Let $\mathrm{f}:[1, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $\mathrm{f}(1)=\frac{1}{3}$ and $3 \int_{1}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{xf}(\mathrm{x})-\frac{\mathrm{x}^{3}}{3}, \mathrm{x} \in[1, \infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $\mathrm{f}(e)$ is

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Differential Equations

Let

f : [1, \infty) \to R

be a differentiable function such that

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

and

3 \int_{1}^{x} f (t) dt = xf (x) - \frac{x ^{3}}{3}, x \in [1, \infty)

. Let

e

denote the base of the natural logarithm. Then the value of

f (e)

Answer: (C)

Differentiating

3 \int_{1}^{x} f d t = x f (x) - \frac{x ^{3}}{3}

gives

3 f = f + x f^{'} - x^{2}

, i.e.

x f^{'} - 2 f = x^{2}

, or

f^{'} - \frac{2}{x} f = x

. With integrating factor

x^{- 2}

this becomes

(x^{- 2} f)^{'} = \frac{1}{x}

, so

x^{- 2} f = ln x + C

and

f (x) = x^{2} ln x + C x^{2}

. The condition

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

gives

C = \frac{1}{3}

, hence

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

and

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

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Single correct option