# MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers

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## Limits and Derivatives Class 11 MCQs Questions with Answers

Students are advised to solve the Limits and Derivatives Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Limits and Derivatives Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Limits and Derivatives Class 11 with answers provided with detailed solutions by looking below.

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 – ……..

Answer

Answer: (d) -x – x² /2 – x³ /3 – ……..
log(1 – x) = -x – x² /2 – x³ /3 – ……..

Question 2.
The value of Limx→a (a × sin x – x × sin a)/(ax² – xa²) is
(a) = (a × cos a + sin a)/a²
(b) = (a × cos a – sin a)/a²
(c) = (a × cos a + sin a)/a
(d) = (a × cos a – sin a)/a

Answer

Answer: (b) = (a × cos a – sin a)/a²
Given,
Limx→a (a × sin x – x × sin a)/(ax² – xa²)
When we put x = a in the expression, we get 0/0 form.
Now apply L. Hospital rule, we get
Limx→a (a × cos x – sin a)/(2ax – a²)
= (a × cos a – sin a)/(2a × a – a²)
= (a × cos a – sin a)/(2a² – a²)
= (a × cos a – sin a)/a²
So, Limx→a (a × sin x – x × sin a)/(ax² – xa²) = (a × cos a – sin a)/a²

Question 3.
Limx→-1 [1 + x + x² + ……….+ x10] is
(a) 0
(b) 1
(c) -1
(d) 2

Answer

Answer: (b) 1
Given, Limx→-1 [1 + x + x² + ……….+ x10]
= 1 + (-1) + (-1)² + ……….+ (-1)10
= 1 – 1 + 1 – ……. + 1
= 1

Question 4.
The value of the limit Limx→0 {log(1 + ax)}/x is
(a) 0
(b) 1
(c) a
(d) 1/a

Answer

Answer: (c) a
Given, Limx→0 {log(1 + ax)}/x
= Limx→0 {ax – (ax)² /2 + (ax)³ /3 – (ax)4 /4 + …….}/x
= Limx→0 {ax – a² x² /2 + a³ x³ /3 – a4 x4 /4 + …….}/x
= Limx→0 {a – a² x /2 + a³ x² /3 – a4 x³ /4 + …….}
= a – 0
= a

Question 5.
The value of the limit Limx→0 (cos x)cot² x is
(a) 1
(b) e
(c) e1/2
(d) e-1/2

Answer

Answer: (d) e-1/2
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

Question 6.
Then value of Limx→1 (1 + log x – x)}/(1 – 2x + x²) is
(a) 0
(b) 1
(c) 1/2
(d) -1/2

Answer

Answer: (d) -1/2
Given, Limx→1 (1 + log x – x)}/(1 – 2x + x²)
= Limx→1 (1/x – 1)}/(-2 + 2x) {Using L. Hospital Rule}
= Limx→1 (1 – x)/{2x(x – 1)}
= Limx→1 (-1/2x)
= -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

Answer

Answer: (d) x × tan x × sec x + sec x
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 (e – cos x)/x² is equals to
(a) 0
(b) 1
(c) 2/3
(d) 3/2

Answer

Answer: (d) 3/2
Given, Limx→0 (e – cos x)/x²
= Limx→0 (e – cos x – 1 + 1)/x²
= Limx→0 {(e – 1)/x² + (1 – cos x)}/x²
= Limx→0 {(e – 1)/x² + Limx→0 (1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2

Question 9.
The expansion of ax is
(a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
(b) ax = 1 – x/1! × (log a) + x² /2! × (log a)² – x³ /3! × (log a)³ + ………..
(c) ax = 1 + x/1 × (log a) + x² /2 × (log a)² + x³ /3 × (log a)³ + ………..
(d) ax = 1 – x/1 × (log a) + x² /2 × (log a)² – x³ /3 × (log a)³ + ………..

Answer

Answer: (a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..

Question 10.
The value of the limit Limn→0 (1 + an)b/n is
(a) ea
(b) eb
(c) eab
(d) ea/b

Answer

Answer: (c) eab
Given, Limn→0 (1 + an)b/n
= eLimn→0(an × b/n)
= eLimn→0(ab)
= eab

Question 11.
The value of Limx→0 cos x/(1 + sin x) is
(a) 0
(b) -1
(c) 1
(d) None of these

Answer

Answer: (c) 1
Given, Limx→0 cos x/(1 + sin x)
= cos 0/(1 + sin 0)
= 1/(1 + 0)
= 1/1
= 1

