Q37 / 57 · JEE Advanced 2009 (Paper 2) · Mathematics
Let $p(x)$ be a polynomial of degree 4 having extremum at $x=1,2$ and $\lim _{x \rightarrow 0}\left[1+\frac{p(x)}{x^2}\right]=2$. Then, the value of $p(2)$ is
Correct answer: 0
Solution
Let $p(x)=a x^4+b x^3+c x^2+d x+e$
$$
\Rightarrow \quad p^{\prime}(x)=4 a x^3+3 b x^2+2 c x+d
$$
$$
\therefore \quad p^{\prime}(1)=4 a+3 b+2 c+d=0
$$
and $p^{\prime}(2)=32 a+12 b+4 c+d=0$
Since, $\lim _{x \rightarrow 0}\left(1+\frac{p(x)}{x^2}\right)=2 \quad$ [given]
$$
\lim _{x \rightarrow 0} \frac{a x^4+b x^3+(c+1) x^2+d x+e}{x^2}=2
$$
$$
\begin{array}{rlrl}
\Rightarrow & & x+1=2, d & =0, e=0 \\
\Rightarrow & c & =1
\end{array}
$$
From Eqs. (i) and (ii)
$$
\begin{array}{rlrl}
& & 4 a+3 b=-2 & \text { and } 32 a+12 b=-4 \\
\Rightarrow & & a=\frac{1}{4} \text { and } b=-1 \\
\therefore & & p(x)=\frac{x^4}{4}-x^3+x^2 \\
\Rightarrow & & p(2)=\frac{16}{4}-8+4
\end{array}
$$
Hence, the correct answer is 0.