CBSE Class 12 Mathematics Application Of Derivative Worksheet Set 04

Very Short Answer Type Questions

Question. Find the values of \( k \) for which \( f(x) = kx^3 - 9kx^2 + 9x + 3 \) is increasing on \( R \).
Answer: We have, \( f(x) = kx^3 - 9kx^2 + 9x + 3 \)
On differentiating w.r.t. \( x \), we get
\( f'(x) = 3kx^2 - 18kx + 9 \)
If \( f(x) \) is increasing on \( R \), then
\( f'(x) \geq 0 \Rightarrow 3kx^2 - 18kx + 9 \geq 0 \)
When a quadratic \( ax^2 + bx + c \geq 0 \)
Then, \( a > 0 \) and \( D \leq 0 \)
So, \( 3k > 0 \) and \( (-18k)^2 - 4 \times 9 \times 3k \leq 0 \)
\( \Rightarrow k > 0 \) and \( 324k^2 - 108k \leq 0 \)
\( \Rightarrow 108k(3k - 1) \leq 0 \)
\( \Rightarrow 0 \leq k \leq \frac{1}{3} \)
So, \( k \in \left(0, \frac{1}{3}\right] \).

Question. Find the interval in which the function \( f(x) = \frac{x}{\sin x} \) is increasing.
Answer: We have, \( f(x) = \frac{x}{\sin x} \)
\( \Rightarrow f'(x) = \frac{\sin x \cdot 1 - x \cdot \cos x}{\sin^2 x} = \frac{\sin x - x \cos x}{\sin^2 x} \)
Clearly, \( \sin^2 x > 0 \), \( \forall x \in \left(0, \frac{\pi}{2}\right) \).
Now, \( f'(x) > 0 \Rightarrow \sin x - x \cos x > 0 \Rightarrow \sin x > x \cos x \)
\( \Rightarrow \tan x > x \), which is true, \( \forall x \in \left(0, \frac{\pi}{2}\right) \)
Thus, \( f'(x) > 0 \), \( \forall x \in \left(0, \frac{\pi}{2}\right) \).
So, \( f(x) \) is increasing on \( \left(0, \frac{\pi}{2}\right) \).

Question. Given that \( f(x) = \frac{\log x}{x} \), find the point of local maximum of \( f(x) \).
Answer: Given, \( f(x) = \frac{\log x}{x} \)
On differentiating both sides w.r.t. \( x \), we get
\( f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} \)
\( \Rightarrow f'(x) = \frac{x \cdot \frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2} \) ...(i)
For local maxima, \( f'(x) = 0 \)
\( \Rightarrow \frac{1 - \log x}{x^2} = 0 \Rightarrow 1 - \log x = 0 \)
\( \Rightarrow \log x = 1 \Rightarrow x = e \)
Again, on differentiating Eq. (i) both sides w.r.t. \( x \), we get
\( f''(x) = \frac{x^2 \frac{d}{dx}(1 - \log x) - (1 - \log x) \frac{d}{dx}(x^2)}{(x^2)^2} \)
\( = \frac{x^2(-1/x) - 2x(1 - \log x)}{x^4} \)
\( = \frac{-x - 2x(1 - \log x)}{x^4} \)
\( = \frac{-1 - 2 + 2\log x}{x^3} = \frac{-3 + 2\log x}{x^3} \)
At \( x = e \), \( f''(e) = \frac{-3 + 2\log e}{e^3} = \frac{-1}{e^3} < 0 \)
\( \therefore x = e \) is point of maxima.
Hence, \( f(x) \) is maximum at \( x = e \).

Question. Find the local maximum and local minimum values of the function \( f(x) = e^{x+2} \).
Answer: We have, \( f(x) = e^{x+2} \)
On differentiating both sides, we get
\( f'(x) = e^{x+2} \)
For local maximum or local minimum, on putting \( f'(x) = 0 \Rightarrow e^{x+2} = 0 \Rightarrow \) No value of \( x \) exist.
Hence, no local maximum and no local minimum value can be determined.

Question. Find the stationary (critical) point of the function \( f(x) = x^x \).
Answer: We have, \( f(x) = x^x \)
Let \( y = x^x \)
On taking log both sides, we get
\( \log y = \log x^x \Rightarrow \log y = x \log x \)
\( \therefore \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \)
\( \Rightarrow \frac{dy}{dx} = (1 + \log x) \cdot x^x \)
For local maxima or local minima, \( \frac{dy}{dx} = 0 \)
\( \Rightarrow (1 + \log x) \cdot x^x = 0 \Rightarrow \log x = -1 \)
\( \Rightarrow \log x = \log e^{-1} \Rightarrow x = e^{-1} \Rightarrow x = \frac{1}{e} \)
Ans. \( f(x) \) has a stationary point at \( x = \frac{1}{e} \).

