}}$. The total volume is $V = 0 + \frac{4}{3}\pi \left(\sqrt{\frac{K}{4\pi}}\right)^3 = \frac{K^{3/2}}{6\sqrt{\pi}}$. This is a boundary case.

The other possibility is $x - 3r = 0$, which implies $x = 3r$. This gives the relationship between $x$ and $r$ at the critical point.

To prove that $x = 3r$ corresponds to a minimum volume, we use the second derivative test. Find $\frac{d^2V}{dx^2}$ by differentiating $\frac{dV}{dx} = 2x^2 - 6rx$ with respect to $x$, treating $r$ as a function of $x$:

$\frac{d^2V}{dx^2} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(6rx)$

$\frac{d^2V}{dx^2} = 4x - 6\left(1 \cdot r + x \frac{dr}{dx}\right)$

$\frac{d^2V}{dx^2} = 4x - 6r - 6x \frac{dr}{dx}$

Substitute $\frac{dr}{dx} = -\frac{3x}{2\pi r}$ from equation (iii):

$\frac{d^2V}{dx^2} = 4x - 6r - 6x \left(-\frac{3x}{2\pi r}\right)$

$\frac{d^2V}{dx^2} = 4x - 6r + \frac{18x^2}{2\pi r} = 4x - 6r + \frac{9x^2}{\pi r}$

Evaluate the second derivative at the critical condition $x = 3r$:

$\frac{d^2V}{dx^2}\Big|_{x=3r} = 4(3r) - 6r + \frac{9(3r)^2}{\pi r}$

$= 12r - 6r + \frac{9(9r^2)}{\pi r}$

$= 6r + \frac{81r^2}{\pi r} = 6r + \frac{81r}{\pi}$ (assuming $r \ne 0$, otherwise $x=0$ which is a boundary case)

$= r\left(6 + \frac{81}{\pi}\right)$

Since $r > 0$ and $6 + \frac{81}{\pi} > 0$, the second derivative $\frac{d^2V}{dx^2}\Big|_{x=3r} > 0$.

By the second derivative test, the sum of the volumes $V$ has a local minimum when $x = 3r$. This local minimum is the global minimum in the relevant domain.

This proves the first part that the sum of volumes is minimum when $x = 3r$.

Now, find the minimum value of the sum of the volumes. Substitute the condition $x = 3r$ into the volume sum formula $V = \frac{2x^3}{3} + \frac{4}{3}\pi r^3$ (equation ii):

$V_{\text{min}} = \frac{2(3r)^3}{3} + \frac{4}{3}\pi r^3$

$V_{\text{min}} = \frac{2(27r^3)}{3} + \frac{4}{3}\pi r^3$

$V_{\text{min}} = 18r^3 + \frac{4}{3}\pi r^3$

$V_{\text{min}} = r^3\left(18 + \frac{4\pi}{3}\right) = r^3\left(\frac{54 + 4\pi}{3}\right)$

To express this minimum volume in terms of the constant $K$, substitute $x = 3r$ into the constraint equation $6x^2 + 4\pi r^2 = K$ (equation i):

$6(3r)^2 + 4\pi r^2 = K$

$6(9r^2) + 4\pi r^2 = K$

$54r^2 + 4\pi r^2 = K$

$r^2(54 + 4\pi) = K \implies r^2 = \frac{K}{54 + 4\pi}$

So, $r = \sqrt{\frac{K}{54 + 4\pi}}$. Then $r^3 = \left(\frac{K}{54 + 4\pi}\right)^{3/2}$.

$V_{\text{min}} = \left(\frac{K}{54 + 4\pi}\right)^{3/2} \left(\frac{54 + 4\pi}{3}\right)$

$V_{\text{min}} = \frac{K^{3/2}}{(54 + 4\pi)^{3/2}} (54 + 4\pi) \frac{1}{3}$

$V_{\text{min}} = \frac{K^{3/2}}{(54 + 4\pi)^{1/2}} \frac{1}{3} = \frac{K^{3/2}}{3\sqrt{54 + 4\pi}}$

We can factor $54 + 4\pi = 2(27 + 2\pi)$.

$V_{\text{min}} = \frac{K^{3/2}}{3\sqrt{2(27 + 2\pi)}} = \frac{K^{3/2}}{3\sqrt{2}\sqrt{27 + 2\pi}}$

The sum of the volumes is minimum when the side length $x$ of the parallelopiped is equal to three times the radius $r$ of the sphere ($x = 3r$).

The minimum value of the sum of their volumes is $\frac{K^{3/2}}{3\sqrt{54 + 4\pi}}$, where $K$ is the constant sum of their surface areas.

Given:

Let $s$ be the side length of the equilateral triangle.

Rate of increase of the side, $\frac{ds}{dt} = 2$ cm/sec.

To Find:

The rate at which the area increases, $\frac{dA}{dt}$, when $s = 10$ cm.

Solution:

The area of an equilateral triangle with side length $s$ is given by the formula:

$A = \frac{\sqrt{3}}{4} s^2$

To find the rate at which the area increases, we differentiate $A$ with respect to time $t$:

$\frac{dA}{dt} = \frac{d}{dt} \left(\frac{\sqrt{3}}{4} s^2\right)$

Using the chain rule, we get:

$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt}$

$\frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}$

We are given that $\frac{ds}{dt} = 2$ cm/sec and we need to find $\frac{dA}{dt}$ when $s = 10$ cm.

Substitute these values into the expression for $\frac{dA}{dt}$:

$\frac{dA}{dt} = \frac{\sqrt{3}}{2} (10)(2)$

$\frac{dA}{dt} = \frac{\sqrt{3}}{\cancel{2}^1} (\cancel{20}^{10})$

$\frac{dA}{dt} = 10\sqrt{3}$

The rate at which the area increases when the side is 10 cm is $10\sqrt{3}$ cm$^2$/sec.

The correct option is (C).

Given:

Let $L$ be the length of the ladder, $L = 5$ m.

Let $y$ be the height of the top of the ladder from the floor.

Let $x$ be the distance of the lower end of the ladder from the wall.

Let $\theta$ be the angle between the floor and the ladder.

Rate at which the top of the ladder slides downwards, $\frac{dy}{dt} = -10$ cm/sec. (Negative sign indicates decreasing height)

Convert the rate to meters per second: $\frac{dy}{dt} = -10 \text{ cm/sec} = -0.1 \text{ m/sec}$.

To Find:

The rate at which the angle $\theta$ is decreasing, i.e., $-\frac{d\theta}{dt}$, when $x = 2$ m.

Solution:

The relationship between $y$, $L$, and $\theta$ is given by trigonometry in the right-angled triangle formed by the ladder, the wall, and the floor:

$\sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{y}{L}$

Substituting the length of the ladder, $L=5$ m:

$\sin \theta = \frac{y}{5}$

Differentiate both sides of the equation with respect to time $t$:

$\frac{d}{dt}(\sin \theta) = \frac{d}{dt}\left(\frac{y}{5}\right)$

Using the chain rule:

$\cos \theta \frac{d\theta}{dt} = \frac{1}{5} \frac{dy}{dt}$

We need to find $\frac{d\theta}{dt}$ when $x = 2$ m. First, we need to find the value of $\cos \theta$ at this moment.

When $x = 2$ m and $L = 5$ m, the relationship between $x$, $L$, and $\theta$ is given by:

$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{x}{L}$

$\cos \theta = \frac{2}{5}$

Now, substitute the values of $\cos \theta = \frac{2}{5}$ and $\frac{dy}{dt} = -0.1$ m/sec into the differentiated equation:

$\left(\frac{2}{5}\right) \frac{d\theta}{dt} = \frac{1}{5} (-0.1)$

$\frac{d\theta}{dt} = \frac{1}{5} (-0.1) \cdot \frac{5}{2}$

$\frac{d\theta}{dt} = \frac{1}{\cancel{5}} (-0.1) \cdot \frac{\cancel{5}}{2}$

$\frac{d\theta}{dt} = \frac{-0.1}{2}$

$\frac{d\theta}{dt} = -0.05$ radian/sec

The rate of change of the angle is $-0.05$ radian/sec. The negative sign indicates that the angle is decreasing.

The rate at which the angle is decreasing is $|-0.05|$ radian/sec.

Rate of decrease $= 0.05$ radian/sec.

Convert the decimal to a fraction:

$0.05 = \frac{5}{100} = \frac{1}{20}$

Thus, the rate at which the angle between the floor and the ladder is decreasing is $\frac{1}{20}$ radian/sec.

The correct option is (B).

Given:

The equation of the curve is $y = x^{\frac{1}{5}}$.

The point is (0, 0).

To Find:

The type of tangent to the curve at the point (0, 0).

Solution:

To find the slope of the tangent to the curve at a point, we first find the derivative $\frac{dy}{dx}$.

$y = x^{\frac{1}{5}}$

Differentiating with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} (x^{\frac{1}{5}})$

$\frac{dy}{dx} = \frac{1}{5} x^{\frac{1}{5}-1}$

$\frac{dy}{dx} = \frac{1}{5} x^{-\frac{4}{5}}$

$\frac{dy}{dx} = \frac{1}{5x^{\frac{4}{5}}}$

Now, we evaluate the derivative at the point (0, 0), which means substituting $x=0$ into the expression for $\frac{dy}{dx}$.

Slope of the tangent at (0, 0) $= \left. \frac{dy}{dx} \right|_{x=0} = \frac{1}{5(0)^{\frac{4}{5}}}$

Since $0^{\frac{4}{5}} = 0$, the expression becomes:

Slope $= \frac{1}{5 \cdot 0} = \frac{1}{0}$

The slope is undefined.

