Question

Let $f\left(x\right)$ be a polynomial of degree $4$ having extreme values at $x=1$ and $x=2$. If $\underset{x\to 0}{lim}(\frac{f\left(x\right)}{x^2}+1)=3$ then $f(-1)$ is equal to :

• A. $\frac{5}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $\frac{9}{2}$

Answer 1

The correct option is D $\frac{9}{2}$

Let $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Given, $\underset{x\to 0}{lim}(\frac{f\left(x\right)}{x^2}+1)=3$
$\Rightarrow \underset{x\to 0}{lim}\left(\frac{f\left(x\right)}{x^2}\right)=2$
$\Rightarrow \underset{x\to 0}{lim}\left(\frac{a{x}^4+b{x}^3+c{x}^2+dx+e}{x^2}\right)=2$
$\Rightarrow \underset{x\to 0}{lim}(a{x}^2+bx+c+\frac{d}{x}+\frac{e}{x^2})=2$
The value of limit can be finite if $d=0,e=0$
Therefore, $\underset{x\to 0}{lim}(a{x}^2+bx+c)=2$
$\Rightarrow c=2$
So, $f\left(x\right)=a{x}^4+b{x}^3+2x^2\because d=0,e=0,c=2$
$\Rightarrow f^{\prime }\left(x\right)=4a{x}^3+3b{x}^2+4x$
As $f\left(x\right)$ has extreme values at $x=1$ and $x=2$ so, $f^{\prime }\left(1\right)=0,f^{\prime }\left(2\right)=0$
$\Rightarrow f^{\prime }\left(1\right)=4a+3b+4=\mathrm{0...}\left(1\right)$
$\Rightarrow f^{\prime }\left(2\right)=8a+3b+2=\mathrm{0...}\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$
$a=\frac{1}{2}$ and $b=-2$
Therefore, $f\left(x\right)=\frac{1}{2}{x}^4-2x^3+2x^2$
$\Rightarrow f(-1)=\frac{9}{2}$