Verified Solution • 2025–26 Session moderate

QJEEM24Ma4c4b: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4c4b Solution: Let the set of all values of , for which does not have any…

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 _________.

Step-by-Step Verified Solution

amp;f(x)=(p2−6p+8)(sin⁡22x−cos⁡22x)+2(2−p)x+7amp;f(x)=−cos⁡4x(p2−6p+8)+2(2−p)x+7amp;f′(x)=4sin⁡4x(p2−6p+8)+2(2−p)≠0amp;2(2−p)+[−4(p2−6p+8),4(p2−6p+8)]amp;⇒[−4p2+24p−32,4p2−24p+32]+(4−2p)amp;[−4p2+22p−28,4p2−26p+36]amp;[(p−2)(−4p+14),(p−2)(4p−18)]amp;⇒(p−2)[(−4p+14),4p−18]⇒p∈(72,92)amp;⇒a=72,b=92amp;⇒16ab=4×63=252\begin{aligned}& f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right) +2(2-p) x+7 \\& f(x)=-\cos 4 x\left(p^2-6 p+8\right)+2(2-p) x+7 \\& f^{\prime}(x)=4 \sin 4 x\left(p^2-6 p+8\right)+2(2-p) \neq 0 \\& 2(2-p)+\left[-4\left(p^2-6 p+8\right), 4\left(p^2-6 p+8\right)\right] \\& \Rightarrow\left[-4 p^2+24 p-32,4 p^2-24 p+32\right]+(4-2 p) \\& {\left[-4 p^2+22 p-28,4 p^2-26 p+36\right]} \\& {[(p-2)(-4 p+14),(p-2)(4 p-18)]} \\& \Rightarrow(p-2)[(-4 p+14), 4 p-18] \Rightarrow p \in\left(\frac{7}{2}, \frac{9}{2}\right) \\& \Rightarrow a=\frac{7}{2}, b=\frac{9}{2} \\& \Rightarrow 16 a b=4 \times 63=252\end{aligned}

Related Learning Resources & Exam Preparation