Question

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

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

Answer 1

Answer: (A)

$\frac{9}{2}$

We have,

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

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

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

Now,

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

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

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

$\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=4$

$\Rightarrow f^{\prime }\left(0\right)=4$

Then,

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

$f^{\prime }\left(x\right)=k\left[x^3-3x^2+2x\right]$

$f\left(x\right)=k\left[3x^2-6x+2\right]$

$f\left(0\right)=2k=4$

$\because k=2$

Now,

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

$f\left(-1\right)=\left[\frac{2}{4}+2+2\right]$

$=4+\frac{1}{2}$

$=\frac{9}{2}$

Hence, the option $\left(A\right)$ is the correct answer.