JEE Advanced 2025 Paper 2 · Mathematics · Sets, Relations and Functions · medium

A function f:mathbb Rsetminus\0\→mathbb R satisfies f(x)+2f (1/x)=3x for all x≠0. Then ∫1^2 f(x) dx equals

Single correct option

A function

f:\mathbb R\setminus\{0\}\to\mathbb R

satisfies

f(x)+2f\!\left(\dfrac1x\right)=3x

for all

x\neq0

. Then

\displaystyle\int_1^2 f(x)\,dx

equals

Show answer and solution

Answer: (A) $2\ln2-\dfrac32$

Replace

x

1/x

f\!\left(\tfrac1x\right)+2f(x)=\dfrac3x

. We now have two linear equations in the unknowns

f(x)

and

f(1/x)

f(x)+2f(1/x)=3x

and

2f(x)+f(1/x)=3/x

. Multiply the second by

2

and subtract the first:

3f(x)=\dfrac6x-3x

, so

f(x)=\dfrac2x-x

. Check:

f(1/x)=2x-\tfrac1x

, and

f(x)+2f(1/x)=\tfrac2x-x+4x-\tfrac2x=3x

. Therefore

\displaystyle\int_1^2\left(\frac2x-x\right)dx=2\ln2-\left[\frac{x^2}{2}\right]_1^2=2\ln2-\dfrac32\approx-0.114

. The option

2\ln2+\tfrac32

comes from a sign error, and

\ln 2-\tfrac32

from forgetting the factor

2

2/x

Original question written in the JEE Advanced 2025 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake