# Class 12 Maths MCQ – Application of Derivatives

This set of Class 12 Maths Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Application of Derivatives”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. What is the slope of the tangent to the curve y = 2x/(x2 + 1) at (0, 0)?
a) 0
b) 1
c) 2
d) 3

Explanation: We have y = 2x/(x2 + 1)
Differentiating y with respect to x, we get
dy/dx = d/dx(2x/(x2 + 1))
= 2 * [(x2 + 1)*1 – x * 2x]/(x2 + 1)2
= 2 * [1 – x2]/(x2 + 1)2
Thus, the slope of tangent to the curve at (0, 0) is
[dy/dx](0, 0) = 2 * [1 – 0]/(0 + 1)2
Thus [dy/dx](0, 0) = 2.

2. The value of f’(x) is -1 at the point P on a continuous curve y = f(x). What is the angle which the tangent to the curve at P makes with the positive direction of x axis?
a) π/2
b) π/4
c) 3π/4
d) 3π/2

Explanation: Let, Φ be the angle which the tangent to the curve y = f(x) at P makes with the positive direction of the x axis.
Then,
tanΦ = [f’(x)]p = -1
= -tan(π/4)
So, it is clear that this can be written as,
= tan(π – π/4)
= tan(3π/4)
So, Φ = 3π/4
Therefore, the required angle which the tangent at P to the curve y = f(x) makes with positive direction of x axis is 3π/4.

3. What will be the differential function of √(x2 + 2)?
a) x√(x2 + 2) dx
b) x/√(x2 + 2) dx
c) x/√(x2 – 2) dx
d) -x/√(x2 + 2) dx

Explanation: Let, y = f(x) = √(x2 + 2)
So, f(x) = (x2 + 2)1/2
On differentiating it we get,
f’(x) = d/dx[(x2 + 2)1/2]
f’(x) = 1/2 * 1/√(x2 + 2) * 2x
So f’(x) = x/√(x2 + 2)
So the differential equation is:
dy = f’(x)dx
Hence, dy = x/√(x2 + 2) dx

4. What will be the differential function of log(x2 + 4)?
a) 2x/(x2 + 4) dx
b) 2x/(x2 – 4) dx
c) -2x/(x2 + 4) dx
d) -2x/(x2 – 4) dx

Explanation: Let, y = f(x) = log(x2 + 4)
So f(x) = log(x2 + 4)
On differentiating it we get,
f’(x) = d/dx[log(x2 + 4)]
So f’(x) = 2x/(x2 + 4)
So the differential equation is:
dy = f’(x)dx
Hence, dy = 2x/(x2 + 4) dx

5. What will be the average rate of change of the function [y = 16 – x2] between x = 3 and x = 4?
a) 7
b) -7
c) 9
d) -9

Explanation: Let, y = f(x) = 16 – x2
If x changes from 3 to 4, then, δx = 4 – 3 = 1
Again f(4) = 16 – 42 = 0
And f(3) = 16 – 32 = 7
Therefore, δy = f(4) – f(3) = 0 – 7 = -7
Hence, the average rate of change of the function between x = 3 and x = 4 is:
δy/δx = -7/1 = -7.
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6. What will be the average rate of change of the function [y = 16 – x2] at x = 4?
a) -8
b) 8
c) -9
d) Depends on the value of x

Explanation: Let, y = f(x) = 16 – x2
dy/dx = -2x
Now, [dy/dx]x = 4 = [-2x]x = 4
So, [dy/dx]x = 4 = -8

7. What will be the value of the co-ordinate whose position of a particle moving along the parabola y2 = 4x at which the rate at of increase of the abscissa is twice the rate of increase of the ordinate?
a) (1, 1)
b) (2, 2)
c) (3, 3)
d) (4, 4)

Explanation: Here, y2 = 4x ……….(1)
Let, (x, y) be the position of the particle moving along the parabola (1) at time t.
Now, differentiating both sides of (1) with respect to t, we get:
2y(dy/dt) = 4(dx/dt)
Or, y(dy/dt) = 2(dy/dt) ……….(2)
By question, dx/dt = 2 * dy/dt ……….(3)
From (2) and (3) we get, y(dy/dt) = 2 * 2 dy/dt
Or, y = 4
Putting y = 4 in (1) we get, 42 = 4x
So, x = 4
Thus, the co-ordinate of the particle is (4, 4).

