On which of the following intervals is the fu | vedclass

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(B) Given the function $f(x) = x^{100} + \sin x - 1$. To determine the intervals where the function is decreasing,we find its derivative: $f'(x) = 100

Vedclass · Accepted Answer

None of these

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(B) Given the function $f(x) = x^{100} + \sin x - 1$. To determine the intervals where the function is decreasing,we find its derivative: $f'(x) = 100x^{99} + \cos x$. $1$. In the interval $(0, 1)$: Since $x \in (0, 1)$,$100x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) > 0$,so the function is strictly increasing. $2$. In the interval $\left(\frac{\pi}{2}, \pi\right)$: Here,$\cos x  \frac{\pi}{2} \approx 1.57$. Specifically,$100x^{99} > 1$ and $|\cos x| \le 1$,so $100x^{99} + \cos x > 0$. Thus,$f'(x) > 0$,and the function is strictly increasing. $3$. In the interval $\left(0, \frac{\pi}{2}\right)$: Here,$100x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) > 0$,and the function is strictly increasing. Since $f'(x) > 0$ in all the given intervals,the function is not decreasing in any of them. Therefore,the correct answer is $B$.

On which of the following intervals is the function $f$ given by $f(x)=x^{100}+\sin x-1$ decreasing?

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