Q26 / 75 · JEE Main 2025 (7 Apr Shift 2) · Mathematics

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a polynomial function of degree four having extreme values at $x=4$ and $x=5$.
If $\lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5$, then $f(2)$ is equal to :

  • A
    12
  • Correct answer
    B
    10
  • C
    8
  • D
    14

Solution

$\begin{aligned}<br/>& \lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5 \\ & \lim _{x \rightarrow 0} \frac{\left.a x^4+b x^3+c x^2+d x+e\right)}{x^2}=5 \\ & c=5 \text { and } d=e=0 \\ & f(x)=a x^4+b x^3+5 x^2 \\ & f^{\prime}(x)=4 a x^3+3 b x^2+10 x \\ & =x\left(4 a x^2+3 b x+10\right)<br/>\end{aligned}$
has extremes at 4 and so $f^{\prime}(4)=0 \& f^{\prime}(5)=0$
so $\mathrm{a}=\frac{1}{8} \& \mathrm{~b}=\frac{-3}{2}$
$\text { so } f(2)=\frac{1}{8} \times 2^4-\frac{3}{2} \times 2^3+5 \times 2^2$
$=2-12+20=10 $

Hence, the correct answer is option: 'B'.

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