f(x)=−(p2−6p+8)cos4n+2(2−p)n+7f1(x)=+4(p2−6p+8)sin4x+(4−2p)=0sin4x=4(p−4)(p−2)2p−4 $\begin{aligned}
& \sin 4 x \neq \frac{2(p-2)}{4(p-4)(p-2)} \
& p \neq 2 \
& \quad \sin 4 x \neq \frac{1}{2(p-4)} \
& \Rightarrow\left|\frac{1}{2(p-4)}\right|>1
\end{aligned}onsolvingweget\therefore \mathrm{p} \in\left(\frac{7}{2}, \frac{9}{2}\right)Hence\mathrm{a}=\frac{7}{2}, \mathrm{~b}=\frac{9}{2}\therefore 16 \mathrm{ab}=252$