Question. Prove that the maximum value of \( \left(\frac{1}{x}\right)^x \) is \( e^{1/e} \).
Answer: Let \( y = \left(\frac{1}{x}\right)^x \). Taking logarithm on both sides, we get:
\( \log y = x \log\left(\frac{1}{x}\right) = -x \log x \)
Differentiating both sides w.r.t. \( x \):
\( \frac{1}{y}\frac{dy}{dx} = - \left(x \cdot \frac{1}{x} + \log x \cdot 1\right) = -(1 + \log x) \)
\( \Rightarrow \frac{dy}{dx} = -\left(\frac{1}{x}\right)^x (1 + \log x) \)
For critical points, put \( \frac{dy}{dx} = 0 \):
\( -\left(\frac{1}{x}\right)^x (1 + \log x) = 0 \Rightarrow 1 + \log x = 0 \Rightarrow \log x = -1 \Rightarrow x = e^{-1} = \frac{1}{e} \)
Now, evaluating \( y \) at \( x = \frac{1}{e} \):
\( y_{\text{max}} = \left(\frac{1}{1/e}\right)^{1/e} = e^{1/e} \). Hence proved.

Question. It is given that at \( x = 1 \), the function \( x^4 - 62x^2 + ax + 9 \) attains maximum value on the interval \( (0, 2) \). Find the value of \( a \).
Answer: Let \( f(x) = x^4 - 62x^2 + ax + 9 \)
Then, \( f'(x) = 4x^3 - 124x + a \)
Since \( f(x) \) attains its maximum value at \( x = 1 \), we must have \( f'(1) = 0 \).
\( \Rightarrow 4(1)^3 - 124(1) + a = 0 \)
\( \Rightarrow 4 - 124 + a = 0 \)
\( \Rightarrow -120 + a = 0 \Rightarrow a = 120 \).

Question. Find the maximum slope of curve \( y = -x^3 + 3x^2 + 9x - 27 \).
Answer: We have, \( y = -x^3 + 3x^2 + 9x - 27 \)
\( \therefore \frac{dy}{dx} = -3x^2 + 6x + 9 = \text{Slope of the curve} \)
Let \( M = \frac{dy}{dx} = -3x^2 + 6x + 9 \)
On differentiating w.r.t. \( x \), we get
\( \frac{dM}{dx} = -6x + 6 = -6(x - 1) \)
and \( \frac{d^2M}{dx^2} = -6 \)
For maxima or minima of slope,
\( \frac{dM}{dx} = 0 \Rightarrow -6(x - 1) = 0 \Rightarrow x = 1 \)
Now, \( \frac{d^2M}{dx^2} = -6 < 0 \)
So, the maximum slope of given curve is at \( x = 1 \).
\( \therefore \left(\frac{dy}{dx}\right)_{x=1} = -3 \cdot 1^2 + 6 \cdot 1 + 9 = 12 \).

Short Answer Type Questions

Question. Find the value(s) of \( x \) for which \( y = [x(x - 2)]^2 \) is an increasing function.
Answer: Given, \( y = [x(x - 2)]^2 = [x^2 - 2x]^2 \)
On differentiating both sides w.r.t. \( x \), we get
\( \frac{dy}{dx} = 2(x^2 - 2x) \cdot \frac{d}{dx}(x^2 - 2x) \)
\( = 2(x^2 - 2x)(2x - 2) \)
\( = 2[2x^3 - 6x^2 + 4x] \)
\( = 4x[x^2 - 3x + 2] \)
\( = 4x(x - 2)(x - 1) \)
On putting \( \frac{dy}{dx} = 0 \), we get
\( 4x(x - 2)(x - 1) = 0 \Rightarrow x = 0, 1 \) and \( 2 \)
These values divide the real line into four disjoint intervals \( (-\infty, 0) \), \( (0, 1) \), \( (1, 2) \) and \( (2, \infty) \).
In each interval, nature of \( y(x) \) is given below:
- In \( (-\infty, 0) \), sign of \( f'(x) \) is \( (-)(-)(-) = \text{-ve} \), strictly decreasing.
- In \( (0, 1) \), sign of \( f'(x) \) is \( (+)(-)(-) = \text{+ve} \), strictly increasing.
- In \( (1, 2) \), sign of \( f'(x) \) is \( (+)(-)(+) = \text{-ve} \), strictly decreasing.
- In \( (2, \infty) \), sign of \( f'(x) \) is \( (+)(+)(+) = \text{+ve} \), strictly increasing.
Therefore, \( y(x) \) is increasing in \( (0, 1) \) and \( (2, \infty) \).
\( \Rightarrow 0 < x < 1 \) and \( x > 2 \).

