Question

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

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

Answer 1

Answer: (D)

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