JEE Advanced 2023 Paper 2 Q01 Mathematics Integration & Differential Equations Differential Equations Hard

JEE Advanced 2023 Paper 2 · Q01 · Differential Equations

Let $f: [1, \infty) \to \mathbb{R}$ be a differentiable function such that $f(1) = \dfrac{1}{3}$ and $3\displaystyle\int_1^x f(t)\,dt = xf(x) - \dfrac{x^3}{3}$ for all $x \in [1, \infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is

  1. A. $\dfrac{e^2 + 4}{3}$
  2. B. $\dfrac{\log_e 4 + e}{3}$
  3. C. $\dfrac{4e^2}{3}$
  4. D. $\dfrac{e^2 - 4}{3}$
Reveal answer + step-by-step solution

Correct answer:C

Solution

Differentiate both sides w.r.t. $x$: $3f(x) = f(x) + xf'(x) - x^2$, so $xf'(x) - 2f(x) = x^2$, i.e. $f'(x) - (2/x)f(x) = x$. Integrating factor $= e^{\int -2/x\,dx} = 1/x^2$. Then $d(f/x^2)/dx = 1/x$, so $f(x)/x^2 = \ln x + C$. With $f(1) = 1/3$: $C = 1/3$. Hence $f(x) = x^2 \ln x + x^2/3$ and $f(e) = e^2 + e^2/3 = 4e^2/3$.

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