JEE Main 2026 — 2 April, Evening Shift — Question 40

Given $\mathrm{f}(\mathrm{x})$ is a polynomial of degree 5 Also, $f^{\prime}(1)=0 ; f^{\prime}(-1)=0$. Also $\because \lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=-5$ (fixed and finite) $\therefore \mathrm{f}(0)=0 ; \mathrm{f}^{\prime}(0)=0 ; \mathrm{f}^{\prime \prime}(0)=0$ $\frac{f^{\prime \prime \prime}(0)}{6}=-5 \Rightarrow f^{\prime \prime \prime}(0)=-30$ Hence, let $\mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{ax}+\mathrm{b})(\mathrm{x}-0)(\mathrm{x}-1)(\mathrm{x}+1)$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{ax}^{4}+\mathrm{bx}^{3}-\mathrm{ax}^{2}-\mathrm{bx}$ $\mathrm{f}^{\prime \prime}(\mathrm{x})=4 \mathrm{ax}^{3}+3 \mathrm{bx}^{2}-2 \mathrm{ax}-\mathrm{b}$ $\because \mathrm{f}^{\prime}(0)=0 \Rightarrow \mathrm{~b}=0$ $f^{\prime \prime \prime}(x)=12 a x^{2}+6 b x-2 a$ $\because f^{\prime \prime \prime}(0)=-2 a=-30$ $\Rightarrow \mathrm{a}=15$ $\therefore \mathrm{f}^{\prime}(\mathrm{x})=15 \mathrm{x} . \mathrm{x} .(\mathrm{x}-1)(\mathrm{x}+1)$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=15 \mathrm{x}^{4}-15 \mathrm{x}^{2}$ $\Rightarrow \mathrm{f}(\mathrm{x})=3 \mathrm{x}^{5}-5 \mathrm{x}^{3}+\mathrm{C}$ $\because \mathrm{f}(0)=0 \Rightarrow \mathrm{C}=0$ $\therefore \mathrm{f}(\mathrm{x})=3 \mathrm{x}^{5}-5 \mathrm{x}^{3}$ $\therefore \mathrm{f}(2)-\mathrm{f}(-2)=112$