A tangent line with an undefined slope is a vertical line, which is parallel to the y-axis.

The correct option is (A).

Given:

Equation of the curve: $3x^2 - y^2 = 8$.

The normal to the curve is parallel to the line: $x + 3y = 8$.

To Find:

The equation of the normal to the curve.

Solution:

First, we find the slope of the tangent to the curve at a general point $(x, y)$. We differentiate the equation of the curve implicitly with respect to $x$:

$\frac{d}{dx}(3x^2 - y^2) = \frac{d}{dx}(8)$

$6x - 2y \frac{dy}{dx} = 0$

$-2y \frac{dy}{dx} = -6x$

$\frac{dy}{dx} = \frac{-6x}{-2y} = \frac{3x}{y}$

The slope of the tangent at a point $(x_1, y_1)$ on the curve is $m_t = \left. \frac{dy}{dx} \right|_{(x_1, y_1)} = \frac{3x_1}{y_1}$.

The slope of the normal to the curve at $(x_1, y_1)$ is the negative reciprocal of the tangent's slope:

$m_n = -\frac{1}{m_t} = -\frac{1}{\frac{3x_1}{y_1}} = -\frac{y_1}{3x_1}$

Now, we find the slope of the given line $x + 3y = 8$. We rewrite it in slope-intercept form $y = mx + c$:

$3y = -x + 8$

$y = -\frac{1}{3}x + \frac{8}{3}$

The slope of the given line is $m_{line} = -\frac{1}{3}$.

Since the normal is parallel to the given line, their slopes are equal:

$m_n = m_{line}$

$-\frac{y_1}{3x_1} = -\frac{1}{3}$

$\frac{y_1}{3x_1} = \frac{1}{3}$

$3y_1 = 3x_1$

$y_1 = x_1$

This means the points where the normal is parallel to the line $x+3y=8$ have equal $x$ and $y$ coordinates. These points must lie on the curve $3x^2 - y^2 = 8$. Substitute $y_1 = x_1$ into the curve equation:

$3x_1^2 - (x_1)^2 = 8$

$3x_1^2 - x_1^2 = 8$

$2x_1^2 = 8$

$x_1^2 = 4$

$x_1 = \pm 2$

Since $y_1 = x_1$, the points on the curve are (2, 2) and (-2, -2).

We use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with $m = m_n = -\frac{1}{3}$.

For the point (2, 2):

$y - 2 = -\frac{1}{3}(x - 2)$

$3(y - 2) = -(x - 2)$

$3y - 6 = -x + 2$

$x + 3y - 8 = 0$

For the point (-2, -2):

$y - (-2) = -\frac{1}{3}(x - (-2))$

$y + 2 = -\frac{1}{3}(x + 2)$

$3(y + 2) = -(x + 2)$

$3y + 6 = -x - 2$

$x + 3y + 8 = 0$

The equations of the normals parallel to the line $x + 3y = 8$ are $x + 3y - 8 = 0$ and $x + 3y + 8 = 0$. These can be written together as $x + 3y \pm 8 = 0$.

The correct option is (C).

Given:

Equation of the first curve: $ay + x^2 = 7$.

Equation of the second curve: $x^3 = y$.

The curves cut orthogonally at the point (1, 1).

To Find:

The value of the constant $a$.

Solution:

Two curves cut orthogonally at a point if their tangents at that point are perpendicular. The product of the slopes of the tangent lines must be -1.

First, find the slope of the tangent for the first curve $ay + x^2 = 7$. Differentiate implicitly with respect to $x$:

$\frac{d}{dx}(ay + x^2) = \frac{d}{dx}(7)$

$a \frac{dy}{dx} + 2x = 0$

$a \frac{dy}{dx} = -2x$

$\frac{dy}{dx} = -\frac{2x}{a}$

The slope of the tangent to the first curve at the point (1, 1) is $m_1$:

$m_1 = \left. \frac{dy}{dx} \right|_{(1, 1)} = -\frac{2(1)}{a} = -\frac{2}{a}$

Next, find the slope of the tangent for the second curve $y = x^3$. Differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(x^3)$

$\frac{dy}{dx} = 3x^2$

The slope of the tangent to the second curve at the point (1, 1) is $m_2$:

$m_2 = \left. \frac{dy}{dx} \right|_{(1, 1)} = 3(1)^2 = 3$

Since the curves cut orthogonally, the product of their tangent slopes at the point of intersection is -1:

$m_1 \cdot m_2 = -1$

$\left(-\frac{2}{a}\right) \cdot (3) = -1$

$-\frac{6}{a} = -1$

$\frac{6}{a} = 1$

$a = 6$

We should also verify that the point (1, 1) lies on the first curve when $a=6$. Substitute $x=1, y=1, a=6$ into $ay + x^2 = 7$:

$6(1) + (1)^2 = 6 + 1 = 7$

This is true, so the point (1, 1) is indeed on the first curve when $a=6$. The point (1, 1) is clearly on the second curve $y=x^3$ since $1^3=1$.

The value of $a$ is 6.

The correct option is (D).

Given:

The function is $y = f(x) = x^4 - 10$.

The initial value of $x$ is $x = 2$.

The new value of $x$ is $x + \Delta x = 1.99$.

To Find:

The change in $y$, denoted by $\Delta y$, when $x$ changes from 2 to 1.99.

Solution:

The change in $x$ is $\Delta x = (x + \Delta x) - x = 1.99 - 2 = -0.01$.

The change in $y$ is $\Delta y = f(x + \Delta x) - f(x)$.

We can approximate the change in $y$ using the differential $dy$, which is given by $dy = \frac{dy}{dx} \Delta x$.

First, find the derivative of $y$ with respect to $x$:

$y = x^4 - 10$

$\frac{dy}{dx} = \frac{d}{dx}(x^4 - 10)$

$\frac{dy}{dx} = 4x^3 - 0 = 4x^3$

Evaluate the derivative at the initial value of $x$, which is $x = 2$:

$\left. \frac{dy}{dx} \right|_{x=2} = 4(2)^3 = 4(8) = 32$

Now, calculate the approximate change in $y$ using the differential:

$dy = \left. \frac{dy}{dx} \right|_{x=2} \cdot \Delta x$

$dy = 32 \cdot (-0.01)$

$dy = -0.32$

The approximate change in $y$ is $-0.32$. Since the options are positive values, the question likely refers to the magnitude of the change.

$|\Delta y| \approx |dy| = |-0.32| = 0.32$

The approximate change in $y$ is 0.32.

The correct option is (A).

Given:

The equation of the curve is $y(1 + x^2) = 2 - x$.

The tangent is required at the point where the curve crosses the x-axis.

To Find:

The equation of the tangent to the curve where it crosses the x-axis.

Solution:

The curve crosses the x-axis when $y = 0$. Substitute $y=0$ into the equation of the curve to find the x-coordinate of the point of intersection:

$0(1 + x^2) = 2 - x$

$0 = 2 - x$

$x = 2$

So, the point where the curve crosses the x-axis is (2, 0).

Now, we need to find the slope of the tangent to the curve at the point (2, 0).

The equation of the curve is $y(1 + x^2) = 2 - x$. We can write it as $y = \frac{2 - x}{1 + x^2}$.

Differentiate $y$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:

Let $u = 2 - x$, so $u' = \frac{du}{dx} = -1$.

Let $v = 1 + x^2$, so $v' = \frac{dv}{dx} = 2x$.

$\frac{dy}{dx} = \frac{(-1)(1 + x^2) - (2 - x)(2x)}{(1 + x^2)^2}$

$\frac{dy}{dx} = \frac{-1 - x^2 - (4x - 2x^2)}{(1 + x^2)^2}$

$\frac{dy}{dx} = \frac{-1 - x^2 - 4x + 2x^2}{(1 + x^2)^2}$

$\frac{dy}{dx} = \frac{x^2 - 4x - 1}{(1 + x^2)^2}$

Evaluate the slope of the tangent at the point (2, 0) by substituting $x=2$:

Slope $m = \left. \frac{dy}{dx} \right|_{(2, 0)} = \frac{(2)^2 - 4(2) - 1}{(1 + (2)^2)^2}$

$m = \frac{4 - 8 - 1}{(1 + 4)^2}$

$m = \frac{-5}{(5)^2}$

$m = \frac{-5}{25} = -\frac{1}{5}$

Now, we use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (2, 0)$ and the slope $m = -\frac{1}{5}$.

$y - 0 = -\frac{1}{5}(x - 2)$

$y = -\frac{1}{5}(x - 2)$

Multiply both sides by 5:

$5y = -(x - 2)$

$5y = -x + 2$

Rearrange the terms to get the standard form:

$x + 5y - 2 = 0$ or $x + 5y = 2$

The equation of the tangent to the curve where it crosses the x-axis is $x + 5y = 2$.

The correct option is (A).

Given:

The equation of the curve is $y = x^3 - 12x + 18$.

The tangents to the curve are parallel to the x-axis.

To Find:

The points on the curve where the tangent is parallel to the x-axis.

Solution:

A tangent line is parallel to the x-axis if its slope is 0. The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

First, find the derivative of $y$ with respect to $x$:

$y = x^3 - 12x + 18$

$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 12x + 18)$

$\frac{dy}{dx} = 3x^2 - 12 + 0$

$\frac{dy}{dx} = 3x^2 - 12$

Set the slope equal to 0 to find the x-coordinates where the tangent is parallel to the x-axis:

$3x^2 - 12 = 0$

$3x^2 = 12$

$x^2 = \frac{12}{3}$

$x^2 = 4$

$x = \pm \sqrt{4}$

$x = \pm 2$

Now, find the corresponding y-coordinates by substituting these x-values back into the original equation of the curve $y = x^3 - 12x + 18$.

