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

Answer 1

Answer: (D)

$\frac{9}{2}$

Given it has extremum values at $x=1$ and $x=2$

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

Given f(x) is a fourth degree polynomial

Let $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+c=0$

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

$\underset{x\to 0}{lim}(\frac{a{x}^4+b{x}^3+c{x}^2+dx+e}{x^2}+1)=3$

$\underset{x\to 0}{lim}(a{x}^2+bx+c+\frac{d}{x}+\frac{e}{x^2}+1)=3$

For limit to have finite value (in this case 3) value of 'd' and 'e' must be 0

$\Rightarrow d=0$ & $e=0$

Substituting x=0 in limit ;

$\Rightarrow$ $c+1=3$

$\Rightarrow$ $c=2$

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

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

$4a\left(1\right)+3b\left(1\right)+2c\left(1\right)+d=0$ $\Rightarrow$ $1$

$4a\left(8\right)+3b\left(4\right)+2c\left(2\right)+d=0$ $\Rightarrow$ $2$

Substituting $c=2$ and $d=0$

$4a+3b+4=0$

$32a+12b+8=0$

Solving two equations, we get $a=\frac{1}{2}andb=-2$

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

$f(-1)=\frac{-1^4}{2}-2(-1{)}^3+2(-1{)}^2$

Hence $f\left(x\right)=\frac{9}{2}$

Therefore correct option is 'D'