8. The time rate of change of the radius of a sphere is 1/2π. When it’s radius is 5cm, what will be the rate of change of the surface of the sphere with time?
a) 10 sq cm
b) 20 sq cm
c) 30 sq cm
d) 40 sq cm

Explanation: Let, r be the radius and s be the area of the surface of the sphere at time t.
By question, dr/dt = 1/2π
Now, s = 4πr2;
Thus, ds/dt = 4π * 2r(dr/dt) = 8πr(dr/dt)
When r = 5cm and dr/dt = 1/2π
Then ds/dt = 8π*5*(1/2π) = 20
Thus, correct answer is 20 sq cm.

9. A solid cube changes its volume such that its shape remains unchanged. For such a cube of unit volume, what will be the value of rate of change of volume?
a) 3/8*(rate of change of area of any face of the cube)
b) 3/4*(rate of change of area of any face of the cube)
c) 3/10*(rate of change of area of any face of the cube)
d) 3/2*(rate of change of area of any face of the cube)

Explanation: Let x be the length of a side of the cube.
If v be the volume and s the area of any face of the cube, then
v = x3 and s = x2
Thus, dv/dt = dx3/dt = 3x2 (dx/dt)
And ds/dt = dx2/dt = 2x(dx/dt)
Now, (dv/dt)/(ds/dt) = 3x/2
Or, dv/dt = (3x/2)(ds/dt)
Now, for a cube of unit volume we have,
v = 1
=>x = 1 [as, x is real]
Therefore, for a cube of unit volume [i.e. for x = 1], we get,
dv/dt = (3/2)(ds/dt)
Thus the rate of change of volume = 3/2*(rate of change of area of any face of the cube)

10. A 5 ft long man walks away from the foot of a 12(½) ft high lamp post at the rate of 3 mph. What will be the rate at which the shadow increases?
a) 0mph
b) 1mph
c) 2mph
d) 3mph

Explanation: Let, AB be the lamp-post whose foot is A, and B is the source of light, and given (AB)’ = 12(½) ft.
Let MN denote the position of the man at time t where (MN)’ = 5ft.
Join BN and produce it to meet AM(produced) at P.
Then the length of man’s shadow= (MP)’
Assume, (AM)’ = x and (MP)’ = y. Then,
(PA)’ = (AM)’ + (MP)’ = x + y
And dx/dt = velocity of the man = 3
Clearly, triangles APB and MPN are similar.
Thus, (PM)’/(MN)’ = (PA)’/(AB)’
Or, y/5 = (x + y)/12(½)
Or, (25/2)y = 5x + 5y
Or, 3y = 2x
Or, y = (2/3)x
Thus, dy/dt = (2/3)(dx/dt)
As, dx/dt = 2,
= 2/3*3 = 2mph

11. A ladder 20 ft long leans against a vertical wall. If the top end slides downwards at the rate of 2ft per second, what will be the rate at which the slope of the ladder changes?
a) -19/54
b) -21/54
c) -23/54
d) -25/54

Explanation: Let the height on the wall be x and laser touches the ground at distance y from the wall. The length of the ladder is 20ft.
By Pythagoras theorem:
x2 + y2 = 400
Differentiating with respect to t:
2x(dx/dt) + 2y(dy/dt) = 0
Dividing throughout by 2:
x (dx/dt) + y (dy/dt) = 0
Now, dx/dt = -2 ft /s. negative because downwards
x(-2) + y (dy/dt) = 0 ………..(1)
When lower end is 12 ft from wall, let us find x:
x2 + 122 = 400
x2 = 400 – 144= 256
x = 16
x(2) + y (dy/dt) = 0 from (1)
16(-2) + 12 (dy/dt) = 0
-32 + 12(dy/dt) = 0
dy/dt = (32/12) = (8/3)
Thus, lower end moves on a horizontal floor when it is 12 ft from the wall at the rate of 8/3 ft/s
Now, assume that the ladder makes an angle θ with the horizontal plane at time t.
If, m be the slope of the ladder at time t, then,
m = tanθ = x/y
Thus, dm/dt = d/dt(x/y) = [y(dx/dt) – x(dx/dt)]/y2
Therefore, the rate of change of slope of ladder is,
[dm/dt]y = 2 = [12*(-2) – 16*(8/3)]/(12)2
Now, putting the value of x = 16, when y = 12 and dx/dt = -2, dy/dt = 8/3
We get, [dm/dt]y = 12 = [12(-2) – 16(8/3)]/(12)2 = -25/54

