The function sin x - cos x is increasing in t | vedclass

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(B) Let $f(x) = \sin x - \cos x$. To find the interval where the function is increasing,we calculate the derivative: $f'(x) = \cos x + \sin x$. The fu

Vedclass · Accepted Answer

$\left[ 0, \frac{3\pi}{4} \right)$

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(B) Let $f(x) = \sin x - \cos x$. To find the interval where the function is increasing,we calculate the derivative: $f'(x) = \cos x + \sin x$. The function $f(x)$ is increasing if $f'(x) > 0$. $f'(x) = \cos x + \sin x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x \right) = \sqrt{2} \cos \left( x - \frac{\pi}{4} \right)$. We require $\sqrt{2} \cos \left( x - \frac{\pi}{4} \right) > 0$,which implies $\cos \left( x - \frac{\pi}{4} \right) > 0$. This holds when $-\frac{\pi}{2} Adding $\frac{\pi}{4}$ to all parts,we get $-\frac{\pi}{4} Considering the standard interval $[0, 2\pi]$,the function is increasing in $[0, \frac{3\pi}{4})$. Thus,option $(B)$ is correct.

The function $\sin x - \cos x$ is increasing in the interval

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