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.
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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:
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
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