Question

lf $log(1+x)-\frac{2x}{2+x}$ is increasing then

• A. $-1<x<\infty$
• B. $-\infty <x<0$
• C. $-\infty <x<\infty$
• D. $1<x<2$

Answer 1

Answer: (A)

$-1<x<\infty$

$f\left(x\right)=\log (1+x)-\frac{2x}{2+x}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{1+x}-\frac{4}{(x+2{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x^2}{(x+2{)}^2(x+1)}$
For f(x) to be increasing ,$f^{\prime }\left(x\right)\ge 0$
$\Rightarrow x+1>0$
$\Rightarrow x>-1$