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

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

Answer 2

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

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

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

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

\(p \neq 2\)

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

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

On solving, we get:

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

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

\(\therefore 16ab = 252\)