CBSE Class 12 Mathematics Application of Derivatives Important Questions Set A

Objective Type Questions:

Question. The points at which the tangents to the curve \( y = x^3 - 12x + 18 \) are parallel to x-axis are
(a) \( (2, -2), (-2, -34) \)
(b) \( (2, 34), (-2, 0) \)
(c) \( (0, 34), (-2, 0) \)
(d) \( (2, 2), (-2, 34) \)
Answer: (d)

Question. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is
(a) \( 10 \text{ cm}^2\text{/s} \)
(b) \( \sqrt{3} \text{ cm}^2\text{/s} \)
(c) \( 10\sqrt{3} \text{ cm}^2\text{/s} \)
(d) \( \frac{10}{3} \text{ cm}^2\text{/s} \)
Answer: (c)

Question. The maximum value of slope of the curve \( y = -x^3 + 3x^2 + 12x - 5 \) is
(a) 15
(b) 12
(c) 9
(d) 0
Answer: (a)

Question. If the function \( f(x) = 2x^2 - kx + 5 \) is increasing on \( [1, 2] \), then k lies in the interval
(a) \( (-\infty, 4) \)
(b) \( (4, \infty) \)
(c) \( (-\infty, 8) \)
(d) \( (8, \infty) \)
Answer: (a)

Question. If the curve \( ay + x^2 = 7 \) and \( x^3 = y \), cut orthogonally at \( (1, 1) \), then the value of a is
(a) 1
(b) 0
(c) -6
(d) 6
Answer: (d)

Question. The approximate value of \( (33)^{1/5} \) is
(a) 2.0125
(b) 2.1
(c) 2.01
(d) none of these
Answer: (a)

Question. The equation of the normal to the curve \( y = x(2 - x) \) at the point \( (2, 0) \) is
(a) \( x - 2y = 2 \)
(b) \( x - 2y + 2 = 0 \)
(c) \( 2x + y = 4 \)
(d) \( 2x + y - 4 = 0 \)
Answer: (a)

Question. The angle of intersection of the parabolas \( y^2 = 4ax \) and \( x^2 = 4ay \) at the origin, is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{\pi}{4} \)
Answer: (c)

Short Answer Questions-I 

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: \( R(x) = 3x^2 + 36x + 5 \)
\( \Rightarrow R'(x) = 6x + 36 \)
\( \therefore \) 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, \( P(x) = 0.005x^3 + 0.02x^2 + 30x \)
\( \therefore P'(x) = 0.015x^2 + 0.04x + 30 \)
\( \Rightarrow [P'(x)]_{x = 3} = 0.015 \times 9 + 0.04 \times 3 + 30 \)
\( = 0.135 + 0.12 + 30 = 30.255 \)

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 \( (A_2) = y^2 = (x - x^2)^2 \)
\( \therefore \frac{dA_2}{dt} = \frac{d}{dt}(x - x^2)^2 = 2(x - x^2) \cdot \left( \frac{dx}{dt} - 2x \cdot \frac{dx}{dt} \right) \)
\( = \frac{dx}{dt}(1 - 2x)2(x - x^2) \)
and \( \frac{dA_1}{dt} = \frac{d}{dx} x^2 \cdot \frac{dx}{dt} = 2x \cdot \frac{dx}{dt} \)
\( \therefore \frac{dA_2}{dA_1} = \frac{dA_2/dt}{dA_1/dt} = \frac{\frac{dx}{dt} \cdot (1 - 2x)(2x - 2x^2)}{2x \cdot \frac{dx}{dt}} \)
\( = \frac{(1 - 2x)2x(1 - x)}{2x} \)
\( = (1 - 2x)(1 - x) = 1 - x - 2x + 2x^2 = 2x^2 - 3x + 1 \)

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 \( \delta x = 0.5 \)
\( \Rightarrow f'(x) = \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{49}} = \frac{1}{14} \)
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 \( f(x) = \tan^{-1} (\sin x + \cos x) \)
\( \Rightarrow f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \times (\cos x - \sin x) \)
\( f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \)
\( \because 1 + (\sin x + \cos x)^2 > 0 \forall x \in R \)
Also, \( x \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \Rightarrow \sin x > \cos x \Rightarrow \cos x - \sin x < 0 \)
\( \therefore f'(x) = \frac{-ve}{+ve} = -ve \) i.e., \( f'(x) < 0 \)
\( \Rightarrow f(x) \) is decreasing in \( \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \)

