Correct answer is : 252
\(f(x)=-\left(p^{2}-6 p+8\right) \cos 4 n+2(2-p) n+7\)
\(f^{1}(x)=+4\left(p^{2}-6 p+8\right) \sin 4 x+(4-2 p) \neq 0\)
\(\sin 4 x \neq \frac{2 p-4}{4(p-4)(p-2)}\)
\(\sin 4 x \neq \frac{2(p-2)}{4(p-4)(p-2)}\)
\(\mathrm{p} \neq 2\)
\(
\begin{aligned}
& \sin 4 x \neq \frac{1}{2(p-4)}
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow & \left|\frac{1}{2(p-4)}\right|>1
\end{aligned}
\)
on solving we get
\(\therefore \mathrm{p} \in\left(\frac{7}{2}, \frac{9}{2}\right)\)
Hence \(\mathrm{a}=\frac{7}{2}, \mathrm{~b}=\frac{9}{2}\)
\(\therefore 16 \mathrm{ab}=252\)
