JEE Advanced 2021 Paper 2 Mathematics: Paragraph 1

Treat the equation as linear: divide by $x^3$ to get $\left(\dfrac{f}{x^2}\right)'=\dfrac{xf'-2f}{x^3}=e^{x}$. Integrating, $f(x)/x^2=e^x+c$, and $f(1)=0$ gives $c=-e$. Hence $f(x)=x^2(e^x-e)$. At $x=2$, $f(2)=4(e^2-e)=4(7.38906-2.71828)=4(4.67077)=18.683$. Therefore $f(2)\approx18.68$.

From $f(x)=x^2(e^x-e)$ we get $f'(x)=x\left[(x+2)e^x-2e\right]$. For $x>0$ the bracket is strictly increasing (its derivative is $(x+3)e^x>0$) and is negative at $x\to0^+$ (value $2-2e<0$) and positive for large $x$, so $f'$ changes sign from negative to positive exactly once at $x_0$ where $(x_0+2)e^{x_0}=2e$. Solving numerically, $x_0\approx0.6999$. Thus $m=f(x_0)=x_0^2(e^{x_0}-e)\approx0.4899\times(2.0136-2.7183)\approx-0.3452$. This is the global minimum on $(0,\infty)$ since $f\to\infty$ as $x\to\infty$. Hence $m\approx-0.35$.