CBSE Class 12 Mathematics Application Of Derivative Worksheet Set 03

Very Short Answer Questions

Question. If the rate of change of volume of a sphere is equal to the rate of change of its radius, find the radius of the sphere.
Answer: We have, \( \frac{dV}{dt} = \frac{dr}{dt} \), where V is the volume and r is the radius of the sphere
\( \implies \) \( \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = \frac{dr}{dt} \)
\( \implies \) \( \frac{4}{3}\pi \times 3r^2 \frac{dr}{dt} = \frac{dr}{dt} \)
\( \implies \) \( 4\pi r^2 = 1 \)
\( \implies \) \( r^2 = \frac{1}{4\pi} \)
\( \therefore r = \frac{1}{2\sqrt{\pi}} \) units

Question. An edge of a variable cube is increasing at the rate of 5 cm per second. How fast is the volume increasing when the side is 15 cm?
Answer: Let \( x \) be the edge of the cube and \( V \) be the volume of the cube at any time \( t \).
Given, \( \frac{dx}{dt} = 5 \) cm/s, \( x = 15 \) cm
Since we know the volume of cube \( = (side)^3 \) i.e., \( V = x^3 \).
\( \implies \) \( \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \)
\( \implies \) \( \frac{dV}{dt} = 3 \cdot (15)^2 \times 5 = 3375 \) cm\(^3\) / sec

Question. Find the slope of the tangent to the curve \( x = t^2 + 3t - 8, y = 2t^2 - 2t - 5 \) at \( t = 2 \).
Answer: Slope of the tangent \( = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t - 2}{2t + 3} \)
\( \implies \) \( \left(\frac{dy}{dx}\right)_{at \: t = 2} = \left(\frac{4t - 2}{2t + 3}\right)_{at \: t = 2} = \frac{4(2) - 2}{2(2) + 3} = \frac{6}{7} \)

Question. If \( y = \log_e x \), then find \( \Delta y \) when \( x = 3 \) and \( \Delta x = 0.03 \).
Answer: We have, \( y = \log_e x \)
\( \therefore \Delta y = \frac{dy}{dx} \cdot \Delta x = \frac{1}{x} \cdot \Delta x = \frac{0.03}{3} = 0.01 \)

Question. Find the rate of change of the area of a circle with respect to its radius 'r' when \( r = 4 \) cm.
Answer: If \( A \) is area and \( r \) is the radius of a circle, then
\( A = \pi r^2 \)
\( \implies \) \( \frac{dA}{dr} = 2\pi r \)
\( \therefore \left[\frac{dA}{dr}\right]_{r = 4} = 8\pi \) cm\(^2\)/cm

Short Answer Questions

Question. The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (marginal revenue). If the total revenue (in rupees) received from the sale of \( x \) units of a product is given by \( R(x) = 3x^2 + 36x + 5 \), find the marginal revenue when \( x = 5 \).
Answer: Given: \( R(x) = 3x^2 + 36x + 5 \)
\( \implies \) \( R'(x) = 6x + 36 \)
\( \therefore \) Marginal revenue (when \( x = 5 \)) \( = [R'(x)]_{x = 5} \)
\( = 6 \times 5 + 36 = Rs. 66 \).

Question. The amount of pollution content added in air in a city due to \( x \)-diesel vehicles is given by \( P(x) = 0.005x^3 + 0.02x^2 + 30x \). Find the marginal increase in pollution content when 3 diesel vehicles are added.
Answer: We have to find \( [P'(x)]_{x = 3} \)
Now, \( P(x) = 0.005x^3 + 0.02x^2 + 30x \)
\( \therefore P'(x) = 0.015x^2 + 0.04x + 30 \)
\( \implies \) \( [P'(x)]_{x = 3} = 0.015 \times 9 + 0.04 \times 3 + 30 \)
\( = 0.135 + 0.12 + 30 = 30.255 \)

Question. The contentment obtained after eating \( x \)-units of a new dish at a trial function is given by the function \( C(x) = x^3 + 6x^2 + 5x + 3 \). If the marginal contentment is defined as rate of change of \( C(x) \) with respect to the number of units consumed at an instant, then find the marginal contentment when three units of dish are consumed.
Answer: Given, \( C(x) = x^3 + 6x^2 + 5x + 3 \)
\( \implies \) \( C'(x) = 3x^2 + 12x + 5 \)
\( \implies \) \( [C'(x)]_{x = 3} = 3 \times 9 + 36 + 5 = 27 + 36 + 5 = 68 \) units
\( \therefore \) Required marginal contentment \( [C'(x)]_{x = 3} = 68 \) units

Question. If \( x \) and \( y \) are the sides of two squares such that \( y = x - x^2 \), then find the rate of change of the area of second square with respect to the area of first square.
Answer: Since, \( x \) and \( y \) are the sides of two squares such that \( y = x - x^2 \).
\( \therefore \) Area of the first square \( (A_1) = x^2 \)
and area of the second square \( (A_2) = y^2 = (x - x^2)^2 \)
\( \therefore \frac{dA_2}{dt} = \frac{d}{dt}(x - x^2)^2 = 2(x - x^2)\left(\frac{dx}{dt} - 2x \cdot \frac{dx}{dt}\right) \)
\( = \frac{dx}{dt}(1 - 2x)2(x - x^2) \)
and \( \frac{dA_1}{dt} = \frac{d}{dx}x^2 = 2x \cdot \frac{dx}{dt} \)
\( \therefore \frac{dA_2}{dA_1} = \frac{dA_2/dt}{dA_1/dt} = \frac{\frac{dx}{dt} \cdot (1 - 2x)(2x - 2x^2)}{2x \cdot \frac{dx}{dt}} \)
\( = \frac{(1 - 2x)2x(1 - x)}{2x} \)
\( = (1 - 2x)(1 - x) = 1 - x - 2x + 2x^2 = 2x^2 - 3x + 1 \)

