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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
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B
0
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C
4
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D
8
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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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