The function f(x) = sin^4 x + cos^4 x increas | vedclass

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(B) Given $f(x) = \sin^4 x + \cos^4 x$. We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$. Since $\sin^2 x + \cos^2 x = 1$,

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$\frac{\pi}{4} < x < \frac{3\pi}{8}$

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(B) Given $f(x) = \sin^4 x + \cos^4 x$. We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$. Since $\sin^2 x + \cos^2 x = 1$,we have $f(x) = 1 - 2\sin^2 x \cos^2 x$. Using the identity $\sin 2x = 2\sin x \cos x$,we get $f(x) = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x$. Using $\sin^2 heta = \frac{1 - \cos 2 heta}{2}$,we have $f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} ight) = 1 - \frac{1}{4} + \frac{1}{4}\cos 4x = \frac{3}{4} + \frac{1}{4}\cos 4x$. For the function to be increasing,$f'(x) > 0$. $f'(x) = \frac{d}{dx} \left( \frac{3}{4} + \frac{1}{4}\cos 4x ight) = -\sin 4x$. Setting $f'(x) > 0$,we get $-\sin 4x > 0$,which implies $\sin 4x The sine function is negative in the interval $(\pi, 2\pi)$. Thus,$\pi Dividing by $4$,we get $\frac{\pi}{4} Comparing this with the given options,the interval $\frac{\pi}{4} < x < \frac{3\pi}{8}$ is a subset of this range,so the function is increasing in this interval.

The function $f(x) = \sin^4 x + \cos^4 x$ increases,if

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