# Let f (x) be a polynomial of degree 4 having extreme values at x=1 and x=2.If $\lim_{x\rightarrow 0}\left ( \frac{f\left ( x \right )}{x^{2}}+1 \right )=3$, then f (−1) is equal to : Option 1) $\frac{9}{2}$ Option 2) $\frac{5}{2}$ Option 3) $\frac{3}{2}$ Option 4) $\frac{1}{2}$

N neha
H Himanshu

As we learned,

Method for maxima or minima -

First and second derivative method :

$Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0$

$Step\:2.\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if\:f'(x)>0$  $and\:local\:minimum\:if\;f'(x)<0.$

$Step\:\:3.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:miximum\:if$  $f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0$

- wherein

$Where\:\:y=f(x)$

$\frac{dy}{dx}=f'(x)$

$\lim_{x\rightarrow 0}\left ( 1+\frac{f\left ( x \right )}{x^{2}} \right )=3$

$\Rightarrow \: \lim_{x\rightarrow 0}\frac{f\left ( x \right )}{x^{2}}=2$

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

$f'\left ( x \right )=4ax^{3}+3bx^{2}+4x$

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

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

on solving $a=\frac{1}{2}$  and b= -2

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

$f\left ( -1 \right )=\frac{1}{2}+2+2=\frac{9}{2}$

Option 1)

$\frac{9}{2}$

Option 2)

$\frac{5}{2}$

Option 3)

$\frac{3}{2}$

Option 4)

$\frac{1}{2}$

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