Multiple Choice Questions on Differentiation Class 11 OP Malhotra Exe-19E ISC Maths Solutions Ch-19. In this article you would learn to solve all type mcq problems on Differentiation. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.

Differentiation Class 11 OP Malhotra  Multiple Choice Questions ISC Maths Solutions Ch-19

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-19	Differentiation
Writer	OP Malhotra
Exe-19(F)	Multiple Choice Questions.

 Multiple Choice Questions on Differentiation

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Limit x → ∞ of (3x³ + 2x² − 7x + 9) / (4x³ + 9x − 2) is equal to

(a) 2/9
(b) 3/4
(c) −9/2
(d) 1/2

Answer: (b) 3/4

Sol:  Divide numerator and denominator by x³ (highest power of x).
= (3 + 2/x − 7/x² + 9/x³) / (4 + 9/x² − 2/x³)
As x → ∞, all terms containing 1/x → 0.
Limit = 3 / 4.

Que-2: Limit x → 0 of (√(1 + 2x) − 1) / x is equal to

(a) 0
(b) −1
(c) 1/2
(d) 1

Answer: (d) 1

Sol:  Rationalize the numerator:
Multiply numerator and denominator by (√(1 + 2x) + 1).
= [(1 + 2x) − 1] / [x(√(1 + 2x) + 1)]
= 2x / [x(√(1 + 2x) + 1)]
Cancel x:
= 2 / (√(1 + 2x) + 1)
Now put x = 0:
= 2 / (1 + 1) = 1.

Que-3: The value of Limit x → 3 of (x⁵ − 3⁵) / (x⁸ − 3⁸) is equal to

(a) 5/8
(b) 5/64
(c) 5/216
(d) 1/27

Answer: (c) 5/216

Sol: Use formula:
Lim x → a (xⁿ − aⁿ) / (xᵐ − aᵐ) = n aⁿ⁻¹ / m aᵐ⁻¹
Here a = 3, n = 5, m = 8
= (5 × 3⁴) / (8 × 3⁷)
= 5 / (8 × 3³)
= 5 / 216.

Que-4: Lim x→2 (x − 2) / (x³ − x² − x − 2) is equal to

(a) 3/5   
(b) 1/5   
(c) 2/7   
(d) 1/7

Answer: (d) 1/7

Sol: Put x = 2, we get 0/0 form. Factor denominator:
x³ − x² − x − 2 = (x − 2)(x² + x + 1)
Cancel (x − 2):
Limit = 1 / (x² + x + 1)
Put x = 2:
= 1 / (4 + 2 + 1) = 1/7

Que-5: Lim x→3 (xⁿ − 3ⁿ)/(x − 3) = 108, find n

(a) 3     (b) 7     (c) 6     (d) 4

Answer: (d) 4

Sol: Using formula:
Lim x→a (xⁿ − aⁿ)/(x − a) = n·aⁿ⁻¹
So:
n·3ⁿ⁻¹ = 108
Try n = 4:
4 × 3³ = 4 × 27 = 108 ✔

Que-6: Lim x→0 (cot 4x / cosec 3x) is equal to

(a) 4/3     (b) 3/4     (c) 2/3     (d) 0

Answer: (b) 3/4

Sol: Write in sine-cosine:
cot 4x = cos4x/sin4x, cosec3x = 1/sin3x
So expression = (cos4x/sin4x) × sin3x
= (sin3x / sin4x) × cos4x
Using limits:
sin3x ≈ 3x, sin4x ≈ 4x
So = (3x / 4x) × 1 = 3/4

Que-7: Lim x→0 (1 − cos x)/x² is

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

Answer: (d) 1/2

Sol: Standard result:
Lim x→0 (1 − cos x)/x² = 1/2

Que-8: Lim x→0 sin(π cos²x)/x² is equal to

(a) 1     (b) −π     (c) π     (d) π/2

Answer: (c) π

Sol: cos²x ≈ 1 − x²
So πcos²x ≈ π(1 − x²)
sin(π − πx²) = sin(πx²)
≈ πx²
Thus limit = πx² / x² = π

Que-9: Lim θ→0 (1 − cos4θ)/(1 − cos6θ)

(a) 9/4     (b) 4/9     (c) 9/3     (d) 3/4

Answer: (b) 4/9

Sol: Use:
1 − cosx ≈ x²/2
So:
(4θ)² / (6θ)²
= 16/36 = 4/9

Que-10: Lim x→4 (√(x² + 9) − 5)/(x − 4)