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

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 \( y = \frac{1}{x^2} \Rightarrow \frac{dy}{dx} = -\frac{2}{x^3} \Rightarrow \frac{dy}{dx} = -\frac{2}{2^3} = -\frac{2}{8} \)
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 \( f(x) = \cos\left( 2x + \frac{\pi}{4} \right) \); is increasing or decreasing in the interval \( \left( \frac{3\pi}{8}, \frac{7\pi}{8} \right) \).
Answer: We have \( f(x) = \cos\left( 2x + \frac{\pi}{4} \right) \)
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, \( f'(x) = - 2 \sin\left( 2x + \frac{\pi}{4} \right) \)
\( \therefore f'(x) > 0 \Rightarrow -2 \sin\left( 2x + \frac{\pi}{4} \right) > 0 \)
\( \Rightarrow \sin\left( 2x + \frac{\pi}{4} \right) < 0 \)
\( \Rightarrow \pi < 2x + \frac{\pi}{4} < 2\pi \) [\( \because \sin x < 0 \forall x \in (\pi, 2\pi) \)]
\( \Rightarrow \pi - \frac{\pi}{4} < 2x < 2\pi - \frac{\pi}{4} \Rightarrow \frac{3\pi}{4} < 2x < \frac{7\pi}{4} \)
\( \Rightarrow \frac{3\pi}{8} < x < \frac{7\pi}{8} \)
Thus \( f'(x) > 0 \forall x \in \left( \frac{3\pi}{8}, \frac{7\pi}{8} \right) \)
\( \Rightarrow f(x) \) is increasing function on \( \left( \frac{3\pi}{8}, \frac{7\pi}{8} \right) \).

Short Answer Questions-II

Question. Prove that the curves \( xy = 4 \) and \( x^2 + y^2 = 8 \) touch each other.
Answer: Given equation of curves are
\( xy = 4 \) ... (i)
and \( x^2 + y^2 = 8 \) ... (ii)
\( \Rightarrow x \cdot \frac{dy}{dx} + y = 0 \)
and \( 2x + 2y \frac{dy}{dx} = 0 \)
\( \Rightarrow \frac{dy}{dx} = \frac{-y}{x} \) and \( \frac{dy}{dx} = \frac{-2x}{2y} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-y}{x} = m_1 \) (say)
and \( \frac{dy}{dx} = \frac{-x}{y} = m_2 \) (say)
\( \therefore \frac{-y}{x} = \frac{-x}{y} \Rightarrow -y^2 = -x^2 \text{ and } x^2 = y^2 \) ... (iii)
Using the value of \( x^2 \) in equation (ii), we get
\( y^2 + y^2 = 8 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2 \)
For \( y = 2, x = \frac{4}{2} = 2 \) and for \( y = -2, x = \frac{4}{-2} = -2 \)
Thus, the required points of intersection are (2, 2) and (-2, -2)
For (2, 2), \( m_1 = \frac{-y}{x} = \frac{-2}{2} = -1 \)
and \( m_2 = \frac{-x}{y} = \frac{-2}{2} = -1 \)
\( \therefore m_1 = m_2 \)
For (-2, -2), \( m_1 = \frac{-y}{x} = \frac{-(-2)}{-2} = -1 \)
and \( m_2 = \frac{-x}{y} = \frac{-(-2)}{-2} = -1 \)
For both the intersection points, we see that slope of both curve are same.

