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
B
32
C
52
D
92

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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