The function f(x) = sin^4 x + cos^4 x is incr | 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 = 1 - \frac{1}{2}(2\sin x \cos x)^

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$\frac{\pi}{4} < x < \frac{\pi}{2}$

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(B) Given $f(x) = \sin^4 x + \cos^4 x$. We can rewrite this as $f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2 2x$. Differentiating with respect to $x$,we get $f'(x) = -\frac{1}{2} \cdot 2\sin 2x \cdot \cos 2x \cdot 2 = -2\sin 2x \cos 2x = -\sin 4x$. For $f(x)$ to be increasing,$f'(x) > 0$,so $-\sin 4x > 0$,which implies $\sin 4x The sine function is negative in the third and fourth quadrants,so $\pi Dividing by $4$,we get $\frac{\pi}{4} < x < \frac{\pi}{2}$.

The function $f(x) = \sin^4 x + \cos^4 x$ is increasing in

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