Question. Using differentials, find the approximate value of \( \sqrt{49.5} \).
Answer: Let \( f(x) = \sqrt{x} \), where \( x = 49 \) and \( \delta x = 0.5 \)
\( \therefore f(x + \delta x) = \sqrt{x + \delta x} = \sqrt{49.5} \)
Now by definition, approximately we can write
\( f'(x) = \frac{f(x + \delta x) - f(x)}{\delta x} \) ...(i)
Here \( f(x) = \sqrt{x} = \sqrt{49} = 7 \) and \( \delta x = 0.5 \)
\( \implies \) \( f'(x) = \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{49}} = \frac{1}{14} \)
Putting these values in (i), we get
\( \frac{1}{14} = \frac{\sqrt{49.5} - 7}{0.5} \)
\( \sqrt{49.5} = \frac{0.5}{14} + 7 = \frac{0.5 + 98}{14} = \frac{98.5}{14} = 7.036 \)

Question. Show that the function \( f \) given by \( f(x) = \tan^{-1} (\sin x + \cos x) \) is decreasing for all \( x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \).
Answer: We have \( f(x) = \tan^{-1}(\sin x + \cos x) \)
\( \implies \) \( f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \times (\cos x - \sin x) \)
\( f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \)
\( \because 1 + (\sin x + \cos x)^2 > 0 \: \forall x \in R \)
Also, \( \forall x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \sin x > \cos x \)
\( \implies \) \( \cos x - \sin x < 0 \)
\( \therefore f'(x) = \frac{-ve}{+ve} = -ve \) i.e., \( f'(x) < 0 \)
\( \implies \) \( f(x) \) is decreasing in \( \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \)

Question. The volume of a cube is increasing at the rate of 9 cm\(^3\)/s. How fast is its surface area increasing when the length of an edge is 10 cm?
Answer: Let \( V \) and \( S \) be the volume and surface area of a cube of side \( x \) cm respectively.
Given \( \frac{dV}{dt} = 9 \) cm\(^3\)/sec
We require \( \left[\frac{dS}{dt}\right]_{x = 10 \text{ cm}} \)
Now \( V = x^3 \)
\( \implies \) \( \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \)
\( \implies \) \( 9 = 3x^2 \frac{dx}{dt} \)
\( \implies \) \( \frac{dx}{dt} = \frac{9}{3x^2} = \frac{3}{x^2} \)
Again, \( \because S = 6x^2 \) [By formula for surface area of a cube]
\( \implies \) \( \frac{dS}{dt} = 12x \cdot \frac{dx}{dt} \)
\( = 12x \cdot \frac{3}{x^2} = \frac{36}{x} \)
\( \implies \) \( \left[\frac{dS}{dt}\right]_{x = 10 \text{ cm}} = \frac{36}{10} = 3.6 \) cm\(^2\)/sec.

Question. Find the approximate change in the value of \( \frac{1}{x^2} \), when \( x \) changes from \( x = 2 \) to \( x = 2.002 \).
Answer: Let \( y = \frac{1}{x^2} \). Let \( \delta x \) be small change in \( x \) and \( \delta y \) the corresponding change in \( y \).
Given \( \delta x = 2.002 - 2 = 0.002 \), where \( x = 2 \).
Now \( y = \frac{1}{x^2} \)
\( \implies \) \( \frac{dy}{dx} = -\frac{2}{x^3} \)
\( \implies \) \( \frac{dy}{dx} = -\frac{2}{2^3} = -\frac{2}{8} \)
We know that, approximately,
\( \delta y = \frac{dy}{dx} \cdot \delta x \)
\( \therefore \delta y = -\frac{2}{8} \times 0.002 = -0.0005 \)

Question. Find whether the function \( f(x) = \cos(2x + \frac{\pi}{4}) \); is increasing or decreasing in the interval \( \frac{3\pi}{8} < x < \frac{5\pi}{8} \).
Answer: We have \( f(x) = \cos(2x + \frac{\pi}{4}) \)
We know that function \( f(x) \) is increasing in \( (a, b) \) if \( f'(x) > 0 \: \forall x \in (a, b) \) & is decreasing if \( f'(x) < 0 \: \forall x \in (a, b) \).
Now, \( f'(x) = -2 \sin(2x + \frac{\pi}{4}) \)
\( \therefore f'(x) > 0 \)
\( \implies \) \( -2 \sin(2x + \frac{\pi}{4}) > 0 \)
\( \implies \) \( \sin(2x + \frac{\pi}{4}) < 0 \)
\( \implies \) \( \pi < 2x + \frac{\pi}{4} < 2\pi \) [\( \because \sin x < 0 \: \forall x \in (\pi, 2\pi) \)]
\( \implies \) \( \pi - \frac{\pi}{4} < 2x < 2\pi - \frac{\pi}{4} \)
\( \implies \) \( \frac{3\pi}{4} < 2x < \frac{7\pi}{4} \)
\( \implies \) \( \frac{3\pi}{8} < x < \frac{7\pi}{8} \)
Thus \( f'(x) > 0 \: \forall x \in \left(\frac{3\pi}{8}, \frac{7\pi}{8}\right) \)
\( \implies f(x) \) is increasing function on \( \left(\frac{3\pi}{8}, \frac{7\pi}{8}\right) \).

