Question

# Let $$f(x)$$ be a polynomial of degree four having extreme value at $$x=1$$ and $$x=2$$.If $$\displaystyle \lim_{x\rightarrow 0}(1+\dfrac{f(x)}{x^2})=3$$,then $$f(2)$$ is equal to ?

A
0
B
4
C
8
D
49

Solution

## The correct option is D $$\dfrac{-4}{9}$$Given $$f'(1)=f'(2)=0$$     $$\displaystyle\lim_{x\rightarrow 0} \bigg(1+\dfrac{f(x)}{x^{2}}\bigg)=3$$           $$\displaystyle\lim _{x\rightarrow 0} \dfrac{f(x)}{x^{2}}=2\implies f(0)=0\ \ \ \ \ (\because$$ for applying L-hospital rule$$)$$     Apply L-Hospital Rule$$\implies \dfrac{f''(0)}{2}=2\implies f''(0)=4$$Let $$f'(x)=(x-1)(x-2)(x-a)$$   $$\implies f''(x)=(x-1)(x-2)+(x-2)(x-a)+(x-1)(x-a)$$    $$f''(0)=2+2{a}+a=4\implies a=\dfrac{2}{3}$$$$\implies f'(x)=(x-1)(x-2)(x-\dfrac{2}{3})=x^{3}-\dfrac{11}{3}x^{2}+4{x}-\dfrac{4}{3}$$$$f(x)=\dfrac{x^{4}}{4}-\dfrac{11}{9}x^{3}+2{x^{2}}-\dfrac{4}{3}x+d$$$$f(0)=0\implies d=0$$$$f(2)=4-\dfrac{88}{9}+8-\dfrac{8}{3}=-\dfrac{4}{9}$$Mathematics

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