(a) 2/5     (b) 8/25     (c) 0     (d) 8/5       (e) 4/5

Answer: (e) 4/5

Sol: Rationalize:
Multiply by conjugate:
= (x² + 9 − 25)/[(x − 4)(√(x²+9)+5)]
= (x² − 16)/[(x − 4)(√(x²+9)+5)]
= (x − 4)(x + 4)/[(x − 4)(√(x²+9)+5)]
Cancel (x−4):
= (x + 4)/(√(x²+9)+5)
Put x = 4:
= 8 / (5 + 5) = 4/5

Que-11: Right hand and left hand limit of f(x) = (e^(1/x) − 1)/(e^(1/x) + 1), x≠0 and f(0)=0

(a) 1 and −1   
(b) −1 and −1   
(c) −1 and 1   
(d) 1 and 1

Answer: (a) 1 and -1

Sol: As x→0⁺, 1/x→∞
⇒ e^(1/x)→∞
So function → (∞−1)/(∞+1) ≈ 1
As x→0⁻, 1/x→−∞ ⇒ e^(1/x)→0
So function → (0−1)/(0+1) = −1
Thus LHL = 1, RHL = 1

Que-12: Limit x→0 (x² cos x)/(1 − cos x) is equal to

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

Answer: (a) 2

Sol: We use standard limits: cos x ≈ 1 and 1 − cos x ≈ x²/2 as x → 0.
So, (x² cos x)/(1 − cos x)
≈ x² / (x²/2) = 2.
Hence the limit is 2.

Que-13: Limit x→1 (x^m − 1)/(xⁿ − 1) is equal to

(a) 1         (b) m/n          (c) −m/n          (d) m²/n²

Answer: (b) m/n

Sol: Using standard limit:
lim (x^k − 1)/(x − 1) = k as x → 1.
So, dividing numerator and denominator by (x−1), we get result = m/n.

Que-14: Limit x→0 (cosec x − cot x)/x is equal to

(a) −1/2     (b) 1       (c) 1/2      (d) 1

Answer: (c) 1/2

Sol: Use identity:
cosec x − cot x = (1 − cos x)/sin x.
So expression becomes (1 − cos x)/(x sin x).
Using limits: 1 − cos x ≈ x²/2 and sin x ≈ x
⇒ result = (x²/2)/(x²) = 1/2.

Que-15: Limit x→0 |sin x|/x is equal to

(a) 1
(b) −1
(c) Does not exist
(d) None of these

Answer: (c) Does not exist

Sol: As x → 0 from right, sin x ≈ x ⇒ limit = 1.
As x → 0 from left, |sin x| = −sin x ⇒ limit = −1.
Since LHL ≠ RHL, limit does not exist.

Que-16: Limit x→0 (tan 2x − x)/(3x − sin x) is equal to

(a) 2     (b) 1/2       (c) −1/2       (d) 1/4

Answer: (b) 1/2

Sol: Use approximations: tan 2x ≈ 2x and sin x ≈ x.
Numerator = 2x − x = x
Denominator = 3x − x = 2x
So limit = x / 2x = 1/2.

Que-17: Let f(x) = x⁶ + 6ˣ, then f'(x) is equal to

(a) 12x
(b) x + 4
(c) 6x⁵ + 6ˣ log 6
(d) 6x⁵ + x·6^(x−1)

Answer: (c)

Sol: Differentiate term-wise:
d/dx(x⁶) = 6x⁵
d/dx(6ˣ) = 6ˣ ln6
So, f'(x) = 6x⁵ + 6ˣ log 6.

Que-18: Differentiate x⁻³ (2 + 7x)

(a) −1/x⁴ (3 + 7x)
(b) −2/x³
(c) 2/x⁴ (2 + 7x)
(d) −2/x⁴ (3 + 7x)

Answer: (d)

Sol: Use product rule:
d/dx [x⁻³(2+7x)] = x⁻³·7 + (2+7x)(−3x⁻⁴)
= 7/x³ − 3(2+7x)/x⁴
Taking LCM and simplifying ⇒ −2/x⁴ (3 + 7x).