Long Answer Questions-I

Question. Find the intervals in which \( f(x) = \sin 3x - \cos 3x, 0 < x < \pi \), is strictly increasing or strictly decreasing.
Answer: Given function is \( f(x) = \sin 3x - \cos 3x \)
\( f'(x) = 3 \cos 3x + 3 \sin 3x \)
For critical points of function \( f(x) \)
\( f'(x) = 0 \)
\( \implies 3 \cos 3x + 3 \sin 3x = 0 \)
\( \implies \cos 3x + \sin 3x = 0 \)
\( \implies \sin 3x = -\cos 3x \)
\( \implies \frac{\sin 3x}{\cos 3x} = -1 \)
\( \implies \tan 3x = - \tan \frac{\pi}{4} \)
\( \implies \tan 3x = \tan (\pi - \frac{\pi}{4}) \)
\( \implies \tan 3x = \tan \frac{3\pi}{4} \)
\( \implies 3x = n\pi + \frac{3\pi}{4} \), where \( n = 0, \pm 1, \pm 2, \dots \)
Putting \( n = 0, \pm 1, \pm 2, \dots \), we get \( x = \frac{\pi}{4}, \frac{7\pi}{12}, \frac{11\pi}{12} \in (0, \pi) \)
Hence, required possible intervals are \( (0, \frac{\pi}{4}), (\frac{\pi}{4}, \frac{7\pi}{12}), (\frac{7\pi}{12}, \frac{11\pi}{12}), (\frac{11\pi}{12}, \pi) \)
For \( (0, \frac{\pi}{4}), f'(x) = +ve \)
For \( (\frac{\pi}{4}, \frac{7\pi}{12}), f'(x) = -ve \)
For \( (\frac{7\pi}{12}, \frac{11\pi}{12}), f'(x) = +ve \)
For \( (\frac{11\pi}{12}, \pi), f'(x) = -ve \)
Hence, given function \( f(x) \) is strictly increasing in \( (0, \frac{\pi}{4}) \cup (\frac{7\pi}{12}, \frac{11\pi}{12}) \) and strictly decreasing in \( (\frac{\pi}{4}, \frac{7\pi}{12}) \cup (\frac{11\pi}{12}, \pi) \).

Question. Find the equation of the normal at the point \( (am^2, am^3) \) for the curve \( ay^2 = x^3 \).
Answer: Given, curve \( ay^2 = x^3 \)
On differentiating, we get
\( 2ay \frac{dy}{dx} = 3x^2 \)
\( \implies \frac{dy}{dx} = \frac{3x^2}{2ay} \)
\( \implies \frac{dy}{dx} \text{ at } (am^2, am^3) = \frac{3 \times a^2m^4}{2a \times am^3} = \frac{3m}{2} \)
\( \therefore \) Slope of normal \( = -\frac{1}{\text{slope of tangent}} = -\frac{1}{\frac{3m}{2}} = -\frac{2}{3m} \)
Equation of normal at the point \( (am^2, am^3) \) is given by
\( \frac{y - am^3}{x - am^2} = -\frac{2}{3m} \)
\( \implies 3my - 3am^4 = -2x + 2am^2 \)
\( \implies 2x + 3my - am^2(2 + 3m^2) = 0 \)
Hence, equation of normal is \( 2x + 3my - am^2(2 + 3m^2) = 0 \)

Question. Show that \( y = \log(1 + x) - \frac{2x}{2 + x}, x > -1 \) is an increasing function of \( x \) throughout its domain.
Answer: Here, \( f(x) = \log(1 + x) - \frac{2x}{2 + x} \) [where \( y = f(x) \)]
\( \implies f'(x) = \frac{1}{1 + x} - 2\left[\frac{(2 + x) \cdot 1 - x}{(2 + x)^2}\right] \)
\( = \frac{1}{1 + x} - \frac{2(2 + x - x)}{(2 + x)^2} = \frac{1}{1 + x} - \frac{4}{(2 + x)^2} \)
\( = \frac{4 + x^2 + 4x - 4 - 4x}{(x + 1)(x + 2)^2} = \frac{x^2}{(x + 1)(x + 2)^2} \)
For \( f(x) \) being increasing function \( f'(x) > 0 \)
\( \implies \frac{x^2}{(x + 1)(x + 2)^2} > 0 \)
\( \implies \frac{1}{x + 1} \cdot \left[\frac{x^2}{(x + 2)^2}\right] > 0 \)
\( \implies \frac{1}{x + 1} > 0 \) [\( \because \frac{x^2}{(x + 2)^2} > 0 \)]
\( \implies x + 1 > 0 \) or \( x > -1 \)
i.e., \( f(x) = \log(1 + x) - \frac{2x}{2 + x} \) is increasing function in its domain \( x > -1 \) i.e., \( (-1, \infty) \)

