In the open interval left(0, frac{pi}{2}right | vedclass

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Question

(B) Let $f(x) = \cos x + x \sin x - 1$. Find the derivative of $f(x)$ with respect to $x$: $f'(x) = -\sin x + (1 \cdot \sin x + x \cos x) = x \cos x$.

Vedclass · Accepted Answer

$\cos x + x \sin x > 1$

Vedclass · Answer

(B) Let $f(x) = \cos x + x \sin x - 1$. Find the derivative of $f(x)$ with respect to $x$: $f'(x) = -\sin x + (1 \cdot \sin x + x \cos x) = x \cos x$. For $x \in \left(0, \frac{\pi}{2}\right)$, both $x > 0$ and $\cos x > 0$. Therefore, $f'(x) = x \cos x > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$. Since $f'(x) > 0$, the function $f(x)$ is strictly increasing on the interval $\left(0, \frac{\pi}{2}\right)$. As $f(x)$ is increasing, for any $x > 0$, we have $f(x) > f(0)$. Calculate $f(0)$: $f(0) = \cos(0) + 0 \cdot \sin(0) - 1 = 1 + 0 - 1 = 0$. Thus, $f(x) > 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$. $\cos x + x \sin x - 1 > 0$ $\cos x + x \sin x > 1$.

In the open interval $\left(0, \frac{\pi}{2}\right)$, which of the following is true for the expression $\cos x + x \sin x$?

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