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. $-4$
• B. $0$
• C. $4$
• D. $-8$

Answer 1

Answer: (B)

$0$

Consider the given limit.

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

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

Since, the limit exists, therefore,

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

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

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

Also,

$f\left(x\right)=0$ at $x=1,2$

Therefore,

$a=\frac{1}{2},b=-2$

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

Therefore,

$f\left(2\right)=8-16+8=0$

Hence, $f\left(2\right)=0$.