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.

Limits and Derivatives Class 11 MCQs Questions with Answers

Question 1.
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
Hint:
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 2.
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

Answer

Answer: (c) 2 cos a
Hint:
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


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

Answer

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


Question 4.
The value of Limx→01 (1/x) × sin-1 {2x/(1 + x²) is
(a) 0
(b) 1
(c) 2
(d) -2

Answer

Answer: (c) 2
Hint:
Given, Limx→0 (1/x) × sin-1 {2x/(1 + x²)
= Limx→0 (2× tan-1 x)/x
= 2 × 1
= 2


Question 5.
Limx→0 log(1 – x) is equals to
(a) 0
(b) 1
(c) 1/2
(d) None of these

Answer

Answer: (a) 0
Hint:
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 6.
Limx→0 {(ax – bx)/ x} is equal to
(a) log a
(b) log b
(c) log (a/b)
(d) log (a×b)

Answer

Answer: (c) log (a/b)
Hint:
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 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
Hint:
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.
Limy→∞ {(x + 6)/(x + 1)}(x+4) equals
(a) e
(b) e³
(c) e5
(d) e6

Answer

Answer: (c) e5
Hint:
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 9.
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

Answer

Answer: (d) A
Hint:
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 10.
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 – ……..
Hint:
log(1 – x) = -x – x²/2 – x³/3 – ……..


Question 11.
If f(x) = x × sin(1/x), x ≠ 0, then Limx→0 f(x) is
(a) 1
(b) 0
(c) -1
(d) does not exist

Answer

Answer: (b) 0
Hint:
Given, f(x) = x × sin(1/x)
Now, Limx→0 f(x) = Limx→0 x × sin(1/x)
⇒ Limx→0 f(x) = 0


Question 12.
The value of Limn→∞ {1² + 2² + 3² + …… + n²}/n³ is
(a) 0
(b) 1
(c) -1
(d) n

Answer

Answer: (a) 0
Hint:
Given, Limn→∞ {1² + 2² + 3² + …… + n²}/n³
= Limn→∞ [{n×(n + 1)×(2n + 1)}/6]/{n(n + 1)/2}²
= Limn→∞ [{n×n×n ×(1 + 1/n)×(2 + 1/n)}/6]/{n × n ×(1 + 1/n)/2}²
= Limn→∞ [{n³ ×(1 + 1/n)×(2 + 1/n)}/6]/{n² ×(1 + 1/n)/2}²
= Limn→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n4 × {(1 + 1/n)/2}²]
⇒ Limn→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n × {(1 + 1/n)/2}²]
= [{(1 + 1/∞)×(2 + 1/∞)}/6]/[∞×{(1 + 1/∞)/2}²
= [{(1 + 0)×(2 + 0)}/6]/∞ {since 1/∞ = 0}
= {(1 × 2)/6}/∞
= (2/6)/∞
= (1/3)/∞
= 0
So, Limn→∞ {1² + 2² + 3² + …… + n²}/n³ = 0


Question 13.
The value of Limn→∞ (sin x/x) is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (a) 0
Hint:
Limn→∞ (sin x/x) = Limy→0 {y × sin (1/y)} = 0


Question 14.
The value of Limx→0 ax is
(a) 0
(b) 1
(c) 1/2
(d) 3/2

Answer

Answer: (b) 1
Hint:
We know that
ax = 1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + ………..
Now,
Limx→0 ax = Limx→0 {1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + …}
⇒ Limx→0 ax = Limx→0 1 + Limx→0 {x/1! × (log a)} + Limx→0 {x² /2! × (log a)²}+ ………
⇒ Limx→0 ax = 1


Question 15.
Let f(x) = cos x, when x ≥ 0 and f(x) = x + k, when x < 0 Find the value of k given that Limx→0 f(x) exists.
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (b) 1
Hint:
Given, Limx→0 f(x) exists
⇒ Limx→0 – f(x) = Limx→0 + f(x)
⇒ Limx→0 (x + k) = Limx→0 cos x
⇒ k = cos 0
⇒ k = 1


Question 16.
The value of Limx→0 (1/x) × sin-1 {2x/(1 + x²) is
(a) 0
(b) 1
(c) 2
(d) -2

Answer

Answer: (c) 2
Hint:
Given, Limx→0 (1/x) × sin-1 {2x/(1 + x²)
= Limx→0 (2 × tan-1 x)/x
= 2 × 1
= 2


Question 17.
Limx→0 sin (ax)/bx is
(a) 0
(b) 1
(c) a/b
(d) b/a

Answer

Answer: (c) a/b
Hint:
Given, Limx→0 sin (ax)/bx
= Limx→0 [{sin (ax)/ax} × (ax/bx)]
⇒ (a/b) Limx→0 sin (ax)/ax
= a/b


Question 18.
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
Hint:
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 19.
If f(x) = (x + 1)/x then df(x)/dx is
(a) 1/x
(b) -1/x
(c) -1/x²
(d) 1/x²

Answer

Answer: (c) -1/x²
Hint:
Given, f(x) = (x + 1)/x
Now, df(x)/dx = d{(x + 1)/x}/dx
= {1 × x – (x + 1)×1}/x²
= (x – x – 1)/x²
= -1/x²


Question 20.
Limx→0 (e – cos x)/x² is equals to
(a) 0
(b) 1
(c) 2/3
(d) 3/2

Answer

Answer: (d) 3/2
Hint:
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


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