Let the set of all values of p , for which f ( x ) = ( p 2 - 6 p + 8 ) ( sin 2 2 x - cos 2 2 x ) + 2 ( 2 - p ) x + 7 does not have any critical point, be the interval ( a , b ) . Then 16 a b is equal to ______ . [2024]

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Solution

Q.
Let the set of all values of $p,$ for which $f (x) = (p^{2} - 6 p + 8) (\sin^{2} 2 x - \cos^{2} 2 x) + 2 (2 - p) x + 7$ does not have any critical point, be the interval $(a, b) .$ Then $16 a b$ is equal to ______ . [2024]

Ans.
(252)

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

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

$f' (x) = 4 \sin 4 x (p^{2} - 6 p + 8) + 2 (2 - p) = 0$ [ $∵$ $f (x)$ has no critical points]

$\Rightarrow \sin 4 x \neq \frac{2 p - 4}{4 (p^{2} - 6 p + 8)}$

$\Rightarrow \sin 4 x \neq \frac{2 (p - 2)}{4 (p - 2) (p - 4)}, p \neq 2 \Rightarrow \sin 4 x \neq \frac{1}{2 (p - 4)}$

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

$\Rightarrow$ $| p - 4 |$ $< \frac{1}{2} \Rightarrow \frac{- 1}{2} < p - 4 < \frac{1}{2}$

$\Rightarrow \frac{7}{2} < p < \frac{9}{2} \Rightarrow a = \frac{7}{2}, b = \frac{9}{2}$

$So, 16 a b = 16 \times \frac{7}{2} \times \frac{9}{2} = 4 \times 62 = 252$

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