Question

The function $f\left(x\right)=\log x-\frac{2x}{2+x}$ is increasing in the interval

• A. $\left(-\infty ,0\right)$
• B. $\left(0,\infty \right)$
• C. $\left(1,\infty \right)$
• D. $\left(-\infty ,1\right)$

Answer 1

The correct option is B $\left(0,\infty \right)$

$f\left(x\right)=logx-\frac{2x}{x+2}$

$f^{\prime }\left(x\right)=\frac{1}{x}-\left(\frac{1\times 2x-2(x+2)}{(x+2{)}^2}\right)$

$\Rightarrow \frac{1}{x}-\left(\frac{2x-2x-4}{x^2+4+4x}\right)\ge 0$

$\Rightarrow \frac{x^2+4x+4+4x}{x(x+2{)}^2}\ge 0$

$\Rightarrow \frac{x^2+8x+4}{x(x+2{)}^2}\ge 0$

so, $xϵ(0,\infty )$