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JEE Advanced Maths Method of Differentiation Previous Year Questions with Solutions

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 + ex/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 + ex/2

f(0) = 0 + e0 = 1

f’(x) = 3x2 + ex/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) – f(2)]/3 ≥ 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) = logex. 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) = logex.

(gof)x = sin2x

F (x) = (hogof)x

= loge sin2x

= 2 loge 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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