MCQ
Let $f:\left[(1, \infty) \rightarrow \mathbb{R}\right.$ be a differentiable function such that $f(1)=\frac{1}{3}$ and $3 \int_1^x f(t) d t=x f(x)-\frac{x^3}{3}, x \in[1, \infty)$.

Let $e$ denote the base of the natural logarithm. Then the value of $\mathrm{f}(e)$ is

  • A
    $\frac{e^2+4}{3}$
  • B
    $\frac{\log _e 4+e}{3}$
  • $\frac{4 e^2}{3}$
  • D
    $\frac{e^2-4}{3}$

Answer

Correct option: C.
$\frac{4 e^2}{3}$
c
$\text { Diff. wr.t } x^{\prime}$

$3 f(x)=f(x)+x f(x)-x^2$

$\frac{d y}{d x}-\left(\frac{2}{x}\right) y=x$

$I F=e^{-2(m x}=\frac{1}{x^2}$

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

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

$\therefore y(1)=\frac{1}{3} \Rightarrow c=\frac{1}{3}$

$y(e)=\frac{4 e^2}{3}$

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