In which of the following intervals does f(x) | vedclass

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Question

(A) To determine where $f(x) = \sin x$ increases less rapidly than $g(x) = \cos x$,we compare their derivatives.$f'(x) = \cos x$ and $g'(x) = -\sin x$

Vedclass · Accepted Answer

$\left( -\frac{\pi}{2}, -\frac{\pi}{4} \right)$

Vedclass · Answer

(A) To determine where $f(x) = \sin x$ increases less rapidly than $g(x) = \cos x$,we compare their derivatives.$f'(x) = \cos x$ and $g'(x) = -\sin x$.We want to find the interval where $f'(x) This is equivalent to $\sin x + \cos x Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x This inequality holds when $\pi Solving for $x$,we get $-\frac{5\pi}{4} Among the given options,the interval $\left( -\frac{\pi}{2}, -\frac{\pi}{4} \right)$ is a subset of this range.Therefore,in the interval $\left( -\frac{\pi}{2}, -\frac{\pi}{4} \right)$,$f(x)$ increases less rapidly than $g(x)$.

In which of the following intervals does $f(x) = \sin x$ increase less rapidly than $g(x) = \cos x$?

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