For $x = 2$:

$y = (2)^3 - 12(2) + 18$

$y = 8 - 24 + 18$

$y = -16 + 18 = 2$

So, one point is (2, 2).

For $x = -2$:

$y = (-2)^3 - 12(-2) + 18$

$y = -8 + 24 + 18$

$y = 16 + 18 = 34$

So, the other point is (-2, 34).

The points at which the tangents are parallel to the x-axis are (2, 2) and (-2, 34).

The correct option is (D).

Given:

The equation of the curve is $y = e^{2x}$.

The point of tangency is (0, 1).

To Find:

The point where the tangent to the curve at (0, 1) meets the x-axis.

Solution:

First, find the slope of the tangent to the curve at the point (0, 1). The slope is given by the derivative $\frac{dy}{dx}$.

Find the derivative of $y$ with respect to $x$:

$y = e^{2x}$

$\frac{dy}{dx} = \frac{d}{dx}(e^{2x})$

Using the chain rule, $\frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x)$:

$\frac{dy}{dx} = e^{2x} \cdot \frac{d}{dx}(2x)$

$\frac{dy}{dx} = e^{2x} \cdot 2 = 2e^{2x}$

Evaluate the slope of the tangent at the point (0, 1) by substituting $x=0$:

Slope $m = \left. \frac{dy}{dx} \right|_{(0, 1)} = 2e^{2(0)} = 2e^0 = 2(1) = 2$

Now, use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (0, 1)$ and the slope $m = 2$.

$y - 1 = 2(x - 0)$

$y - 1 = 2x$

The equation of the tangent line is $y = 2x + 1$.

To find where the tangent line meets the x-axis, set $y = 0$ in the equation of the tangent line:

$0 = 2x + 1$

$2x = -1$

$x = -\frac{1}{2}$

The y-coordinate is 0.

The tangent line meets the x-axis at the point $\left(-\frac{1}{2}, 0\right)$.

The correct option is (B).

Given:

The parametric equations of the curve are $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$.

The point on the curve is (2, -1).

To Find:

The slope of the tangent to the curve at the point (2, -1).

Solution:

The slope of the tangent to a parametric curve at a point is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

First, find the derivative of $x$ with respect to $t$:

$x = t^2 + 3t - 8$

$\frac{dx}{dt} = \frac{d}{dt}(t^2 + 3t - 8)$

$\frac{dx}{dt} = 2t + 3$

Next, find the derivative of $y$ with respect to $t$:

$y = 2t^2 - 2t - 5$

$\frac{dy}{dt} = \frac{d}{dt}(2t^2 - 2t - 5)$

$\frac{dy}{dt} = 4t - 2$

Now, find the value of the parameter $t$ that corresponds to the point (2, -1). Substitute the coordinates into the given equations:

For the x-coordinate:

$2 = t^2 + 3t - 8$

$t^2 + 3t - 10 = 0$

Factor the quadratic equation:

$(t + 5)(t - 2) = 0$

Possible values for $t$ are $t = -5$ or $t = 2$.

For the y-coordinate:

$-1 = 2t^2 - 2t - 5$

$2t^2 - 2t - 4 = 0$

Divide by 2:

$t^2 - t - 2 = 0$

Factor the quadratic equation:

$(t - 2)(t + 1) = 0$

Possible values for $t$ are $t = 2$ or $t = -1$.

The common value of $t$ that satisfies both equations is $t = 2$. This is the parameter value corresponding to the point (2, -1).

Now, evaluate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ at $t = 2$:

$\left. \frac{dx}{dt} \right|_{t=2} = 2(2) + 3 = 4 + 3 = 7$

$\left. \frac{dy}{dt} \right|_{t=2} = 4(2) - 2 = 8 - 2 = 6$

Finally, calculate the slope of the tangent $\frac{dy}{dx}$ at $t = 2$ (which corresponds to the point (2, -1)):

$\left. \frac{dy}{dx} \right|_{(2, -1)} = \left. \frac{dy/dt}{dx/dt} \right|_{t=2} = \frac{6}{7}$

The slope of the tangent to the curve at the point (2, -1) is $\frac{6}{7}$.

The correct option is (B).

Given:

The equations of the two curves are:

Curve 1: $x^3 - 3xy^2 + 2 = 0$

Curve 2: $3x^2y - y^3 - 2 = 0$

To Find:

The angle of intersection between the two curves.

Solution:

First, find the point(s) of intersection. Adding the two equations:

$(x^3 - 3xy^2 + 2) + (3x^2y - y^3 - 2) = 0$

$x^3 + 3x^2y - 3xy^2 - y^3 = 0$

This equation can be factored as $(x-y)(x^2+4xy+y^2) = 0$.

One possibility is $x-y = 0$, which means $y=x$.

Substitute $y=x$ into the first equation:

$x^3 - 3x(x)^2 + 2 = 0$

$x^3 - 3x^3 + 2 = 0$

$-2x^3 + 2 = 0$

$-2x^3 = -2$

$x^3 = 1$

$x = 1$

Since $y=x$, $y=1$. Thus, (1, 1) is a point of intersection.

We should also check if (1, 1) satisfies the second equation:

$3(1)^2(1) - (1)^3 - 2 = 3 - 1 - 2 = 0$. It does.

Other intersection points are given by $x^2+4xy+y^2=0$ and $y=x$. Substituting $y=x$ into $x^2+4xy+y^2=0$ gives $x^2+4x(x)+x^2 = x^2+4x^2+x^2=6x^2=0$, which implies $x=0$. If $x=0$, then $y=0$. Substituting (0, 0) into the original equations gives $0^3 - 3(0)(0)^2 + 2 = 2 \neq 0$ and $3(0)^2(0) - 0^3 - 2 = -2 \neq 0$. So (0, 0) is not an intersection point. The only real intersection point is (1, 1).

To find the angle of intersection, we find the slopes of the tangents to each curve at (1, 1).

For Curve 1: $x^3 - 3xy^2 + 2 = 0$. Differentiate implicitly with respect to $x$:

$3x^2 - 3(y^2 + x \cdot 2y \frac{dy}{dx}) = 0$

$3x^2 - 3y^2 - 6xy \frac{dy}{dx} = 0$

$-6xy \frac{dy}{dx} = 3y^2 - 3x^2$

$\frac{dy}{dx} = \frac{3y^2 - 3x^2}{-6xy} = \frac{y^2 - x^2}{-2xy}$

The slope of the tangent to Curve 1 at (1, 1) is $m_1$:

$m_1 = \left. \frac{dy}{dx} \right|_{(1, 1)} = \frac{(1)^2 - (1)^2}{-2(1)(1)} = \frac{1 - 1}{-2} = \frac{0}{-2} = 0$

For Curve 2: $3x^2y - y^3 - 2 = 0$. Differentiate implicitly with respect to $x$:

$3(2xy + x^2 \frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0$

$6xy + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0$

$(3x^2 - 3y^2) \frac{dy}{dx} = -6xy$

$\frac{dy}{dx} = \frac{-6xy}{3x^2 - 3y^2} = \frac{-2xy}{x^2 - y^2}$

The slope of the tangent to Curve 2 at (1, 1) is $m_2$:

$m_2 = \left. \frac{dy}{dx} \right|_{(1, 1)} = \frac{-2(1)(1)}{(1)^2 - (1)^2} = \frac{-2}{1 - 1} = \frac{-2}{0}$

The slope $m_2$ is undefined, which means the tangent to Curve 2 at (1, 1) is a vertical line (parallel to the y-axis).

The slope of the tangent to Curve 1 at (1, 1) is $m_1 = 0$, which means the tangent is a horizontal line (parallel to the x-axis).

Since the tangent lines at the point of intersection (1, 1) are horizontal and vertical, they are perpendicular to each other.

The angle between two perpendicular lines is $\frac{\pi}{2}$ (or $90^\circ$).

The angle of intersection between the two curves at the point (1, 1) is $\frac{\pi}{2}$.

The correct option is (C).

Given:

The function is $f(x) = 2x^3 + 9x^2 + 12x - 1$.

To Find:

The interval on which the function is decreasing.

Solution:

A function $f(x)$ is decreasing on an interval if its derivative $f'(x)$ is less than or equal to 0 on that interval.

First, find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(2x^3 + 9x^2 + 12x - 1)$

$f'(x) = 2(3x^2) + 9(2x) + 12(1) - 0$

$f'(x) = 6x^2 + 18x + 12$

To find where the function is decreasing, we need to find the interval(s) where $f'(x) \leq 0$.

$6x^2 + 18x + 12 \leq 0$

Divide the inequality by 6:

$x^2 + 3x + 2 \leq 0$

Find the roots of the quadratic equation $x^2 + 3x + 2 = 0$ to determine the critical points.

Factor the quadratic expression:

$(x + 1)(x + 2) = 0$

The roots are $x = -1$ and $x = -2$.

These roots divide the number line into three intervals: $(-\infty, -2]$, $[-2, -1]$, and $[-1, \infty)$.

The quadratic expression $x^2 + 3x + 2$ represents a parabola that opens upwards (since the coefficient of $x^2$ is positive, which is 1). A parabola that opens upwards is less than or equal to zero between its roots (inclusive).

Therefore, the inequality $x^2 + 3x + 2 \leq 0$ is satisfied for values of $x$ between the roots -2 and -1, inclusive.

The interval where $f'(x) \leq 0$ is $[-2, -1]$.

Thus, the function $f(x)$ is decreasing on the interval $[-2, -1]$.

