Question

Let $f\left(x\right)$ be a polynomial of degree four having extreme value at $x=1$ and $x=2$.If $\underset{x\to 0}{lim}(1+\frac{f\left(x\right)}{x^2})=3$,then $f\left(2\right)$ is equal to ?

• A. $0$
• B. $4$
• C. $-8$
• D. $\frac{-4}{9}$

Answer 1

The correct option is D $\frac{-4}{9}$

Given $f^{\prime }\left(1\right)=f^{\prime }\left(2\right)=0$

$\underset{x\to 0}{lim}(1+\frac{f\left(x\right)}{x^2})=3$

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^2}=2⟹f\left(0\right)=0(\because$ for applying L-hospital rule$)$

Apply L-Hospital Rule

$⟹\frac{f^{\prime \prime }\left(0\right)}{2}=2⟹f^{\prime \prime }\left(0\right)=4$

Let $f^{\prime }\left(x\right)=(x-1)(x-2)(x-a)$

$⟹f^{\prime \prime }\left(x\right)=(x-1)(x-2)+(x-2)(x-a)+(x-1)(x-a)$

$f^{\prime \prime }\left(0\right)=2+2a+a=4⟹a=\frac{2}{3}$

$⟹f^{\prime }\left(x\right)=(x-1)(x-2)(x-\frac{2}{3})=x^3-\frac{11}{3}{x}^2+4x-\frac{4}{3}$

$f\left(x\right)=\frac{x^4}{4}-\frac{11}{9}{x}^3+2x^2-\frac{4}{3}x+d$

$f\left(0\right)=0⟹d=0$

$f\left(2\right)=4-\frac{88}{9}+8-\frac{8}{3}=-\frac{4}{9}$