Question. Show that \( f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x) \) is increasing in \( R \).
Answer: We have \( f'(x) = \frac{1 + 2x^2}{1 + x^2} - \frac{1}{1 + x^2} - \frac{1}{\sqrt{1 + x^2}} = \frac{1 + 2x^2 - \sqrt{1 + x^2}}{1 + x^2} \)
Now, \( f'(x) \geq 0 \)
\( \Rightarrow \frac{1 + 2x^2 - \sqrt{1 + x^2}}{1 + x^2} \geq 0 \Rightarrow 1 + 2x^2 \geq \sqrt{1 + x^2} \)
\( \Rightarrow 1 + 4x^2 + 4x^4 \geq 1 + x^2 \)
\( \Rightarrow 4x^4 + 3x^2 \geq 0 \), which is true for all \( x \in R \).
So, \( f'(x) \geq 0 \), \( \forall x \in R \).

Question. Show that the function \( f \) given by \( f(x) = \tan^{-1}(\sin x + \cos x) \), \( x > 0 \) is always an strictly increasing function in \( \left(0, \frac{\pi}{4}\right) \).
Answer: Here, \( f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \times (\cos x - \sin x) \)
\( = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \)
In the interval \( \left(0, \frac{\pi}{4}\right) \), \( \cos x > \sin x > 0 \)
\( \therefore \cos x - \sin x > 0 \) or \( \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} > 0 \)
\( \Rightarrow f'(x) > 0 \), \( \forall x \in \left(0, \frac{\pi}{4}\right) \)
Hence, \( f(x) \) is strictly increasing in \( \left(0, \frac{\pi}{4}\right) \).

Question. Find the local maxima and local minima and the corresponding local maximum and local minimum values of the function \( f(x) = \frac{2}{x} + \frac{x}{2} \), \( x > 0 \).
Answer: Given, \( f(x) = \frac{2}{x} + \frac{x}{2} \)
On differentiating twice w.r.t. \( x \), we get
\( f'(x) = -\frac{2}{x^2} + \frac{1}{2} \)
and \( f''(x) = \frac{4}{x^3} + 0 \)
For local maxima or local minima, put \( f'(x) = 0 \)
\( \Rightarrow -\frac{2}{x^2} + \frac{1}{2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \)
Since, it is given that \( x > 0 \), so we consider \( x = 2 \) only.
Now, \( f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0 \)
Thus, \( f(x) \) is local minima at \( x = 2 \).
So, local minimum value of \( f(x) \) at \( x = 2 \) is
\( f(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2 \).

Question. Find the maximum profit that a company can make, if the profit function is given by \( P(x) = 51 + 42x - x^2 \), where \( x \) is the number of units and \( P \) is the profit in rupees.
Answer: Given, \( P(x) = 51 + 42x - x^2 \)
On differentiating w.r.t. \( x \), we get
\( P'(x) = 42 - 2x \)
For maximum profit, we put \( P'(x) = 0 \)
\( \Rightarrow 42 - 2x = 0 \)
\( \Rightarrow x = 21 \)
Now, \( P''(x) = -2 \)
\( \because P''(x) < 0 \)
\( \therefore P(x) \) is maximum at \( x = 21 \).
\( \therefore \) The maximum value of \( P(x) \) is at \( x = 21 \).
\( \Rightarrow P(21) = 51 + 42(21) - (21)^2 \)
\( = 51 + 882 - 441 = 492 \)
Thus, the maximum profit is ₹ \( 492 \).

Question. Check whether the function \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = x^3 + x \), has any critical point(s) or not? If yes, then find the point(s).
Answer: Given, \( f(x) = x^3 + x, \forall x \in \mathbb{R} \).
On differentiating w.r.t. \( x \), we get:
\( f'(x) = 3x^2 + 1 \).
Since \( x^2 \ge 0, \forall x \in \mathbb{R} \), we have \( 3x^2 + 1 \ge 1 > 0 \).
Therefore, \( f'(x) \) can never be zero or undefined for any \( x \in \mathbb{R} \).
\( \Rightarrow f'(x) > 0 \).
Hence, no critical point exists.

