Very Short Answer on Differentiation Class 11 OP Malhotra Exe-19E ISC Maths Solutions Ch-19. In this article you would learn to solve all type 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 Very Short Answer ISC Maths Solutions Ch-19

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-19	Differentiation
Writer	OP Malhotra
Exe-19(E)	Very Short Answer Type Questions.

Very Short Answer on Differentiation

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Limit x→-1 of (x−2)(x+5) / (x²−5x+6)

Sol: Factor denominator: (x−2)(x−3).
Cancel (x−2).
Limit = (x+5)/(x−3) at x = -1
⇒ (-1+5)/(-1−3)
= 4/(-4) = -1

Que-2: Limit x→5 of f(x), where f(x) = |x| − 5

Sol: For x→5, |x| = x.
So f(x) = x−5
⇒ limit = 5−5 = 0

Que-3: Limit x→0 of [(x+1)⁵ − 1] / x

Sol: Use expansion: (1+x)⁵ = 1 + 5x + …
So numerator ≈ 5x
Limit = 5x/x = 5

Que-4: Limit x→0 of sin(px)/x

Sol: Standard result: sin(ax)/x = a
So answer = p

Que-5: Limit x→1 of (x^(1/3) − 1)/(x^(1/6) − 1)

Sol: Put x = t⁶
⇒ numerator = t² −1,
denominator = t−1
Limit = (t²−1)/(t−1)
= t+1 → at t=1
⇒ 2

Que-6: Limit x→π of tan x / (x − π)

Sol: tan x ≈ (x−π) near π
So limit = 1

Que-7: Limit x→3 of x / [x]

Sol: [x] = greatest integer
As x→3⁻ ⇒ [x]=2 ⇒ value = 3/2
As x→3⁺ ⇒ [x]=3 ⇒ value = 3/3 =1
Left ≠ Right ⇒ limit does not exist

Que-8: d/dx (4x³ − 7x + 20)

Sol: Differentiate termwise
= 12x² − 7

Que-9: d/dx [(x−1)(x+2)]

Sol: Expand: x² + x −2
Derivative = 2x + 1

Que-10: d/dx (x + 1/x)

Sol: = 1 − 1/x²

Que-11: d/dx [(x−2)/(x−3)], x ≠ 3

Sol: Quotient rule
= [(x−3) − (x−2)]/(x−3)²
= −1/(x−3)²

Que-12: d/dx (x cos x)

Sol: Product rule
= cos x − x sin x

Que-13: d/dx (sin x cos x)

Sol: = cos²x − sin²x
= cos 2x

Que-14: Derivative of sin(3x+5) − 7cos x + 12

Sol: = 3cos(3x+5) + 7sin x

Que-15: Derivative of 5 tan(1/5 x + 7)

Sol: = 5 sec²(1/5 x +7) × (1/5)
= sec²(1/5 x +7)

Evaluate the following limits

Que-16: Limit x→1/2 of (4x² −1)/(2x−1)

Sol: Factor: (2x−1)(2x+1)/(2x−1)
Cancel ⇒ 2x+1 = 2(1/2)+1 = 2

Que-17: Limit x→0 of (x sec x)

Sol: sec x →1, so x·1 → 0

Que-18: Limit x→0 of cos x / (π − x)

Sol: cos 0 =1 ⇒ 1/π

Que-19: Limit x→π of sin x /(π − x)

Sol: sin x ≈ (x−π)(−1)
So limit = 1

Que-20: Limit x→0 of sin(ax)/(bx)

Sol: Multiply and divide by x:
(sin(ax)/ax) × (a/b)
We know standard limit:
lim x→0 (sin x / x) = 1
So,
= 1 × (a/b) = a/b

Que-21: Limit x→∞ of (2x² + 7x + 5)/(4x² + 3x −1)

Sol: Divide numerator and denominator by x²:
= (2 + 7/x + 5/x²) / (4 + 3/x − 1/x²)
As x → ∞, all terms with x in denominator → 0
= 2/4 = 1/2

Que-22: Limit h→0 of (√(x+h) − √x)/h

Sol: Rationalize numerator:
= [(√(x+h) − √x)(√(x+h) + √x)] / [h(√(x+h) + √x)]
= [(x+h − x)] / [h(√(x+h) + √x)]
= h / [h(√(x+h) + √x)]
Cancel h:
= 1 / (√(x+h) + √x)
Now h → 0:
= 1 / (2√x)

