IIT-JEE 2009 · Paper 2 · Mathematics Q34

MathematicsInteger+4 / -1Question 34 of 57

Let $p(x)$ be a polynomial of degree 4 having extremum at $x=1,2$ and
$\lim_{x\to0}\left(1+\frac{p(x)}{x^2}\right)=2.$
Then the value of $p(2)$ is

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Answer and worked solution

Official answer: 0

Form of p. $\frac{p(x)}{x^2}\to1$ as $x\to0$ , so $p(x)=x^2+bx^3+cx^4$ .

Extrema at 1 and 2. $p'(x)=x(2+3bx+4cx^2)$ , and $4cx^2+3bx+2=4c(x-1)(x-2)$ : $8c=2$ , $c=\frac14$ , and $3b=-12c$ , $b=-1$ .

Value. $p(2)=4-8+4=0$ .

Answer: 0

IIT-JEE 2009 · Paper 2, Mathematics question 34