Question

Let p(x) 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(2) is ___

Answer 1

Let $p\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Now $\underset{x\to 0}{lim}[1+\frac{p\left(x\right)}{x^2}]=2$
$\Rightarrow \underset{x\to 0}{lim}\frac{p\left(x\right)}{x^2}=1\cdots \left(i\right)$
$\Rightarrow p\left(0\right)=0\Rightarrow e=0$
Applying L 'Hospital's rule to $e{q}^n\left(1\right)$, we get
$\underset{x\to 0}{lim}\frac{p^{\prime }\left(x\right)}{2x}=1\Rightarrow p^{\prime }\left(0\right)=0$
$\Rightarrow d=0$
Again applying 'Hospital's' rule, we get
$\underset{x\to 0}{lim}\frac{p^{\prime \prime }\left(x\right)}{2}=1\Rightarrow p^{\prime \prime }\left(0\right)=2$
$\Rightarrow 2c=2orc=1$
$\therefore p\left(x\right)=a{x}^4+b{x}^3+x^2$
$\Rightarrow p^{\prime }\left(x\right)=4a{x}^3+3b{x}^2+2x$
As p(x) has extremum at x = 1 and 2
$\therefore p^{\prime }\left(1\right)=0$ and $p^{\prime }\left(2\right)=0$
$\Rightarrow 4a+3b+2=0\cdots \left(i\right)$
$\Rightarrow 32a+12b+4=0$ or $8a+3b+1=0\cdots \left(ii\right)$
Solving eq's (i) and (ii) we get $a=\frac{1}{4}$ and $b=-1$
$\therefore p\left(x\right)=\frac{1}{4}{x}^4-x^3+x^2$
So, that $p\left(2\right)=\frac{16}{4}-8+4=0$