Question. A person wants to plant some trees in his community park. The local nursery has to perform this task. It charges the cost of planting trees by the following formula: \( C(x) = x^3 - 45x^2 + 600x \), where \( x \) is the number of trees and \( C(x) \) is the cost of planting \( x \) trees in rupees. The local authority has imposed a restriction that it can plant 10 to 20 trees in one community park for a fair distribution. For how many trees should the person place the order so that he has to spend the least amount? Use calculus to answer these questions.
Answer: We have, \( C(x) = x^3 - 45x^2 + 600x \) [where \( x \) is number of trees and \( C(x) \) be the cost of planting \( x \) trees, also, \( x \in [10, 20] \)]
\( \implies C'(x) = 3x^2 - 90x + 600 \)
For critical value \( C'(x) = 0 \)
\( 3x^2 - 90x + 600 = 0 \)
\( \implies 3(x^2 - 30x + 200) = 0 \)
\( \implies 3\{x^2 - 20x - 10x + 200\} = 0 \)
\( \implies 3\{x(x - 20) - 10(x - 20)\} = 0 \)
\( \implies 3\{(x - 20)(x - 10)\} = 0 \)
\( \implies x = 10, 20 \).
i.e., maximum or minimum value will occur at \( x = 10 \) and \( x = 20 \).
Now, \( C(10) = 10^3 - 45 \times 100 + 600 \times 10 \)
\( = 1000 - 4500 + 6000 = 2500 \)
\( C(20) = 20^3 - 45 \times 400 + 600 \times 20 = 8000 - 18,000 + 12,000 = 2000 \)
Hence, the person must place the order for 20 trees to spend least amount of Rs. 2000.

Question. Show that \( f(x) = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x) \) is increasing in R.
Answer: We have, \( f(x) = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x) \)
\( f'(x) = 2 + \left(\frac{-1}{1 + x^2}\right) + \frac{1}{(\sqrt{1 + x^2} - x)} \cdot \left(\frac{1}{2\sqrt{1 + x^2}} \cdot 2x - 1\right) \)
\( = 2 - \frac{1}{1 + x^2} + \frac{1}{(\sqrt{1 + x^2} - x)} \cdot \frac{(x - \sqrt{1 + x^2})}{\sqrt{1 + x^2}} = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{1 + x^2}} \)
\( = \frac{2 + 2x^2 - 1 - \sqrt{1 + x^2}}{1 + x^2} = \frac{1 + 2x^2 - \sqrt{1 + x^2}}{1 + x^2} \)
For increasing function, \( f'(x) \geq 0 \)
\( \frac{1 + 2x^2 - \sqrt{1 + x^2}}{1 + x^2} \geq 0 \)
\( \implies 1 + 2x^2 \geq \sqrt{1 + x^2} \)
\( \implies (1 + 2x^2)^2 \geq 1 + x^2 \)
\( \implies 1 + 4x^4 + 4x^2 \geq 1 + x^2 \)
\( \implies 4x^4 + 3x^2 \geq 0 \)
\( \implies x^2 (4x^2 + 3) \geq 0 \)
It is true for any real value of \( x \).
Hence, \( f(x) \) is increasing in R.

Question. Find the points on the curve \( y = x^3 \) at which the slope of the tangent is equal to the \( y \)-coordinate of the point.
Answer: Let \( P(x_1, y_1) \) be the required point on the curve \( y = x^3 \) ...(i)
\( \implies \frac{dy}{dx} = 3x^2 \)
\( \implies \left[\frac{dy}{dx}\right]_{(x_1, y_1)} = 3x_1^2 \)
\( \implies \) Slope of tangent at \( (x_1, y_1) = 3x_1^2 \)
According to the question, \( 3x_1^2 = y_1 \) ...(ii)
Also \( (x_1, y_1) \) lies on (i) \( y_1 = x_1^3 \) ...(iii)
From (ii) and (iii), we get \( 3x_1^2 = x_1^3 \)
\( \implies x_1^3 - 3x_1^2 = 0 \)
\( \implies x_1^2(x_1 - 3) = 0 \)
\( \implies x_1 = 0 \) or \( x_1 = 3 \)
\( \implies y_1 = 0 \) or \( y_1 = 27 \)
Hence, required points are (0, 0) and (3, 27).

Question. The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing when the side of the triangle is 20 cm?
Answer: Let 'A' be the area and 'a' be the side of an equilateral triangle.
\( A = \frac{\sqrt{3}}{4} a^2 \)
Differentiating with respect to \( t \), we get
\( \frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2a \cdot \frac{da}{dt} \)
\( \implies \frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2a \times 2 \) [Given \( \frac{da}{dt} = 2 \) cm/sec]
\( \implies \frac{dA}{dt} = \sqrt{3} a \)
\( \implies \left[\frac{dA}{dt}\right]_{a = 20 \text{ cm}} = 20\sqrt{3} \) sq cm/s

