Question

Class 11 Maths MCQ Questions of Limits and Derivatives with Answers?

Answer 1

To attempt these Class 11 Maths MCQ Questions of Limits and Derivatives with Answers students should be comfortable with the essential concept. They should to have ideas regarding the many subjects which may be tested in MCQ Questions. To help students in preparing for objective-type questions, Sarthaks eConnect has arranged Chapter-wise MCQ Questions for class 11 with answers.

This series of questions will allow class 11 students a chance to check their insight into ideas and assist them with getting prepared for the approaching CBSE Exam. All of the Class 11 Maths MCQ Questions of Limits and Derivatives with Answers are given precise Answers.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The value of the limit Lim_x→0 (cos x)\(^{cot^{2}x}\) is

(a) 1
(b) e
(c) e^1/2
(d) e^-1/2

2. The value of limit Lim_x→0 {sin (a + x) – sin (a – x)}/x is

(a) 0
(b) 1
(c) 2 cos a
(d) 2 sin a

3. Lim_x→-1 [1 + x + x² + ……….+ x¹⁰] is

(a) 0
(b) 1
(c) -1
(d) 2

4. Lim_x→0 ( \(e^{x^2}\)– cos x)/x² is equals to

(a) 0
(b) 1
(c) 2/3
(d) 3/2

5. Lim_x→0 log(1 – x) is equals to

(a) 0
(b) 1
(c) 1/2
(d) None of these

6. Lim_x→0 {(a^x – b^x)/ x} is equal to

(a) log a
(b) log b
(c) log (a/b)
(d) log (a×b)

7. 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

8. Lim _y→∞ {(x + 6)/(x + 1)}(x+4) equals

(a) e
(b) e³
(c) e⁵
(d) e⁶

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

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 – ……..

11. If f(x) = x × sin(1/x), x ≠ 0, then Lim_x→0 f(x) is

(a) 1
(b) 0
(c) -1
(d) does not exist

12. The value of Lim_n→∞ {1² + 2² + 3²+ …… + n²}/n² is

(a) 0
(b) 1
(c) -1
(d) n

13. The value of Lim_x→0 ax is

(a) 0
(b) 1
(c) 1/2
(d) 3/2

14. Let f(x) = cos x, when x ≥ 0 and f(x) = x + k, when x < 0 Find the value of k given that Lim_x→0 f(x) exists.

(a) 0
(b) 1
(c) -1
(d) None of these

15. The value of Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²) is

(a) 0
(b) 1
(c) 2
(d) -2

16. Lim_x→0 sin (ax)/bx is

(a) 0
(b) 1
(c) a/b
(d) b/a

17. lim _x→0 |Sinx | /x  is      

(a) 1
(b) –1
(c) does not exist
(d) None of these

18. The derivative of x² cos x is

(a) 2x sin x – x² sin x
(b) 2x cos x – x² sin x
(c) 2x sin x – x² cos x
(d) cosx – x² sin x cos x

19. If f(x) = x sin x, then f′(π/2) is equal to

(a) 0
(b) 1
(c) –1
(d) 1/2

20. If f(x) = x¹⁰⁰ + x⁹⁹ + … + x + 1, then f′(1) is equal to

(a) 5050
(b) 5049
(c) 5051
(d) 50051

Answer 2

 1. Answer: (d) e^-1/2

Explanation: Given, Lim_x→0 (cos x)cot²

= Lim_x→0 (1 + cos x – 1)cot²x

= eLim_x→0 (cos x – 1) × cot² x

= eLim^x→0 (cos x – 1)/tan² x

= e^-1/2

2. Answer: (c) 2 cos a

Explanation: Given, Lim_x→0 {sin (a + x) – sin (a – x)}/x

= Lim_x→0 {2 × cos a × sin x}/x

= 2 × cos a × Lim_x→0 sin x/x

= 2 cos a

3. Answer: (b) 1

Explanation: Given, Limx→-1 [1 + x + x² + ……….+ x¹⁰]

= 1 + (-1) + (-1)² + ……….+ (-1)¹⁰

= 1 – 1 + 1 – ……. + 1

= 1

4. Answer: (d) 3/2

Explanation: Lim_x→0 (\(e^{x^2}\) – cos x)/x²

= Lim_x→0 (\(e^{x^2}\) – cos x -1 + 1)/x²

= Lim_x→0 {(\(e^{x^2}\) – 1)/x² + (1 – cos x)}/x²

= Lim_x→0 {(\(e^{x^2}\) – 1)/x² + Lim_x→0 (1 – cos x)}/x²

= 1 + 1/2

= (2 + 1)/2

= 3/2

5. Answer: (a) 0

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

Now,

Lim_x→0 log(1 – x) = Lim_x→0 {-x – x²/2 – x³/3 – ……..}

⇒ Lim_x→0 log(1 – x) = Lim_x→0 {-x} – Lim_x→0 {x²/2} – Lim_x→0 {x³/3} – ……..