Question 12.
Lim tanx→π/4 tan 2x × tan(π/4 – x) is
(a) 0
(b) 1
(c) 1/2
(d) 3/2

Answer

Answer: (c) 1/2
Given, Lim tanx→π/4 tan 2x × tan(π/4 – x)
= Lim tanh→0 tan 2(π/4 – x) × tan(-h)
= Lim tanh→0 -cot 2h/(-cot h)
= Lim tanh→0 tan h/tan 2h
= (1/2) × Lim tanh→0 (tan h/h)/(2h/tan 2h)
= (1/2) × {Lim tanh→0 (tan h/h)}/{Lim tanh→0 (2h/tan 2h)}
= (1/2) × 1
= 1/2

Question 13.
Limx→2 (x³ – 6x² + 11x – 6)/(x² – 6x + 8) =
(a) 0
(b) 1
(c) 1/2
(d) Limit does not exist

Answer

Answer: (c) 1/2
When x = 2, the expression
(x³ – 6x² + 11x – 6)/(x² – 6x + 8) assumes the form 0/0
Now,
Limx→2 (x³ – 6x² + 11x – 6)/(x² – 6x + 8) = Limx→2 {(x – 1) × (x – 2) × (x – 3)}/{(x – 2) × (x – 4)}
= Limx→2 {(x – 1) × (x – 3)}/(x – 4)
= {(2 – 1) × (2 – 3)}/(2 – 4)
= 1/2

Question 14.
The value of the limit Limx→2 (x – 2)/√(2 – x) is
(a) 0
(b) 1
(c) -1
(d) 2

Answer

Answer: (a) 0
Given, Limx→2 (x – 2)/√(2 – x)
= Limx→2 -(2 – x)/√(2 – x)
= Limx→2 -{√(2 – x) × √(2 – x)}/√(2 – x)
= Limx→2 -√(2 – x)
= -√(2 – 2)
= 0

Question 15.
The derivative of the function f(x) = 3x³ – 2x³ + 5x – 1 at x = -1 is
(a) 0
(b) 1
(c) -18
(d) 18

Answer

Answer: (d) 18
Given, function f(x) = 3x³ – 2x² + 5x – 1
Differentiate w.r.t. x, we get
df(x)/dx = 3 × 3 × x² – 2 × 2 × x + 5
⇒ df(x)/dx = 9x² – 4x + 5
⇒ {df(x)/dx}x =-1 = 9 × (-1)² – 4 × (-1) + 5
⇒ {df(x)/dx}x =-1 = 9 + 4 + 5
⇒ {df(x)/dx}x =-1 = 18

Question 16.
Limx→0 sin²(x/3)/x² is equals to
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/9

Answer

Answer: (d) 1/9
Given, Limx→0 sin² (x/3)/ x²
= Limx→0 [sin² (x/3)/ (x/3)² × {(x/3)² /x²}]
= Limx→0 [{sin (x/3)/ (x/3)}² × {(x² /9)/x²}]
= 1 × 1/9
= 1/9

Question 17.
The expansion of ax is
(a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
(b) ax = 1 – x/1! × (log a) + x² /2! × (log a)² – x³ /3! × (log a)³ + ………..
(c) ax = 1 + x/1 × (log a) + x² /2 × (log a)² + x³ /3 × (log a)³ + ………..
(d) ax = 1 – x/1 × (log a) + x² /2 × (log a)² – x³ /3 × (log a)³ + ………..

Answer

Answer: (a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..

Question 18.
Differentiation of cos √x with respect to x is
(a) sin x /2√x
(b) -sin x /2√x
(c) sin √x /2√x
(d) -sin √x /2√x

Answer

Answer: (d) -sin √x /2√x
Let y = cos √x
Put u = √x
du/dx = 1/2√x
Now, y = cos u
dy/du = -sin u
dy/dx = (dy/du) × (du/dx)
= -sin u × (1/2√x)
= -sin √x /2√x

Question 19.
Differentiation of log(sin x) is
(a) cosec x
(b) cot x
(c) sin x
(d) cos x

Answer

Answer: (b) cot x
Let y = log(sin x)
Again let u = sin x
du/dx = cos x
Now, y = log u
dy/du = 1/u = 1/sin x
Now, dy/dx = (dy/du) × (du/dx)
⇒ dy/dx = (1/sin x) × cos x
⇒ dy/dx = cos x/sin x
⇒ dy/dx = cot x

Question 20.
Limx→∞ {(x + 5)/(x + 1)}x equals
(a) e²
(b) e4
(c) e6
(d) e8

Answer

Answer: (c) e6
Given, Limx→∞ {(x + 5)/(x + 1)}x
= Limx→∞ {1 + 6/(x + 1)}x
= eLimx→∞6x/(x + 1)
= eLimx→∞ 6/(1 + 1/x)
= e6/(1 + 1/∞)
= e6/(1 + 0)
= e6

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