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Q.
On which of the following intervals is the function $x ^{100}+\sin\, x -1$ decreasing?
Application of Derivatives
Solution:
$f(x)=x^{100}+\sin \,x-1 $
$\Rightarrow f'(x)=100 x^{99}+\cos \,x$.
If $0 < x < \pi / 2$, then $f '( x )>0$,
therefore $f ( x )$ is increasing on $(0, \pi / 2)$.
If $0 < x < 1$, then $100 x^{99}>0$ and $\cos \,x > 0$
$[\because x$ lies between 0 and 1 radian $]$
$\Rightarrow f '( x )=100 x ^{99}+\cos\, x > 0$
$ \Rightarrow f ( x )$ is increasing on $(0,1)$
If $\pi / 2 < x < \pi$, then $100 x^{99} > > 100$
$\left[\because x > 1, \therefore x^{99} ; >1\right]$
$\Rightarrow 100 x^{99}+\cos \,x > 0 $
$[\because \cos\, x \geq-1, \left.\therefore 100 x^{99}+\cos \,x > 99\right]$
$\Rightarrow f '( x ) > 0 $
$\Rightarrow f ( x )$ is increasing on $(\pi / 2, \pi)$