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 ab is equal to .

Mathematics · Application of Derivatives

JEE Main 2024 — 9 April, Shift 2 — Question 22

Let the set of all values of $p$ , for which $f(x)=\left(p^{2}-6 p+8\right)\left(\sin ^{2} 2 x-\cos ^{2} 2 x\right)+2(2-p) x+7$ does not

have any critical point, be the interval $(a, b)$ . Then 16 ab is equal to $\qquad$ .

Numerical answer — enter this value.

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

$f^{1}(x)=+4\left(p^{2}-6 p+8\right) \sin 4 x+(4-2 p) \neq 0$

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

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

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

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

on solving we get

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

$\mathrm{a}=\frac{7}{2}, \mathrm{~b}=\frac{9}{2}$

$\therefore 16 a b=252$

Answer key and solution verified before publishing.

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