If x lies in the interval (0, pi/2),then the | vedclass

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(B) Given function: $f(x) = x \sin x + \cos x + \cos^2 x$ Step $1$: Find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(x \sin x) + \frac{d}{dx}(\cos x

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Decreasing

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(B) Given function: $f(x) = x \sin x + \cos x + \cos^2 x$ Step $1$: Find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(x \sin x) + \frac{d}{dx}(\cos x) + \frac{d}{dx}(\cos^2 x)$ $f'(x) = (\sin x + x \cos x) - \sin x - 2 \cos x \sin x$ $f'(x) = x \cos x - 2 \sin x \cos x$ $f'(x) = \cos x (x - 2 \sin x)$ Step $2$: Analyze the sign of $f'(x)$ in the interval $(0, \pi/2)$. In the interval $(0, \pi/2)$,$\cos x > 0$. We know that for $x \in (0, \pi/2)$,$\sin x > x$ is false; rather,$\sin x  0$. However,comparing $x$ and $2 \sin x$: Let $g(x) = x - 2 \sin x$. Then $g'(x) = 1 - 2 \cos x$. $g'(x) = 0$ at $x = \pi/3$. For $x \in (0, \pi/3)$,$g'(x) For $x \in (\pi/3, \pi/2)$,$g(x)$ increases,but at $x = \pi/2$,$g(\pi/2) = \pi/2 - 2 Thus,$(x - 2 \sin x) Step $3$: Conclusion. Since $\cos x > 0$ and $(x - 2 \sin x) Therefore,$f(x)$ is a decreasing function.

If $x$ lies in the interval $(0, \pi/2)$,then the function $f(x) = x \sin x + \cos x + \cos^2 x$ is:

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