Question

$f\left(x\right)=log(1+x)-\frac{2x}{x+2}(x>0)$ is increasing in interval

• A. $(1,3)$
• B. $(0,\infty )$
• C. $(0,1)$
• D. $(1,\infty )$

Answer 1

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

We need to differentiate the function.

So,$\frac{df}{dx}=\frac{1}{1+x}-\frac{2x+4-2x}{(x+2{)}^2}$

$=\frac{1}{1+x}-\frac{4}{(x+2{)}^2}$

$=\frac{x^2+4x+4-4x-4}{(x+1)\times (x+2{)}^2}$

$=\frac{x^2}{(x+1)\times (x+2{)}^2}$

This is greater than 0 for $x\in (-1,\infty )$

But in the question, it is given that $x>0$.

Hence $x\in (0,\infty )$