JEE Advanced 2024 Paper 2 · Mathematics · Differential Equations · medium

Let f:(0,∞)→mathbbR be the differentiable function satisfying x f'(x)+2f(x)=x^2 for all x>0 and f(1)=1.

One or more options correct

Let

f:(0,\infty)\to\mathbb{R}

be the differentiable function satisfying

x f'(x)+2f(x)=x^2

for all

x>0

and

f(1)=1

. Which of the following is/are correct?

A. $f(2)=\dfrac{7}{4}$
B. $f$ attains its least value at $x=3^{1/4}$ , and this least value is $\dfrac{\sqrt3}{2}$
C. $\displaystyle\int_1^2 f(x)\,dx=\dfrac{25}{24}$
D. $\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{x^2}=\dfrac12$

Show answer and solution

Answer: B

Multiply the equation by

x

\dfrac{d}{dx}\left(x^2 f\right)=x^3

, so

x^2f=\dfrac{x^4}{4}+C

and

f(1)=1

gives

C=\dfrac34

. Hence

f(x)=\dfrac{x^2}{4}+\dfrac{3}{4x^2}

. Then

f(2)=1+\dfrac{3}{16}=\dfrac{19}{16}

, so A is false.

f'(x)=\dfrac x2-\dfrac{3}{2x^3}=0

gives

x^4=3

, and

f''>0

there, so the minimum is

f(3^{1/4})=\dfrac{\sqrt3}{4}+\dfrac{3}{4\sqrt3}=\dfrac{\sqrt3}{2}

; B is true.

\int_1^2 f\,dx=\dfrac{7}{12}+\dfrac38=\dfrac{23}{24}\ne\dfrac{25}{24}

, so C is false.

f(x)/x^2\to\dfrac14

, so D is false.

Original question written in the JEE Advanced 2024 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