Answer:
Option C
Explanation:
Central Idea. Any function have extreme values (maximum or minimum) at its critical points , where f " (x)=0
Since , the function have extreme values at x=1 and x=2
$\therefore$ $f'(x)=0$ = 0 at x=1 and x=2
$\Rightarrow$ $f'(1)=0$ and $f'(2)=0$
Also it is given that
$\lim_{x \rightarrow 0}\left[ 1+\frac{f(x)}{x^{2}}\right]=3$
$\Rightarrow$ $1+\lim_{x\rightarrow 0}\frac{f(x)}{x^{2}}=3$
$\Rightarrow$ $\lim_{x\rightarrow 0}\frac{f(x)}{x^{2}}=2$
$\Rightarrow$ f(x) will be of the form
$ax^{4}+bx^{2}+2x^{2}$
[ $\therefore$ f(x) is four degree polynomial]
Let $f(x)=ax^{4}+bx^{2}+2x^{2}$
$\Rightarrow$ $f'(x)=4ax^{3}+3bx^{2}+4x$
$\Rightarrow$ $f'(1)=4a+3b+4=0$ ........(i)
and $f'(2)=32a+12b+8=0$
$\Rightarrow$ 8a+3b+2=0 ......(ii)
On solving Eqs(i) and (ii) , we get
$a=\frac{1}{2},b=-2$
$\therefore$ $f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$
$\Rightarrow$ f(2)= 8-16+8=0
