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JEE Main 7-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 7-April-2025 Paper - Shift 24 Marks
MCQ
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a polynomial function of degree four having extreme values at $\mathrm{x}=4$ and $\mathrm{x}=5$.
If $\lim _{x \rightarrow 0} \frac{f(\mathrm{x})}{\mathrm{x}^{2}}=5$, then $f(2)$ is equal to :
  • A
    12
  • B
    10
  • C
    8
  • D
    14
✓

Answer

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

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