Question. Find the intervals in which the function \( f(x) = -3 \log(1 + x) + 4 \log(2 + x) - \frac{4}{2 + x} \) is strictly increasing or strictly decreasing.
Answer: Given \( f(x) = -3 \log(1 + x) + 4 \log(2 + x) - \frac{4}{2 + x} \)
\( \implies f'(x) = \frac{-3}{1 + x} + \frac{4}{2 + x} + \frac{4}{(2 + x)^2} = \frac{-3(2 + x)^2 + 4(1 + x)(2 + x) + 4(1 + x)}{(1 + x)(2 + x)^2} \)
\( = \frac{-3(4 + 4x + x^2) + 4(2 + x + 2x + x^2) + 4 + 4x}{(1 + x)(2 + x)^2} \)
\( = \frac{-12 - 12x - 3x^2 + 8 + 12x + 4x^2 + 4 + 4x}{(1 + x)(2 + x)^2} \)
\( f'(x) = \frac{x(x + 4)}{(1 + x)(2 + x)^2} \)
Now, \( f'(x) = 0 \)
\( \implies \frac{x(x + 4)}{(1 + x)(2 + x)^2} = 0 \implies x(x + 4) = 0 \)
\( \implies x = 0 \) [\( \because x \neq -4 \) as \( f(x) \) is defined on \( (-1, \infty) \)]
Hence, required intervals are \( (-1, 0) \) and \( (0, \infty) \).
For \( (-1, 0) \)
\( f'(x) = \frac{(-ve) \times (+ve)}{(+ve) \times (+ve)} = -ve \implies f(x) \) is strictly decreasing in \( (-1, 0) \)
For \( (0, \infty) \)
\( f'(x) = \frac{(+ve) \times (+ve)}{(+ve) \times (+ve)} = +ve \implies f(x) \) is strictly increasing in \( (0, \infty) \).
i.e., \( f(x) \) is strictly decreasing on \( (-1, 0) \) and strictly increasing on \( (0, \infty) \).

Question. Find the condition that curves \( 2x = y^2 \) and \( 2xy = k \) intersect orthogonally.
Answer: Given, equation of curves are \( 2x = y^2 \) ...(i) and \( 2xy = k \) ...(ii)
\( \implies y = \frac{k}{2x} \)
From equation (i) \( 2x = \left(\frac{k}{2x}\right)^2 \)
\( \implies 8x^3 = k^2 \)
\( \implies x^3 = \frac{1}{8}k^2 \)
\( \implies x = \frac{1}{2}k^{2/3} \)
\( \implies y = \frac{k}{2x} = \frac{k}{2 \cdot \frac{1}{2}k^{2/3}} = k^{1/3} \)
Thus, we get point of intersection of curves which is from equations (i) and (ii).
\( 2 = 2y \frac{dy}{dx} \) and \( 2\left[x \cdot \frac{dy}{dx} + y \cdot 1\right] = 0 \)
\( \implies \frac{dy}{dx} = \frac{1}{y} \) and \( \left(\frac{dy}{dx}\right) = \frac{-2y}{2x} = -\frac{y}{x} \)
\( \implies \left(\frac{dy}{dx}\right)_{(\frac{1}{2}k^{2/3}, k^{1/3})} = \frac{1}{k^{1/3}} \) [say \( m_1 \)]
\( \implies \left(\frac{dy}{dx}\right)_{(\frac{1}{2}k^{2/3}, k^{1/3})} = \frac{-k^{1/3}}{\frac{1}{2}k^{2/3}} = -2k^{-1/3} \) [say \( m_2 \)]
Since, the curves intersect orthogonally.
i.e., \( m_1 \cdot m_2 = -1 \)
\( \implies \frac{1}{k^{1/3}} \cdot (-2k^{-1/3}) = -1 \implies -2k^{-2/3} = -1 \)
\( \implies \frac{2}{k^{2/3}} = 1 \implies k^{2/3} = 2 \)
\( \therefore k^2 = 8 \) which is the required condition.

Long Answer Questions-II

Question. Prove that the surface area of a solid cuboid, of square base and given volume, is minimum when it is a cube.
Answer: Let \( x \) be the side of square base of cuboid and other side be \( y \).
Then volume of cuboid with square base, \( V = x \cdot x \cdot y = x^2 y \)
As volume of cuboid is given so volume is taken constant throughout the question, therefore,
\( y = \frac{V}{x^2} \) ...(i)
In order to show that surface area is minimum when the given cuboid is cube, we have to show \( S'' > 0 \) and \( x = y \).
Let \( S \) be the surface area of cuboid, then
\( S = x^2 + xy + xy + xy + xy + x^2 = 2x^2 + 4xy \) ...(ii)
\( \implies = 2x^2 + 4x \cdot \frac{V}{x^2} \implies S = 2x^2 + \frac{4V}{x} \) ...(iii)
\( \implies \frac{dS}{dx} = 4x - \frac{4V}{x^2} \) ...(iv)
For maximum/minimum value of \( S \), we have \( \frac{dS}{dx} = 0 \)
\( \implies 4x - \frac{4V}{x^2} = 0 \implies 4V = 4x^3 \implies V = x^3 \) ...(v)
Putting \( V = x^3 \) in (i), we have
\( y = \frac{x^3}{x^2} = x \)
Here, \( y = x \implies \) cuboid is a cube.
Differentiating (iv) w.r.t \( x \), we get
\( \frac{d^2S}{dx^2} = \left(4 + \frac{8V}{x^3}\right) > 0 \)
Hence, surface area is minimum when given cuboid is a cube.

