CBSE Class 11 Maths - MCQ and Online Tests - Unit 13 - Limits and Derivatives

CBSE Class 11 Maths  – MCQ and Online Tests – Unit 13 – Limits and Derivatives

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.

 

CBSE Class 11 Maths  – MCQ and Online Tests – Unit 13 – Limits and Derivatives

Question 1.
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

Answer

Answer: (b) 1
Hint:
Given, Lim_x?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

---

Question 2.
The value of Lim_x?0 (1/x) × sin^-1 {2x/(1 + x²) is
(a) 0
(b) 1
(c) 2
(d) -2

Answer

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

---

Question 3.
The value of the limit Lim_x?0 {log(1 + ax)}/x is
(a) 0
(b) 1
(c) a
(d) 1/a

Answer

Answer: (c) a
Hint:
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

---

Question 4.
Lim_x?-1 [1 + x + x² + ……….+ x¹⁰] is
(a) 0
(b) 1
(c) -1
(d) 2

Answer

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

---

Question 5.
The value of Lim_x?01 (1/x) × sin^-1 {2x/(1 + x²) is
(a) 0
(b) 1
(c) 2
(d) -2

Answer

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

---

Question 6.
Lim_x?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,
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

---

Question 7.
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)

Answer

Answer: (c) log (a/b)
Hint:
Given, Lim_x?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)

---

Question 8.
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

Answer

Answer: (d) x × tan x × sec x + sec x
Hint:
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

---

Question 9.
Lim_y?8 {(x + 6)/(x + 1)}^(x+4) equals
(a) e
(b) e³
(c) e⁵
(d) e⁶

Answer

Answer: (c) e⁵
Hint:
Given, Lim_y?8 {(x + 6)/(x + 1)}(x + 4)
= Lim_y?8 {1 + 5/(x + 1)}(x + 4)
= e^{Lim_y?8 5(x + 4)/(x + 1)}
= e^{Lim_y?8 5(1 + 4/x)/(1 + 1/x)}
= e^{5(1 + 4/8)/(1 + 1/8)}
= e^{5/(1 + 0)}
= e⁵

---

Question 10.
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 11.
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

Answer

Answer: (d) e^-1/2
Hint:
Given, Lim_x?0 (cos x)^{cot² x}
= Lim_x?0 (1 + cos x – 1)^{cot² x}
= eLim_x?0 (cos x – 1) × cot² x
= eLim_x?0 (cos x – 1)/tan² x
= e^-1/2

---

Question 12.
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

Answer

Answer: (c) 2 cos a
Hint:
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

---

Question 13.
The value of Lim_n?8 {1² + 2² + 3² + …… + n²}/n³ is
(a) 0
(b) 1
(c) -1
(d) n

Answer

Answer: (a) 0
Hint:
Given, Lim_n?8 {1² + 2² + 3² + …… + n²}/n³
= Lim_n?8 [{n×(n + 1)×(2n + 1)}/6]/{n(n + 1)/2}²
= Lim_n?8 [{n×n×n ×(1 + 1/n)×(2 + 1/n)}/6]/{n × n ×(1 + 1/n)/2}²
= Lim_n?8 [{n³ ×(1 + 1/n)×(2 + 1/n)}/6]/{n² ×(1 + 1/n)/2}²
= Lim_n?8 [{(1 + 1/n)×(2 + 1/n)}/6]/[n⁴ × {(1 + 1/n)/2}²]
? Lim_n?8 [{(1 + 1/n)×(2 + 1/n)}/6]/[n × {(1 + 1/n)/2}²]
= [{(1 + 1/8)×(2 + 1/8)}/6]/[8×{(1 + 1/8)/2}²
= [{(1 + 0)×(2 + 0)}/6]/8 {since 1/8 = 0}
= {(1 × 2)/6}/8
= (2/6)/8
= (1/3)/8
= 0
So, Lim_n?8 {1² + 2² + 3² + …… + n²}/n³ = 0

---

Question 14.
The value of Lim_n?8 (sin x/x) is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (a) 0
Hint:
Lim_n?8 (sin x/x) = Lim_y?0 {y × sin (1/y)} = 0

---

Question 15.
The value of Lim_x?0 ax is
(a) 0
(b) 1
(c) 1/2
(d) 3/2

Answer

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

---

Question 16.
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 17.
Lim_x?0 (e^x² – cos x)/x² is equals to
(a) 0
(b) 1
(c) 2/3
(d) 3/2

Answer

Answer: (d) 3/2
Hint:
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

---

Question 18.
Lim_x?0 sin (ax)/bx is
(a) 0
(b) 1
(c) a/b
(d) b/a

Answer

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

---

Question 19.
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 20.
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

Answer

Answer: (b) 0
Hint:
Given, 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

---