Question with Solution

Let $\mathrm{P}(\mathrm{x})={\mathrm{a}}_0{\mathrm{x}}^4+\dots ..+{\mathrm{a}}_4$ by hypothesis, ${\mathrm{P}}^1(1)=0$ and ${\mathrm{P}}^1(2)=0$ $\Rightarrow 4{\mathrm{a}}_0+3{\mathrm{a}}_1+2{\mathrm{a}}_2+{\mathrm{a}}_3=0$ and $32{\mathrm{a}}_0+12{\mathrm{a}}_1+4{\mathrm{a}}_2+{\mathrm{a}}_3=0$

Also, $\underset{x\to 0}{\mathrm{Lt}}\frac{P(x)}{x^2}=1\Rightarrow a_4=0$ and $a_3=0$ hence $\underset{x\to 0}{\mathrm{Lt}}\left(a_0{x}^3+a_{1}x+a_2\right)=1\Rightarrow a_2=1$

Solving, we get, ${\mathrm{a}}_0=\frac{1}{4},{\mathrm{a}}_1=-1,{\mathrm{a}}_2=1,{\mathrm{a}}_3=0,{\mathrm{a}}_4=0$

$\therefore P(x)=\frac{1}{4}{x}^4-x^3+x^2\Rightarrow P(2)=0$