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 ________

$$f(x)=(p^2-6p+8)(\sin^22x-\cos^22x)+2(2-p)x+7$$

$$\sin^22x-\cos^22x=-\cos4x$$
$$\Rightarrow f'(x)=4(p^2-6p+8)\sin4x+2(2-p)$$
For no critical point:
$$4(p^2-6p+8)\sin4x+2(2-p)\ne0$$

$$\Rightarrow\left|\frac{2(p-2)}{4(p^2-6p+8)}\right|>1$$
$$\Rightarrow|p-2|>2|p^2-6p+8|$$

$$(p^2-6p+8)=(p-2)(p-4)$$
$$\Rightarrow|p-2|>2|p-2||p-4|$$

$$For(p\ne2):$$
$$1>2|p-4|\Rightarrow|p-4|<\frac{1}{2}$$

$$\Rightarrow p\in\left(\frac{7}{2},\frac{9}{2}\right)$$

16ab=16$$\cdot$$ $$\frac{7}{2}\cdot$$ $$\frac{9}{2}$$=252