CBSE Class 12 Mathematics Application Of Derivatives MCQs Set 09

Question. Edge of a variable cube increases at the rate of \( 5\text{ cm/s} \). The rate at which the surface area of the cube increases when the edge is \( 2\text{ cm} \) long is
(a) \( 24\text{ cm}^2/\text{s} \)
(b) \( 120\text{ cm}^2/\text{s} \)
(c) \( 12\text{ cm}^2/\text{s} \)
(d) \( 5\text{ cm}^2/\text{s} \)
Answer: (b) \( 120\text{ cm}^2/\text{s} \)

Question. For \( 0 < \theta < \frac{\pi}{2} \), the value of \( \theta \), if it increases twice as fast as its sine, is
(a) \( \frac{\pi}{2} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{6} \)
(d) None of the options
Answer: (b) \( \frac{\pi}{3} \)

Question. Given a curve \( y = 7x - x^3 \) and \( x \) increases at the rate of \( 2 \) units per sec. The rate at which the slope of the curve is changing when \( x = 5 \) is
(a) \( -60 \) units/sec
(b) \( 60 \) units/sec
(c) \( -70 \) units/sec
(d) \( -140 \) units/sec
Answer: (a) \( -60 \) units/sec

Question. If the radius of a circle is increasing at the rate of \( 0.5\text{ cm/s} \), then the rate of increase of its circumference is
(a) \( \frac{2\pi}{3}\text{ cm/s} \)
(b) \( \pi\text{ cm/s} \)
(c) \( \frac{4\pi}{3}\text{ cm/s} \)
(d) \( 2\pi\text{ cm/s} \)
Answer: (b) \( \pi\text{ cm/s} \)

Question. The function \( f(x) = x^3 + 3x \) is increasing in interval
(a) \( (-\infty, 0) \)
(b) \( (0, \infty) \)
(c) \( \mathbb{R} \)
(d) \( (0, 1) \)
Answer: (c) \( \mathbb{R} \)

Question. The interval in which the function \( f(x) = 2x^3 + 9x^2 + 12x - 1 \) is decreasing is
(a) \( (-1, \infty) \)
(b) \( (-2, -1) \)
(c) \( (-\infty, -2) \)
(d) \( (-1, 1) \)
Answer: (b) \( (-2, -1) \)

Question. If \( f : \mathbb{R} \to \mathbb{R} \) is defined as \( f(x) = 2x - \sin x \), then \( f \) is
(a) a decreasing function
(b) an increasing function
(c) maximum at \( x = \frac{\pi}{2} \)
(d) maximum at \( x = 0 \)
Answer: (b) an increasing function

Question. The real function \( f(x) = 2x^3 - 3x^2 - 36x + 7 \) is
(a) strictly increasing in \( (-\infty, -2) \) and strictly decreasing in \( (-2, \infty) \)
(b) strictly decreasing in \( (-2, 3) \)
(c) strictly decreasing in \( (-\infty, 3) \) and strictly increasing in \( (3, \infty) \)
(d) strictly decreasing in \( (-\infty, -2) \cup (3, \infty) \)
Answer: (b) strictly decreasing in \( (-2, 3) \)

Question. The value of \( b \) for which the function \( f(x) = x + \cos x + b \) is strictly decreasing over \( \mathbb{R} \) is
(a) \( b < 1 \)
(b) no value of \( b \) exist
(c) \( b \le 1 \)
(d) \( b \ge 1 \)
Answer: (b) no value of \( b \) exist

Question. The function \( y = x^2 e^{-x} \) is decreasing in the interval
(a) \( (0, 2) \)
(b) \( (2, \infty) \)
(c) \( (-\infty, 0) \)
(d) \( (-\infty, 0) \cup (2, \infty) \)
Answer: (d) \( (-\infty, 0) \cup (2, \infty) \)

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) \( (-\infty, 4) \)

