Question

The function $\mathrm{f}\left(\mathrm{x}\right)=\log \left(1+\mathrm{x}\right)-\left(\frac{2\mathrm{x}}{\left(2+\mathrm{x}\right)}\right)$ is increasing for all values of $\mathrm{x}$, then

• A. $\left(0,\infty \right)$
• B. $\left(-\infty ,0\right)$
• C. $\left(-\infty ,\infty \right)$
• D. None of these

Answer 1

The correct option is A

$\left(0,\infty \right)$

Explanation for correct option

Step 1. Determine the interval of the function

The given function is, $\mathrm{f}\left(\mathrm{x}\right)=\log \left(1+\mathrm{x}\right)-\left(\frac{2\mathrm{x}}{\left(2+\mathrm{x}\right)}\right)$

Differentiate with respect to $\mathrm{x}$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\log \left(1+\mathrm{x}\right)-\left(\frac{2}{\left(2+\mathrm{x}\right)}\right)\right)$

Use sum/difference rule $\left(\mathrm{f}\pm \mathrm{g}\right)\text{'}=\mathrm{f}\text{'}\pm \mathrm{g}\text{'}$, where $\mathrm{f}$ and $\mathrm{g}$ are the function.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\ln \left(1+\mathrm{x}\right)\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{2+\mathrm{x}}\right)$

Use chain rule

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{1}{1+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{x}\right)-\frac{{\displaystyle \frac{1}{2\mathrm{x}}}}{2+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{2+\mathrm{x}}\right)$

Derivative of constant $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{a}\right)=0$ and common derivative $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)=1$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\left(\frac{1}{1+\mathrm{x}}\times 1\right)-2\frac{1\left(2+\mathrm{x}\right)-1\times \mathrm{x}}{{(2+\mathrm{x})}^2}$

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

Step 2 simplify the equation

We will simplify the equation $\frac{1}{1+\mathrm{x}}-\frac{4}{{\left(2+\mathrm{x}\right)}^2}$.

$\Rightarrow$ $\mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{{\left(\mathrm{x}+2\right)}^2-4\left(\mathrm{x}+1\right)}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^2}$

$=\frac{\left({\mathrm{x}}^2+4\mathrm{x}+4\right)}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^2}$

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

$=\frac{{\mathrm{x}}^2}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^2}$

Therefore, $\mathrm{f}\text{'}\left(\mathrm{x}\right)>0$

By substituting the values,

$\frac{{\mathrm{x}}^2}{\left(\mathrm{x}+1\right){\left(\mathrm{x}+2\right)}^2}>0$, where $\left(1+\mathrm{x}\right)>0$

Thus, $\mathrm{f}\left(\mathrm{x}\right)$ is increasing function on $\left(0,\infty \right)$ for all $\mathrm{x}>-1$.

Hence, option $\mathrm{A}$ is correct answer.