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If $f$ be twice differentiable function such that $f^{\prime \prime}(x)<0 \forall x \in R$ , then $h(x)=f\left(\sin ^2 x\right)+f\left(\cos ^2 x\right)$ where $|x| \leq \frac{\pi}{2}$ , increases in
Option: 1
$\left[0, \frac{\pi}{4}\right]$

Option: 2
$\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$

Option: 3
$\left[-\frac{\pi}{2}, 0\right]$

Option: 4
None of these

Answers (1)

best_answer

$h^{\prime}(x)=\sin 2 x\left[f^{\prime}\left(\sin ^2 x\right)-f^{\prime}\left(\cos ^2 x\right)\right]$

$\Rightarrow x=0, \frac{\pi}{4},-\frac{\pi}{4},-\frac{\pi}{2}, \frac{\pi}{2} \\$
$\sin ^2 x=\cos ^2 x \Rightarrow \tan ^2 x=1 \Rightarrow x= \pm \frac{\pi}{4}$

When $0 \leq x \leq \frac{\pi}{4} \\$

$\sin 2 x \geq 0 \\$

$\sin ^2 x<\cos ^2 x \\$

$\Rightarrow f^{\prime}\left(\sin ^2 x\right)>f^{\prime}\left(\cos ^2 x\right)$ ( since $f^{\prime}$ is decreasing function)
$h^{\prime}(x) \geq 0 \quad\left[0, \frac{\pi}{4}\right] \text {. } \\$

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