12. A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 6t2 – 15t + 18. What will be the velocity of the particle at the end of 2 seconds?
a) 20cm/sec
b) 22cm/sec
c) 21cm/sec
d) 23cm/sec

Explanation: We have, x = t3 + 6t2 – 15t + 18
Let, v be the velocity of the particle at the end of t seconds. Then,
v = dx/dt = d/dt(t3 + 6t2 – 15t + 18)
So, v = 3t2 + 12t – 15
Thus, velocity of the particle at the end of 2 seconds is,
[dx/dt]t = 2 = 3(2)2 + 12(2) – 15 = 21cm/sec.

13. A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 6t2 – 15t + 18. What will be the acceleration of the particle at the end of 2 seconds?
a) 22cm/sec2
b) 23cm/sec2
c) 24cm/sec2
d) 25cm/sec2

Explanation: We have, x = t3 + 6t2 – 15t + 18
Let, ‘v’ be the velocity of the particle and ‘a’ be the acceleration of the particle at the end of t seconds. Then,
v = dx/dt = d/dt(t3 + 6t2 – 15t + 18)
So, v = 3t2 + 12t – 15
Therefore, a = dv/dt = d/dt(3t2 + 12t – 15)
So, a = 6t + 12
Thus, acceleration of the particle at the end of 2 seconds is,
[dv/dt]t = 2 = 6(2) + 12 = 24cm/sec2.

14. A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 6t2 – 15t + 18. When does the particle stop?
a) 1/4 second
b) 1/3 second
c) 1 second
d) 1/2 second

Explanation: We have, x = t3 + 6t2 – 15t + 18
The particle stops when dx/dt = 0
And, dx/dt = d/dt(t3 + 6t2 – 15t + 18)
3t2 + 12t – 15 = 0
=>t2 + 4t –5 = 0
=>(t – 1)(t + 5)= 0
Thus, t = 1 or t = -5
Hence, the particle stops at the end of 1 second.

15. Water is flowing into a right circular conical vessel, 45 cm deep and 27 cm in diameter at the rate of 11 cc per minute. How fast is the water level rising when the water is 30 cm deep?
a) 0.033cm/minute
b) 0.043cm/minute
c) 0.053cm/minute
d) 0.045cm/minute

Explanation: Let ‘r’ be the radius and ‘h’ be the height of the water level at time t.
Then the volume of water level ‘V’ at time t is given by,
V = 1/3 (πr2h) ……….(1)
Given the radius of the base of the cone is = (OA)’ = 27/2 cm and its height = (OB)’ = 45cm.
Again at time t, the radius of the water level = r = (CD)’ and its height = h = (CB)’
Clearly, the triangle OAB and CBD are similar.
Therefore, (CD)’/(CB)’ = (OA)’/(OB)’
Or, r/h = (22/7)/45 = 3/10
or, r = 3h/10
Thus, from (1) we get,
V = 1/3 π (3h/10)2 h = (3π/100)h3
Thus, dV/dt = (3π/100)*3h2(dh/dt)
= (9πh2/100)(dh/dt)
When, h = 30cm, then,
11 = (9π/100) (30)2 (dh/dt) [as for all the values of t we have, dV/dt = 11]
Or, dh/dt = (11/(9π*9)) = 0.043
Thus, the rising rate of rising is 0.043 cm/minute.

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