The correct option is (B).

Given:

The function is $f(x) = 2x + \cos x$.

The domain of the function is R (all real numbers).

To Determine:

Whether the function has a minimum/maximum or if it is increasing/decreasing.

Solution:

To determine if the function has local extrema or if it is increasing/decreasing, we examine its first derivative $f'(x)$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(2x + \cos x)$

$f'(x) = 2 + (-\sin x)$

$f'(x) = 2 - \sin x$

Now, analyze the sign of $f'(x)$. The range of the sine function is $[-1, 1]$, i.e., $-1 \leq \sin x \leq 1$ for all real values of $x$.

Consider the expression for $f'(x)$: $2 - \sin x$.

Since $-1 \leq \sin x \leq 1$, we can find the range of $2 - \sin x$:

Multiply the inequality by -1 and reverse the inequality signs:

$-1 \leq -\sin x \leq 1$

Add 2 to all parts of the inequality:

$2 - 1 \leq 2 - \sin x \leq 2 + 1$

$1 \leq 2 - \sin x \leq 3$

So, the range of $f'(x) = 2 - \sin x$ is $[1, 3]$.

This means that $f'(x) \geq 1$ for all $x \in R$.

Since $f'(x) \geq 1$ for all $x \in R$, it implies that $f'(x) > 0$ for all $x \in R$.

A function is strictly increasing on an interval if its derivative is positive on that interval.

Since $f'(x) > 0$ for all $x \in R$, the function $f(x) = 2x + \cos x$ is strictly increasing on R.

A strictly increasing function does not have any local maximum or local minimum points.

The function $f(x)$ is an increasing function.

The correct option is (D).

Given:

The function is $y = f(x) = x (x - 3)^2$.

To Find:

The interval on which the function is decreasing.

Solution:

A function $f(x)$ is decreasing on an interval if its derivative $f'(x)$ is less than or equal to 0 on that interval.

First, find the derivative of $y$ with respect to $x$ using the product rule $\frac{d}{dx}(uv) = u'v + uv'$:

$y = x (x - 3)^2$

Let $u = x$ and $v = (x - 3)^2$.

$u' = \frac{du}{dx} = 1$

$v' = \frac{dv}{dx} = 2(x - 3) \cdot \frac{d}{dx}(x - 3) = 2(x - 3) \cdot 1 = 2(x - 3)$

$f'(x) = (1)(x - 3)^2 + x(2(x - 3))$

$f'(x) = (x - 3)^2 + 2x(x - 3)$

Factor out the common term $(x - 3)$:

$f'(x) = (x - 3)[(x - 3) + 2x]$

$f'(x) = (x - 3)[x - 3 + 2x]$

$f'(x) = (x - 3)(3x - 3)$

$f'(x) = 3(x - 3)(x - 1)$

To find where the function is decreasing, we need to find the interval(s) where $f'(x) \leq 0$.

$3(x - 3)(x - 1) \leq 0$

Divide the inequality by the positive constant 3:

$(x - 3)(x - 1) \leq 0$

The roots of the quadratic expression $(x - 3)(x - 1)$ are $x = 3$ and $x = 1$. These roots divide the number line into three intervals: $(-\infty, 1]$, $[1, 3]$, and $[3, \infty)$.

The expression $(x - 3)(x - 1)$ represents a parabola that opens upwards (since the coefficient of $x^2$ after expanding is positive). A parabola that opens upwards is less than or equal to zero between its roots (inclusive).

Therefore, the inequality $(x - 3)(x - 1) \leq 0$ is satisfied for values of $x$ between the roots 1 and 3, inclusive.

The interval where $f'(x) \leq 0$ is $[1, 3]$.

Thus, the function $y = x(x - 3)^2$ decreases on the interval $[1, 3]$.

Looking at the options, the interval $1 < x < 3$ is a subset of $[1, 3]$ and represents the open interval where the function is strictly decreasing ($f'(x) < 0$). At the endpoints $x=1$ and $x=3$, $f'(x)=0$. By convention, decreasing intervals are usually given as closed intervals if the derivative is $\leq 0$. However, among the given options, only $1 < x < 3$ falls within the decreasing range.

The function decreases for the values of $x$ in the interval $1 < x < 3$.

The correct option is (A).

Given:

The function is $f(x) = 4 \sin^3 x - 6 \sin^2 x + 12 \sin x + 100$.

To Determine:

The interval where the function is strictly increasing or decreasing.

Solution:

To determine where the function is strictly increasing or decreasing, we find its derivative $f'(x)$ and analyze its sign.

Let $u = \sin x$, so $f(x)$ can be viewed as a polynomial in $u$: $g(u) = 4u^3 - 6u^2 + 12u + 100$.

Using the chain rule, $f'(x) = \frac{dg}{du} \cdot \frac{du}{dx}$.

Find the derivative of $g(u)$ with respect to $u$:

$\frac{dg}{du} = \frac{d}{du}(4u^3 - 6u^2 + 12u + 100)$

$\frac{dg}{du} = 12u^2 - 12u + 12$

Find the derivative of $u = \sin x$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$

Now, combine these to find $f'(x)$:

$f'(x) = (12u^2 - 12u + 12) \cos x$

Substitute back $u = \sin x$:

$f'(x) = (12 \sin^2 x - 12 \sin x + 12) \cos x$

$f'(x) = 12 (\sin^2 x - \sin x + 1) \cos x$

Consider the quadratic expression in $\sin x$: $\sin^2 x - \sin x + 1$. Let $s = \sin x$. The expression is $s^2 - s + 1$. This is a quadratic in $s$. Its discriminant is $\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3$. Since the discriminant is negative and the coefficient of $s^2$ is positive (which is 1), the quadratic $s^2 - s + 1$ is always positive for all real values of $s$.

Since $s = \sin x$ and the range of $\sin x$ is $[-1, 1]$, which are real values, the expression $\sin^2 x - \sin x + 1$ is always positive for all $x$.

So, $12 (\sin^2 x - \sin x + 1)$ is always positive.

The sign of $f'(x) = 12 (\sin^2 x - \sin x + 1) \cos x$ depends solely on the sign of $\cos x$.

The function $f(x)$ is strictly increasing when $f'(x) > 0$, which happens when $\cos x > 0$.

The function $f(x)$ is strictly decreasing when $f'(x) < 0$, which happens when $\cos x < 0$.

Let's examine the sign of $\cos x$ in the intervals given in the options:

(A) $\left( π, \frac{3π}{2} \right)$: In this interval, $\cos x < 0$. So, $f'(x) < 0$. The function is strictly decreasing.

(B) $\left( \frac{π}{2} , π \right)$: In this interval, $\cos x < 0$. So, $f'(x) < 0$. The function is strictly decreasing.

(C) $\left[\frac{−π}{2} , \frac{π}{2} \right]$: In this interval, $\cos x \geq 0$ (positive for $(-\frac{\pi}{2}, \frac{\pi}{2})$, zero at endpoints). So, $f'(x) \geq 0$. The function is increasing (strictly increasing on $(-\frac{\pi}{2}, \frac{\pi}{2})$).

(D) $\left[ 0, \frac{π}{2} \right]$: In this interval, $\cos x \geq 0$ (positive for $(0, \frac{\pi}{2})$, zero at endpoints). So, $f'(x) \geq 0$. The function is increasing (strictly increasing on $(0, \frac{\pi}{2})$).

Let's re-examine the question and options. It asks for where the function is strictly increasing or decreasing.

Strictly increasing when $\cos x > 0$, i.e., in intervals like $(-\frac{\pi}{2} + 2n\pi, \frac{\pi}{2} + 2n\pi)$.

Strictly decreasing when $\cos x < 0$, i.e., in intervals like $(\frac{\pi}{2} + 2n\pi, \frac{3\pi}{2} + 2n\pi)$.

Let's check the options again based on strict inequality:

(A) $\left( π, \frac{3π}{2} \right)$: $\cos x < 0$. $f'(x) < 0$. Strictly decreasing.

(B) $\left( \frac{π}{2} , π \right)$: $\cos x < 0$. $f'(x) < 0$. Strictly decreasing.

(C) $\left[\frac{−π}{2} , \frac{π}{2} \right]$: $\cos x \geq 0$. $f'(x) \geq 0$. Increasing, but not strictly increasing at endpoints.

(D) $\left[ 0, \frac{π}{2} \right]$: $\cos x \geq 0$. $f'(x) \geq 0$. Increasing, but not strictly increasing at endpoint $\frac{\pi}{2}$.

There seems to be a discrepancy between the options and the strict definition. However, option (A) and (B) both indicate intervals where the function is strictly decreasing. Option (C) and (D) indicate intervals where it is increasing (possibly strictly increasing within the open interval).

Let's re-read the options carefully. Option (A) says strictly increasing, but our analysis shows it's decreasing there. Option (B) says strictly decreasing in $(\frac{\pi}{2}, \pi)$, which is correct as $\cos x < 0$ in this open interval.

Option (C) says decreasing in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. In the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\cos x > 0$, so it's strictly increasing. At the endpoints, $\cos x = 0$, so $f'(x) = 0$. The function is increasing on the closed interval, not decreasing.

Option (D) says decreasing in $[0, \frac{\pi}{2}]$. In the open interval $(0, \frac{\pi}{2})$, $\cos x > 0$, so it's strictly increasing. At endpoints, $\cos x \geq 0$. The function is increasing on the closed interval, not decreasing.

Let's check option (B) again. Is it possible that the question intended to ask about increasing instead of decreasing for options (A), (C), and (D)? Assuming the options as written are correct, option (B) states the function is decreasing in $(\frac{\pi}{2}, \pi)$. In this interval, $\cos x < 0$, so $f'(x) < 0$. This means the function is indeed strictly decreasing in $(\frac{\pi}{2}, \pi)$.

