Question

Let $p\left(x\right)$ be a polynomial of degree 4 having extremum at $x=1,2$ and $\underset{x\to 0}{lim}(1+\frac{p\left(x\right)}{x^2})=2$ Then the value of $p\left(2\right)$ is

Answer 1

Let $P\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$

When $x\to 0$ then, for existence of $\frac{P\left(x\right)}{x^2}$ $d=e=0$ and $\underset{x\to 0}{lim}(1+\frac{P\left(x\right)}{x^2})=2$

So $c=1.$

$P^{\prime }\left(1\right)=P^{\prime }\left(2\right)=04a+3b+2c+d=04a+3b+2=0.........eq.14a.8+3b.4+2c.2+d=032a+12b+4c+d=032a+12b+4=0......eq.2$

By solving $eq.1$ and $eq.2$ we get $a=\frac{1}{4};b=-1$

So, $P\left(x\right)=\frac{1}{4}{x}^4-x^3+x^2.$

$P\left(2\right)=4-8+4=0$