# MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives 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: (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: (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

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

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

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

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: (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

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: (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

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

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

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

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

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

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

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: (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

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

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