If the question intended to ask about strictly increasing intervals, then the function is strictly increasing where $\cos x > 0$. One such interval is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Neither (C) nor (D) is exactly this open interval, and they state "decreasing".

Given the options, the most consistent answer based on our derivative analysis is that the function is strictly decreasing in $(\frac{\pi}{2}, \pi)$.

The function is strictly decreasing in $\left( \frac{π}{2} , π \right)$.

The correct option is (B).

To Find:

The function among the given options that is decreasing on the interval $\left( 0, \frac{π}{2} \right)$.

Solution:

A function $f(x)$ is decreasing on an interval if its derivative $f'(x) \leq 0$ on that interval. For strict decrease, $f'(x) < 0$. Let's find the derivative of each function and check its sign on the interval $\left( 0, \frac{π}{2} \right)$.

For $x \in \left( 0, \frac{π}{2} \right)$, we know that $\sin x > 0$ and $\cos x > 0$.

Option (A): $f(x) = \sin(2x)$

$f'(x) = \frac{d}{dx}(\sin(2x)) = 2 \cos(2x)$

For $x \in \left( 0, \frac{π}{2} \right)$, $2x \in (0, \pi)$. In this interval, $\cos(2x)$ is positive for $2x \in \left( 0, \frac{\pi}{2} \right)$ (i.e., $x \in \left( 0, \frac{\pi}{4} \right)$) and negative for $2x \in \left( \frac{\pi}{2}, \pi \right)$ (i.e., $x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)$).

So, $f'(x)$ changes sign in $\left( 0, \frac{π}{2} \right)$. This function is not decreasing on the entire interval.

Option (B): $f(x) = \tan x$

$f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}$

For $x \in \left( 0, \frac{π}{2} \right)$, $\cos x > 0$, so $\cos^2 x > 0$. Thus, $\frac{1}{\cos^2 x} > 0$.

$f'(x) > 0$ on $\left( 0, \frac{π}{2} \right)$. This function is strictly increasing.

Option (C): $f(x) = \cos x$

$f'(x) = \frac{d}{dx}(\cos x) = -\sin x$

For $x \in \left( 0, \frac{π}{2} \right)$, $\sin x > 0$. Thus, $f'(x) = -\sin x < 0$.

$f'(x) < 0$ on $\left( 0, \frac{π}{2} \right)$. This function is strictly decreasing.

Option (D): $f(x) = \cos(3x)$

$f'(x) = \frac{d}{dx}(\cos(3x)) = -3 \sin(3x)$

For $x \in \left( 0, \frac{π}{2} \right)$, $3x \in \left( 0, \frac{3\pi}{2} \right)$. In this interval, $\sin(3x)$ is positive for $3x \in (0, \pi)$ (i.e., $x \in \left( 0, \frac{\pi}{3} \right)$) and negative for $3x \in \left( \pi, \frac{3\pi}{2} \right)$ (i.e., $x \in \left( \frac{\pi}{3}, \frac{\pi}{2} \right)$).

So, $f'(x) = -3 \sin(3x)$ is negative for $x \in \left( 0, \frac{\pi}{3} \right)$ and positive for $x \in \left( \frac{\pi}{3}, \frac{\pi}{2} \right)$.

$f'(x)$ changes sign in $\left( 0, \frac{π}{2} \right)$. This function is not decreasing on the entire interval.

Based on the analysis of the derivatives, only $f(x) = \cos x$ has a derivative that is negative throughout the interval $\left( 0, \frac{π}{2} \right)$.

The function $\cos x$ is decreasing on $\left( 0, \frac{π}{2} \right)$.

The correct option is (C).

Given:

The function is $f(x) = \tan x - x$.

To Determine:

Whether the function is increasing or decreasing.

Solution:

To determine the increasing or decreasing nature of the function, we find its first derivative $f'(x)$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\tan x - x)$

$f'(x) = \frac{d}{dx}(\tan x) - \frac{d}{dx}(x)$

$f'(x) = \sec^2 x - 1$

Recall the trigonometric identity: $\sec^2 x = 1 + \tan^2 x$. Substitute this into the expression for $f'(x)$:

$f'(x) = (1 + \tan^2 x) - 1$

$f'(x) = \tan^2 x$

For any real value of $x$ where $\tan x$ is defined, $\tan^2 x$ is always non-negative, i.e., $\tan^2 x \geq 0$.

The function $f(x) = \tan x - x$ is defined for all real numbers $x$ except $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

So, $f'(x) = \tan^2 x \geq 0$ for all $x$ in the domain of $f(x)$.

The derivative $f'(x) = 0$ only when $\tan x = 0$, which occurs at isolated points where $x = n\pi$ for some integer $n$.

In any interval where the function is defined and does not contain points of the form $n\pi$, $f'(x) = \tan^2 x > 0$.

Since $f'(x) \geq 0$ throughout its domain, and $f'(x)=0$ only at isolated points, the function is strictly increasing on each interval of its domain.

For example, on the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $f'(x) = \tan^2 x \geq 0$, and $f'(x)=0$ only at $x=0$. The function is strictly increasing on this interval.

The same applies to all intervals of the form $\left(n\pi - \frac{\pi}{2}, n\pi + \frac{\pi}{2}\right)$.

Therefore, the function $f(x) = \tan x - x$ always increases on its domain.

The correct option is (A).

Given:

The function is $f(x) = x^2 - 8x + 17$.

The variable $x$ is real.

To Find:

The minimum value of the function $f(x)$.

Solution:

To find the minimum value of a function, we can use calculus. We first find the critical points by setting the first derivative equal to zero.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^2 - 8x + 17)$

$f'(x) = 2x - 8$

Set the derivative equal to zero to find critical points:

$f'(x) = 0$

$2x - 8 = 0$

$2x = 8$

$x = 4$

The critical point is $x = 4$. To confirm that this corresponds to a minimum, we can use the second derivative test.

Find the second derivative of $f(x)$:

$f''(x) = \frac{d}{dx}(2x - 8)$

$f''(x) = 2$

Since $f''(4) = 2 > 0$, the function has a local minimum at $x = 4$. As this is a quadratic function opening upwards (coefficient of $x^2$ is positive), this local minimum is also the global minimum.

Now, evaluate the function $f(x)$ at $x = 4$ to find the minimum value:

$f(4) = (4)^2 - 8(4) + 17$

$f(4) = 16 - 32 + 17$

$f(4) = -16 + 17$

$f(4) = 1$

The minimum value of the function is 1.

Alternate Solution (Completing the Square):

We can rewrite the quadratic expression by completing the square:

$f(x) = x^2 - 8x + 17$

$f(x) = (x^2 - 8x + (4)^2) - (4)^2 + 17$

$f(x) = (x - 4)^2 - 16 + 17$

$f(x) = (x - 4)^2 + 1$

Since $(x - 4)^2$ is a square, its minimum value is 0, which occurs when $x - 4 = 0$, i.e., $x = 4$.

Therefore, the minimum value of $f(x)$ is $0 + 1 = 1$.

The minimum value of the function is 1.

The correct option is (C).

Given:

The polynomial is $f(x) = x^3 - 18x^2 + 96x$.

The interval is [0, 9].

To Find:

The smallest value (absolute minimum) of the polynomial on the closed interval [0, 9].

Solution:

To find the absolute minimum of a continuous function on a closed interval, we need to evaluate the function at the critical points within the interval and at the endpoints of the interval.

First, find the critical points by setting the first derivative equal to zero.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(x^3 - 18x^2 + 96x)$

$f'(x) = 3x^2 - 36x + 96$

Set the derivative equal to zero:

$3x^2 - 36x + 96 = 0$

Divide the equation by 3:

$x^2 - 12x + 32 = 0$

Factor the quadratic equation:

We look for two numbers that multiply to 32 and add up to -12. These numbers are -4 and -8.

$(x - 4)(x - 8) = 0$

The critical points are $x = 4$ and $x = 8$.

Now, check if these critical points lie within the given interval [0, 9]. Both $x=4$ and $x=8$ are in the interval [0, 9].

Evaluate the function $f(x)$ at the critical points ($x=4$, $x=8$) and the endpoints of the interval ($x=0$, $x=9$).

Evaluate at $x = 0$ (endpoint):

$f(0) = (0)^3 - 18(0)^2 + 96(0) = 0 - 0 + 0 = 0$

Evaluate at $x = 4$ (critical point):

$f(4) = (4)^3 - 18(4)^2 + 96(4)$

$f(4) = 64 - 18(16) + 384$

$f(4) = 64 - 288 + 384$

$f(4) = -224 + 384 = 160$

Evaluate at $x = 8$ (critical point):

$f(8) = (8)^3 - 18(8)^2 + 96(8)$

$f(8) = 512 - 18(64) + 768$

$f(8) = 512 - 1152 + 768$

$f(8) = -640 + 768 = 128$

Evaluate at $x = 9$ (endpoint):

$f(9) = (9)^3 - 18(9)^2 + 96(9)$

$f(9) = 729 - 18(81) + 864$

$f(9) = 729 - 1458 + 864$

$f(9) = -729 + 864 = 135$

Compare the values of the function at the critical points and endpoints:

$f(0) = 0$

$f(4) = 160$

$f(8) = 128$

$f(9) = 135$

The smallest value among these is 0.

The smallest value of the polynomial $x^3 - 18x^2 + 96x$ in the interval [0, 9] is 0.

The correct option is (B).

