Question

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

• A. $(0,1)$
• B. $(\frac{\pi }{2},\pi )$
• C. $(0,\frac{\pi }{2})$
• D. None of these

Answer 1

Answer: (D)

None of these

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

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

Option A,

For $0<x<1$ , both $100x^9$ and $\cos x$ are greater than $0$.

So,$f^{\prime }\left(x\right)>0$

Hence, $f\left(x\right)$ is strictly increasing on $(0,1)$

Option C,

For $x\in (\frac{\pi }{2},\pi ]$,

$\cos \left(\frac{\pi }{2}\right)=0,\cos \left(\pi \right)=-1$

i.e $\cos x$ decreases from $0$ to $-1$ in $(0,\frac{\pi }{2})$

At $x=\frac{\pi }{2}$, $100(\frac{\pi }{2}{)}^{99}$ and $x=\pi$ ,$100(\pi {)}^{99}$

i.e. $100x^{99}$ increases from $(\frac{\pi }{2},\pi ]$

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

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

Hence, $f\left(x\right)$ is strictly increasing on $(\frac{\pi }{2},\pi )$