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Question

Let $$f(x)$$ be a polynomial of degree $$4$$ having extreme values at $$x = 1$$ and $$x = 2$$.
If $$\displaystyle \lim_{x\rightarrow 0}\left (\dfrac {f(x)}{x^{2}} + 1\right ) = 3$$ then $$f(-1)$$ is equal to


A
12
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B
32
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C
52
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D
92
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Solution

The correct option is C $$\dfrac {9}{2}$$
Given it has extremum values at $$x=1$$ and $$x=2$$

$$\Rightarrow f'\left ( 1 \right )=0$$  and  $$ f'\left ( 2 \right )=0$$

Given f(x) is a fourth degree polynomial 

Let $$f\left ( x \right )=ax^{4}+bx^{3}+cx^{2}+dx+c=0$$

Given $$\displaystyle\lim_{x\to0} \left ( \frac{f(x) }{x^{2}}+1 \right )$$=3

$$\displaystyle\lim_{x\to0} \left ( \frac{ax^{4}+bx^{3}+cx^{2}+dx+e }{x^{2}}+1 \right )=3$$

$$\displaystyle\lim_{x\to0} \left ( ax^{2}+bx+c+\frac{d}{x}+\frac{e}{x^{2}}+1 \right )=3$$

For limit to have finite value (in this case 3) value of 'd' and 'e' must be 0

$$\Rightarrow d=0$$  & $$ e=0$$

Substituting x=0 in limit ;

$$\Rightarrow$$ $$c+1=3$$
$$\Rightarrow$$ $$ c=2$$

$$f'(x) =4ax^{3}+3bx^{2}+2cx+d$$
Applying $$ f'\left ( 1 \right )=0 , f'\left ( 2 \right )=0$$

$$4a(1)+3b(1)+2c(1)+d=0$$     $$\Rightarrow$$ $$1$$
$$4a(8)+3b(4)+2c(2)+d=0$$    $$\Rightarrow$$ $$2$$

Substituting $$c=2$$ and $$d=0$$

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

Solving two equations, we get $$a=\dfrac{1}{2}\  and\   b= -2$$

$$f\left ( x \right )=\dfrac{x^{4}}{2}-2x^{3}+2x^{2} $$

$$f\left ( - 1\right )=\dfrac{-1^{4}}{2}-2(-1)^{3}+2(-1)^{2} $$

Hence $$f(x) =\dfrac{9}{2}$$

Therefore correct option is 'D' 



Mathematics

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