On which the following intervals is the function $x^{100}+\sin x-1$ decreasing?

$f\left(x\right)=x^{100}+\sin x-1\therefore f\left(x\right)=100x^{99}+\cos x$

$\text{ if }0<x<\frac{\pi }{2}\text{, then }{f}^{\mathrm{\prime }}\left(x\right)>0\text{. Therefore, }$

$f\left(x\right)\text{ is increasing on }(0,\pi /2)\text{. }$

$\text{ if }0<x<1\text{, Then }100x^{99}>0\text{ and }\cos x>0$

$[\because x\text{ lies between }0\text{ and }1\text{ radian }]$

$\therefore f^{\mathrm{\prime }}\left(x\right)=100x^{99}+\cos x>0$

$\text{ Thus, }f\left(x\right)\text{ is increasing on }(0,1)\text{ if }\frac{\pi }{2}<x<\pi \text{, }$

$\text{ then }100x^{99}>100\left(\because x>1\Rightarrow x^{99}>1\right)$

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

$\left(\because \cos x\ge -1\Rightarrow 100x^{99}+\cos x>99\right)$

$f^{\mathrm{\prime }}\left(x\right)>0\text{ implies }f\left(x\right)\text{ is increasing in }(\pi /2,\pi ).$