JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 27)
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 $$16 a b$$ is equal to _________.
Answer
252
Explanation
$$\begin{aligned}
& f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right) +2(2-p) x+7 \\
& f(x)=-\cos 4 x\left(p^2-6 p+8\right)+2(2-p) x+7 \\
& f^{\prime}(x)=4 \sin 4 x\left(p^2-6 p+8\right)+2(2-p) \neq 0 \\
& 2(2-p)+\left[-4\left(p^2-6 p+8\right), 4\left(p^2-6 p+8\right)\right] \\
& \Rightarrow\left[-4 p^2+24 p-32,4 p^2-24 p+32\right]+(4-2 p) \\
& {\left[-4 p^2+22 p-28,4 p^2-26 p+36\right]} \\
& {[(p-2)(-4 p+14),(p-2)(4 p-18)]} \\
& \Rightarrow(p-2)[(-4 p+14), 4 p-18] \Rightarrow p \in\left(\frac{7}{2}, \frac{9}{2}\right) \\
& \Rightarrow a=\frac{7}{2}, b=\frac{9}{2} \\
& \Rightarrow 16 a b=4 \times 63=252
\end{aligned}$$
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