Question. If the function \( f \) is given by \( f(x) = x^3 - 3x^2 + 4x, x \in \mathbb{R} \), then
(a) \( f \) is strictly increasing on \( \mathbb{R} \)
(b) \( f \) is decreasing on \( \mathbb{R} \)
(c) \( f \) is neither increasing nor decreasing on \( \mathbb{R} \)
(d) \( f \) is strictly decreasing on \( \mathbb{R} \)
Answer: (a) \( f \) is strictly increasing on \( \mathbb{R} \)

Question. The minimum value of \( f(x) = x \log_e x \) is equal to
(a) \( e \)
(b) \( \frac{1}{e} \)
(c) \( -\frac{1}{e} \)
(d) \( 2e \)
Answer: (c) \( -\frac{1}{e} \)

Question. If the function \( f \) is given by \( f(x) = x^3 - 3x + 3 \), then
(a) \( x = \pm 2 \) are the only critical points for local maxima or local minima.
(b) \( x = 3 \) is a point of local minima.
(c) local minimum value is 2.
(d) local maximum value is 5.
Answer: (d) local maximum value is 5.

Question. The smallest value of polynomial \( x^3 - 18x^2 + 96x \) in \( [0, 9] \) is
(a) \( 126 \)
(b) \( 0 \)
(c) \( 135 \)
(d) \( 160 \)
Answer: (b) \( 0 \)

Assertion-Reason Based Questions

Question. Assertion (A) The rate of change of area of a circle with respect to its radius \( r \) when \( r = 3\text{ cm} \), is \( 6\pi\text{ cm}^2/\text{cm} \).
Reason (R) Rate of change of area of a circle with respect to its radius \( r \) is \( \frac{dA}{dr} \), where \( A \) is the area of the circle.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) A balloon, which always remains spherical, has a variable radius. The rate, at which its volume is increasing with the radius when the radius is \( 10\text{ cm} \), is \( 400\pi\text{ cm}^3/\text{cm} \).
Reason (R) Rate of change of volume \( V \) of balloon with respect to radius \( r \) is \( \frac{dV}{dr} = \left(\frac{2}{3}\pi\right) \cdot 3r^2 \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A) The side of an equilateral triangle is increasing at the rate of \( 0.5\text{ cm/s} \), then the rate of increase of its perimeter is \( 2.5\text{ cm/s} \).
Reason (R) Rate of change in side of a triangle with side \( a \) is \( \frac{da}{dt} \), then rate of change in the perimeter of triangle \( = \frac{dP}{dt} \), where \( P = \text{perimeter} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A) If the total revenue (in rupees) received from the sale of \( x \) units of an article is given by \( R(x) = 3x^2 + 36x + 5 \), then the marginal revenue when \( x = 15 \) is \( 126 \).
Reason (R) If \( R(x) \) is the revenue function for \( x \) units sold, then marginal revenue \( \text{MR} = \frac{d}{dx}\{R(x)\} \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) The function \( f(x) = 4x + 3, x \in \mathbb{R} \) is an increasing function.
Reason (R) Let \( f(x) \) be a function. Then, \( f(x) \) is increasing function, if \( f'(x) > 0 \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) Function \( f(x) = \tan x - x \) always increases.
Reason (R) Any function \( y = f(x) \) is increasing, if \( \frac{dy}{dx} < 0 \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (c) A is correct; R is incorrect.

