Let p(x) be a polynomial of degree 4 having extremum at x=1,2 and _...

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) $\begin{array}{rlrl} & & 4 a+3 b=-2 & \text{ and } 32 a+12 b=-4 \\ \Rightarrow & & a=\frac{1}{4} \text{ and } b=-1 \\ \therefore & & p(x)=\frac{x^4}{4}-x^3+x^2 \\ \Rightarrow & & p(2)=\frac{16}{4}-8+4 \end{array}$