JEE Advanced 2023 Paper 2Question 1
Let $f:[1,\infty)\to\mathbb{R}$ be a differentiable function such that $f(1)=\frac{1}{3}$ and $3\int_1^x f(t)\,dt = x f(x)-\frac{x^3}{3},\ x\in[1,\infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is
Differential Equation
Solution
Concept. Differentiate: $xf'-2f=x^2$, i.e. $\left(\frac f{x^2}\right)'=\frac1x$. Formulas used. (1) $\frac{d}{dx}\int_1^xf(t)\,dt=f(x)$ and $\left(\dfrac f{x^2}\right)'=\dfrac{xf'-2f}{x^3}$Step 1 - Differentiate. $3f=f+xf'-x^2$, so $xf'-2f=x^2$. Step 2 - Integrate. By (1), $\frac{f}{x^2}=\ln x+C$. Step 3 - Constant. $f(1)=\frac13$ gives $C=\frac13$. Step 4 - At $x=e$. $f(e)=e^2\left(1+\frac13\right)=\dfrac{4e^2}{3}$. Answer: (C)