Question. Assertion (A) The function is given by \( f(x) = x^4 \) is decreasing in the interval \( (0, \infty) \).
Reason (R) Any function \( y = f(x) \) is decreasing, if \( \frac{dy}{dx} < 0 \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A) The function \( f(x) = x^2 - 4x + 6 \) is strictly increasing in the interval \( (2, \infty) \).
Reason (R) The function \( f(x) = x^2 - 4x + 6 \) is strictly decreasing in the interval \( (-\infty, 2) \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Consider the information given below. The function \( f \) is given by \( f(x) = 2x^3 - 3x^2 - 36x + 7 \).
Assertion (A) The given function \( f \) is strictly increasing in intervals \( (-\infty, +2) \) and \( (-3, \infty) \).
Reason (R) The given function \( f \) is strictly decreasing in interval \( (-2, 3) \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A) The slope of the curve \( y = -x^3 + 3x^2 + 9x - 27 \) is maximum at the point \( (1, -16) \).
Reason (R) Let \( f \) be a function defined on an interval \( I \) and \( c \in I \) and also, if \( f \) be twice differentiable at \( c \), then \( x = c \) is a point of local maxima if \( f'(c) = 0 \) and \( f''(c) < 0 \) and the value \( f(c) \) is local maximum value of \( f \).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Case Study Based Questions

A rectangular visiting card is to contain 24 sq cm of printed matter. The margins at the top and bottom of the card are to be 1 cm and the margins on the left and right are to be \( 1\frac{1}{2}\text{ cm} \) as shown below.

Question. Write the expression for the area of the visiting card in terms of \( x \).
Answer: Let the length of the printed part be \( x\text{ cm} \) and width of the printed part be \( y\text{ cm} \). Given, area of printed part \( = 24\text{ cm}^2 \Rightarrow xy = 24 \Rightarrow y = \frac{24}{x} \). From given condition, length of the card \( = (x + 3)\text{ cm} \) and width of the card \( = (y + 2)\text{ cm} \). Area of card, \( A = (x + 3)(y + 2) = (x + 3)\left(\frac{24}{x} + 2\right) = 24 + 2x + \frac{72}{x} + 6 = 30 + 2x + \frac{72}{x}\text{ cm}^2 \).

Question. Obtain the dimensions of the card of minimum area.
Answer: We have, \( A = 30 + 2x + \frac{72}{x} \Rightarrow \frac{dA}{dx} = 2 - \frac{72}{x^2} \). For maxima or minima, \( \frac{dA}{dx} = 0 \Rightarrow 2x^2 - 72 = 0 \Rightarrow x^2 = 36 \Rightarrow x = 6 \) (since \( x \) cannot be negative). Now, \( \frac{d^2A}{dx^2} = \frac{144}{x^3} \). At \( x = 6 \), \( \frac{d^2A}{dx^2} > 0 \). Thus, area is minimum when \( x = 6 \). Then, \( y = \frac{24}{6} = 4 \). Dimensions of the card are: length of the card \( = x + 3 = 6 + 3 = 9\text{ cm} \), and width of the card \( = y + 2 = 4 + 2 = 6\text{ cm} \).

Case Study Based Questions - I

In order to set up a rainwater harvesting system, a tank to collect rainwater is to be dug. The tank should have a square base and a capacity of \( 250\text{ m}^3 \). The cost of land is ₹ 5000 per \( \text{m}^2 \) and cost of digging increases with depth and for the whole tank, it is \( 40000h^2 \), where \( h \) is the depth of the tank in metres. \( x \) is the side of the square base of the tank in metres.

Question. Find the total cost \( C \) of digging the tank in terms of \( x \).
Answer: We have, volume of the tank \( = 250\text{ m}^3 \Rightarrow x^2 h = 250 \Rightarrow h = \frac{250}{x^2} \). Total cost of digging tank \( = 40000h^2 + 5000x^2 \). Thus, \( C(x) = 40000 \left(\frac{250}{x^2}\right)^2 + 5000x^2 = \frac{2500000000}{x^4} + 5000x^2 \).

Question. Find \( \frac{dC}{dx} \).
Answer: Differentiating \( C(x) \) with respect to \( x \), we get \( \frac{dC}{dx} = 40000 \times 62500 \times \left(-\frac{4}{x^5}\right) + 10000x = \frac{-10000000000}{x^5} + 10000x \).