Question. If the straight line \( x \cos \alpha + y \sin \alpha = p \) touches the curve \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), then prove that \( a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2 \).
Answer: We know that, if a line \( y = mx + c \) touches ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) then the required condition is \( c^2 = a^2m^2 + b^2 \).
Here, given equation of the line is \( x \cos \alpha + y \sin \alpha = p \).
\( \implies y = \frac{p - x \cos \alpha}{\sin \alpha} = -x \cot \alpha + \frac{p}{\sin \alpha} \)
\( \implies c = \frac{p}{\sin \alpha} \)
and \( m = -\cot \alpha \)
\( \therefore \left(\frac{p}{\sin \alpha}\right)^2 = a^2 (-\cot \alpha)^2 + b^2 \implies \frac{p^2}{\sin^2 \alpha} = a^2 \frac{\cos^2 \alpha}{\sin^2 \alpha} + b^2 \)
\( \implies p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha \)

Question. If the sum of hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is \( \frac{\pi}{3} \).
Answer: Let \( h \) and \( x \) be the length of hypotenuse and one side of a right triangle and \( y \) is length of the third side.
If \( A \) be the area of triangle, then
\( A = \frac{1}{2}xy = \frac{1}{2}x\sqrt{h^2 - x^2} \) [also given \( h + x = k \) (constant) \( \therefore h = k - x \)]
\( A = \frac{1}{2}x\sqrt{(k - x)^2 - x^2} = \frac{1}{2}x\sqrt{k^2 - 2kx + x^2 - x^2} \)
\( A^2 = \frac{x^2}{4}(k^2 - 2kx) \implies A^2 = \frac{1}{4}(k^2x^2 - 2kx^3) \)
Differentiating with respect to \( x \) we get
\( \frac{d(A^2)}{dx} = \frac{1}{4}(2k^2x - 6kx^2) \) ...(i)
For maxima or minima of \( A^2 \)
\( \frac{d(A^2)}{dx} = 0 \implies \frac{1}{4}(2k^2x - 6kx^2) = 0 \)
\( 2k^2x - 6kx^2 = 0 \implies 2kx(k - 3x) = 0 \)
\( k - 3x = 0 \); \( 2kx \neq 0 \)
\( x = \frac{k}{3} \)
Differentiating (i) again with respect to \( x \), we get
\( \frac{d^2(A^2)}{dx^2} = \frac{1}{4}(2k^2 - 12kx) \)
\( \left[\frac{d^2(A^2)}{dx^2}\right]_{x = k/3} = \frac{1}{4}(2k^2 - 12k \cdot \frac{k}{3}) < 0 \)
Hence, \( A^2 \) is maximum when \( x = \frac{k}{3} \) and \( h = k - \frac{k}{3} = \frac{2k}{3} \).
i.e., \( A \) is maximum when \( x = \frac{k}{3}, h = \frac{2k}{3} \)
\( \therefore \cos \theta = \frac{x}{h} = \frac{k}{3} \times \frac{3}{2k} = \frac{1}{2} \implies \cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3} \)

Question. Find the shortest distance of the point (0, c) from the parabola \( y = x^2 \), where \( 1 \leq c \leq 5 \).
Answer: Let \( P(\alpha, \beta) \) be required point on parabola \( y = x^2 \) such that the distance of \( P \) to given point \( Q(0, c) \) is shortest.
Let \( PQ = D \)
\( D = \sqrt{(\alpha - 0)^2 + (\beta - c)^2} \implies D^2 = \alpha^2 + (\beta - c)^2 \)
\( D^2 = \alpha^2 + (\alpha^2 - c)^2 \) [\( \because (\alpha, \beta) \) lie on \( y = x^2 \implies \beta = \alpha^2 \)] ...(i)
Now, \( \frac{d(D^2)}{d\alpha} = 2\alpha + 2(\alpha^2 - c) \cdot 2\alpha = 2\alpha(1 + 2\alpha^2 - 2c) = 2\alpha + 4\alpha^3 - 4\alpha c \)
For extremum value of \( D \) or \( D^2 \)
\( \frac{d(D^2)}{d\alpha} = 0 \implies 2\alpha(1 + 2\alpha^2 - 2c) = 0 \)
\( \implies \alpha = 0, \) or \( 1 + 2\alpha^2 - 2c = 0 \implies \alpha = 0 \) or \( \alpha = \pm \sqrt{\frac{2c - 1}{2}} \)
Again \( \frac{d^2(D^2)}{d\alpha^2} = 2 + 12\alpha^2 - 4c \implies \left[\frac{d^2(D^2)}{d\alpha^2}\right]_{\alpha = 0} = 2 - 4c = -ve \) [\( \because 1 \leq c \leq 5 \)]
\( \left[\frac{d^2(D^2)}{d\alpha^2}\right]_{\alpha = \pm \sqrt{\frac{2c - 1}{2}}} = 2 + 12\left(\frac{2c - 1}{2}\right) - 4c = 2 + 12c - 6 - 4c = 8c - 4 > 0 \) [\( \because 1 \leq c \leq 5 \)]
i.e., for \( \alpha = \pm \sqrt{\frac{2c - 1}{2}} \), \( D^2 \) i.e., \( D \) is minimum (shortest)
Now, the shortest distance \( D \) is
\( D = \sqrt{\alpha^2 + (\alpha^2 - c)^2} = \sqrt{\alpha^4 + \alpha^2 + c^2 - 2\alpha^2c} \) [From (i)]
\( = \sqrt{\left(\frac{2c - 1}{2}\right)^2 + \frac{2c - 1}{2} + c^2 - 2c\left(\frac{2c - 1}{2}\right)} \) [\( \because \alpha^2 = \frac{2c - 1}{2} \)]
\( = \sqrt{\frac{(2c - 1)^2 + 4c^2 + 4c - 2 - 4c(2c - 1)}{4}} \)
\( = \frac{1}{2}\sqrt{4c^2 + 1 - 4c + 4c^2 + 4c - 2 - 8c^2 + 4c} = \frac{1}{2}\sqrt{4c - 1} \)
Hence, required shortest distance is \( \frac{1}{2}\sqrt{4c - 1} \).

