If f(x) = sin x - cos x,0 leq x leq 2pi,then | vedclass

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(D) Given $f(x) = \sin x - \cos x$.To find the intervals where $f(x)$ is strictly decreasing,we find the derivative $f'(x)$.$f'(x) = \cos x + \sin x$.

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None of these

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(D) Given $f(x) = \sin x - \cos x$.To find the intervals where $f(x)$ is strictly decreasing,we find the derivative $f'(x)$.$f'(x) = \cos x + \sin x$.For $f(x)$ to be strictly decreasing,we must have $f'(x) $\cos x + \sin x $\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x ight) $\sqrt{2} \cos \left( x - \frac{\pi}{4} ight) Since $\sqrt{2} > 0$,we need $\cos \left( x - \frac{\pi}{4} ight) We know that $\cos 	heta Substituting $	heta = x - \frac{\pi}{4}$,we get $\frac{\pi}{2} Adding $\frac{\pi}{4}$ to all parts: $\frac{\pi}{2} + \frac{\pi}{4} $\frac{3\pi}{4} Comparing this with the given options,none of the intervals provided are subsets of $\left( \frac{3\pi}{4}, \frac{7\pi}{4} ight)$ except for the possibility of none of these being correct.

If $f(x) = \sin x - \cos x$,$0 \leq x \leq 2\pi$,then $f(x)$ is strictly decreasing in the interval:

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