Class 12 Maths Chapter 6 Application of Derivatives MCQs

Class 12 Maths Chapter 6 Application of Derivatives MCQs are available online to assist students in improving their test performance. These multiple-choice questions are provided with answers and detailed explanations. The questions are written in accordance with the CBSE curriculum (2022–2023) and NCERT guidelines. To obtain all chapter-by-chapter MCQs for class 12 Maths, click here.

MCQs on Class 12 Maths Chapter 6 Application of Derivatives

Check out the multiple-choice questions (MCQs) for Class 12 Maths Chapter 6 Application of Derivatives. Each MCQ is provided with four options, out of which only one option is accurate. Students must choose the proper option and compare their answers to the solutions provided on our page. Also, check important questions for class 12 Maths.

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1) The side of an equilateral triangle is increasing at the rate of 2 cm/s. The rate at which area increases when the side is 10 is

1. 10 cm²/s
2. 10/3 cm²/s
3. √3 cm²/s
4. 10√3 cm²/s

Answer: (d) 10√3 cm²/s

Explanation:

Assume that x be the side of an equilateral triangle and A be its area.

Hence, A = (√3/4)x² square units

Differentiate both sides with respect to t, we get

dA/dt = (√3/4)2x(dx/dt) …(1)

Given that x = 10 cm and dx/dt = 2 cm/s

Now, substitute the values in (1), we get

dA/dt = (√3/4)2(10)(2)

Therefore, dA/dt = 10√3 cm²/s

Hence, option (d) is the correct answer.

2) The equation of the normal to the curve y = sin x at (0, 0) is

1. x = 0
2. y = 0
3. x+y =0
4. x-y =0

Answer: (c) x+y = 0

Explanation:

Given: y = sin x

Hence, dy/dx = cos x

So, (dy/dx) at (0, 0) = cos 0 = 1

Thus, the slope of the normal = -1/(dy/dx) = -1/1 = -1.

Therefore, the equation of the normal at (0, 0) is

y-0 =-1(x-0)

y=-x

Hence, x+y =0 is the correct answer.

3) If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is

1. 1%
2. 2%
3. 3%
4. 4%

Answer: (a) 1%

We know that the formula for the time period of a pendulum is T = 2π × √(l/g)

dT/dl = π/√gl

Given that the percentage error in measuring the length “l” = 2%

Δl/l = 2/100

Δl = 2l/100

Therefore, the approximate error in T = dT = (dT/dl)Δl

dT = T/100 = 1% of T.

Hence, the percentage error in T is 1%.

4) The absolute maximum value of y = x³ – 3x + 2 in 0 ≤ x ≤ 2 is

1. 0
2. 2
3. 4
4. 6

Answer: (c) 4

Explanation:

Given: y = x³ – 3x + 2

Therefore, y’ = 3x²-3.

For a point of absolute maximum or minimum, y’ = 0

Hence, x = ±1

Let y = f(x)

Therefore, f(0) = 0³-3(0) +2 = 2

f(1) = 1³– 3(1)+2 = 0

f(2) = 2³ – 3(2) +2 = 4

Hence, f(x) achieves a maximum value of 4 when x = 2.

Hence, the correct answer is an option (c) 4.

5) The line y = x + 1 is a tangent to the curve y² = 4x at the point

1. (1, 2)
2. (2, 1)
3. (-1, 2)
4. (1, -2)

Answer: (a) (1, 2)

Explanation:

y = x+1…(1)

y²= 4x …(2)

Substitute (1) in (2), we get

(x+1)² = 4x

x²+1+2x = 4x

x²-2x+1 = 0, which is equal to (x-1)²=0

⇒ x = 1

Now, substitute x = 1 in y=x+1, we get

y = 1+1 = 2.

Hence, the line y = x+1 is a tangent to the curve y² = 4x at the point (1, 2).

Therefore, option (a) (1, 2) is the correct answer.

6) The function f(x) = x + cos x is

1. Always increasing
2. Always decreasing
3. Increasing for a certain range of x
4. None of these

Answer: (a) Always increasing

Explanation:

Given:

f(x) = x+cos x

f’(x) = 1 – sinx

f’(x)>0 for all vaues of x.

Since sin x is lying between -1 and +1, f(x) is always increasing

7) Let the f: R → R be defined by f(x) = 2x + cos x, then f

1. has a maximum, at x = 0
2. has a minimum at x = 3t
3. is an increasing function
4. is a decreasing function

Answer: (c) is an increasing function

Explanation:

Given: f(x) = 2x + cos x

f’(x) = 2 – sin x

Now, f’(x) will be positive for all values of x

It means, f’ (x)>0 ∀ x∈R

Hence, the function is increasing for all values of x and it does not have any defined maximum and minimum. Hence, option (c) is the correct answer.

8) The point(s) on the curve y = x², at which y-coordinate is changing six times as fast as x-coordinate is/are

1. (6, 2)
2. (2, 4)
3. (3, 9)
4. (3, 9), (9, 3)

Answer: (c) (3, 9)

Explanation:

dy/dt = (2x)(dx/dt)

⇒6. (dx/dt) = 2x.(dx/dt)

⇒x = 3

Now, substitute x = 3 in y=x², we get

y = (3)² = 9

Therefore, y = 9

Hence, the coordinate is (3, 9)

Therefore, the correct answer is an option (c).

9) If y = x³ + x² + x + 1, then y

1. has a local minimum
2. has a local maximum
3. neither has a local minimum nor local maximum
4. None of the above

Answer: (c) neither has a local minimum nor local maximum

If y = x³ + x² + x + 1, then y neither has a local minimum nor local maximum.

10) The tangent to the curve y=e^2x at the point (0,1) meet the x-axis at

1. (0, 1)
2. (2, 0)
3. (-2, 0)
4. (-½, 0)

Answer: (d) (-½, 0)

Explanation:

Given: y=e^2x

dy/dx = 2e^2x

Hence, dy/dx at (0, 1) is = 2e⁰ = 2(1) = 2

Therefore, the equation of a tangent is y-1 = 2(x-0)

⇒2x-y+1 = 0, if it meet the x-axis, then y=0

Hence, x = -½.

Hence, the point is (-½, 0).

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