Given:

The function is $f(x) = 2x^3 - 3x^2 - 12x + 4$.

To Determine:

The number and type of local extrema (maxima and minima) of the function.

Solution:

Local extrema occur at critical points, where the first derivative $f'(x)$ is equal to zero or undefined. For a polynomial function, the derivative is always defined.

First, find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 4)$

$f'(x) = 2(3x^2) - 3(2x) - 12(1) + 0$

$f'(x) = 6x^2 - 6x - 12$

Set the derivative equal to zero to find critical points:

$f'(x) = 0$

$6x^2 - 6x - 12 = 0$

Divide the equation by 6:

$x^2 - x - 2 = 0$

Factor the quadratic equation:

We look for two numbers that multiply to -2 and add up to -1. These numbers are -2 and 1.

$(x - 2)(x + 1) = 0$

The critical points are $x = 2$ and $x = -1$.

To determine if these critical points correspond to a local maximum or minimum, we can use the second derivative test.

Find the second derivative of $f(x)$:

$f''(x) = \frac{d}{dx}(6x^2 - 6x - 12)$

$f''(x) = 12x - 6$

Evaluate the second derivative at each critical point:

At $x = 2$:

$f''(2) = 12(2) - 6 = 24 - 6 = 18$

Since $f''(2) = 18 > 0$, the function has a local minimum at $x = 2$.

At $x = -1$:

$f''(-1) = 12(-1) - 6 = -12 - 6 = -18$

Since $f''(-1) = -18 < 0$, the function has a local maximum at $x = -1$.

Thus, the function has one local maximum (at $x=-1$) and one local minimum (at $x=2$).

We can also find the values of the function at these points, although it's not required by the question:

Local maximum value at $x = -1$:

$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 4$

$f(-1) = 2(-1) - 3(1) + 12 + 4$

$f(-1) = -2 - 3 + 12 + 4 = 11$

Local minimum value at $x = 2$:

$f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 4$

$f(2) = 2(8) - 3(4) - 24 + 4$

$f(2) = 16 - 12 - 24 + 4$

$f(2) = 4 - 24 + 4 = -16$

The function has one local maximum and one local minimum.

The correct option is (C).

Given:

The function is $f(x) = \sin x \cos x$.

To Find:

The maximum value of the function $f(x)$.

Solution:

We can use a trigonometric identity to simplify the function.

Recall the double angle identity for sine:

$\sin(2x) = 2 \sin x \cos x$

We can rewrite the given function using this identity:

$f(x) = \sin x \cos x = \frac{1}{2} (2 \sin x \cos x)$

$f(x) = \frac{1}{2} \sin(2x)$

Now, we need to find the maximum value of $f(x) = \frac{1}{2} \sin(2x)$.

The range of the sine function, $\sin \theta$, for any real value of $\theta$, is the interval $[-1, 1]$.

This means:

$-1 \leq \sin(2x) \leq 1$

To find the range of $\frac{1}{2} \sin(2x)$, multiply the inequality by $\frac{1}{2}$:

$\frac{1}{2} (-1) \leq \frac{1}{2} \sin(2x) \leq \frac{1}{2} (1)$

$-\frac{1}{2} \leq \frac{1}{2} \sin(2x) \leq \frac{1}{2}$

So, the range of the function $f(x) = \sin x \cos x$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$.

The maximum value of the function is the upper bound of this range.

Maximum value $= \frac{1}{2}$.

This maximum value is attained when $\sin(2x) = 1$, which occurs for values of $x$ such as $2x = \frac{\pi}{2}, \frac{5\pi}{2}, \dots$ or $x = \frac{\pi}{4}, \frac{5\pi}{4}, \dots$

The maximum value of $\sin x \cos x$ is $\frac{1}{2}$.

The correct option is (B).

Given:

The function is $f(x) = 2 \sin 3x + 3 \cos 3x$.

The point is $x = \frac{5\pi}{6}$.

To Determine:

Whether the function has a maximum, minimum, zero, or neither at $x = \frac{5\pi}{6}$.

Solution:

To determine if the function has a local maximum or minimum at a point, we first find the first derivative $f'(x)$ and evaluate it at the given point. If $f'(x) \neq 0$ at that point, it is neither a local maximum nor a local minimum.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(2 \sin 3x + 3 \cos 3x)$

$f'(x) = 2 \frac{d}{dx}(\sin 3x) + 3 \frac{d}{dx}(\cos 3x)$

Using the chain rule:

$\frac{d}{dx}(\sin 3x) = \cos 3x \cdot \frac{d}{dx}(3x) = 3 \cos 3x$

$\frac{d}{dx}(\cos 3x) = -\sin 3x \cdot \frac{d}{dx}(3x) = -3 \sin 3x$

So, $f'(x) = 2(3 \cos 3x) + 3(-3 \sin 3x)$

$f'(x) = 6 \cos 3x - 9 \sin 3x$

Now, evaluate $f'(x)$ at $x = \frac{5\pi}{6}$:

We need to evaluate $3x$ at this point:

$3x = 3 \cdot \frac{5\pi}{6} = \frac{15\pi}{6} = \frac{5\pi}{2}$

Substitute $3x = \frac{5\pi}{2}$ into the expression for $f'(x)$:

$f'(\frac{5\pi}{6}) = 6 \cos \left(\frac{5\pi}{2}\right) - 9 \sin \left(\frac{5\pi}{2}\right)$

We know the values of $\cos$ and $\sin$ at $\frac{5\pi}{2}$ (which is coterminal with $\frac{\pi}{2}$):

$\cos \left(\frac{5\pi}{2}\right) = \cos \left(2\pi + \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{2}\right) = 0$

$\sin \left(\frac{5\pi}{2}\right) = \sin \left(2\pi + \frac{\pi}{2}\right) = \sin \left(\frac{\pi}{2}\right) = 1$

Substitute these values into $f'(\frac{5\pi}{6})$:

$f'(\frac{5\pi}{6}) = 6(0) - 9(1)$

$f'(\frac{5\pi}{6}) = 0 - 9 = -9$

Since the derivative $f'(\frac{5\pi}{6}) = -9$ is not equal to zero, $x = \frac{5\pi}{6}$ is not a critical point of the function. A function can only have a local maximum or minimum at a critical point where the derivative is zero or undefined.

Therefore, at $x = \frac{5\pi}{6}$, the function has neither a local maximum nor a local minimum.

We can also evaluate the function value at $x = \frac{5\pi}{6}$ to check option (C):

$f(\frac{5\pi}{6}) = 2 \sin \left(3 \cdot \frac{5\pi}{6}\right) + 3 \cos \left(3 \cdot \frac{5\pi}{6}\right)$

$f(\frac{5\pi}{6}) = 2 \sin \left(\frac{5\pi}{2}\right) + 3 \cos \left(\frac{5\pi}{2}\right)$

$f(\frac{5\pi}{6}) = 2(1) + 3(0) = 2 + 0 = 2$

Since $f(\frac{5\pi}{6}) = 2 \neq 0$, the function is not zero at this point.

At $x = \frac{5\pi}{6}$, the function $f(x)$ is neither maximum nor minimum.

The correct option is (D).

Given:

The equation of the curve is $y = -x^3 + 3x^2 + 9x - 27$.

To Find:

The maximum value of the slope of the tangent to the curve.

Solution:

The slope of the tangent to the curve at any point $x$ is given by the first derivative $\frac{dy}{dx}$.

Find the first derivative of $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(-x^3 + 3x^2 + 9x - 27)$

$\frac{dy}{dx} = -3x^2 + 6x + 9$

Let the slope function be $m(x) = -3x^2 + 6x + 9$. We need to find the maximum value of this quadratic function.

The function $m(x)$ is a quadratic function of the form $ax^2 + bx + c$ with $a = -3$, $b = 6$, and $c = 9$. Since the coefficient of $x^2$ ($a = -3$) is negative, the parabola opens downwards, which means it has a maximum value at its vertex.

The x-coordinate of the vertex of a parabola $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$.

For $m(x) = -3x^2 + 6x + 9$, the x-coordinate of the vertex is:

$x = -\frac{6}{2(-3)} = -\frac{6}{-6} = 1$

The maximum slope occurs at $x = 1$. Now, substitute this value of $x$ back into the slope function $m(x)$ to find the maximum slope value:

Maximum slope $= m(1) = -3(1)^2 + 6(1) + 9$

Maximum slope $= -3(1) + 6 + 9$

Maximum slope $= -3 + 6 + 9$

Maximum slope $= 3 + 9 = 12$

Alternatively, using calculus to find the maximum of $m(x)$: Find the derivative of $m(x)$ with respect to $x$ (which is the second derivative of $y$):

$m'(x) = \frac{d}{dx}(-3x^2 + 6x + 9)$

$m'(x) = -6x + 6$

Set $m'(x) = 0$ to find the critical point(s) of the slope function:

$-6x + 6 = 0$

$-6x = -6$

$x = 1$

To confirm this is a maximum for $m(x)$, find the second derivative of $m(x)$ (which is the third derivative of $y$):

$m''(x) = \frac{d}{dx}(-6x + 6) = -6$

Since $m''(1) = -6 < 0$, the slope function $m(x)$ has a local maximum at $x = 1$. Since $m(x)$ is a downward-opening parabola, this local maximum is also the global maximum.

The maximum slope is $m(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12$.

The maximum slope of the curve is 12.

The correct option is (B).

Given:

The function is $f(x) = x^x$.

The domain of the function $x^x$ for real $x$ is typically considered for $x > 0$.

To Find:

The value of $x$ where the function has a stationary point.

Solution:

A stationary point of a function occurs where the first derivative is equal to zero.