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 \( f(x) \)
\( f'(x) = 0 \)
\( \Rightarrow 3 \cos 3x + 3 \sin 3x = 0 \Rightarrow \cos 3x + \sin 3x = 0 \)
\( \Rightarrow \sin 3x = - \cos 3x \Rightarrow \frac{\sin 3x}{\cos 3x} = -1 \)
\( \Rightarrow \tan 3x = - \tan \frac{\pi}{4} \Rightarrow \tan 3x = \tan\left( \pi - \frac{\pi}{4} \right) \)
\( \Rightarrow \tan 3x = \tan \frac{3\pi}{4} \)
\( \Rightarrow 3x = n\pi + \frac{3\pi}{4} \), where \( n = 0, \pm 1, \pm 2, \dots \)
Putting \( n = 0, \pm 1, \dots \), we get
\( x = \frac{\pi}{4}, \frac{7\pi}{12}, \frac{11\pi}{12} \in (0, \pi) \)
Hence, required possible intervals are \( \left( 0, \frac{\pi}{4} \right), \left( \frac{\pi}{4}, \frac{7\pi}{12} \right), \left( \frac{7\pi}{12}, \frac{11\pi}{12} \right), \left( \frac{11\pi}{12}, \pi \right) \)
For \( \left( 0, \frac{\pi}{4} \right), f'(x) = +ve \)
For \( \left( \frac{\pi}{4}, \frac{7\pi}{12} \right), f'(x) = -ve \)
For \( \left( \frac{7\pi}{12}, \frac{11\pi}{12} \right), f'(x) = +ve \)
For \( \left( \frac{11\pi}{12}, \pi \right), f'(x) = -ve \)
Hence, given function \( f(x) \) is strictly increasing in \( \left( 0, \frac{\pi}{4} \right) \cup \left( \frac{7\pi}{12}, \frac{11\pi}{12} \right) \) and strictly decreasing in \( \left( \frac{\pi}{4}, \frac{7\pi}{12} \right) \cup \left( \frac{11\pi}{12}, \pi \right) \).

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
\( 2ay \frac{dy}{dx} = 3x^2 \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ay} \)
\( \Rightarrow \frac{dy}{dx} \text{ at } (am^2, am^3) = \frac{3 \times a^2 m^4}{2a \times am^3} = \frac{3m}{2} \)
\( \therefore \) Slope of normal \( = - \frac{1}{\text{slope of tangent}} = - \frac{1}{3m/2} = - \frac{2}{3m} \)
Equation of normal at the point \( (am^2, am^3) \) is given by
\( \frac{y - am^3}{x - am^2} = - \frac{2}{3m} \Rightarrow 3my - 3am^4 = - 2x + 2am^2 \)

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
\( 2ay \frac{dy}{dx} = 3x^2 \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ay} \)
\( \Rightarrow \frac{dy}{dx} \text{ at } (am^2, am^3) = \frac{3 \times a^2 m^4}{2a \times am^3} = \frac{3m}{2} \)
\( \therefore \) Slope of normal \( = -\frac{1}{\text{slope of tangent}} = -\frac{1}{3m/2} = -\frac{2}{3m} \)
Equation of normal at the point \( (am^2, am^3) \) is given by
\( \frac{y - am^3}{x - am^2} = -\frac{2}{3m} \Rightarrow 3my - 3am^4 = -2x + 2am^2 \)
\( \Rightarrow 2x + 3my - am^2(2 + 3m^2) = 0 \)
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 \( y = f(x) \)]
\( \Rightarrow f'(x) = \frac{1}{1 + x} - 2 \left[ \frac{(2 + x) \cdot 1 - x}{(2 + x)^2} \right] \)
\( = \frac{1}{1 + x} - \frac{2(2 + x - x)}{(2 + x)^2} = \frac{1}{1 + x} - \frac{4}{(2 + x)^2} \)
\( = \frac{4 + x^2 + 4x - 4 - 4x}{(x + 1)(x + 2)^2} = \frac{x^2}{(x + 1)(x + 2)^2} \)
For \( f(x) \) being increasing function
\( f'(x) > 0 \)
\( \Rightarrow \frac{x^2}{(x + 1)(x + 2)^2} > 0 \Rightarrow \frac{1}{x + 1} \cdot \frac{x^2}{(x + 2)^2} > 0 \)
\( \Rightarrow \frac{1}{x + 1} > 0 \) \( \left[ \because \frac{x^2}{(x + 2)^2} > 0 \right] \)
\( \Rightarrow x + 1 > 0 \) or \( x > -1 \)
i.e., \( f(x) = y = \log (1 + x) - \frac{2x}{2 + x} \) is increasing function in its domain \( x > -1 \) i.e., \( (-1, \infty) \).

