JEE Advanced 2023 · Paper 2 · Mathematics Q1

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MathematicsSingle correct+3 / -1Question 1 of 51

Let $f : [1,\infty) \to \mathbb{R}$ be a differentiable function such that $f(1) = \dfrac13$ and $3\displaystyle\int_1^x f(t)\,dt = x\,f(x) - \dfrac{x^3}{3}$ , $x \in [1,\infty)$ . Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is

\dfrac{e^2+4}{3}

\dfrac{\log_e 4 + e}{3}

\dfrac{4e^2}{3}

\dfrac{e^2-4}{3}

Answer and worked solution

Official answer: C

Differentiate the given equation:
$3f(x)=f(x)+xf'(x)-x^2\implies f'(x)-\frac2xf(x)=x.$

Integrating factor $x^{-2}$ :
$\left(\frac{f}{x^2}\right)'=\frac1x\implies\frac{f(x)}{x^2}=\ln x+C.$

Constant. $f(1)=\frac13$ gives $C=\frac13$ , so $f(x)=x^2\left(\ln x+\frac13\right)$ .

$f(e)=e^2\cdot\frac43=\frac{4e^2}{3}.$

Topics:Linear first-order equations Fundamental theorem of calculus

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JEE Advanced 2023 · Paper 2, Mathematics question 1