Question

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 _____ .

Answer 1

$f(x) = - (p^2 - 6p + 8) \cos 4x + 2(2 - p)x + 7$

$f'(x) = +4 \left( p^2 - 6p + 8 \right) \sin 4x + (4 - 2p) \neq 0$

$\sin 4x \neq \frac{2p - 4}{4(p-4)(p-2)}$

$\sin 4x \neq \frac{2(p-2)}{4(p-4)(p-2)}$

$p \neq 2$

$\sin 4x = \frac{1}{2(p-4)}$

$\implies \left| \frac{1}{2(p-4)} \right| > 1$

On solving, we get:

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

Hence $a = \frac{7}{2}$, $b = \frac{9}{2}$.

$\therefore 16ab = 252$