Question. Show that \( f(x) = 2x + \cot^{-1}x + \log(\sqrt{1+x^2} - x) \) is increasing in R.
Answer: We have,
\( f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x) \)
\( f'(x) = 2 + \left( \frac{-1}{1 + x^2} \right) + \frac{1}{\sqrt{1 + x^2} - x} \left( \frac{1}{2\sqrt{1 + x^2}} \cdot 2x - 1 \right) \)
\( = 2 - \frac{1}{1 + x^2} + \frac{1}{\sqrt{1 + x^2} - x} \cdot \frac{(x - \sqrt{1 + x^2})}{\sqrt{1 + x^2}} = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{1 + x^2}} \)
\( = \frac{2 + 2x^2 - 1 - \sqrt{1 + x^2}}{1 + x^2} = \frac{1 + 2x^2 - \sqrt{1 + x^2}}{1 + x^2} \)
For increasing function, \( f'(x) \ge 0 \)
\( \frac{1 + 2x^2 - \sqrt{1 + x^2}}{1 + x^2} \ge 0 \Rightarrow 1 + 2x^2 \ge \sqrt{1 + x^2} \)
\( \Rightarrow (1 + 2x^2)^2 \ge 1 + x^2 \Rightarrow 1 + 4x^4 + 4x^2 \ge 1 + x^2 \)
\( \Rightarrow 4x^4 + 3x^2 \ge 0 \Rightarrow x^2 (4x^2 + 3) \ge 0 \)
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)
\( \Rightarrow \frac{dy}{dx} = 3x^2 \Rightarrow \left[ \frac{dy}{dx} \right]_{(x_1, y_1)} = 3x_1^2 \)
\( \Rightarrow \) 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)
\( \Rightarrow y_1 = x_1^3 \) ...(iii)
From (ii) and (iii), we get
\( 3x_1^2 = x_1^3 \)
\( \Rightarrow x_1^3 - 3x_1^2 = 0 \Rightarrow x_1^2 (x_1 - 3) = 0 \)
\( \Rightarrow x_1 = 0 \) or \( x_1 = 3 \)
\( \Rightarrow y_1 = 0 \) or \( y_1 = 27 \)
Hence, required points are \( (0, 0) \) and \( (3, 27) \).

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 \( f(x) = -3 \log (1 + x) + 4 \log(2 + x) - \frac{4}{2 + x} \)
\( \Rightarrow f'(x) = \frac{-3}{1 + x} + \frac{4}{2 + x} + \frac{4}{(2 + x)^2} = \frac{-3(2 + x)^2 + 4(1 + x)(2 + x) + 4(1 + x)}{(1 + x)(2 + x)^2} \)
\( = \frac{-3(4 + 4x + x^2) + 4(2 + x + 2x + x^2) + 4 + 4x}{(1 + x)(2 + x)^2} \)
\( = \frac{-12 - 12x - 3x^2 + 8 + 12x + 4x^2 + 4 + 4x}{(1 + x)(2 + x)^2} \)
\( f'(x) = \frac{x(x + 4)}{(1 + x)(2 + x)^2} \)
Now, \( f'(x) = 0 \Rightarrow \frac{x(x + 4)}{(1 + x)(2 + x)^2} = 0 \Rightarrow x(x + 4) = 0 \)
\( \Rightarrow x = 0 \) [\( \because x \ne -4 \) as \( f(x) \) is defined on \( (-1, \infty) \)]
Hence, required intervals are \( (-1, 0) \) and \( (0, \infty) \).
For \( (-1, 0) \)
\( f'(x) = \frac{(-ve) \times (+ve)}{(+ve) \times (+ve)} = -ve \Rightarrow f(x) \) is strictly decreasing in \( (-1, 0) \)
For \( (0, \infty) \)
\( f'(x) = \frac{(+ve) \times (+ve)}{(+ve) \times (+ve)} = +ve \Rightarrow f(x) \) 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)
\( \Rightarrow y = \frac{k}{2x} \)
From equation (i), \( 2x = \left( \frac{k}{2x} \right)^2 \Rightarrow 8x^3 = k^2 \Rightarrow x^3 = \frac{1}{8}k^2 \Rightarrow x = \frac{1}{2}k^{2/3} \)
\( \Rightarrow y = \frac{k}{2 \cdot \frac{1}{2}k^{2/3}} \Rightarrow y = k^{1/3} \)
Thus, we get point of intersection of curves which is \( \left( \frac{1}{2}k^{2/3}, k^{1/3} \right) \)
From (i) and (ii), differentiating:
\( 2 = 2y \frac{dy}{dx} \) and \( 2 \left[ x \cdot \frac{dy}{dx} + y \cdot 1 \right] = 0 \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{y} \) and \( \frac{dy}{dx} = -\frac{y}{x} \)
\( \Rightarrow \left( \frac{dy}{dx} \right)_{\left( \frac{1}{2}k^{2/3}, k^{1/3} \right)} = \frac{1}{k^{1/3}} \) [say \( m_1 \)]
\( \Rightarrow \left( \frac{dy}{dx} \right)_{\left( \frac{1}{2}k^{2/3}, k^{1/3} \right)} = \frac{-k^{1/3}}{\frac{1}{2}k^{2/3}} = -2k^{-1/3} \) [say \( m_2 \)]
Since, the curves intersect orthogonally.
i.e., \( m_1 m_2 = -1 \)
\( \Rightarrow \frac{1}{k^{1/3}} \cdot (-2k^{-1/3}) = -1 \Rightarrow -2k^{-2/3} = -1 \)
\( \Rightarrow \frac{2}{k^{2/3}} = 1 \Rightarrow k^{2/3} = 2 \)
\( \therefore k^2 = 8 \)
which is the required condition.