To find the derivative of $y = x^x$, we can use logarithmic differentiation.

Let $y = x^x$.

Take the natural logarithm of both sides:

$\log y = \log (x^x)$

Using the logarithm property $\log (a^b) = b \log a$:

$\log y = x \log x$

Now, differentiate both sides implicitly with respect to $x$:

$\frac{d}{dx}(\log y) = \frac{d}{dx}(x \log x)$

Using the chain rule on the left side and the product rule on the right side:

$\frac{1}{y} \frac{dy}{dx} = (1) \log x + x \left(\frac{1}{x}\right)$

$\frac{1}{y} \frac{dy}{dx} = \log x + 1$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = y (\log x + 1)$

Substitute $y = x^x$ back into the equation:

$f'(x) = \frac{dy}{dx} = x^x (\log x + 1)$

To find stationary points, set the derivative equal to zero:

$f'(x) = 0$

$x^x (\log x + 1) = 0$

Since the domain is $x > 0$, $x^x$ is always positive and never zero.

Therefore, for the product to be zero, the other factor must be zero:

$\log x + 1 = 0$

$\log x = -1$

To solve for $x$, exponentiate both sides with base $e$:

$e^{\log x} = e^{-1}$

$x = e^{-1}$

$x = \frac{1}{e}$

The function $f(x) = x^x$ has a stationary point at $x = \frac{1}{e}$.

The correct option is (B).

Given:

The function is $f(x) = \left(\frac{1}{x}\right)^x$.

For the function to be well-defined for real $x$, we require $\frac{1}{x} > 0$, which implies $x > 0$.

To Find:

The maximum value of the function $f(x)$.

Solution:

To find the maximum value, we first find the critical points by setting the first derivative equal to zero.

Let $y = \left(\frac{1}{x}\right)^x$. We can write $\frac{1}{x} = x^{-1}$.

$y = (x^{-1})^x = x^{-x}$

Use logarithmic differentiation to find the derivative.

Take the natural logarithm of both sides:

$\log y = \log (x^{-x})$

Using the logarithm property $\log (a^b) = b \log a$:

$\log y = -x \log x$

Now, differentiate both sides implicitly with respect to $x$:

$\frac{d}{dx}(\log y) = \frac{d}{dx}(-x \log x)$

Using the chain rule on the left side and the product rule on the right side:

$\frac{1}{y} \frac{dy}{dx} = -(1 \cdot \log x + x \cdot \frac{1}{x})$

$\frac{1}{y} \frac{dy}{dx} = -(\log x + 1)$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = -y (\log x + 1)$

Substitute $y = x^{-x}$ back into the equation:

$f'(x) = \frac{dy}{dx} = -x^{-x} (\log x + 1)$

To find stationary points, set the derivative equal to zero:

$f'(x) = 0$

$-x^{-x} (\log x + 1) = 0$

Since $x > 0$, $x^{-x} = \frac{1}{x^x}$ is always positive and never zero. The factor $-x^{-x}$ is always negative and never zero.

Therefore, for the product to be zero, the other factor must be zero:

$\log x + 1 = 0$

$\log x = -1$

To solve for $x$, exponentiate both sides with base $e$:

$e^{\log x} = e^{-1}$

$x = e^{-1}$

$x = \frac{1}{e}$

The critical point is $x = \frac{1}{e}$. To determine if this is a maximum, we can use the first or second derivative test. Let's use the first derivative test by checking the sign of $f'(x)$ around $x = \frac{1}{e}$.

$f'(x) = -x^{-x} (\log x + 1)$

The term $-x^{-x}$ is always negative for $x > 0$. The sign of $f'(x)$ is determined by the sign of $\log x + 1$.

$\log x + 1 > 0 \implies \log x > -1 \implies x > e^{-1} = \frac{1}{e}$

$\log x + 1 < 0 \implies \log x < -1 \implies x < e^{-1} = \frac{1}{e}$

So, for $0 < x < \frac{1}{e}$, $\log x + 1 < 0$, which means $f'(x) = \text{(negative)} \times \text{(negative)} > 0$. The function is increasing.

For $x > \frac{1}{e}$, $\log x + 1 > 0$, which means $f'(x) = \text{(negative)} \times \text{(positive)} < 0$. The function is decreasing.

Since the function changes from increasing to decreasing at $x = \frac{1}{e}$, there is a local maximum at $x = \frac{1}{e}$. For this function, this local maximum is also the global maximum.

Now, evaluate the function $f(x)$ at $x = \frac{1}{e}$ to find the maximum value:

$f\left(\frac{1}{e}\right) = \left(\frac{1}{\frac{1}{e}}\right)^{\frac{1}{e}}$

$f\left(\frac{1}{e}\right) = (e)^{\frac{1}{e}}$

$f\left(\frac{1}{e}\right) = e^{\frac{1}{e}}$

The maximum value of $\left( \frac{1}{x} \right)^x$ is $e^{\frac{1}{e}}$.

The correct option is (C).

Given:

The equations of the two curves are:

Curve 1: $y_1 = 4x^2 + 2x - 8$

Curve 2: $y_2 = x^3 - x + 13$

To Find:

The point where the two curves touch each other.

Solution:

For two curves to touch at a point $(x_0, y_0)$, they must satisfy two conditions at that point:

1. They must intersect: $y_1(x_0) = y_2(x_0)$

2. They must have the same slope: $y_1'(x_0) = y_2'(x_0)$

First, find the derivatives of both functions with respect to $x$:

$y_1' = \frac{d}{dx}(4x^2 + 2x - 8) = 8x + 2$

$y_2' = \frac{d}{dx}(x^3 - x + 13) = 3x^2 - 1$

Set the derivatives equal to each other to find the x-values where the tangent lines are parallel:

$8x + 2 = 3x^2 - 1$

Rearrange the terms to form a quadratic equation:

$3x^2 - 8x - 1 - 2 = 0$

$3x^2 - 8x - 3 = 0$

Factor the quadratic equation:

We look for two numbers that multiply to $(3)(-3) = -9$ and add up to -8. These numbers are -9 and 1.

$3x^2 - 9x + x - 3 = 0$

$3x(x - 3) + 1(x - 3) = 0$

$(3x + 1)(x - 3) = 0$

The possible x-values where the slopes are equal are $x = 3$ and $x = -\frac{1}{3}$.

Now, we check if the curves intersect at these x-values by evaluating the original functions at $x = 3$ and $x = -\frac{1}{3}$.

For $x = 3$:

$y_1(3) = 4(3)^2 + 2(3) - 8 = 4(9) + 6 - 8 = 36 + 6 - 8 = 34$

$y_2(3) = (3)^3 - (3) + 13 = 27 - 3 + 13 = 24 + 13 = 37$

Since $y_1(3) = 34$ and $y_2(3) = 37$, $y_1(3) \neq y_2(3)$. The curves do not intersect at $x = 3$.

For $x = -\frac{1}{3}$:

$y_1(-\frac{1}{3}) = 4(-\frac{1}{3})^2 + 2(-\frac{1}{3}) - 8 = 4(\frac{1}{9}) - \frac{2}{3} - 8 = \frac{4}{9} - \frac{6}{9} - \frac{72}{9} = \frac{4 - 6 - 72}{9} = -\frac{74}{9}$

$y_2(-\frac{1}{3}) = (-\frac{1}{3})^3 - (-\frac{1}{3}) + 13 = -\frac{1}{27} + \frac{1}{3} + 13 = -\frac{1}{27} + \frac{9}{27} + \frac{351}{27} = \frac{-1 + 9 + 351}{27} = \frac{359}{27}$

Since $y_1(-\frac{1}{3}) = -\frac{74}{9} = -\frac{222}{27}$ and $y_2(-\frac{1}{3}) = \frac{359}{27}$, $y_1(-\frac{1}{3}) \neq y_2(-\frac{1}{3})$. The curves do not intersect at $x = -\frac{1}{3}$.

Based on the given equations, there is no point where the curves intersect and have the same slope simultaneously. This means the curves do not touch each other as written.

However, in problems of this type, it is often intended that the curves touch at one of the points where the slopes are equal. Assuming there is a minor typo in one of the equations such that they touch at $x=3$ (which yielded simpler coordinates), the required y-coordinate for the touching point would be $y_1(3) = 34$. If the second curve were meant to pass through (3, 34) with the calculated slope, its equation might have had a different constant term (e.g., $y=x^3-x+10$, since $3^3-3+10=34$).

Assuming the intended question corresponds to a scenario where the curves touch at $x=3$, the point of touching is (3, 34).

Answer for the blank: (3, 34)

Given:

The equation of the curve is $y = \tan x$.

The point on the curve is (0, 0).

To Find:

The equation of the normal to the curve at the point (0, 0).

Solution:

First, we need to find the slope of the tangent to the curve at the point (0, 0). The slope of the tangent is given by the derivative $\frac{dy}{dx}$.

Find the derivative of $y$ with respect to $x$:

$y = \tan x$

$\frac{dy}{dx} = \frac{d}{dx}(\tan x) = \sec^2 x$

Evaluate the slope of the tangent at the point (0, 0) by substituting $x=0$:

Slope of tangent $m_t = \left. \frac{dy}{dx} \right|_{x=0} = \sec^2 (0)$

Since $\sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1$, we have:

$m_t = (1)^2 = 1$

The slope of the normal to the curve at (0, 0) is the negative reciprocal of the slope of the tangent, provided the tangent slope is not zero.

Slope of normal $m_n = -\frac{1}{m_t} = -\frac{1}{1} = -1$

Now, use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (0, 0)$ and the slope $m = m_n = -1$.

