Let the set of all values of p, for which f(x)=(p^2-6 p+8)(sin ^2 2 x-cos ^2 2 x)+2(2-p) x+7 does not have any critical point, be the interval (a, b). Then 16 a b is equal to _________.

f(x)=(p^2-6 p+8)(sin ^2 2 x-cos ^2 2 x) +2(2-p) x+7 \ f(x)=-cos 4 x(p^2-6 p+8)+2(2-p) x+7 \ f^prime(x)=4 sin 4 x(p^2-6 p+8)+2(2-p) ≠ 0 \ 2(2-p)+[-4(p^2-6 p+8), 4(p^2-6 p+8)] \ => [-4 p^2+24 p-32,4 p^2-24 p+32]+(4-2 p) \ [-4 p^2+22 p-28,4 p^2-26 p+36] \ [(p-2)(-4 p+14),(p-2)(4 p-18)] \ => (p-2)[(-4 p+14), 4 p-18] => p in ((7 / 2), (9 / 2)) \ => a=(7 / 2), b=(9 / 2) \ => 16 a b=4 x 63=252 \

Let the set of all values of p for which fxp26 p8sin 2 ... | (Mathematics)

$\begin{aligned} & f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right) +2(2-p) x+7 \\ & f(x)=-\cos 4 x\left(p^2-6 p+8\right)+2(2-p) x+7 \\ & f^{\prime}(x)=4 \sin 4 x\left(p^2-6 p+8\right)+2(2-p) \neq 0 \\ & 2(2-p)+\left[-4\left(p^2-6 p+8\right), 4\left(p^2-6 p+8\right)\right] \\ & \Rightarrow\left[-4 p^2+24 p-32,4 p^2-24 p+32\right]+(4-2 p) \\ & {\left[-4 p^2+22 p-28,4 p^2-26 p+36\right]} \\ & {[(p-2)(-4 p+14),(p-2)(4 p-18)]} \\ & \Rightarrow(p-2)[(-4 p+14), 4 p-18] \Rightarrow p \in\left(\frac{7}{2}, \frac{9}{2}\right) \\ & \Rightarrow a=\frac{7}{2}, b=\frac{9}{2} \\ & \Rightarrow 16 a b=4 \times 63=252 \end{aligned}$

Explanation

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