Question. Find the least value of the function \( f(x) = ax + \frac{b}{x} \) (\( a > 0, b > 0, x > 0 \)).
Answer: We have, \( f(x) = ax + \frac{b}{x} \) where \( a > 0, b > 0, x > 0 \).
Differentiating w.r.t. \( x \), we get:
\( f'(x) = a - \frac{b}{x^2} \) and \( f''(x) = \frac{2b}{x^3} \).
For maxima or minima of \( f(x) \), put \( f'(x) = 0 \):
\( \Rightarrow a - \frac{b}{x^2} = 0 \Rightarrow x^2 = \frac{b}{a} \).
\( \Rightarrow x = \sqrt{\frac{b}{a}} \) [since \( x > 0 \) (given)].
Again, evaluating the second derivative at this point:
\( f''\left(\sqrt{\frac{b}{a}}\right) = \frac{2b}{\left(\frac{b}{a}\right)^{3/2}} = \frac{2a^{3/2}}{\sqrt{b}} > 0 \) [since \( a > 0, b > 0 \)].
So, \( f(x) \) has its least value at \( x = \sqrt{\frac{b}{a}} \).
\( \therefore \) Least value of \( f(x) = f\left(\sqrt{\frac{b}{a}}\right) = a\sqrt{\frac{b}{a}} + \frac{b}{\sqrt{\frac{b}{a}}} = \sqrt{ab} + \sqrt{ab} = 2\sqrt{ab} \).

Question. Find the points of local maxima, local minima and the points of inflection of the function \( f(x) = x^5 - 5x^4 + 5x^3 - 1 \). Also, find the corresponding local maximum and local minimum values.
Answer: Given, \( f(x) = x^5 - 5x^4 + 5x^3 - 1 \).
On differentiating w.r.t. \( x \), we get:
\( f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2 - 4x + 3) = 5x^2(x-1)(x-3) \).
For local maxima or local minima, put \( f'(x) = 0 \):
\( 5x^2(x-1)(x-3) = 0 \Rightarrow x = 0, 1, 3 \).
Now, differentiating again to get the second derivative:
\( f''(x) = 20x^3 - 60x^2 + 30x \).

At \( x = 1 \):
\( f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 < 0 \).
Thus, \( x = 1 \) is a point of local maxima.
Local maximum value is \( f(1) = (1)^5 - 5(1)^4 + 5(1)^3 - 1 = 1 - 5 + 5 - 1 = 0 \).

At \( x = 3 \):
\( f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 540 - 540 + 90 = 90 > 0 \).
Thus, \( x = 3 \) is a point of local minima.
Local minimum value is \( f(3) = (3)^5 - 5(3)^4 + 5(3)^3 - 1 = 243 - 405 + 135 - 1 = -28 \).

At \( x = 0 \):
\( f''(0) = 0 \).
Let's examine the third derivative:
\( f'''(x) = 60x^2 - 120x + 30 \).
At \( x = 0 \):
\( f'''(0) = 30 \neq 0 \).
Since \( f''(0) = 0 \) and \( f'''(0) \neq 0 \), \( x = 0 \) is a point of inflection.

Question. Prove that the largest rectangle with a given perimeter is a square.
Answer: Let \( x \) be the length and \( y \) be the breadth of the rectangle.
Given that the perimeter \( P \) is constant, we have:
\( 2(x+y) = P \Rightarrow y = \frac{P}{2} - x \).
The area \( A \) of the rectangle is given by:
\( A = xy = x\left(\frac{P}{2} - x\right) = \frac{Px}{2} - x^2 \).
Differentiating w.r.t. \( x \):
\( \frac{dA}{dx} = \frac{P}{2} - 2x \).
For maximum area, we put \( \frac{dA}{dx} = 0 \):
\( \Rightarrow \frac{P}{2} - 2x = 0 \Rightarrow x = \frac{P}{4} \).
Now, differentiating again:
\( \frac{d^2A}{dx^2} = -2 < 0 \).
Since the second derivative is negative, the area is maximum when \( x = \frac{P}{4} \).
Substituting \( x = \frac{P}{4} \) into the perimeter equation:
\( y = \frac{P}{2} - \frac{P}{4} = \frac{P}{4} \).
Since \( x = y \), the rectangle of maximum area is a square.

Question. Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m, BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that \( RP^2 + RQ^2 \) is minimum.
Answer: Let \( AR = x \text{ m} \). Then \( BR = (20 - x) \text{ m} \).
In right-angled triangles \( \Delta PAR \) and \( \Delta QBR \), using Pythagoras theorem:
\( RP^2 = AP^2 + AR^2 = 16^2 + x^2 = 256 + x^2 \)
\( RQ^2 = BQ^2 + BR^2 = 22^2 + (20-x)^2 = 484 + (20-x)^2 \)
Let \( Z = RP^2 + RQ^2 \)
\( \Rightarrow Z = 256 + x^2 + 484 + (20-x)^2 = x^2 + (20-x)^2 + 740 \)
On differentiating w.r.t. \( x \):
\( \frac{dZ}{dx} = 2x + 2(20-x)(-1) = 2x - 40 + 2x = 4x - 40 \)
For minimum value, put \( \frac{dZ}{dx} = 0 \):
\( 4x - 40 = 0 \Rightarrow x = 10 \).
Now, differentiating again:
\( \frac{d^2Z}{dx^2} = 4 > 0 \).
Since the second derivative is positive, \( RP^2 + RQ^2 \) is minimum at \( x = 10 \text{ m} \).
Hence, the distance of point R from point A is \( 10 \text{ m} \).

