Question

Let $f\left(x\right)$ be a polynomial of degree four having extreme values 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. $-8$
• B. $-4$
• C. $0$
• D. $4$

Answer 1

The correct option is C $0$

Given that:

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

To find:

$f\left(2\right)=?$

Solution:

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

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

since limit exits hence, $x^2+f\left(x\right)=a{x}^4+b{x}^3+3x^2$

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

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

Also $f^{\prime }\left(x\right)=0$ at $x=1,2$

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

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

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

Hence, C is the correct option.