Let the set of all values of p,for which f(x) | vedclass

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(B) Given $f(x) = (p^2 - 6p + 8)(\sin^2 2x - \cos^2 2x) + 2(2 - p)x + 7$.Using $\cos 4x = \cos^2 2x - \sin^2 2x$,we have $f(x) = -(p^2 - 6p + 8)\cos 4

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$252$

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(B) Given $f(x) = (p^2 - 6p + 8)(\sin^2 2x - \cos^2 2x) + 2(2 - p)x + 7$.Using $\cos 4x = \cos^2 2x - \sin^2 2x$,we have $f(x) = -(p^2 - 6p + 8)\cos 4x + 2(2 - p)x + 7$.For $f(x)$ to have no critical points,$f'(x) 
eq 0$ for all $x \in \mathbb{R}$.$f'(x) = 4(p^2 - 6p + 8)\sin 4x + 2(2 - p)$.Setting $f'(x) = 0$,we get $4(p - 4)(p - 2)\sin 4x = 2(p - 2)$.If $p = 2$,$f'(x) = 0$ for all $x$,so $p 
eq 2$.For $p 
eq 2$,$\sin 4x = \frac{2(p - 2)}{4(p - 4)(p - 2)} = \frac{1}{2(p - 4)}$.For no critical points,the equation $\sin 4x = \frac{1}{2(p - 4)}$ must have no solution.This happens if $\left| \frac{1}{2(p - 4)} ight| > 1$.$|2(p - 4)| Thus,$a = 3.5 = \frac{7}{2}$ and $b = 4.5 = \frac{9}{2}$.$16ab = 16 	imes \frac{7}{2} 	imes \frac{9}{2} = 4 	imes 7 	imes 9 = 252$.

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

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