Long Answer Questions 

Question. Find the minimum value of \( (ax + by) \), where \( xy = c^2 \).
Answer: Let \( z = ax + by \) ...(i)
Given \( xy = c^2 \Rightarrow y = \frac{c^2}{x} \)
Putting \( y = \frac{c^2}{x} \) in equation (i), we have
\( z = ax + \frac{bc^2}{x} \)
For \( z \) to be maximum or minimum
\( \frac{dz}{dx} = a - \frac{bc^2}{x^2} = 0 \Rightarrow a = \frac{bc^2}{x^2} \Rightarrow x^2 = \frac{bc^2}{a} \Rightarrow x = \pm c \sqrt{\frac{b}{a}} \)
Now, \( \frac{d^2z}{dx^2} = \frac{2bc^2}{x^3} \)
\( \therefore \) at \( x = c \sqrt{\frac{b}{a}}, \frac{d^2z}{dx^2} = \frac{2bc^2}{\left( c \sqrt{\frac{b}{a}} \right)^3} > 0 \)
\( \therefore z \) will be minimum at \( x = c \sqrt{\frac{b}{a}} \)
\( y = \frac{c^2}{x} = \frac{c^2}{c \sqrt{\frac{b}{a}}} = c \sqrt{\frac{a}{b}} \)
\( \therefore \) Minimum value of \( z = ax + by \)
\( = a \times c\sqrt{\frac{b}{a}} + b \times c\sqrt{\frac{a}{b}} = c\sqrt{ab} + c\sqrt{ab} = 2c\sqrt{ab} \)

