JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 9)
Let $$p(x)$$ be a polynomial of degree $$4$$ having extremum at
$$x = 1,2$$ and $$\mathop {\lim }\limits_{x \to 0} \left( {1 + {{p\left( x \right)} \over {{x^2}}}} \right) = 2$$.
$$x = 1,2$$ and $$\mathop {\lim }\limits_{x \to 0} \left( {1 + {{p\left( x \right)} \over {{x^2}}}} \right) = 2$$.
Then the value of $$p (2)$$ is
Answer
0
Explanation
Let us consider
$$P(x) = a{x^4} + b{x^3} + c{x^2} + dx + e$$
$$P'(1) = P'(2) = 0$$
$$\mathop {\lim }\limits_{x \to 0} \left( {{{{x^2} + P(x)} \over {{x^2}}}} \right) = 2$$
$$ \Rightarrow P(0) = 0 \Rightarrow e = 0$$
$$\mathop {\lim }\limits_{x \to 0} \left( {{{2x + P'(x)} \over {2x}}} \right) = 2$$
$$ \Rightarrow P'(0) = 0 \Rightarrow d = 0$$
$$\mathop {\lim }\limits_{x \to 0} \left( {{{2 + P''(x)} \over 2}} \right) = 2$$
$$ \Rightarrow c = 1$$
On solving, we get $$a = 1/4,b = - 1$$. Thus,
$$P(x) = {{{x^4}} \over 4} - {x^3} + {x^2} \Rightarrow P(2) = 0$$
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