Que-23: Limit x→0 of (3²ˣ −1)/(2³ˣ −1)

Sol: Use formula: lim (a^x − 1)/x = ln a
Let numerator = 3^2x = (3²)^x = 9^x
So,
(3^2x − 1)/(2^3x − 1)
= (9^x − 1)/(8^x − 1)
= (ln 9)/(ln 8) = (2 ln 3)/(3 ln 2)

Que-24: Limit x→0 of [(x+2)^(1/3) − 2^(1/3)]/x

Sol: This is of form f(x) − f(a) / (x − a), so use derivative at x = 0
Let f(x) = (x+2)^1/3
f'(x) = (1/3)(x+2)^-2/3
At x = 0:
f'(0) = 1 / [3 · 2^2/3]

Differentiate the following w.r.t. x,

Que-25: Differentiate (√x + 1/√x)²

Sol: First expand:
(√x + 1/√x)² = x + 2 + 1/x
Now differentiate:
d/dx (x + 2 + 1/x) = 1 − 1/x²

Que-26: Differentiate x⁻⁴(3 − 4x⁻⁵)

Sol: Using Product Rule: (uv)’ = u’v + uv’
u = x^-4,   v = (3 − 4x^-5)
u’ = −4x^-5
v’ = 20x^-6
∴ dy/dx = −4x^-5(3 − 4x^-5) + x^-4(20x^-6)
= −12x^-5 + 16x^-10 + 20x^-10
= −12x^-5 + 36x^-10

Que-27: Differentiate sin⁴x

Sol: Let y = (sin x)⁴
Using Chain Rule:
dy/dx = 4(sin x)³ × cos x
= 4 sin³x cos x

Que-28: Differentiate [(3x+4)/(4x+3)] (x − 3/4)

Sol: Using Product Rule:
Let u = (3x+4)/(4x+3),   v = (x − 3/4)
u’ = [ (4x+3)(3) − (3x+4)(4) ] / (4x+3)²
= (12x+9 − 12x−16)/(4x+3)²
= −7/(4x+3)²
v’ = 1
∴ dy/dx = u’v + uv’
= [−7/(4x+3)²](x − 3/4) + (3x+4)/(4x+3)

Que-29: √(x² + 1)

Sol: Differentiate using chain rule. Let y = √(x² + 1) = (x² + 1)^1/2.
dy/dx = (1/2)(x² + 1)^-1/2 · 2x = x / √(x² + 1).

Que-30: (3x + 5)(1 + tan x)

Sol: Use product rule: (uv)’ = u’v + uv’.
u = 3x + 5 ⇒ u’ = 3
v = 1 + tan x ⇒ v’ = sec²x
dy/dx = 3(1 + tan x) + (3x + 5)sec²x.

Que-31: x² sin x + cos 2x

Sol: Differentiate term-wise.
d/dx (x² sin x) = x² cos x + 2x sin x
d/dx (cos 2x) = -2 sin 2x
Final answer: x² cos x + 2x sin x – 2 sin 2x.

Que-32: 1 / (ax² + bx + c)

Sol: Use derivative of 1/f(x):
d/dx [1/f(x)] = -f'(x)/[f(x)]²
Here f(x) = ax² + bx + c
⇒ f'(x) = 2ax + b
So derivative = -(2ax + b)/(ax² + bx + c)².

Que-33: If f(x) = x|x|, then f'(x)

Sol: Use definition of modulus:
If x ≥ 0, f(x) = x² ⇒ f'(x) = 2x
If x < 0, f(x) = -x² ⇒ f'(x) = -2x
Thus, f'(x) = 2|x|.

Que-34: If f(x) = 2|x| + 3 sin x + 6, find right hand derivative at x = 0

Sol: For right-hand derivative, take x > 0 ⇒ |x| = x
So f(x) = 2x + 3 sin x + 6
Differentiate: f'(x) = 2 + 3 cos x
At x = 0: f'(0) = 2 + 3(1) = 5.

Que-35: If f(x) = sin 2x − cos 2x, find f'(π/6)

Sol: Differentiate:
d/dx (sin 2x) = 2 cos 2x
d/dx (cos 2x) = -2 sin 2x
So f'(x) = 2 cos 2x + 2 sin 2x
At x = π/6:
2x = π/3
f'(π/6) = 2 cos(π/3) + 2 sin(π/3)
= 2(1/2) + 2(√3/2)
= 1 + √3.

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

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
Thanks
Please share with your friends