QJEEM23Ma98b1: JEE Main 2023 Mathematics - Class 12 JEEM23Ma98b1 Solution: Let be a differentiable function such that , . Then is equal to :

Question

Verified Solution • 2025–26 Session moderate

Let $f$ be a differentiable function such that ${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt}$ , $f(1) = {2 \over 3}$ . Then $18f(3)$ is equal to :

Accepted Answer

Given that

x^2 f(x)-x=4 \int_0^x t f(t) d t

On differentiating both sides with respect to

x

, we get

\begin{array}{rlrl} &amp; 2 x f(x)+x^2 f^{\prime}(x)-1 =4 x f(x) \\\\&amp;\Rightarrow x^2 f^{\prime}(x)-2 x f(x)-1 =0 \\\\&amp;\Rightarrow x^2 \frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\&amp;\frac{d y}{d x}-\frac{2}{x} y =\frac{1}{x^2}\end{array}

On comparing above equation with

\frac{d y}{d x}+P y=Q

, where

P=-\frac{2}{x}, Q=\frac{1}{x^2}

Now, IF

=e^{\int(-2 / x) d x}=e^{-2 \log x}=1 / x^2

Solution is

y\left(\frac{1}{x^2}\right)=\int\left(\frac{1}{x^2}\right) \frac{1}{x^2} d x

\begin{array}{ll}&amp;\Rightarrow \frac{y}{x^2}=\int \frac{1}{x^4} d x \Rightarrow \frac{y}{x^2}=-\frac{1}{3 x^3}+C \\\\&amp;\Rightarrow y =-\frac{1}{3 x}+C x^2\end{array}

Given,

f(1)=\frac{2}{3}

.

So,

\frac{2}{3}=-\frac{1}{3}+C \Rightarrow C=1

Thus,

y=f(x)=-\frac{1}{3 x}+x^2

\therefore 18 f(3)=18\left\{9-\frac{1}{9}\right\}=18 \times \frac{80}{9}=160

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