Que-19: Let f(x) = x² + bx + 7. If f'(5) = 2f'(7/2), find b

(a) 4       (b) 3       (c) −4        (d) −3

Answer: (c) -4

Sol: f'(x) = 2x + b
So f'(5) = 10 + b
f'(7/2) = 7 + b
Given: 10 + b = 2(7 + b)
⇒ 10 + b = 14 + 2b
⇒ b = −4.

Que-20: If f(x) = x − 1/x, then f'(−1) is

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

Answer: (c) 2

Sol: f(x) = x − x⁻¹
So f'(x) = 1 + 1/x²
At x = −1 ⇒ f'(−1) = 1 + 1 = 2.

Que-21: d/dx (sin x°) is equal to

(a) cos x°
(b) (π/180) cos x
(c) (π/180) cos x°
(d) None of these

Answer: (c)

Sol: sin(x°) = sin(xπ/180)
Using chain rule:
d/dx[sin(xπ/180)] = cos(xπ/180) × (π/180)
= (π/180) cos(x°).

Que-22: Let f(x) = (x − 4)/(2√x), then f'(1) is equal to

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

Answer: (d) none of these

Sol: f(x) = (x−4)/(2√x) = (1/2)(x−4)x^(-1/2).
Using product rule:
f'(x) = (1/2)[ x^(-1/2) + (x−4)(−1/2)x^(-3/2) ]
Put x = 1:
= (1/2)[1 + (−3)(−1/2)] = (1/2)[1 + 3/2] = (1/2)(5/2) = 5/4.
Since 5/4 is not in options ⇒ Answer = None of these.

Que-23: If y = (cos x + sin x)/(cos x − sin x), then dy/dx is equal to

(a) sec²(π/4 + x)
(b) −2/(cos x − sin x)²
(c) −2 sin x/(cos x − sin x)²
(d) None of these

Answer: (d) none of these

Sol: Using quotient rule:
Let u = cos x + sin x, v = cos x − sin x
u’ = −sin x + cos x, v’ = −sin x − cos x
dy/dx = (u’v − uv’)/v²
Simplifying numerator gives 2 ⇒ dy/dx = 2/(cos x − sin x)²
Not matching options ⇒ Answer = (d).

Que-24: If y = (1 − x)² / x², then dy/dx is

(a) −2/x² + 2/x³
(b) −1/x² + 2/x³
(c) −2/x³ + 2/x²
(d) 2/x² − 1/x³

Answer: (c)

Sol: y = (1−x)² x⁻²
Using product rule:
dy/dx = 2(1−x)(−1)x⁻² + (1−x)²(−2x⁻³)
Simplify ⇒ dy/dx = −2/x³ + 2/x².

Que-25: If y = √x + 1/√x, then dy/dx at x = 1 is equal to

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

Answer: (d) 0

Sol: y = x^(1/2) + x^(−1/2)
dy/dx = (1/2)x^(−1/2) − (1/2)x^(−3/2)
At x=1 ⇒ (1/2 − 1/2) = 0.

Que-26: If y = (1 + 1/x²)/(1 − 1/x²), then dy/dx is equal to

(a) −4x/(x² − 1)²
(b) −4x/(x² − 1)
(c) (1 − x²)/(4x)
(d) 4x/(x² − 1)

Answer: (a)

Sol: Write y = (x²+1)/(x²−1).
Using quotient rule:
dy/dx = [(2x)(x²−1) − (x²+1)(2x)]/(x²−1)²
= [2x(x²−1−x²−1)]/(x²−1)²
= −4x/(x²−1)².

Que-27: If y = sin(x+9)/cos x, then dy/dx at x = 0 is equal to

(a) cos 9       (b) sin 9        (c) 0        (d) 1

Answer: (a) cos 9

Sol: y = sin(x+9)/cos x
Using quotient rule:
dy/dx = [cos(x+9)cos x + sin(x+9)sin x]/cos²x
At x=0 ⇒ cos9·1 + sin9·0 = cos9.

Que-28: Find the derivative of f(x) = 1 + x + x² + x³ + … + x⁵⁰ at x = 1.

(a) 1275        (b) 1200        (c) 1326       (d) 1542

Answer: (a) 1275

Sol: f(x) = Σ xⁿ from n=0 to 50
f'(x) = Σ n·xⁿ⁻¹
At x=1 ⇒ f'(1) = 1 + 2 + 3 + … + 50
Sum = 50×51/2 = 1275.

–: End Differentiation Class 11 OP Malhotra Exe-19F ISC Maths Ch-19 Solutions :–

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