Question. An Apache helicopter of enemy is flying along the curve given by \( y = x^2 + 7 \). A soldier placed at \( (3,7) \) wants to shoot down the helicopter, when it is nearest to him. Find the nearest distance.
Answer: Let \( A(h,k) \) be any point on the curve \( y = x^2 + 7 \) representing the helicopter and let \( B(3,7) \) be the position of the soldier.
The distance between A and B is given by:
\( AB = \sqrt{(h-3)^2 + (k-7)^2} \)
Since the point \( A(h,k) \) lies on the curve \( y = x^2 + 7 \):
\( k = h^2 + 7 \Rightarrow k-7 = h^2 \)
Substituting this into the distance formula:
\( AB = \sqrt{(h-3)^2 + (h^2)^2} \)
Let \( f(h) = AB^2 = (h-3)^2 + h^4 \)
Differentiating \( f(h) \) w.r.t. \( h \):
\( f'(h) = 2(h-3) + 4h^3 = 4h^3 + 2h - 6 \)
For maximum or minimum, put \( f'(h) = 0 \):
\( 4h^3 + 2h - 6 = 0 \Rightarrow 2h^3 + h - 3 = 0 \)
By inspection, \( h = 1 \) is a root since \( 2(1)^3 + (1) - 3 = 0 \).
Factorizing the equation, we get:
\( (h-1)(2h^2 + 2h + 3) = 0 \)
Since \( 2h^2 + 2h + 3 = 0 \) has no real roots (as its discriminant \( D = 2^2 - 4(2)(3) = -20 < 0 \)), \( h = 1 \) is the only real critical point.
Now, finding the second derivative:
\( f''(h) = 12h^2 + 2 \)
At \( h = 1 \):
\( f''(1) = 12(1)^2 + 2 = 14 > 0 \)
Since the second derivative is positive, \( f(h) \) is minimum at \( h = 1 \).
Substituting \( h = 1 \) back into \( f(h) \):
\( f(1) = (1-3)^2 + (1)^4 = 4 + 1 = 5 \).
So, the nearest distance is \( AB = \sqrt{f(1)} = \sqrt{5} \text{ units} \).

Long Answer Type Questions

Question. Prove that the function f defined by \( f(x) = x^2 - x + 1 \) is neither increasing nor decreasing in \( (-1, 1) \). Hence, find the interval in which \( f(x) \) is (i) strictly increasing, (ii) strictly decreasing.
Answer: Given, \( f(x) = x^2 - x + 1 \).
Differentiating w.r.t. \( x \):
\( f'(x) = 2x - 1 \).
For critical points, put \( f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \).
Now, in the interval \( (-1, 1) \), let's examine the behavior of \( f'(x) \):
- For \( x \in \left(-1, \frac{1}{2}\right) \), \( f'(x) < 0 \), meaning the function is strictly decreasing.
- For \( x \in \left(\frac{1}{2}, 1\right) \), \( f'(x) > 0 \), meaning the function is strictly increasing.
Since \( f(x) \) is decreasing in one part of \( (-1, 1) \) and increasing in the other part, the function \( f(x) \) is neither increasing nor decreasing in \( (-1, 1) \).
Hence:
(i) \( f(x) \) is strictly increasing in \( \left(\frac{1}{2}, 1\right) \).
(ii) \( f(x) \) is strictly decreasing in \( \left(-1, \frac{1}{2}\right) \).

