$\begin{aligned}
& \text { (a) Let } P(x)=a x^3+b x^2+c x+d \\
& \therefore \quad P^{\prime}(x)=3 a x^2+2 b x+c
\end{aligned}$
Since, $P(x)$ has extreme value at $x=1$


Now, we have
$\lim _{x \rightarrow 0}\left(\frac{P(x)+4}{x^2}+2\right)=6$
$\begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{a x^3+b x^2+c x+d+4}{x^2}=4 \\ & \Rightarrow \quad \lim _{x \rightarrow 0} a x+b+\frac{c}{x}+\frac{d+4}{x^2}=4\end{aligned}$
Since, value of limit is finite
$\begin{aligned}
& \therefore \quad c=0 \text { and } d+4=0 \Rightarrow c=0, d=-4 \\
& \therefore \quad \lim _{x \rightarrow 0} a x+b=4 \Rightarrow b=4 \\
&
\end{aligned}$
Putting $b=4, c=0$ in Eq. (i), we get
$3 a+8=0$
$\begin{aligned} & \Rightarrow \quad a=\frac{-8}{3} \quad \therefore \quad P(x)=\frac{-8}{3} x^3+4 x^2-4 \\ & \Rightarrow \quad \frac{d P(x)}{d x}=-8 x^2+8 x \\ & \left.\therefore \quad \frac{d P(x)}{d x}\right|_{x=1 / 2}=\frac{-8}{4}+\frac{8}{2}=2\end{aligned}$