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The minimum value of \(f(x)=\sin ^4 x+\cos ^4 x\) in the interval \(\left(0, \frac{\pi}{2}\right)\) is
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Verified Answer
The correct answer is:
\(\frac{1}{2}\)
Let \(y=\sin ^4 x+\cos ^4 x\)
\(\begin{aligned}
& \frac{d y}{d x}=4 \sin ^3 x \cos x+4 \cos ^3 x(-\sin x) \\
& =4 \sin x \operatorname{cox}\left(\sin ^2 x-\cos ^2 x\right) \\
& =(2 \sin 2 x)(-\cos 2 x)=-\sin 4 x \\
& \therefore \frac{d y}{d x}=0 \Rightarrow \sin 4 x=0 \Rightarrow 4 x=0, \pi, 2 \pi, 3 \pi
\end{aligned}\)
or \(\mathrm{x}=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \ldots \ldots \ldots, \Rightarrow \mathrm{x}=\frac{\pi}{4}\)
\(\begin{aligned}
& \frac{d y}{d x}=4 \sin ^3 x \cos x+4 \cos ^3 x(-\sin x) \\
& =4 \sin x \operatorname{cox}\left(\sin ^2 x-\cos ^2 x\right) \\
& =(2 \sin 2 x)(-\cos 2 x)=-\sin 4 x \\
& \therefore \frac{d y}{d x}=0 \Rightarrow \sin 4 x=0 \Rightarrow 4 x=0, \pi, 2 \pi, 3 \pi
\end{aligned}\)
or \(\mathrm{x}=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \ldots \ldots \ldots, \Rightarrow \mathrm{x}=\frac{\pi}{4}\)
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