JEE Advanced 2023 — Paper 2 — MATH-I solution

JEE Advanced2023Paper 2MATH-I

Let $f : [1, \infty) \to R$ 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 $f (e)$ is

A
$\frac{e ^{2} + 4}{3}$
B
$\frac{lo g _{e} 4 + e}{3}$
C
$\frac{4 e ^{2}}{3}$
D
$\frac{e ^{2} - 4}{3}$

Solution

Differentiating both sides of $3 \int_{1}^{x} f (t) d t = x f (x) - \frac{x ^{3}}{3}$ with respect to $x$ gives $3 f (x) = f (x) + x f^{'} (x) - x^{2}$ , which rearranges to the linear ODE

$\frac{d y}{d x} - \frac{2}{x} y = x, y = f (x) .$

The integrating factor is $e^{- 2 l n x} = \frac{1}{x ^{2}}$ , so

$\frac{d}{d x} (\frac{y}{x ^{2}}) = \frac{1}{x} \Rightarrow \frac{y}{x ^{2}} = ln x + C .$

Hence $f (x) = x^{2} ln x + C x^{2}$ . Using $f (1) = \frac{1}{3}$ gives $C = \frac{1}{3}$ , so $f (x) = x^{2} ln x + \frac{x ^{2}}{3}$ .

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

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