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 ≥ 1f(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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