Question. Prove that \( y = \frac{4\sin\theta}{2+\cos\theta} - \theta \) is an increasing function of \( \theta \) on \( \left[0, \frac{\pi}{2}\right] \).
Answer: Let \( y = f(\theta) = \frac{4\sin\theta}{2+\cos\theta} - \theta \).
On differentiating w.r.t. \( \theta \), we get:
\( f'(\theta) = \frac{(2+\cos\theta)(4\cos\theta) - (4\sin\theta)(-\sin\theta)}{(2+\cos\theta)^2} - 1 \)
\( \Rightarrow f'(\theta) = \frac{8\cos\theta + 4\cos^2\theta + 4\sin^2\theta}{(2+\cos\theta)^2} - 1 \)
\( \Rightarrow f'(\theta) = \frac{8\cos\theta + 4(\cos^2\theta + \sin^2\theta)}{(2+\cos\theta)^2} - 1 \)
\( \Rightarrow f'(\theta) = \frac{8\cos\theta + 4}{(2+\cos\theta)^2} - 1 \) [since \( \sin^2\theta + \cos^2\theta = 1 \)]
\( \Rightarrow f'(\theta) = \frac{8\cos\theta + 4 - (4 + \cos^2\theta + 4\cos\theta)}{(2+\cos\theta)^2} \)
\( \Rightarrow f'(\theta) = \frac{4\cos\theta - \cos^2\theta}{(2+\cos\theta)^2} = \frac{\cos\theta(4 - \cos\theta)}{(2+\cos\theta)^2} \).
In the interval \( \theta \in \left[0, \frac{\pi}{2}\right] \):
- \( \cos\theta \ge 0 \)
- \( 4 - \cos\theta > 0 \) (since \( \cos\theta \in [0, 1] \))
- \( (2+\cos\theta)^2 > 0 \)
Therefore, \( f'(\theta) \ge 0 \) for all \( \theta \in \left[0, \frac{\pi}{2}\right] \).
Hence, \( y \) is an increasing function of \( \theta \) on \( \left[0, \frac{\pi}{2}\right] \).

Question. Find the maximum and minimum values of the function given by \( f(x) = 5 + \sin 2x \).
Answer: Given, \( f(x) = 5 + \sin 2x \).
We know that for any real value of \( x \):
\( -1 \le \sin 2x \le 1 \)
Adding 5 to all parts of the inequality:
\( 5 - 1 \le 5 + \sin 2x \le 5 + 1 \)
\( \Rightarrow 4 \le f(x) \le 6 \).
Hence, the maximum value of the function is 6 and the minimum value of the function is 4.

Question. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \( \frac{4r}{3} \). Also, find the maximum volume in terms of volume of the sphere.
Answer: Let \( R \) be the radius and \( h \) be the height of the right circular cone inscribed in a sphere of radius \( r \).
Let \( O \) be the center of the sphere. The distance from \( O \) to the base of the cone is \( OA = h - r \).
In the right-angled triangle \( \Delta OAB \), using Pythagoras theorem:
\( r^2 = R^2 + (h-r)^2 \)
\( \Rightarrow r^2 = R^2 + h^2 + r^2 - 2rh \)
\( \Rightarrow R^2 = 2rh - h^2 \) ... (i)
The volume \( V \) of the cone is:
\( V = \frac{1}{3}\pi R^2 h \)
Substituting \( R^2 \) from Eq. (i):
\( V = \frac{1}{3}\pi (2rh - h^2)h = \frac{1}{3}\pi (2rh^2 - h^3) \) ... (ii)
Differentiating \( V \) w.r.t. \( h \):
\( \frac{dV}{dh} = \frac{1}{3}\pi (4rh - 3h^2) \) ... (iii)
For maximum volume, put \( \frac{dV}{dh} = 0 \):
\( \frac{1}{3}\pi (4rh - 3h^2) = 0 \Rightarrow h(4r - 3h) = 0 \).
Since \( h \neq 0 \), we have:
\( h = \frac{4r}{3} \).
Differentiating Eq. (iii) again w.r.t. \( h \):
\( \frac{d^2V}{dh^2} = \frac{1}{3}\pi (4r - 6h) \).
At \( h = \frac{4r}{3} \):
\( \frac{d^2V}{dh^2} = \frac{1}{3}\pi \left(4r - 6 \times \frac{4r}{3}\right) = -\frac{4\pi r}{3} < 0 \).
Since the second derivative is negative, the volume is maximum when the altitude \( h = \frac{4r}{3} \).

Now, the maximum volume of the cone is:
\( V = \frac{1}{3}\pi \left[ 2r\left(\frac{4r}{3}\right)^2 - \left(\frac{4r}{3}\right)^3 \right] = \frac{1}{3}\pi \left[ \frac{32r^3}{9} - \frac{64r^3}{27} \right] \)
\( V = \frac{1}{3}\pi \left( \frac{96r^3 - 64r^3}{27} \right) = \frac{32\pi r^3}{81} \).
We can write this as:
\( V = \frac{8}{27} \left(\frac{4}{3}\pi r^3\right) \).
Since \( \frac{4}{3}\pi r^3 \) is the volume of the sphere, we have:
\( \text{Maximum Volume} = \frac{8}{27} \times (\text{Volume of sphere}) \).

