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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Application of Derivatives

Let f: R arrow R be a polynomial function of degree four having...

Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R} be a polynomial function of degree four having extreme values at x=4x=4 and x=5x=5.
If lim⁡x→0f(x)x2=5\lim _{x \rightarrow 0} \frac{f(x)}{x^2}=5, then f(2)f(2) is equal to :

Options

  1. A12
  2. B10
  3. C8
  4. D14

Correct answer

B. 10

Step-by-step solution

lim⁡x→0f(x)x2=5lim⁡x→0ax4+bx3+cx2+dx+e)x2=5c=5 and  d=e=0f(x)=ax4+bx3+5x2f′(x)=4ax3+3bx2+10x=x(4ax2+3bx+10)\begin{aligned}& \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{\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)\end{aligned}has extremes at 4 and so f′(4)=0&f′(5)=0f^{\prime}(4)=0 \& f^{\prime}(5)=0so a=18&b=−32 so f(2)=18×24−32×23+5×22=2−12+20=10\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
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