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

\[ f(x) = - (p^2 - 6p + 8)\cos 4n + 2(2 - p)n + 7 \] \[ f'(x) = +4(p^2 - 6p + 8)\sin 4x + (4 - 2p) eq 0 \] \[ \sin 4x eq \frac{2p - 4}{4(p - 4)(p - 2)} \] \[ \sin 4x eq \frac{2(p - 2)}{4(p - 4)(p - 2)} \] \[ p eq 2 \] \[ \sin 4x eq \frac{1}{2(p - 4)} \] \[ \Rightarrow \left| \frac{1}{2(p - 4)} \right| > 1 \] Solving yields \[ \therefore p \in \left( \frac{7}{2}, \frac{9}{2} \right) \] Thus, \[ a = \frac{7}{2}, \quad b = \frac{9}{2} \] \[ \therefore 16ab = 252 \]