CBSE Class 11 Mathematics Chapter 13 Limits and Derivatives Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Limits and Derivatives with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Limits and Derivatives MCQs with Answers to know their preparation level.
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MCQ Class 11 Mathematics Limits and Derivatives with Answerss - Set - 1
Question 1:
The expansion of log(1 – x) is
(a) x – x²/2 + x³/3 – ……..
(b) x + x²/2 + x³/3 + ……..
(c) -x + x²/2 – x³/3 + ……..
(d) -x – x²/2 – x³/3 – ……..
Correct Answer – (D)
Question 2 :
Limy→∞ {(x + 6)/(x + 1)}(x+4) equals
(a) e
(b) e³
(c) e5
(d) e6
Correct Answer – (C)
Given, Limy→∞ {(x + 6)/(x + 1)}(x + 4)
= Limy→∞ {1 + 5/(x + 1)}(x + 4)
= eLimy→∞ 5(x + 4)/(x + 1)
= eLimy→∞ 5(1 + 4/x)/(1 + 1/x)
= e5(1 + 4/∞)/(1 + 1/∞)
= e5/(1 + 0)
= e5
Question 3 :
Limx→0 {(ax – bx)/ x} is equal to
(a) log a
(b) log b
(c) log (a/b)
(d) log (a×b)
Correct Answer – (C)
Given, Limx→0 {(ax – bx)/ x}
= Limx→0 {(ax – bx – 1 + 1)/ x}
= Limx→0 {(ax – 1) – (bx – 1)}/ x
= Limx→0 {(ax – 1)/x – (bx – 1)/x}
= Limx→0 (ax – 1)/x – Limx→0 (bx – 1)/x
= log a – log b
= log (a/b)
Question 4 :
The value of Limx→01 (1/x) × sin-1 {2x/(1 + x²) is
(a) 0
(b) 1
(c) 2
(d) -2
Correct Answer – (C)
Given, Limx→0 (1/x) × sin-1 {2x/(1 + x²)
= Limx→0 (2× tan-1 x)/x
= 2 × 1
= 2
Question 5 :
The value of limit Limx→0 {sin (a + x) – sin (a – x)}/x is
(a) 0
(b) 1
(c) 2 cos a
(d) 2 sin a
Correct Answer – (C)
Given, Limx→0 {sin (a + x) – sin (a – x)}/x
= Limx→0 {2 × cos a × sin x}/x
= 2 × cos a × Limx→0 sin x/x
= 2 cos a
MCQ Class 11 Mathematics Limits and Derivatives with Answers
Question 6 :
The derivative of [1+(1/x)] /[1-(1/x)] is
(a) 1/(x-1)²
(b) -1/(x-1)²
(c) 2/(x-1)²
(d) -2/(x-1)²
Correct Answer – (D)
Let y = [1+(1/x)] /[1-(1/x)]
then dy/dx = [{1-(1/x)}*(-1/x²)]/[{1+(1/x)}*(1/x²)]
= (1/x²) [(1/x) -1 – 1 – (1/x)]/[1-(1/x)]²
= [-2/x²]/[(x-1)/x]²
= -2/(x-1)²
Question 7 :
The value of limy→0 {(x + y) × sec (x + y) – x × sec x}/y is
(a) x × tan x × sec x
(b) x × tan x × sec x + x × sec x
(c) tan x × sec x + sec x
(d) x × tan x × sec x + sec x
Correct Answer – (D)
Given, limy→0 {(x + y) × sec (x + y) – x×sec x}/y
= limy→0 {x sec (x + y) + y sec (x + y) – x×sec x}/y
= limy→0 [x{ sec (x + y) – sec x} + y sec (x + y)]/y
= limy→0 x{ sec (x + y) – sec x}/y + limy→0 {y sec (x + y)}/y
= limy→0 x{1/cos (x + y) – 1/cos x}/y + limy→0 {y sec (x + y)}/y
= limy→0 [{cos x – cos (x + y)} × x/{y×cos (x + y)×cos x}] + limy→0 {y sec (x + y)}/y
= limy→0 [{2sin (x + y/2) × sin(y/2)} × 2x/{2y×cos (x + y)×cos x}] + limy→0 {y sec (x + y)}/y
= limy→0 {sin (x + y/2) × limy→0 {sin(y/2)/(2y/2)} × limy→0 { x/{y×cos (x + y)×cos x}] + sec x
= sin x × 1 × x/cos² x + sec x
= x × tan x × sec x + sec x
So, limy→0 {(x + y) × sec (x + y) – x×sec x}/y = x × tan x × sec x + sec x
Question 8 :
Limx→0 log(1 – x) is equals to
(a) 0
(b) 1
(c) 1/2
(d) None of these
Correct Answer – (A)
We know that
log(1 – x) = -x – x²/2 – x³/3 – ……..
Now,
Limx→0 log(1 – x) = Limx→0 {-x – x²/2 – x³/3 – ……..}
⇒ Limx→0 log(1 – x) = Limx→0 {-x} – Limx→0 {x²/2} – Limx→0 {x³/3} – ……..
⇒ Limx→0 log(1 – x) = 0
Question 9 :
Limx→-1 [1 + x + x² + ……….+ x10] is
(a) 0
(b) 1
(c) -1
(d) 2
Correct Answer – (B)
Given, Limx→-1 [1 + x + x² + ……….+ x10]
= 1 + (-1) + (-1)² + ……….+ (-1)10
= 1 – 1 + 1 – ……. + 1
= 1
Question 10 :
The value of the limit Limx→0 (cos x)cot2 x is
(a) 1
(b) e
(c) e1/2
(d) e-1/2
Correct Answer – (D)
Given, Limx→0 (cos x)cot² x
= Limx→0 (1 + cos x – 1)cot² x
= eLimx→0 (cos x – 1) × cot² x
= eLimx→0 (cos x – 1)/tan² x
= e-1/2
- NCERT Solutions Class 11 Mathematics Limits and Derivatives with Answers : Exercise 13.1
- NCERT Solutions Class 11 Mathematics Limits and Derivatives with Answers : Exercise 13.2
- NCERT Solutions Class 11 Mathematics Limits and Derivatives with Answers : Exercise 13 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