$y - 0 = -1(x - 0)$

$y = -x$

Rearrange the terms to get the standard form:

$x + y = 0$

The equation of the normal to the curve $y = \tan x$ at (0, 0) is $x + y = 0$.

Answer for the blank: $x + y = 0$

Given:

The function is $f(x) = \sin x - ax + b$.

The function increases on R (all real numbers).

To Find:

The values of the constant $a$ for which the function is increasing on R.

Solution:

A function $f(x)$ is increasing on an interval if its derivative $f'(x) \geq 0$ on that interval. For strict increase, $f'(x) > 0$. Since the question asks for increasing on R, we need $f'(x) \geq 0$ for all $x \in R$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(\sin x - ax + b)$

$f'(x) = \cos x - a + 0$

$f'(x) = \cos x - a$

For the function to be increasing on R, we must have $f'(x) \geq 0$ for all $x \in R$.

$\cos x - a \geq 0$

$\cos x \geq a$

This inequality must hold for all real values of $x$. The range of the cosine function is $[-1, 1]$, which means $-1 \leq \cos x \leq 1$ for all $x \in R$.

For $\cos x \geq a$ to be true for all $x$, the value of $a$ must be less than or equal to the minimum possible value of $\cos x$.

The minimum value of $\cos x$ is -1.

Therefore, we must have $a \leq -1$.

If $a < -1$, then $\cos x \geq a$ is always true since $\cos x \geq -1$ and $-1 > a$. In this case, $\cos x - a > 0$ for all $x$, so $f'(x) > 0$, and the function is strictly increasing.

If $a = -1$, then $\cos x \geq -1$ is always true. In this case, $f'(x) = \cos x - (-1) = \cos x + 1$. $f'(x) \geq 0$ for all $x \in R$, and $f'(x) = 0$ at isolated points where $\cos x = -1$ (i.e., $x = \pi + 2n\pi$). Since the derivative is non-negative and is zero only at isolated points, the function is increasing on R.

So, the values of $a$ for which the function increases on R are $a \leq -1$. This can be written in interval notation as $(-\infty, -1]$.

The values of a for which the function f (x) = sin x – ax + b increases on R are $a \leq -1$.

Answer for the blank: $a \leq -1$ or $(-\infty, -1]$

Given:

The function is $f(x) = \frac{2x^2 - 1}{x^4}$ for $x > 0$.

To Find:

The interval on which the function is decreasing.

Solution:

A function $f(x)$ is decreasing on an interval if its derivative $f'(x) \leq 0$ on that interval. Since the domain is $x > 0$, we will analyze the sign of $f'(x)$ for $x > 0$.

First, simplify the function by splitting the fraction:

$f(x) = \frac{2x^2}{x^4} - \frac{1}{x^4} = 2x^{-2} - x^{-4}$

Now, find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(2x^{-2} - x^{-4})$

$f'(x) = 2(-2x^{-3}) - (-4x^{-5})$

$f'(x) = -4x^{-3} + 4x^{-5}$

$f'(x) = -\frac{4}{x^3} + \frac{4}{x^5}$

Combine the terms into a single fraction:

$f'(x) = \frac{-4x^2 + 4}{x^5} = \frac{4(1 - x^2)}{x^5}$

To find where the function is decreasing, we set $f'(x) \leq 0$:

$\frac{4(1 - x^2)}{x^5} \leq 0$

We are given that $x > 0$. For $x > 0$, the denominator $x^5$ is always positive.

So, the sign of $f'(x)$ depends only on the sign of the numerator $4(1 - x^2)$. Since 4 is positive, the sign depends on $(1 - x^2)$.

We need $1 - x^2 \leq 0$ for $f'(x) \leq 0$.

$1 - x^2 \leq 0$

$1 \leq x^2$

$x^2 \geq 1$

Taking the square root of both sides, and considering $x > 0$:

$\sqrt{x^2} \geq \sqrt{1}$

$|x| \geq 1$

Since $x > 0$, $|x| = x$.

$x \geq 1$

So, the function is decreasing when $x \geq 1$. Since the domain is $x > 0$, the interval where the function is decreasing is $[1, \infty)$. For strict decrease ($f'(x) < 0$), the interval is $(1, \infty)$. At $x=1$, $f'(1)=0$. Typically, the interval of decrease is given as a closed interval if the derivative is $\leq 0$. Considering the options usually presented in multiple-choice questions, the intended interval for decrease might be where $f'(x) \leq 0$. Without specific multiple-choice options, we state the interval where $f'(x) \leq 0$. If the blank requires an open interval for strict decrease, it would be $(1, \infty)$. Given the context of fill-in-the-blank, providing the interval where $f'(x) \leq 0$ is standard for decreasing intervals unless "strictly decreasing" is specified.

The function decreases in the interval $[1, \infty)$.

Answer for the blank: $[1, \infty)$

Given:

The function is $f(x) = ax + \frac{b}{x}$.

Constraints: $a > 0$, $b > 0$, $x > 0$.

To Find:

The least value (minimum value) of the function $f(x)$.

Solution (Using Calculus):

To find the minimum value, we first find the critical points by setting the first derivative equal to zero.

Rewrite the function: $f(x) = ax + bx^{-1}$.

Find the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx}(ax + bx^{-1})$

$f'(x) = a(1) + b(-1x^{-2})$

$f'(x) = a - \frac{b}{x^2}$

Set the derivative equal to zero to find critical points:

$f'(x) = 0$

$a - \frac{b}{x^2} = 0$

$a = \frac{b}{x^2}$

$ax^2 = b$

$x^2 = \frac{b}{a}$

Since $x > 0$ and $a, b > 0$, we take the positive square root:

$x = \sqrt{\frac{b}{a}}$

This is the critical point. To confirm that this corresponds to a minimum, we can use the second derivative test.

Find the second derivative of $f(x)$:

$f'(x) = a - bx^{-2}$

$f''(x) = \frac{d}{dx}(a - bx^{-2}) = 0 - b(-2x^{-3})$

$f''(x) = 2bx^{-3} = \frac{2b}{x^3}$

Evaluate the second derivative at the critical point $x = \sqrt{\frac{b}{a}}$:

$f''\left(\sqrt{\frac{b}{a}}\right) = \frac{2b}{\left(\sqrt{\frac{b}{a}}\right)^3}$

Since $b > 0$ and $x > 0$, $\frac{2b}{x^3} > 0$. Thus, $f''\left(\sqrt{\frac{b}{a}}\right) > 0$.

Since the second derivative is positive at the critical point, the function has a local minimum at $x = \sqrt{\frac{b}{a}}$. Since this is the only critical point in the domain ($x>0$), this local minimum is the global minimum (least value).

Now, evaluate the function $f(x)$ at $x = \sqrt{\frac{b}{a}}$ to find the minimum value:

$f\left(\sqrt{\frac{b}{a}}\right) = a\left(\sqrt{\frac{b}{a}}\right) + \frac{b}{\sqrt{\frac{b}{a}}}$

$f\left(\sqrt{\frac{b}{a}}\right) = a \frac{\sqrt{b}}{\sqrt{a}} + b \frac{\sqrt{a}}{\sqrt{b}}$

$f\left(\sqrt{\frac{b}{a}}\right) = \frac{a \sqrt{b} \sqrt{a}}{\sqrt{a}} + \frac{b \sqrt{a} \sqrt{b}}{\sqrt{b}}$

$f\left(\sqrt{\frac{b}{a}}\right) = \sqrt{a^2 b} + \sqrt{b^2 a}$

$f\left(\sqrt{\frac{b}{a}}\right) = a\sqrt{b} + b\sqrt{a}$

Let's simplify differently:

$f\left(\sqrt{\frac{b}{a}}\right) = a\sqrt{\frac{b}{a}} + \frac{b}{\sqrt{\frac{b}{a}}}$

$f\left(\sqrt{\frac{b}{a}}\right) = \sqrt{a^2 \cdot \frac{b}{a}} + b \cdot \sqrt{\frac{a}{b}}$

$f\left(\sqrt{\frac{b}{a}}\right) = \sqrt{ab} + \sqrt{b^2 \cdot \frac{a}{b}}$

$f\left(\sqrt{\frac{b}{a}}\right) = \sqrt{ab} + \sqrt{ab}$

$f\left(\sqrt{\frac{b}{a}}\right) = 2\sqrt{ab}$

Alternate Solution (Using AM-GM Inequality):

Since $a > 0$, $b > 0$, and $x > 0$, the terms $ax$ and $\frac{b}{x}$ are both positive.

By the Arithmetic Mean - Geometric Mean (AM-GM) inequality, for any two non-negative numbers $u$ and $v$, we have $\frac{u+v}{2} \geq \sqrt{uv}$, with equality if and only if $u = v$.

Let $u = ax$ and $v = \frac{b}{x}$. Both are positive for $x > 0, a > 0, b > 0$.

$\frac{ax + \frac{b}{x}}{2} \geq \sqrt{(ax)\left(\frac{b}{x}\right)}$

$\frac{ax + \frac{b}{x}}{2} \geq \sqrt{ab}$

$ax + \frac{b}{x} \geq 2\sqrt{ab}$

So, $f(x) \geq 2\sqrt{ab}$. The minimum value is $2\sqrt{ab}$.

Equality holds when $ax = \frac{b}{x}$, which gives $ax^2 = b$, or $x^2 = \frac{b}{a}$. Since $x > 0$, $x = \sqrt{\frac{b}{a}}$. This confirms that the minimum is attained within the domain.

The least value of the function $f(x) = ax + \frac{b}{x}$ is $2\sqrt{ab}$.

Answer for the blank: $2\sqrt{ab}$