Question. Find the absolute maximum and absolute minimum values of the function \( f \) given by \( f(x) = \sin^2 x - \cos x, x \in [0, \pi] \).
Answer: Here, \( f(x) = \sin^2 x - \cos x \)
\( f'(x) = 2 \sin x \cos x + \sin x \implies f'(x) = \sin x (2 \cos x + 1) \)
For critical point: \( f'(x) = 0 \)
\( \implies \sin x (2 \cos x + 1) = 0 \implies \sin x = 0 \) or \( \cos x = -\frac{1}{2} \)
\( \implies x = 0 \) or \( \cos x = \cos \frac{2\pi}{3} \implies x = 0 \) or \( x = 2n\pi \pm \frac{2\pi}{3} \), where \( n = 0, \pm 1, \pm 2 \dots \)
\( \implies x = 0 \) or \( x = \frac{2\pi}{3} \) other values does not belong to \( [0, \pi] \).
For absolute maximum or minimum values:
\( f(0) = \sin^2 0 - \cos 0 = 0 - 1 = -1 \)
\( f(\frac{2\pi}{3}) = \sin^2 \frac{2\pi}{3} - \cos \frac{2\pi}{3} = \left(\frac{\sqrt{3}}{2}\right)^2 - (-\frac{1}{2}) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} \)
\( f(\pi) = \sin^2 \pi - \cos \pi = 0 - (-1) = 1 \)
Hence, absolute maximum value \( = \frac{5}{4} \) and absolute minimum value \( = -1 \).

Question. If the function \( f(x) = 2x^3 - 9mx^2 + 12m^2x + 1 \), where \( m > 0 \) attains its maximum and minimum at \( p \) and \( q \) respectively such that \( p^2 = q \), then find the value of \( m \).
Answer: Given, \( f(x) = 2x^3 - 9mx^2 + 12m^2x + 1 \)
\( \implies f'(x) = 6x^2 - 18mx + 12m^2 \)
For extremum value of \( f(x), f'(x) = 0 \)
\( \implies 6x^2 - 18mx + 12m^2 = 0 \implies x^2 - 3mx + 2m^2 = 0 \)
\( \implies x^2 - 2mx - mx + 2m^2 = 0 \implies x(x - 2m) - m(x - 2m) = 0 \)
\( \implies (x - m)(x - 2m) = 0 \implies x = m \) or \( x = 2m \)
Now, \( f''(x) = 12x - 18m \)
And, \( f''(x) \) at \( [x = m] = f''(m) = 12m - 18m = -6m < 0 \)
And, \( f''(x) \) at \( [x = 2m] = f''(2m) = 24m - 18m = 6m > 0 \)
Hence, \( f(x) \) attains maximum and minimum value at \( m \) and \( 2m \) respectively.
\( \implies m = p \) and \( 2m = q \)
But, \( p^2 = q \) [Given]
\( \therefore m^2 = 2m \implies m^2 - 2m = 0 \)
\( \implies m(m - 2) = 0 \implies m = 0 \) or \( m = 2 \)
\( \implies m = 2 \) as \( m > 0 \)

Question. The sum of the surface areas of a cuboid with sides \( x, 2x \) and \( \frac{x}{3} \) and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if \( x \) is equal to three times the radius of sphere. Also find the minimum value of the sum of their volumes.
Answer: Let \( r \) be the radius of sphere and S, V be the sum of surface area and volume of cuboid and sphere.
Now \( V = (x \cdot 2x \cdot \frac{x}{3}) + \frac{4}{3}\pi r^3 \)
\( \implies V = \frac{2}{3}x^3 + \frac{4}{3}\pi r^3 \)
\( \implies V = \frac{2}{3} \left[ \left(\frac{S - 4\pi r^2}{6}\right)^{3/2} + 2\pi r^3 \right] \dots S = 2\left[x \cdot 2x + x \cdot \frac{x}{3} + \frac{x}{3} \cdot 2x\right] + 4\pi r^2 \)
\( \implies S = \frac{18x^2}{3} + 4\pi r^2 = 6x^2 + 4\pi r^2 \)
\( \implies x^2 = \frac{S - 4\pi r^2}{6} \implies x^3 = \left(\frac{S - 4\pi r^2}{6}\right)^{3/2} \)
\( \implies \frac{dV}{dr} = \frac{2}{3} \left[ \frac{3}{2} \left(\frac{S - 4\pi r^2}{6}\right)^{1/2} \cdot \left(\frac{1}{6}\right) \cdot (-8\pi r) + 6\pi r^2 \right] \)
For maximum or minimum value
\( \frac{dV}{dr} = 0 \)
\( \implies \frac{2}{3} \left\{ -2\pi r \left(\frac{S - 4\pi r^2}{6}\right)^{1/2} + 6\pi r^2 \right\} = 0 \)
\( \implies \left(\frac{S - 4\pi r^2}{6}\right)^{1/2} = \frac{6\pi r^2}{2\pi r} \)
\( \implies \left(\frac{S - 4\pi r^2}{6}\right)^{1/2} = 3r \implies r = \frac{1}{3} \left(\frac{S - 4\pi r^2}{6}\right)^{1/2} \)
Obviously, \( \frac{d^2V}{dr^2}\bigg|_{r = \frac{1}{3}(\frac{S - 4\pi r^2}{6})^{1/2}} = +ve \)
V is minimum when \( r = \frac{1}{3} \left(\frac{S - 4\pi r^2}{6}\right)^{1/2} \)
\( \implies 3r = \left(\frac{S - 4\pi r^2}{6}\right)^{1/2} \implies 9r^2 = \left(\frac{S - 4\pi r^2}{6}\right) \implies 54r^2 = S - 4\pi r^2 \)
\( \implies 54r^2 = 6x^2 + 4\pi r^2 - 4\pi r^2 \) [\( \because S = 6x^2 + 4\pi r^2 \)]
\( \implies x^2 = 9r^2 \implies x = 3r \)
i.e., \( x \) is equal to three times the radius of sphere.
Now, minimum value of V (sum of volume) \( = \frac{2}{3} \left[ x^3 + 2\pi (\frac{x}{3})^3 \right] \)
\( = \frac{2}{3} \left[ x^3 + \frac{2\pi x^3}{27} \right] = \frac{2x^3}{81} (27 + 2\pi) \) cubic unit.

