Question

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

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

Answer 1

Answer: (D)

$f(x)=x^{100}+\sin x-1$
$\Rightarrow f^{\prime }(x)=100x^{99}+\cos x$.
If $0<x<\pi /2$, then $f^{\prime }(x)>0$,
therefore $f(x)$ is increasing on $(0,\pi /2)$.
If $0<x<1$, then $100x^{99}>0$ and $\cos x>0$
$[\because x$ lies between 0 and 1 radian $]$
$\Rightarrow f^{\prime }(x)=100x^{99}+\cos x>0$
$\Rightarrow f(x)$ is increasing on $(0,1)$
If $\pi /2<x<\pi$, then $100x^{99}>>100$
$\left[\because x>1,\therefore x^{99};>1\right]$
$\Rightarrow 100x^{99}+\cos x>0$
$[\because \cos x\ge -1,\therefore 100x^{99}+\cos x>99]$
$\Rightarrow f^{\prime }(x)>0$
$\Rightarrow f(x)$ is increasing on $(\pi /2,\pi )$