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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Application of Derivatives

Let the set of all values of p , for which f(x)=(p^2-6 p+8)( ^2 2...

Let the set of all values of pp, for which f(x)=(p2−6p+8)(sin⁡22x−cos⁡22x)+2(2−p)x+7f(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)(a, b). Then 16ab16 a b is equal to _______

Correct answer

0

Step-by-step solution

f(x)=−(p2−6p+8)cos⁡4n+2(2−p)n+7f1(x)=+4(p2−6p+8)sin⁡4x+(4−2p)≠0sin⁡4x≠2p−44(p−4)(p−2)\begin{aligned} & 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)}\end{aligned}sin⁡4x≠2(p−2)4(p−4)(p−2)p≠2sin⁡4x≠12(p−4)⇒∣12(p−4)∣>1\begin{aligned} & \sin 4 x \neq \frac{2(p-2)}{4(p-4)(p-2)} \\ & p \neq 2 \\ & \quad \sin 4 x \neq \frac{1}{2(p-4)} \\ & \Rightarrow\left|\frac{1}{2(p-4)}\right|>1 \end{aligned}onsolvingwegeton solving we get\therefore \mathrm{p} \in\left(\frac{7}{2}, \frac{9}{2}\right)HenceHence\mathrm{a}=\frac{7}{2}, \mathrm{~b}=\frac{9}{2} \therefore 16 \mathrm{ab}=252$
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