⇒ Lim_x→0 log(1 – x)

= 0

6. Answer: (c) log (a/b)

Explanation: Limx→0 {(a^x – b^x)/ x}

= Lim_x→0 {(a^x – b^x – 1 + 1)/ x}

= Lim_x→0 {(a^x – 1) – (b^x – 1)}/ x

= Lim_x→0 {(a^x – 1)/x – (b^x – 1)/x}

= Lim_x→0 (a^x – 1)/x – Lim_x→0 (b^x – 1)/x

= log a – log b

= log (a/b)

7. Answer: (d) x × tan x × sec x + sec x

Explanation: 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

8. Answer: (c) e⁵

Explanation: Limy→∞ {(x + 6)/(x + 1)}(x + 4)

= Lim_y→∞ {1 + 5/(x + 1)}(x + 4)

= eLim_y→∞ 5(x + 4)/(x + 1)

= eLim_y→∞ 5(1 + 4/x)/(1 + 1/x)

= e⁵(1 + 4/∞)/(1 + 1/∞)

= e⁵/(1 + 0)

= e⁵

9. Answer: (d) -2/(x-1)²

Explanation: 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

10. Answer: (d) -x – x²/2 – x³/3 – ……..

Explanation: log(1 – x)

= -x – x²/2 – x³/3 – ……..

Answer 3

 11. Answer: (b) 0

Explanation: f(x) = x × sin(1/x)

Now, Lim_x→0 f(x) = Lim_x→0 x × sin(1/x)

⇒ Lim_x→0 f(x) = 0

12. Answer: (a) 0

Explanation: Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³

= Lim_n→∞ [{n×(n + 1)×(2n + 1)}/6]/{n(n + 1)/2}²

= Lim_n→∞ [{n×n×n ×(1 + 1/n)×(2 + 1/n)}/6]/{n × n ×(1 + 1/n)/2}²

= Lim_n→∞ [{n³ ×(1 + 1/n)×(2 + 1/n)}/6]/{n² ×(1 + 1/n)/2}²

= Lim_n→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n⁴ × {(1 + 1/n)/2}²]

⇒ Lim_n→∞ [{(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, Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³ = 0

13. Answer: (b) 1

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

Now,

Lim_x→0 ax = Lim_x→0 {1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + …}

⇒ Lim_x→0 ax = Lim_x→0 1 + Lim_x→0 {x/1! × (log a)} + Limx→0 {x²/2! × (log a)²}+ ………

⇒ Lim_x→0 ax = 1

14. Answer: (b) 1

Explanation: Limx→0 f(x) exists

⇒ Lim_x→0 – f(x) = Lim^x→0 + f(x)

⇒ Lim_x→0 (x + k) = Lim_x→0 cos x

⇒ k = cos 0

⇒ k = 1

15. Answer: (c) 2

Explanation: Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²)

= Lim_x→0 (2 × tan^-1 x)/x

= 2 × 1

= 2

16. Answer: (c) a/b

Explanation: Lim_x→0 sin (ax)/bx

= Lim_x→0 [{sin (ax)/ax} × (ax/bx)]

⇒ (a/b) Lim_x→0 sin (ax)/ax

= a/b

17. Answer: (c) does not exist

Explanation: \(\underset{x\to 0}{lim}\frac{sin|x|}{x}\)

LHL= −1,RHL= 1 

Limit does not exist.

18. Answer: (b) 2x cos x – x² sin x

Explanation: d/dx(x² cos x)

Using the formula d/dx [f(x) g(x)] = f(x) [d/dx g(x)] + g(x) [d/dx f(x)]

d/dx(x² cos x) = x² [d/dx (cos x)] + cos x [d/dx x²]

= x²(-sin x) + cos x (2x)

= 2x cos x – x² sin x

19. Answer: (b) 1

Explanation: f(x) = x sin x

f'(x) = x[d/dx sin x] + sin x [d/dx (x)]

= x cos x + sin x

Now,

f′(π/2) = (π/2) cos π/2 + sin π/2

= (π/2) (0) + 1

= 1

20. Answer: (a) 50500

Explanation: f(x) = x¹⁰⁰ + x99 + … + x + 1

f′(x) = 100x⁹⁹ + 99x⁹⁸ + …. + 1 + 0

f′(1) = 100(1)⁹⁹ + 99(1)98 + ….+ 1

= 100 + 99 + …. + 1

This is an AP with common difference -1, a = 100, n = 100 and l = 1.

So, the sum of this AP = (100/2)[100 + 1]

= 50(101)

= 5050

Therefore, f′(1) = 5050

Click here to practice MCQ Questions for Limits and Derivatives class 11