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

0

The correct answer is option (1) : 0

It is given that f(x) is a fourth degree polynomial such that

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

$⇒\lim\limits_{x→0}\frac{f(x)}{x^2}$ is finite

$⇒ f (x) has a repeated root at x = 0.

Let $ f(x) = x^2 (ax^2 + bx + c) .$ Then,

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

$⇒\lim\limits_{x→0}(1+ax^2+bx+c)=3$

$⇒x+1=3$

$⇒c=2$

It is given that f(x) has extreme values at x = 1 and x= 2.

$∴f'(1) = 0 $ and $f'(2) = 0 $

$⇒4a+3b + 2c =0\, 32 a + 12 b + 4c = 0 $   $[∵f'(x) = 4ax^3+3bx^2 + 2cx]$

$⇒4a + 3b + 4=0 $ and $  32 a + 12b + 8 = 0 $

$⇒a = 1/2, b = - 2 $

$∴f(x) =x^2(\frac{1}{2}x^2 -2x+2)$

$⇒ f(2) + 4 (2 - 4 + 2) = 0 $