On which of the following intervals is the function

Solution:

We have,

$f(x)=x^{100}+\sin x-1$

$\therefore f^{\prime}(x)=100 x^{99}+\cos x$

In interval $(0,1), \cos x>0$ and $100 x^{99}>0$

$\therefore f^{\prime}(x)>0$

Thus, function f is strictly increasing in interval (0, 1).

In interval $\left(\frac{\pi}{2}, \pi\right), \cos x<0$ and $100 x^{99}>0 .$ Also, $100 x^{99}>\cos x$

$\therefore f^{\prime}(x)>0$ in $\left(\frac{\pi}{2}, \pi\right)$.

Thus, function $f$ is strictly increasing in interval $\left(\frac{\pi}{2}, \pi\right)$.

In interval $\left(0, \frac{\pi}{2}\right), \cos x>0$ and $100 x^{99}>0$.'

$\therefore 100 x^{99}+\cos x>0$

$\Rightarrow f^{\prime}(x)>0$ on $\left(0, \frac{\pi}{2}\right)$

$\therefore f$ is strictly increasing in interval $\left(0, \frac{\pi}{2}\right)$.

Hence, function f is strictly decreasing in none of the intervals.

The correct answer is D.