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)
\( = 2x^2 + 4x \cdot \frac{V}{x^2} \Rightarrow S = 2x^2 + \frac{4V}{x} \) ...(iii)
\( \Rightarrow \frac{dS}{dx} = 4x - \frac{4V}{x^2} \) ...(iv)
For maximum/minimum value of \( S \), we have \( \frac{dS}{dx} = 0 \)
\( \Rightarrow 4x - \frac{4V}{x^2} = 0 \Rightarrow 4V = 4x^3 \Rightarrow V = x^3 \) ...(v)
Putting \( V = x^3 \) in (i), we have \( y = \frac{x^3}{x^2} = x \)
Here, \( y = x \Rightarrow \) 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. 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, \( f(x) = \sin^2 x - \cos x \)
\( f'(x) = 2\sin x.\cos x + \sin x \Rightarrow f'(x) = \sin x(2\cos x + 1) \)
For critical point: \( f'(x) = 0 \)
\( \Rightarrow \sin x(2\cos x + 1) = 0 \Rightarrow \sin x = 0 \) or \( \cos x = -\frac{1}{2} \)
\( \Rightarrow x = 0 \) or \( \cos x = \cos \frac{2\pi}{3} \Rightarrow x = 0 \) or \( x = 2n\pi \pm \frac{2\pi}{3} \), where \( n = 0, \pm 1, \pm 2, \dots \)
\( \Rightarrow x = 0 \) or \( x = \frac{2\pi}{3} \) (other values do not belong to \( [0, \pi] \)).
For absolute maximum or minimum values:
\( f(0) = \sin^2 0 - \cos 0 = 0 - 1 = -1 \)
\( f\left(\frac{2\pi}{3}\right) = \sin^2 \frac{2\pi}{3} - \cos \frac{2\pi}{3} = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(-\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} \)
\( f(\pi) = \sin^2 \pi - \cos \pi = 0 - (-1) = 1 \)
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, \( f(x) = 2x^3 - 9mx^2 + 12m^2x + 1 \)
\( \Rightarrow f'(x) = 6x^2 - 18mx + 12m^2 \)
For extremum value of \( f(x), f'(x) = 0 \)
\( \Rightarrow 6x^2 - 18mx + 12m^2 = 0 \Rightarrow x^2 - 3mx + 2m^2 = 0 \)
\( \Rightarrow x^2 - 2mx - mx + 2m^2 = 0 \Rightarrow x(x - 2m) - m(x - 2m) = 0 \)
\( \Rightarrow (x - m)(x - 2m) = 0 \Rightarrow x = m \) or \( x = 2m \)
Now, \( f''(x) = 12x - 18m \)
\( \Rightarrow f''(x) \) 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.
\( \Rightarrow m = p \) and \( 2m = q \)
But, \( p^2 = q \) [Given]
\( \therefore m^2 = 2m \Rightarrow m^2 - 2m = 0 \Rightarrow m(m - 2) = 0 \)
\( \Rightarrow m = 2 \) 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 \( V = \left( x \cdot 2x \cdot \frac{x}{3} \right) + \frac{4}{3}\pi r^3 \)
\( \Rightarrow V = \frac{2}{3}x^3 + \frac{4}{3}\pi r^3 \Rightarrow V = \frac{2}{3}(x^3 + 2\pi r^3) \)
\( \because S = 2\left[ x \cdot 2x + x \cdot \frac{x}{3} + \frac{x}{3} \cdot 2x \right] + 4\pi r^2 \)
\( \Rightarrow S = 2\left[ 2x^2 + \frac{x^2}{3} + \frac{2x^2}{3} \right] + 4\pi r^2 = 2[3x^2] + 4\pi r^2 = 6x^2 + 4\pi r^2 \)
\( \Rightarrow x^2 = \frac{S - 4\pi r^2}{6} \Rightarrow x^3 = \left( \frac{S - 4\pi r^2}{6} \right)^{3/2} \)
\( V = \frac{2}{3}\left[ \left( \frac{S - 4\pi r^2}{6} \right)^{3/2} + 2\pi r^3 \right] \)
\( \frac{dV}{dr} = \frac{2}{3} \left[ \frac{3}{2} \left( \frac{S - 4\pi r^2}{6} \right)^{1/2} \cdot \left( -\frac{8\pi r}{6} \right) + 6\pi r^2 \right] \)
For maximum or minimum value \( \frac{dV}{dr} = 0 \)
\( \Rightarrow \frac{2}{3} \left[ -2\pi r \left( \frac{S - 4\pi r^2}{6} \right)^{1/2} + 6\pi r^2 \right] = 0 \Rightarrow \left( \frac{S - 4\pi r^2}{6} \right)^{1/2} = \frac{6\pi r^2}{2\pi r} = 3r \)
\( \Rightarrow \frac{S - 4\pi r^2}{6} = 9r^2 \Rightarrow S - 4\pi r^2 = 54r^2 \)
\( \because S = 6x^2 + 4\pi r^2 \Rightarrow 54r^2 = 6x^2 + 4\pi r^2 - 4\pi r^2 \Rightarrow 54r^2 = 6x^2 \)
\( \Rightarrow x^2 = 9r^2 \Rightarrow x = 3r \)
i.e., \( x \) is equal to three times the radius of sphere.
Now, minimum value of \( V \) (sum of volume) \( = \frac{2}{3} \left\{ x^3 + 2\pi \left( \frac{x}{3} \right)^3 \right\} \)
\( = \frac{2}{3} \left\{ x^3 + \frac{2\pi}{27} x^3 \right\} = \frac{2}{81} x^3 (27 + 2\pi) \) 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 \( f(x) = \sec x + \log \cos^2 x \)
\( f'(x) = \sec x \cdot \tan x + \frac{1}{\cos^2 x} \cdot 2\cos x(-\sin x) = \sec x \cdot \tan x - 2 \tan x = \tan x (\sec x - 2) \)
For critical point \( f'(x) = 0 \)
\( \Rightarrow \tan x (\sec x - 2) = 0 \Rightarrow \tan x = 0 \) or \( \sec x - 2 = 0 \)
\( \Rightarrow x = n\pi \) or \( \sec x = 2 \Rightarrow \cos x = \frac{1}{2} \)
\( \Rightarrow x = n\pi \) or \( x = 2n\pi \pm \frac{\pi}{3} \), \( n = 0, \pm 1, \pm 2 \dots \)
Possible values of \( x \) in interval \( 0 < x < 2\pi \) are \( x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \).
Now, \( f\left(\frac{\pi}{3}\right) = \sec \frac{\pi}{3} + \log \cos^2 \frac{\pi}{3} = 2 + \log \left( \frac{1}{2} \right)^2 \)
\( = 2 + 2(\log 1 - \log 2) = 2 - 2 \log 2 = 2(1 - \log 2) \) [\( \because \log 1 = 0 \)]
\( f(\pi) = \sec \pi + \log \cos^2 \pi = -1 + \log (-1)^2 = -1 \)
\( f\left(\frac{5\pi}{3}\right) = \sec \left( 2\pi - \frac{\pi}{3} \right) + 2 \log \cos \left( 2\pi - \frac{\pi}{3} \right) = \sec \frac{\pi}{3} + 2 \log \cos \frac{\pi}{3} \)
\( = \sec \frac{\pi}{3} + 2 \log \frac{1}{2} = 2 + 2 \log \frac{1}{2} \)
\( = 2 + 2(\log 1 - \log 2) = 2 - 2 \log 2 = 2(1 - \log 2) \)
Hence, maximum value of \( f(x) = 2(1 - \log 2) \)
minimum value of \( f(x) = -1 \)

