JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q73

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MathematicsSingle correct+4 / -1Question 73 of 90

Let $f(x)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2$ . If $\displaystyle\lim_{x\to0}\left[1+\frac{f(x)}{x^2}\right]=3$ , then $f(2)$ is equal to :

−8

−4

Answer and worked solution

Official answer: C

The limit condition gives $f(x)=x^2(ax^2+bx+2)$ . Then $f'(x)=x(4ax^2+3bx+4)$ vanishes at 1 and 2:

$4a+3b+4=0$ ;
$16a+6b+4=0$ .

So $a=\frac12$ , $b=-2$ , and $f(2)=4(2-4+2)=0$ .

Topics:Maxima and minima Limits

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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 73