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

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

Answer: (b) = (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

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

Answer: (c) 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

Answer: (d) 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

Answer: (d) -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

Answer: (d) 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

Answer: (d) 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

Answer: (a) 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 + ………..

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

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

Answer: (c) 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

Answer: (c) 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

Answer: (a) 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

Answer: (d) 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

Answer: (d) 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

Answer: (a) 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 + ………..

Answer: (d) -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

Answer: (b) 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

Answer: (c) 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