Question. AB is the diameter of a circle and C is any point on the circle. Show that the area of \( \Delta ABC \) is maximum, when it is an isosceles triangle.
Answer: Let \( AC = x \), \( BC = y \) and \( r \) be the radius of the circle.
Since \( AB \) is the diameter, the angle in the semi-circle is \( \angle C = 90^\circ \).
In right-angled \( \Delta ABC \), by Pythagoras theorem:
\( AB^2 = AC^2 + BC^2 \)
\( \Rightarrow (2r)^2 = x^2 + y^2 \Rightarrow 4r^2 = x^2 + y^2 \) ... (i)
The area of \( \Delta ABC \), \( A \), is given by:
\( A = \frac{1}{2}xy \).
Squaring both sides:
\( A^2 = \frac{1}{4}x^2y^2 \).
Let \( S = A^2 \):
\( S = \frac{1}{4}x^2(4r^2 - x^2) = \frac{1}{4}(4r^2x^2 - x^4) \) [from Eq. (i)].
Differentiating \( S \) w.r.t. \( x \):
\( \frac{dS}{dx} = \frac{1}{4}(8r^2x - 4x^3) \).
For maxima or minima, put \( \frac{dS}{dx} = 0 \):
\( 8r^2x - 4x^3 = 0 \Rightarrow x^2 = 2r^2 \Rightarrow x = \sqrt{2}r \).
Substituting \( x = \sqrt{2}r \) into Eq. (i):
\( y^2 = 4r^2 - 2r^2 = 2r^2 \Rightarrow y = \sqrt{2}r \).
Since \( x = y = \sqrt{2}r \), the triangle is an isosceles triangle.
Differentiating \( \frac{dS}{dx} \) again:
\( \frac{d^2S}{dx^2} = \frac{1}{4}(8r^2 - 12x^2) = 2r^2 - 3x^2 \).
At \( x = \sqrt{2}r \):
\( \frac{d^2S}{dx^2} = 2r^2 - 3(2r^2) = -4r^2 < 0 \).
Since the second derivative is negative, the area is maximum when \( AC = BC \), i.e., when \( \Delta ABC \) is an isosceles triangle.

Question. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is \( \cos^{-1}(1/\sqrt{3}) \).
Answer: Let \( \theta \) be the semi-vertical angle of the cone, where \( \theta \in \left(0, \frac{\pi}{2}\right) \).
Let \( r \), \( h \), and \( l \) be the radius, height, and slant height of the cone, respectively.
Since the slant height \( l \) is given, it is constant.
From the geometry of the cone:
\( r = l\sin\theta \) and \( h = l\cos\theta \).
The volume \( V \) of the cone is:
\( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l\sin\theta)^2 (l\cos\theta) = \frac{1}{3}\pi l^3 \sin^2\theta \cos\theta \).
Differentiating \( V \) w.r.t. \( \theta \):
\( \frac{dV}{d\theta} = \frac{1}{3}\pi l^3 \left[ \sin^2\theta(-\sin\theta) + \cos\theta(2\sin\theta\cos\theta) \right] \)
\( \Rightarrow \frac{dV}{d\theta} = \frac{1}{3}\pi l^3 \left( 2\sin\theta\cos^2\theta - \sin^3\theta \right) \).
For maxima or minima, put \( \frac{dV}{d\theta} = 0 \):
\( 2\sin\theta\cos^2\theta - \sin^3\theta = 0 \Rightarrow \sin\theta(2\cos^2\theta - \sin^2\theta) = 0 \).
Since \( \theta \in \left(0, \frac{\pi}{2}\right) \), \( \sin\theta \neq 0 \), so:
\( 2\cos^2\theta = \sin^2\theta \Rightarrow \tan^2\theta = 2 \Rightarrow \tan\theta = \sqrt{2} \).
Differentiating again to find the second derivative:
\( \frac{d^2V}{d\theta^2} = \frac{1}{3}\pi l^3 \left( 2\cos^3\theta - 7\sin^2\theta\cos\theta \right) \).
At \( \tan^2\theta = 2 \) (i.e., \( \sin^2\theta = 2\cos^2\theta \)):
\( \frac{d^2V}{d\theta^2} = \frac{1}{3}\pi l^3 \left( 2\cos^3\theta - 14\cos^3\theta \right) = -4\pi l^3 \cos^3\theta < 0 \).
Since the second derivative is negative, the volume is maximum when \( \tan\theta = \sqrt{2} \).
Now, \( \cos\theta = \frac{1}{\sqrt{1 + \tan^2\theta}} = \frac{1}{\sqrt{1+2}} = \frac{1}{\sqrt{3}} \).
\( \Rightarrow \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).
Hence, the semi-vertical angle of the cone of maximum volume is \( \cos^{-1}\left(1/\sqrt{3}\right) \).

