JEE Main 2025 — 7 April, Evening Shift — Question 40

$f(x)=a x^{4}+b x^{3}+c x^{2}+d x+e$

$$f^{\prime}(4)=0, f^{\prime}(5)=0$$

Also $\lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=5$

$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{a x^{4}+b x^{3}+c x^{2}+d x+e}{x^{2}}\right)=5$

$\Rightarrow d=e=0$ and $c=5$

$\therefore f(x)=a x^{4}+b x^{3}+5 x^{2}$

$f^{\prime}(x)=4 a x^{3}+3 b x^{2}+10 x$

$$\begin{gathered} f^{\prime}(4)=256 a+48 b+40=0 …(i) \end{gathered}$$ $$\begin{gathered} f^{\prime}(5)=500 a+75 b+50=0 …(ii) \end{gathered}$$

Solving equation (i) and (ii)

We get $a=\frac{1}{8}, b=\frac{-3}{2}$

$\therefore f(x)=\frac{1}{8} x^{4}-\frac{3}{2} x^{3}+5 x^{2}$

$f(2)=\frac{16}{8}-\frac{3}{2}(2)^{3}+5(2)^{2}$

$f(2)=2-12+20$

$f(2)=10$