QJEEAdv23Ma249f: JEE Advanced 2023 Mathematics - Class 12 JEEAdv23Ma249f Solution: Let be a differentiable function such that and . Let denote the …

1. We are given a relationship between the integral of the function $f(t)$ and the function $f(x)$ :

$3 \int\limits_1^x f(t) d t=x f(x)-\frac{x^3}{3}, x \in[1, \infty)$

2. Differentiate both sides of the above equation with respect to $x$ :

$3(f(x) \cdot 1-0)=1 . f(x)+x f^{\prime}(x)-\frac{3 x^2}{3}$

$\Rightarrow$ $3f(x) = f(x) + xf'(x) - x^2$

$\Rightarrow$ $2f(x) = xf'(x) - x^2$

3. We can rewrite this as a first order linear differential equation:

$xf'(x) - 2f(x) = x^2$

4. The general form of a first order linear differential equation is:

$y'(x) + p(x)y = q(x)$

Here $y = f(x)$ , $p(x) = -\frac{2}{x}$ , and $q(x) = x$ .

5. The integrating factor is obtained by exponentiating the integral of $p(x)$ with respect to $x$ :

$e^{\int p(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = \frac{1}{x^2}$

6. $\therefore$ Solution of D.E :

$\begin{gathered}y \cdot\left(\frac{1}{x^2}\right)=\int(x) \frac{1}{x^2} d x \\\\\frac{y}{x^2}=\ln x+c \\\\y=x^2 ln x+c x^2\end{gathered}$

$\therefore$ The general solution is :

$y(x) = x^2(\ln x + c)$

Here, $y(x)$ is the same as $f(x)$ , i.e. the function we're trying to find.

7. Given that $f(1) = \frac{1}{3}$ , let's use this condition to find the constant $c$ :

$\frac{1}{3} = 1^2(\ln 1 + c) \Rightarrow c = \frac{1}{3}$

Therefore, the function $f(x)$ we are looking for is:

$f(x) = x^2(\ln x + \frac{1}{3})$

We want to find $f(e)$ , so substituting $x=e$ into the function, we get :

$f(e) = e^2(\ln e + \frac{1}{3}) = e^2(1 + \frac{1}{3}) = \frac{4e^2}{3}$

So, the correct answer is

Option C : $\frac{4e^2}{3}$