Fill in the blanks.

Question. The equation of normal to the curve \( 2y + x^2 = 3 \) at point (1, 1) is _____________ .
Answer: \( x - y = 0 \)

Question. The maximum value of \( \sin x \cdot \cos x \) is _____________ .
Answer: \( \frac{1}{2} \)

Solve the following questions.

Question. The maximum and minimum value of the function \( f(x) = |x + 2| - 1 \).
Answer: Min value = – 1 & maximum value does not exist

Question. Show that \( f(x) = e^x \) do not have maxima or minima.
Answer: For \( f(x) = e^x \), \( f'(x) = e^x \). Since \( e^x \) is never zero for any real value of \( x \), the function has no critical points and thus no maxima or minima.

Question. Show that the tangent to the curve \( y = 7x^3 + 11 \) are parallel at the points \( x = 2 \) and \( x = -2 \).
Answer: Differentiating \( y = 7x^3 + 11 \) with respect to \( x \), we get \( \frac{dy}{dx} = 21x^2 \). At \( x = 2 \), slope \( = 21(2)^2 = 84 \). At \( x = -2 \), slope \( = 21(-2)^2 = 84 \). Since the slopes are equal, the tangents are parallel.

Question. Find two numbers whose sum is 24 and whose product is as large as possible.
Answer: Both numbers are same and is 12.

Question. Find the least value of \( \lambda \) such that the function \( (x^2 + \lambda x + 1) \) is increasing on [1, 2].
Answer: \( \lambda = -2 \)

Question. Find the value of a for which the function \( f(x) = \sin x - ax + b \) increasing on R.
Answer: \( (-\infty, -1) \)

Question. Find the intervals in which the function f given by \( f(x) = \frac{4 \sin x - 2x - x \cos x}{2 + \cos x} \) is
(i) increasing (ii) decreasing.
Answer: (i) \( f(x) \) is increasing in the interval \( \left( 0, \frac{\pi}{2} \right) \) and \( \left( \frac{3\pi}{2}, 2\pi \right) \) (ii) decreasing in the interval \( \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \)

Question. Find the equation of tangent to the curve \( y = \sqrt{3x - 2} \), which is parallel to the line \( 4x - 2y + 5 = 0 \).
Answer: \( 48x - 24y - 23 = 0 \)

Question. The fuel cost for running a train is proportional to the square of the speed generated in km per hour. If the fuel costs Rs 48 per hour at speed 16 km per hour and the fixed charges amount to Rs 1200 per hour then find the most economical speed of train, when total distance covered by train is 5 km.
Answer: \( v = 50 \text{ km/hour} \)

Question. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. Find the maximum volume of the box.
Answer: \( 432 \text{ cm}^3 \)