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
Answer & explanation
Correct answer: option 1
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 $