MCQ
The function $f(x) = {\sin ^4}x + {\cos ^4}x$ increases, if
- A$0 < x < {\pi \over 8}$
- ✓${\pi \over 4} < x < {{3\pi } \over 8}$
- C${{3\pi } \over 8} < x < {{5\pi } \over 8}$
- D${{5\pi } \over 8} < x < {{3\pi } \over 4}$
$ = {({\sin ^2}x + {\cos ^2}x)^2} - 2{\sin ^2}x{\cos ^2}x$
$ = 1 - \frac{{4{{\sin }^2}x{{\cos }^2}x}}{2} = 1 - \frac{{{{\sin }^2}2x}}{2}$
$ = 1 - \frac{1}{4}(2{\sin ^2}2x)$
$ = 1 - \left( {\frac{{1 - \cos 4x}}{4}} \right) = \frac{3}{4} + \frac{1}{4}\cos 4x$
Hence function $ f(x)$ is increasing when $f'(x) > 0$
$f'(x) = - \sin 4x > 0 \Rightarrow \sin 4x < 0$
Hence $\pi < 4x < \frac{{3\pi }}{2}$ or $\frac{\pi }{4} < x < \frac{{3\pi }}{8}$.
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