Question. (a) Find the value of \( x \) for which cost \( C \) is minimum. Or (b) Check whether the cost function \( C(x) \) expressed in terms of \( x \) is increasing or not, where \( x > 0 \).
Answer: (a) For maximum or minimum cost \( C(x) \), \( \frac{dC}{dx} = 0 \Rightarrow \frac{-10000000000}{x^5} + 10000x = 0 \Rightarrow x^6 = 1000000 \Rightarrow x = 10 \). Since \( \frac{d^2C}{dx^2} > 0 \) at \( x = 10 \), \( C(x) \) is minimum at \( x = 10\text{ m} \).
(b) For \( C(x) \) to be increasing, \( C'(x) > 0 \Rightarrow 10000x - \frac{10000000000}{x^5} > 0 \Rightarrow x^6 > 1000000 \Rightarrow x > 10 \). So, for \( C(x) \) to be increasing, \( x > 10 \), and it is decreasing for \( 0 < x < 10 \). Hence, \( C(x) \) is not increasing for all \( x > 0 \).

Case Study Based Questions - II

A channel company in a town has 500 subscribers on its list and collects fixed charges of ₹ 300 per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of ₹ 1, one subscriber will discontinue the service.

Question. If \( x \) is the increment in annual subscription, then find the total revenue of the company after increment, (in terms of \( x \)).
Answer: New subscription cost \( = 300 + x \). New number of subscribers \( = 500 - x \). Total revenue after increment, \( R(x) = (300 + x)(500 - x) = 150000 + 200x - x^2 \).

Question. How much fee the company should increase to have maximum revenue?
Answer: We have, \( R(x) = 150000 + 200x - x^2 \Rightarrow \frac{dR}{dx} = 200 - 2x \). For critical points, \( \frac{dR}{dx} = 0 \Rightarrow 200 - 2x = 0 \Rightarrow x = 100 \). Since \( \frac{d^2R}{dx^2} = -2 < 0 \), \( R \) is maximum at \( x = 100 \). Hence, the company should increase the fee by ₹ 100.

Question. What will be the maximum revenue of the company?
Answer: Maximum revenue, \( R(100) = 150000 + 200(100) - (100)^2 = 150000 + 20000 - 10000 = \text{₹ } 160000 \).

Question. How many subscriber will be there when revenue is maximum?
Answer: New number of subscribers \( = 500 - x = 500 - 100 = 400 \).

Case Study Based Questions - III

The petrol cost per hour for running a car is proportional to the square of the speed, it generates in km per hour. If the petrol cost ₹ 48 per hour at speed of 16 km/h and the fixed charges to run the car amounts to ₹ 1200 per hour. Assume the speed of car as \( v\text{ km/h} \).

Question. If it is given that the petrol cost per hour is \( K \) times the square of the speed the car generates in km/h, then find the value of \( K \).
Answer: Let \( f \) be the cost of petrol per hour. Then, \( f \propto v^2 \Rightarrow f = K \cdot v^2 \). Given, \( f = 48 \) when \( v = 16 \). So, \( 48 = K \cdot (16)^2 \Rightarrow K = \frac{48}{256} = \frac{3}{16} \).

Question. If the car has travelled a distance of 1000 km, then find the function representing the total cost running the car.
Answer: Let \( X \) be the total cost of running the car. Cost per hour \( = K \cdot v^2 + 1200 = \frac{3}{16}v^2 + 1200 \). Time taken to travel 1000 km, \( t = \frac{1000}{v} \). Total cost, \( X = \left(\frac{3}{16}v^2 + 1200\right)\frac{1000}{v} = \frac{375}{2}v + \frac{1200000}{v} \).

Question. Find the most economical speed to run the car.
Answer: For minimum cost, \( \frac{dX}{dv} = 0 \Rightarrow \frac{375}{2} - \frac{1200000}{v^2} = 0 \Rightarrow v^2 = \frac{2400000}{375} = 6400 \Rightarrow v = 80\text{ km/h} \). Since \( \frac{d^2X}{dv^2} = \frac{2400000}{v^3} > 0 \) when \( v = 80 \), the most economical speed to run the car is \( 80\text{ km/h} \).

