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 (ex² – 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 (ex² – cos x)/x²
= Limx→0 (ex² – cos x -1 + 1)/x²
= Limx→0 {(ex² – 1)/x² + (1 – cos x)}/x²
= Limx→0 {(ex² – 1)/x² + Limx→0 (1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2
We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Limits and Derivatives MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.
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