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

Correct Answer: 252

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

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

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

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

\(p \neq 2\)

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

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

On solving, we get:

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

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

\(\therefore 16ab = 252\)