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Let the set of all values of $p$, for which $f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right)+2(2-p) x+7$ does not have any critical point, be the interval $(a, b)$. Then $16 a b$ is equal to _______
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Verified Answer
The correct answer is:
252
$\begin{aligned} & f(x)=-\left(p^2-6 p+8\right) \cos 4 n+2(2-p) n+7 \\ & f^1(x)=+4\left(p^2-6 p+8\right) \sin 4 x+(4-2 p) \neq 0 \\ & \sin 4 x \neq \frac{2 p-4}{4(p-4)(p-2)}\end{aligned}$
$\begin{aligned}
& \sin 4 x \neq \frac{2(p-2)}{4(p-4)(p-2)} \\
& p \neq 2 \\
& \quad \sin 4 x \neq \frac{1}{2(p-4)} \\
& \Rightarrow\left|\frac{1}{2(p-4)}\right|>1
\end{aligned}$
on solving we get
$\therefore \mathrm{p} \in\left(\frac{7}{2}, \frac{9}{2}\right)$
Hence $\mathrm{a}=\frac{7}{2}, \mathrm{~b}=\frac{9}{2}$
$\therefore 16 \mathrm{ab}=252$
$\begin{aligned}
& \sin 4 x \neq \frac{2(p-2)}{4(p-4)(p-2)} \\
& p \neq 2 \\
& \quad \sin 4 x \neq \frac{1}{2(p-4)} \\
& \Rightarrow\left|\frac{1}{2(p-4)}\right|>1
\end{aligned}$
on solving we get
$\therefore \mathrm{p} \in\left(\frac{7}{2}, \frac{9}{2}\right)$
Hence $\mathrm{a}=\frac{7}{2}, \mathrm{~b}=\frac{9}{2}$
$\therefore 16 \mathrm{ab}=252$
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