Class 11 Maths Chapter 13 Limits and Derivatives MCQs

Class 11 Maths Chapter 13 Limits and Derivatives MCQs are given here for the students of Class 11 to help them in their preparation for exams 2022-23. MCQs of Class 11 Maths Chapter 13 are provided here, along with detailed explanations for the right options. Here, we have covered all the important concepts of the NCERT curriculum for Chapter 13 Limits and Derivatives.

Get MCQs for all the chapters of Class 11 Maths here.

The objective-type questions given for Chapter 13 of Class 11 Maths will help the students to practise and verify the solutions at any time. Practising all these MCQs will help the students to improve their application skills and problem-solving skills.

Download PDF – Chapter 13 Limits and Derivatives MCQs

Also, check:

MCQs for Class 11 Maths Chapter 13 Limits and Derivatives with Answers

1. The derivative of x2 cos x is

(a) 2x sin x – x2 sin x

(b) 2x cos x – x2 sin x

(c) 2x sin x – x2 cos x

(d) cosx – x2 sin x cos x

Correct option: (b) 2x cos x – x2 sin x

Solution:

d/dx(x2 cos x)

Using the formula d/dx [f(x) g(x)] = f(x) [d/dx g(x)] + g(x) [d/dx f(x)]

d/dx(x2 cos x) = x2 [d/dx (cos x)] + cos x [d/dx x2]

= x2(-sin x) + cos x (2x)

= 2x cos x – x2 sin x

2. limx→0 | sin x|/x is equal to

(a) 1

(b) –1

(c) 0

(d) does not exist

Correct option: (d) does not exist

Solution:

Right hand side limit, R.H.S = limx→0+ |sin x|/ x = limx→0+ sin x/x = 1

Left hand side limit, L.H.S = limx→0- |sin x|/ x = limx→0- -sin x/x = -1

R.H.S ≠ L.H.S

Therefore, the solution does not exist.

3. If f(x) = x sin x, then f′(π/2) is equal to

(a) 0

(b) 1

(c) –1

(d) 1/2

Correct option: (b) 1

Solution:

Given,

f(x) = x sin x

f'(x) = x[d/dx sin x] + sin x [d/dx (x)]

= x cos x + sin x

Now,

f′(π/2) = (π/2) cos π/2 + sin π/2

= (π/2) (0) + 1

= 1

4. limx→0 (cosec x – cot x)/x is

(a) -1/2

(b) 1

(c) 1/2

(d) 1

Correct option: (c) 1/2

Solution:

\(\begin{array}{l}\displaystyle \lim_{x \to 0} \frac{cosec\;x-cot\;x}{x}=\frac{\frac{1}{sin\;x}-\frac{cos\;x}{sin\;x}}{x}\end{array} \)

\(\begin{array}{l}=\displaystyle \lim_{x \to 0} \frac{1-cos\;x}{x\;sin\;x}=\displaystyle \lim_{x \to 0}\frac{1-1+2\:sin^{2}(x/2)}{2x\;sin\;(x/2)cos(x/2)}\end{array} \)

\(\begin{array}{l}=\displaystyle \lim_{x \to 0} \frac{sin\left ( \frac{x}{2} \right )}{x\:cos\;\left ( \frac{x}{2} \right )}=\displaystyle \lim_{x/2 \to 0}\frac{tan\left ( \frac{x}{2} \right )}{\frac{x}{2}}\times \frac{1}{2}\end{array} \)

\(\begin{array}{l}=\frac{1}{2}\times 1=\frac{1}{2}\end{array} \)

5. If f(x) = x100 + x99 + … + x + 1, then f′(1) is equal to

(a) 5050

(b) 5049

(c) 5051

(d) 50051

Correct option: (a) 5050

Solution:

f(x) = x100 + x99 + … + x + 1

f′(x) = 100x99 + 99x98 + …. + 1 + 0

f′(1) = 100(1)99 + 99(1)98 + ….+ 1

= 100 + 99 + …. + 1

This is an AP with common difference -1, a = 100, n = 100 and l = 1.

So, the sum of this AP = (100/2)[100 + 1]

= 50(101)

= 5050

Therefore, f′(1) = 5050

6. limx→0 x sin(1/x) is equal to

(a) 0

(b) 1

(c) ½

(d) does not exist

Correct option: (a) 0

Solution:

We know that,

limx→0 x = 0

And

-1 ≤ sin 1/x ≤ 1

By Sandwich theorem,

limx→0 x sin(1/x) = 0

7. limx→π (sin x)/(x – π) is equal to

(a) 1

(b) 2

(c) -1

(d) -2

Correct option: (c) -1

Solution:

limx→π (sin x)/(x – π) = limx→π [sin(π – x)])/(x – π)

We know that, limx→0 (sin x)/x = 1

When π – x → 0

x → π

Therefore,

limx→π [sin(π – x)])/(x – π) = limx→π -[sin(π – x)])/(π – x) = -1

8. Let f(x) = x – [x]; ∈ R, then f′(1/2) is

(a) 3/2

(b) 1

(c) 0

(d) -1

Correct option: (b) 1

Solution:

Given,

f(x) = x – [x]

f′(x) = 1 – 0 {[x] = integer less than or equal to x}

f′(1/2) = 1

9. If y = (sin x + cos x)/(sin x – cos x), dy/dx at x = 0 is

(a) -2

(b) 0

(c) ½

(d) does not exist

Correct option: (a) -2

Solution:

Given,

y = (sin x + cos x)/(sin x – cos x)

Dividing the numerator and denominator by cos x,

y = (tan x + 1)/(tan x – 1)

y = (1 + tan x)/ [-(1 – tan x)]

We know that tan π/4 = 1,

y = -(tan π/4 + tan x)/(1 – tan π/4 tan x)

y = -tan(π/4 + x)

dy/dx = -d/dx tan(π/4 + x)

= -sec2(π/4 + x) {since d/dx tan x = sec2x}

(dy/dx)x = 0 = -sec2(π/4 + 0)

= -sec2(π/4)

= -(√2)2

= -2

10. The positive integer n so that limx→3 (xn – 3n)/(x – 3) = 108 is

(a) 3

(b) 4

(c) -2

(d) 1

Correct option: (b) 4

Solution:

We know that,

limx→3 (xn – 3n)/(x – 3) = n(3)n-1

Thus, n(3)n-1 = 108 {from the given}

n(3)n-1 = 4(27) = 4(33) = 4(3)4-1

Therefore, n = 4

Comments

Leave a Comment

Your Mobile number and Email id will not be published.

*

*