Question

Let $$f(x)$$ be a polynomial of degree four having extreme value at $$x=1$$ and $$x=2$$. If $$\displaystyle\lim _{ x\rightarrow 0 }{ \left[ 1+\dfrac { f(x) }{ { x }^{ 2 } } \right] } =3$$, then $$f(2)$$ is equal to:

A
4
B
0
C
4
D
8

Solution

The correct option is A $$0$$Consider the given limit. $$\displaystyle\underset{x\to 0}{\mathop{\lim }}\,\left[ 1+\dfrac{f\left( x \right)}{{{x}^{2}}} \right]=3$$ $$\displaystyle\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{x+f\left( x \right)}{{{x}^{2}}} \right]=3$$   Since, the limit exists, therefore,  $${{x}^{2}}+f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+3{{x}^{2}}$$  $$f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+2{{x}^{2}}$$  $$f'\left( x \right)=4a{{x}^{3}}+3b{{x}^{2}}+4x$$   Also, $$f\left( x \right)=0$$ at $$x=1,2$$   Therefore, $$a=\dfrac{1}{2},b=-2$$ $$f\left( x \right)=\dfrac{{{x}^{4}}}{2}-2{{x}^{3}}+2{{x}^{2}}$$   Therefore, $$f\left( 2 \right)=8-16+8=0$$   Hence, $$f\left( 2 \right)=0$$.Mathematics

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