Question

${(1+x)}^p\le 1+x^p$, where

• A. $p>1$
• B. $0\le p\le 1$
• C. $x>0$
• D. $x<0$

Answer 1

Answer: (B), (C)

$0\le p\le 1$
$x>0$

Let $f\left(x\right)=1+x^p-{(1+x)}^p$

$\Rightarrow f^{\prime }\left(x\right)=p{x}^p-p{(1+x)}^{p-1}=p(\frac{1}{x^{1-p}}-\frac{1}{{(1+x)}^{1-p}})\ge 0$

if $p\ge 0,1-p\ge 0,x>0$

$\therefore f\left(x\right)$ increases when $x>0$ and $0\le p\le 1.$

Since $x>0,f\left(x\right)>f\left(0\right)$

$\therefore 1+x^p-{(1+x)}^p>0(\because f\left(0\right)=0)$

$\Rightarrow 1+x^p>{(1+x)}^p$ if $0\le p\le 1$ and $x>0$