Question. If the length of three sides of a trapezium other than the base are each equal to 10 cm, then find the area of the trapezium, when it is maximum.
Answer: Let \( ABCD \) be the trapezium in which \( AD = BC = CD = 10 \text{ cm} \).
Let \( AP = x \text{ cm} \) and \( QB = x \text{ cm} \).
In the right-angled triangle \( \Delta APD \), by Pythagoras theorem:
\( DP = \sqrt{AD^2 - AP^2} = \sqrt{10^2 - x^2} = \sqrt{100 - x^2} \).
The parallel sides of the trapezium are \( CD = 10 \text{ cm} \) and \( AB = AP + PQ + QB = x + 10 + x = 2x + 10 \text{ cm} \).
The area \( A \) of the trapezium is:
\( A = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height} \)
\( A = \frac{1}{2} \times (2x + 10 + 10) \times \sqrt{100 - x^2} = (x+10)\sqrt{100-x^2} \) ... (i)
Differentiating \( A \) w.r.t. \( x \):
\( \frac{dA}{dx} = (x+10)\frac{-2x}{2\sqrt{100-x^2}} + \sqrt{100-x^2} \)
\( \Rightarrow \frac{dA}{dx} = \frac{-x(x+10) + (100-x^2)}{\sqrt{100-x^2}} = \frac{-2x^2 - 10x + 100}{\sqrt{100-x^2}} \) ... (ii)
For maximum area, put \( \frac{dA}{dx} = 0 \):
\( -2x^2 - 10x + 100 = 0 \Rightarrow x^2 + 5x - 50 = 0 \)
\( \Rightarrow (x+10)(x-5) = 0 \Rightarrow x = 5 \) or \( x = -10 \).
Since length cannot be negative, we have \( x = 5 \text{ cm} \).
Differentiating Eq. (ii) again, it can be shown that \( \frac{d^2A}{dx^2} < 0 \) at \( x = 5 \).
Therefore, the area is maximum when \( x = 5 \text{ cm} \).
The maximum area of the trapezium is:
\( A_{\text{max}} = (5 + 10)\sqrt{100 - 5^2} = 15\sqrt{75} = 15(5\sqrt{3}) = 75\sqrt{3} \text{ cm}^2 \).

Question. Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface area is \( \cot^{-1}\sqrt{2} \).
Answer: Let \( r \) be the radius of the base, \( h \) be the height, \( V \) be the volume, and \( S \) be the curved surface area of the cone.
The volume of the cone is given by:
\( V = \frac{1}{3}\pi r^2 h \Rightarrow h = \frac{3V}{\pi r^2} \) ... (i)
The curved surface area of the cone is:
\( S = \pi r l = \pi r \sqrt{r^2 + h^2} \).
Squaring both sides:
\( S^2 = \pi^2 r^2 (r^2 + h^2) \).
Substituting \( h^2 \) from Eq. (i):
\( S^2 = \pi^2 r^2 \left( r^2 + \frac{9V^2}{\pi^2 r^4} \right) = \pi^2 r^4 + \frac{9V^2}{r^2} \) ... (ii)
Let \( Z = S^2 \). For \( S \) to be minimum, \( Z \) must be minimum.
Differentiating \( Z \) w.r.t. \( r \):
\( \frac{dZ}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3} \) ... (iii)
For minima, put \( \frac{dZ}{dr} = 0 \):
\( 4\pi^2 r^3 = \frac{18V^2}{r^3} \Rightarrow V^2 = \frac{2\pi^2 r^6}{9} \) ... (iv)
Differentiating Eq. (iii) again w.r.t. \( r \):
\( \frac{d^2Z}{dr^2} = 12\pi^2 r^2 + \frac{54V^2}{r^4} > 0 \).
Since the second derivative is positive, \( S^2 \) (and thus \( S \)) is minimum.
From Eq. (iv), substituting \( V = \frac{1}{3}\pi r^2 h \):
\( \left(\frac{1}{3}\pi r^2 h\right)^2 = \frac{2\pi^2 r^6}{9} \)
\( \Rightarrow \frac{1}{9}\pi^2 r^4 h^2 = \frac{2}{9}\pi^2 r^6 \)
\( \Rightarrow h^2 = 2r^2 \Rightarrow h = \sqrt{2}r \Rightarrow \frac{h}{r} = \sqrt{2} \).
In a right circular cone, \( \dots \cot\theta = \frac{h}{r} \) where \( \theta \) is the semi-vertical angle.
\( \therefore \cot\theta = \sqrt{2} \Rightarrow \theta = \cot^{-1}\sqrt{2} \).
Hence, the semi-vertical angle of the right circular cone of given volume and least curved surface area is \( \cot^{-1}\sqrt{2} \).