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

Limits and Derivatives Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Limits and Derivatives Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 13 Quiz

Class 11 Maths Chapter 13 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Limits and Derivatives Class 11 Maths MCQ online test

Q1.	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 – ……..

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

Q2.	The value of Lim_x→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

Ans: = (a × cos a – sin a)/a²
Given,
Lim_x→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
Lim_x→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, Lim_x→a (a × sin x – x × sin a)/(ax² – xa²) = (a × cos a – sin a)/a²

Q3.	Lim_x→-1 [1 + x + x² + ……….+ x¹⁰] is
	A.0
	B.1
	C.-1
	D.2

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

Q4.	The value of the limit Lim_x→0 {log(1 + ax)}/x is
	A.0
	B.1
	C.a
	D.1/a

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

Q5.	The value of the limit Lim_x→0 (cos x)^{cot² x} is
	A.1
	B.e
	C.e^1/2
	D.e^-1/2

Ans: e^-1/2
Given, Lim_x→0 (cos x)^{cot² x}
= Lim_x→0 (1 + cos x – 1)^{cot² x}
= e^Lim_x→0 (cos x – 1) × cot² x
= e^Lim_x→0 (cos x – 1)/tan² x
= e^-1/2

Q6.	Then value of Lim_x→1 (1 + log x – x)}/(1 – 2x + x²) is
	A.0
	B.1
	C.1/2
	D.-1/2

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

Q7.	The value of lim_y→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

Ans: x × tan x × sec x + sec x
Given, lim_y→0 {(x + y) × sec (x + y) – x × sec x}/y
= lim_y→0 {x sec (x + y) + y sec (x + y) – x × sec x}/y
= lim_y→0 [x{ sec (x + y) – sec x} + y sec (x + y)]/y
= lim_y→0 x{ sec (x + y) – sec x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 x{1/cos (x + y) – 1/cos x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{cos x – cos (x + y)} × x/{y × cos (x + y) × cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{2sin (x + y/2) × sin(y/2)} × 2x/{2y × cos (x + y) × cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 {sin (x + y/2) × lim_y→0 {sin(y/2)/(2y/2)} × lim_y→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, lim_y→0 {(x + y) × sec (x + y) – x × sec x}/y = x × tan x × sec x + sec x

Q8.	Lim_x→0 (e^x² – cos x)/x² is equals to
	A.0
	B.1
	C.2/3
	D.3/2

Ans: 3/2
Given, Lim_x→0 (e^x² – cos x)/x²
= Lim_x→0 (e^x² – cos x – 1 + 1)/x²
= Lim_x→0 {(e^x² – 1)/x² + (1 – cos x)}/x²
= Lim_x→0 {(e^x² – 1)/x² + Lim_x→0 (1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2

Q9.	The expansion of a^x is
	A.a^x = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
	B.a^x = 1 – x/1! × (log a) + x² /2! × (log a)² – x³ /3! × (log a)³ + ………..
	C.a^x = 1 + x/1 × (log a) + x² /2 × (log a)² + x³ /3 × (log a)³ + ………..
	D.a^x = 1 – x/1 × (log a) + x² /2 × (log a)² – x³ /3 × (log a)³ + ………..

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

Q10.	The value of the limit Lim_n→0 (1 + an)^b/n is
	A.e^a
	B.e^b
	C.e^ab
	D.e^a/b

Ans: e^ab
Given, Lim_n→0 (1 + an)^b/n
= e^{Lim_n→0(an × b/n)}
= e^{Lim_n→0(ab)}
= e^ab

Q11.	The value of Lim_x→0 cos x/(1 + sin x) is
	A.0
	B.-1
	C.1
	D.None of these

Ans: 1
Given, Lim_x→0 cos x/(1 + sin x)
= cos 0/(1 + sin 0)
= 1/(1 + 0)
= 1/1
= 1

Q12.	Lim tan_x→π/4 tan 2x × tan(π/4 – x) is
	A.0
	B.1
	C.1/2
	D.3/2

Ans: 1/2
Given, Lim tan_x→π/4 tan 2x × tan(π/4 – x)
= Lim tan_h→0 tan 2(π/4 – x) × tan(-h)
= Lim tan_h→0 -cot 2h/(-cot h)
= Lim tan_h→0 tan h/tan 2h
= (1/2) × Lim tan_h→0 (tan h/h)/(2h/tan 2h)
= (1/2) × {Lim tan_h→0 (tan h/h)}/{Lim tan_h→0 (2h/tan 2h)}
= (1/2) × 1
= 1/2

Q13.	Lim_x→2 (x³ – 6x² + 11x – 6)/(x² – 6x + 8) =
	A.0
	B.1
	C.1/2
	D.Limit does not exist

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

Q14.	The value of the limit Lim_x→2 (x – 2)/√(2 – x) is
	A.0
	B.1
	C.-1
	D.2

Ans: 0
Given, Lim_x→2 (x – 2)/√(2 – x)
= Lim_x→2 -(2 – x)/√(2 – x)
= Lim_x→2 -{√(2 – x) × √(2 – x)}/√(2 – x)
= Lim_x→2 -√(2 – x)
= -√(2 – 2)
= 0

Q15.	The derivative of the function f(x) = 3x³ – 2x³ + 5x – 1 at x = -1 is
	A.0
	B.1
	C.-18
	D.18

Ans: 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

Q16.	Lim_x→0 sin²(x/3)/x² is equals to
	A.1/2
	B.1/3
	C.1/4
	D.1/9

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

Q17.	The expansion of ax is
	A.a^x = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
	B.a^x = 1 – x/1! × (log a) + x² /2! × (log a)² – x³ /3! × (log a)³ + ………..
	C.a^x = 1 + x/1 × (log a) + x² /2 × (log a)² + x³ /3 × (log a)³ + ………..
	D.a^x = 1 – x/1 × (log a) + x² /2 × (log a)² – x³ /3 × (log a)³ + ………..

Ans: a^x = 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 + ………..

Q18.	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

Ans: -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

Q19.	Differentiation of log(sin x) is
	A.cosec x
	B.cot x
	C.sin x
	D.cos x

Ans: 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

Q20.	Lim_x→∞ {(x + 5)/(x + 1)}^x equals
	A.e²
	B.e⁴
	C.e⁶
	D.e⁸

Ans: e⁶
Given, Lim_x→∞ {(x + 5)/(x + 1)}x
= Lim_x→∞ {1 + 6/(x + 1)}x
= e^{Lim_x→∞6x/(x + 1)}
= e^{Lim_x→∞ 6/(1 + 1/x)}
= e^{6/(1 + 1/∞)}
= e^{6/(1 + 0)}
= e⁶