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(D) To determine where the function $f(x) = x^{100} + \sin x - 1$ is strictly increasing,we find its derivative:$f'(x) = 100x^{99} + \cos x$For the in

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All of the above

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(D) To determine where the function $f(x) = x^{100} + \sin x - 1$ is strictly increasing,we find its derivative:$f'(x) = 100x^{99} + \cos x$For the interval $(0, 1)$:Since $x \in (0, 1)$,$x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) = 100x^{99} + \cos x > 0$.For the interval $(0, \pi / 2)$:Since $x \in (0, \pi / 2)$,$x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) = 100x^{99} + \cos x > 0$.For the interval $(\pi / 2, \pi )$:Here,$x > \pi / 2 \approx 1.57$,so $x^{99}$ is very large. $100x^{99} > 100(1.57)^{99}$,while $\cos x$ lies between $-1$ and $0$. Therefore,$100x^{99} + \cos x > 0$.Since the derivative is positive in all the given intervals,the function is strictly increasing in all of them.Therefore,the correct option is $D$.

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

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