The JEE Advanced Maths method of differentiation previous year questions and solutions are given on this page. These are solved by our top JEE subject experts. Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables. These solutions are given in a detailed manner so that students can easily understand the solutions.

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Question 1: The slope of the tangent to the curve (y-x⁵)² = x(1+x²)² at the point (1, 3) is

(a) 2

(b) 6

(c) 3

(d) 8

Solution:

Given (y – x⁵)² = x(1+x²)²

Differentiate with respect to x.

2(y – x⁵)(dy/dx – 5x⁴) = (1+x²)² + 2x(1+x²)×2x

At (1, 3)

2(3 – 1)(dy/dx – 5) = (1+1)² + 2(1+1)×2

4(dy/dx – 5) = 4 + 8

(dy/dx – 5) = 12/4 = 3

dy/dx = 3 + 5

= 8

The slope of the tangent is 8.

Hence option d is the answer.

Question 2: If the function f(x) = x³ + e^x/2 and g(x) = f^-1(x), then the value of g’(1) is

(a) 1

(b) -1

(c) 2

(d) 4

Solution:

Given f(x) = x³ + e^x/2

f(0) = 0 + e⁰ = 1

f’(x) = 3x² + e^x/2×1/2

Given g(x) = f^-1(x)

=> g(f(x)) = x

Differentiate w.r.t.x

g’(f(x)). f’(x) = 1

f’(0) = 0 + ½ e⁰

= 1/2

g’(f(0)). f’(0) = 1

g’(1).½ = 1

=> g’(1) = 2

Hence option c is the answer.

Question 3: If x = (1-√y)/(1+√y), then dy/dx is equal to

(a) 4/(x+1)²

(b) 4(x-1)/(x+1)³

(c) (x-1)/(x+1)³

(d) 4/(x+1)³

Solution:

x = (1-√y)/(1+√y)

Cross multiply

x(1+√y) = (1-√y)

x+x√y = 1-√y

x√y +√y = 1-x

√y (1+x) = 1-x

√y = (1-x)/(1+x)

y = (1-x)²/(1+x)²

Differentiate with respect to x

dy/dx = [(1+x)²2(1-x)(-1) – (1-x)²(1+x)2]/(1+x)⁴

= [(1+x)²2(x-1) – (x-1)²(1+x)2]/(1+x)⁴

= [2(1+x)(x-1){(1+x)-(x-1)}]/(1+x)⁴

= [2(1+x)(x-1)(1+1)]/(1+x)⁴

= 4(x-1)/(1+x)³

Hence option (b) is the answer.

Question 4: The derivative of f(tan x) with respect to g(sec x) at x = π/4, where f ‘(1) = 2 and g'(√2) = 4, is

(a) 1/√2

(b) √2

(c) 1

(d) None of these

Solution:

Let u = f(tan x), v = g(sec x)

du/dx = f’(tan x) sec² x

dv/dx = g’(sec x) sec x tan x

du/dv = f’(tan x) sec² x/ g’(sec x) sec x tan x

= f’(tan x) sec x/ g’(sec x) tan x

Put x = π/4,

du/dv = f’(1)√2 / g’(√2) ×1

Given f ‘(1) = 2 and g'(√2) = 4

So du/dv = 2√2 /4 ×1

= 1/√2

Hence option (a) is the answer.

Question 5: Let f be a twice differentiable function on (1, 6). If f(2) = 8, f’(2) = 5, f’(x) ≥ 1 and f’’(x) ≥ 4, for all x belongs to (1, 6), then

(a) f(5) + f’(5) ≥ 28

(b) f’(5) + f’’(5) ≤ 20

(c) f(5) ≤ 10

(d) f(5) + f’(5) ≤ 26

Solution:

Let f be a twice differentiable function.