Question. Find the maximum and minimum values of \( f(x) = \sec x + \log \cos^2 x, 0 < x < 2\pi \).
Answer: We have \( f(x) = \sec x + \log \cos^2 x \)
\( f'(x) = \sec x \tan x + \frac{1}{\cos^2 x} \cdot 2 \cos x (-\sin x) = \sec x \tan x - 2 \tan x = \tan x (\sec x - 2) \)
For critical point
\( f'(x) = 0 \)
\( \tan x (\sec x - 2) = 0 \implies \tan x = 0 \) or \( \sec x - 2 = 0 \)
\( x = n\pi \) or \( \sec x = 2 \implies x = n\pi \) or \( \cos x = \frac{1}{2} \)
\( x = n\pi \) or \( \cos x = \cos \frac{\pi}{3} \implies x = n\pi \) or \( x = 2n\pi \pm \frac{\pi}{3}, n = 0, \pm 1, \pm 2 \dots \)
Thus possible value of \( x \) in interval \( 0 < x < 2\pi \) are \( x = \pi, \frac{\pi}{3}, \frac{5\pi}{3} \)
Now, \( f(\frac{\pi}{3}) = \sec \frac{\pi}{3} + \log \cos^2 \frac{\pi}{3} = 2 + \log (\frac{1}{2})^2 \)
\( = 2 + 2(\log 1 - \log 2) = 2 - 2 \log 2 = 2(1 - \log 2) \) [\( \because \log 1 = 0 \)]
\( f(\pi) = \sec \pi + \log \cos^2 \pi = -1 + \log (-1)^2 = -1 \)
\( f(\frac{5\pi}{3}) = \sec \frac{5\pi}{3} + \log \cos^2 \frac{5\pi}{3} = \sec (2\pi - \frac{\pi}{3}) + 2 \log \cos (2\pi - \frac{\pi}{3}) \)
\( = \sec \frac{\pi}{3} + 2 \log \cos \frac{\pi}{3} = 2 + 2 \log \frac{1}{2} \)
\( = 2 + 2(\log 1 - \log 2) = 2 - 2 \log 2 = 2(1 - \log 2) \)
Hence, maximum value of \( f(x) = 2(1 - \log 2) \)
minimum value of \( f(x) = -1 \)

Question. Show that the normal at any point \( \theta \) to the curve \( x = a \cos \theta + a\theta \sin \theta, y = a \sin \theta - a\theta \cos \theta \) is at a constant distance from the origin.
Answer: Given \( x = a \cos \theta + a\theta \sin \theta \) and \( y = a \sin \theta - a\theta \cos \theta \)
\( \therefore \frac{dx}{d\theta} = -a \sin \theta + a(\theta \cos \theta + \sin \theta) = a\theta \cos \theta \)
and \( \frac{dy}{d\theta} = a \cos \theta - a(-\theta \sin \theta + \cos \theta) = a\theta \sin \theta \)
\( \implies \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\theta \sin \theta}{a\theta \cos \theta} = \tan \theta \)
\( \therefore \) Slope of tangent at \( \theta = \tan \theta \)
\( \implies \) Slope of normal at \( \theta = -\frac{1}{\tan \theta} = -\cot \theta \)
Hence, equation of normal at \( \theta \) is given by
\( \frac{y - (a \sin \theta - a\theta \cos \theta)}{x - (a \cos \theta + a\theta \sin \theta)} = -\cot \theta \)
\( \implies y - a \sin \theta + a\theta \cos \theta + x \cot \theta - \cot \theta (a \cos \theta + a\theta \sin \theta) = 0 \)
\( \implies y - a \sin \theta + a\theta \cos \theta + x \frac{\cos \theta}{\sin \theta} - a \frac{\cos^2 \theta}{\sin \theta} - a\theta \cos \theta = 0 \)
\( \implies x \cos \theta + y \sin \theta - a = 0 \)
Distance from origin (0, 0) to (i) \( = \left| \frac{0 \cdot \cos \theta + 0 \cdot \sin \theta - a}{\sqrt{\cos^2 \theta + \sin^2 \theta}} \right| = a \)