Let f(x) and g(x) be twice differentiable functions satisfying f''(x) = g''(x) for all x ∈ R , f'(1) = 2g'(1) = 4 and g(2) = 3f(2) = 9 . Then f(25) - g(25) is equal to :

JEE Main 20265 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 5 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differentiation

JEE Main 2026 · 5 April · Evening Shift · Mathematics · Differentiation

Let f(x) and g(x) be twice differentiable functions satisfying...

Let

f(x)

and

g(x)

be twice differentiable functions satisfying

f''(x) = g''(x)

for all

x \in \mathbf{R}

f'(1) = 2g'(1) = 4

and

g(2) = 3f(2) = 9

. Then

f(25) - g(25)

is equal to :

Options

A $20$
B $40$
C $-20$
D $-40$

Correct answer

B. $40$

Step-by-step solution

Given

f''(x) = g''(x)

for all

x \in \mathbf{R}

.

Let

h(x) = f(x) - g(x)

.

Taking the second derivative,

h''(x) = f''(x) - g''(x) = 0

.

Integrating with respect to

x

h'(x) = c_1

.

Given

f'(1) = 4

and

g'(1) = 2

h'(1) = f'(1) - g'(1) = 4 - 2 = 2

.

Therefore,

c_1 = 2

, which gives

h'(x) = 2

.

Integrating again with respect to

x

h(x) = 2x + c_2

.

Given

3f(2) = 9 \Rightarrow f(2) = 3

and

g(2) = 9

h(2) = f(2) - g(2) = 3 - 9 = -6

.

Substituting

x = 2

h(x)

h(2) = 2(2) + c_2 = -6 \Rightarrow c_2 = -10

.

Thus,

h(x) = 2x - 10

.

Substituting

x = 25

h(25) = f(25) - g(25) = 2(25) - 10 = 40

.

Answer:

40

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