Question. Find the cost of petrol when car is running 1000 km at economical speed.
Answer: Cost of petrol per hour, \( f = \frac{3}{16}v^2 \). At economical speed \( v = 80\text{ km/h} \), petrol cost per hour \( = \frac{3}{16}(80)^2 = 1200 \). Time to travel 1000 km is \( t = \frac{1000}{80} = 12.5\text{ hours} \). Total cost of petrol \( = 1200 \times 12.5 = \text{₹ } 15000 \).

Case Study Based Questions - IV

A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its height is 2 m and volume is \( 8\text{ m}^3 \). If building of tank cost ₹ 700 per square metre for the base and ₹ 450 per square metre for sides. Let \( x \) and \( y \) be the length and breadth of the least expensive tank.

Question. Find the total cost in terms of \( x \) and \( y \).
Answer: Area of base \( = xy\text{ m}^2 \). Cost of base \( = 700xy \). Area of four walls \( = 2h(x + y) = 4(x + y)\text{ m}^2 \). Cost of walls \( = 450 \times 4(x + y) = 1800(x + y) \). Total cost \( = 700xy + 1800(x + y) \).

Question. Find the cost of constructing the tank depends on \( x \) only.
Answer: Volume \( V = x \cdot y \cdot 2 = 8 \Rightarrow xy = 4 \Rightarrow y = \frac{4}{x} \). Substituting \( y = \frac{4}{x} \) in the total cost function, we get: \( T(x) = 700(4) + 1800\left(x + \frac{4}{x}\right) = 2800 + 1800\left(x + \frac{4}{x}\right) \).

Question. When cost is minimum, what will be the length?
Answer: Differentiating \( T(x) \) with respect to \( x \), we get \( \frac{dT}{dx} = 1800\left(1 - \frac{4}{x^2}\right) \). For critical points, \( \frac{dT}{dx} = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2\text{ m} \) (rejecting negative value). Since \( \frac{d^2T}{dx^2} = 1800\left(\frac{8}{x^3}\right) > 0 \) at \( x = 2 \), \( T \) is minimum at \( x = 2\text{ m} \). Thus, length is \( 2\text{ m} \).

Question. Find the minimum cost of the tank in rupees.
Answer: Minimum cost \( = 2800 + 1800\left(2 + \frac{4}{2}\right) = 2800 + 7200 = \text{₹ } 10000 \).

Case Study Based Questions - V

The lifespan of a certain flowering plant is around 6 yr. After \( t \) years, the sapling is planted, the plant produces \( r \) grams of flowers each day. The relation between \( r \) and \( t \) can be approximated as \( r = \frac{t^3}{3} - 6t^2 + 32t, 0 \le t \le 6 \).

Question. What will be the yield per day, 3 yr after planting the sapling?
Answer: After 3 years, the yield per day \( r \) is obtained by putting \( t = 3 \): \( r = \frac{3^3}{3} - 6(3)^2 + 32(3) = 9 - 54 + 96 = 51\text{ grams} \).

Question. When will the maximum yield per day? Show your steps with valid reasons.
Answer: We have \( r = \frac{t^3}{3} - 6t^2 + 32t \). Differentiating with respect to \( t \), we get \( \frac{dr}{dt} = t^2 - 12t + 32 \). For critical points, \( \frac{dr}{dt} = 0 \Rightarrow (t - 4)(t - 8) = 0 \Rightarrow t = 4 \) or \( t = 8 \). Since \( 0 \le t \le 6 \), we take \( t = 4 \). Differentiating again, \( \frac{d^2r}{dt^2} = 2t - 12 \). At \( t = 4 \), \( \frac{d^2r}{dt^2} = 2(4) - 12 = -4 < 0 \). Hence, the maximum yield per day is obtained at \( t = 4 \) years.