Given f(2) = 8 and f’(2) = 5

Since f’(x) ≥1

f(5) ≥ 3 + f(2)

f(5) ≥ 3 + 8

=> f(5) ≥ 11

Given f’’(x) ≥ 4

(f’(5) – f’(2))/(5-2) ≥ 4

=> f’(5) ≥ 12 + f’(2)

=> f’(5) ≥ 12 + 5

=> f’(5) ≥ 17

So f(5) + f’(5) ≥ 28

Hence option (a) is the answer.

Question 6: Let f be a twice differentiable function such that g’(x) = -f(x) and f’(x) = g(x), h(x) = {f(x)}² + {g(x)}². If h(5) = 11, then h(10) is equal to

(a) 22

(b) 11

(c) 0

(d) 20

Solution:

Given g’(x) = -f(x) and f’(x) = g(x)

h(x) = {f(x)}² + {g(x)}²

Differentiate w.r.t x

h’(x) = 2f(x) f’(x) + 2g(x)g’(x)

= 2f(x) g(x) + 2g(x)(-f(x))

= 0

So h(x) is a constant.

h(5) = 11

So h(10) = 11

Hence option (b) is the answer.

Question 7: Let f and g be differentiable functions on R such that fog is the identity function. If for some a, b ∈ R, g'(a) = 5 and g(a) = b, then f'(b) is equal to :

(a) 2/5

(b) 1

(c) 1/5

(d) 5

Solution:

Given f and g be differentiable functions and fog is identity function.

=> (fog)(x) = x

=> f(g(x)) = x

Differentiate w.r.t.x

f’(g(x)). g’(x) = 1

Put x = a

f’(g(a)). g’(a) = 1

f’(b). 5 = 1

f’(b) = 1/5

Hence option (c) is the answer.

Question 8: The derivative of sin^-1 (2x√(1-x²) with respect to sin^-1 (3x- 4x³)

(a) 2/3

(b) 3/2

(c) 1/2

(d) 1

Solution:

Let u = sin^-1 (2x√(1-x²)

Put x = sin θ

Then u = sin^-1 (2 sin θ√(1- sin² θ)

= sin^-1 (2 sin θ√cos² θ

= sin^-1 2 sin θ cos θ

= sin^-1 sin 2 θ

= 2 θ

= 2 sin^-1x

du/dx = 2/√(1-x²)

Let v = sin^-1 (3x – 4x³)

Put x = sin θ

v = sin^-1 (3sin θ – 4sin³ θ)

= sin^-1 sin 3θ

= 3θ

= 3 sin^-1 x

dv/dt = 3/√(1-x²)

du/dv = [ 2/√(1-x²) ]/[ 3/√(1-x²) ]

= 2/3

Hence option (a) is the answer.

Question 9: The second-order derivative of a sin³ t with respect to a cos³ t at t = π/4 is

(a) 2

(b) 1/12a

(c) 4√2/3a

(d) 3a/4√2

Solution:

Let u = a sin³ t

du/dt = 3 a sin²t cos t

Let v = a cos³t

dv/dt = -3a cos²t sin t

du/dv = 3a sin²t cos t/-3a cos²t sin t

= – tan t

d²u/dv² = (d/dt)(du/dv)(dt/dv)

= -sec²t (dt/dv)

= -sec²t /-3a cos²t sin t

= 1/3a cos⁴t sin t

Put t = π/4

d²u/dv² = 1/3a cos⁴(π/4) sin (π/4)

= 1/3a (1/√2)⁴(1/√2)

= 4√2/3a

Hence option (c) is the answer.

Question 10: If f (x) = sin x, g (x) = x² and h (x) = log_ex. If F (x) = (hogof) (x), then F ” (x) is equal to

(a) a cosec³ x

(b) 2 cot x² – 4x² cosec² x²

(c) 2 cot x²

(d) – 2 cosec² x

Solution:

f (x) = sin x

g (x) = x²

h (x) = log_ex.

(gof)x = sin²x

F (x) = (hogof)x

= log_e sin²x

= 2 log_e sin x

F’(x) = (2/sinx)cos x

= 2 cot x

F’’(x) = -2 cosec²x

Hence option (d) is the answer.

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