The function $f\left( x \right) = \log \left( {1 + x} \right) - \left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)$ is increasing on
A) $\left( {0,\infty } \right)$
B) $\left( { - \infty ,0} \right)$
C) $\left( { - \infty ,\infty } \right)$
D) None of these

Hint: In order to find the status of the function $f\left( x \right) = \log \left( {1 + x} \right) - \left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)$ given above, we need to find the range of the function for which the function is increasing. For that we need to differentiate the function and check for the values for which the values will be greater than zero.

Formula used:
1. $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$
2. $\dfrac{{dx}}{{dx}} = 1$
3. $\dfrac{{d\left( {{\text{constant}}} \right)}}{{dx}} = 0$
4. ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
5. $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$

Complete step by step answer:
We are given with the function $f\left( x \right) = \log \left( {1 + x} \right) - \left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)$, we need to find the range of the function, for that we are initiating with differentiating the function.
So, Differentiating the function $f\left( x \right) = \log \left( {1 + x} \right) - \left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)$ , with respect to $x$ and we get:
$f'\left( x \right) = \dfrac{d}{{dx}}\left( {\log \left( {1 + x} \right) - \left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)} \right)$
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {\log \left( {1 + x} \right)} \right) - \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)$ …….(1)
Since, we know that $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$ , similarly, we have $\log \left( {1 + x} \right)$, so comparing $\log \left( {1 + x} \right)$ with $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$, we get:
$ \Rightarrow \dfrac{{d\left( {\log \left( {1 + x} \right)} \right)}}{{dx}} = \dfrac{1}{{1 + x}}$ ……(2)
For the other operand $\dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)$, we can compare the function with the quotient rule, that is $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ and we get:
$u = 2x$
$v = 2 + x$
Substituting u, v in the equation $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right) = \dfrac{{\left( {2 + x} \right)\dfrac{{d\left( {2x} \right)}}{{dx}} - 2x\dfrac{{d\left( {2 + x} \right)}}{{dx}}}}{{{{\left( {2 + x} \right)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right) = \dfrac{{2\left( {2 + x} \right)\dfrac{{d\left( x \right)}}{{dx}} - 2x\dfrac{{d\left( {2 + x} \right)}}{{dx}}}}{{{{\left( {2 + x} \right)}^2}}}$
Since, we know that $\dfrac{{dx}}{{dx}} = 1$ and $\dfrac{{d\left( {{\text{constant}}} \right)}}{{dx}} = 0$, substituting it in the above equation, we get:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right) = \dfrac{{2\left( {2 + x} \right) \times 1 - 2x \times 1}}{{{{\left( {2 + x} \right)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right) = \dfrac{{4 + 2x - 2x}}{{{{\left( {2 + x} \right)}^2}}}$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right) = \dfrac{4}{{{{\left( {2 + x} \right)}^2}}}$ …..(3)
Substituting equation 2 and equation 3 in equation 1 and we get:
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {\log \left( {1 + x} \right)} \right) - \dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\left[ {2 + x} \right]}}} \right)$
$ \Rightarrow f'\left( x \right) = \dfrac{1}{{\left( {1 + x} \right)}} - \dfrac{4}{{{{\left( {2 + x} \right)}^2}}}$
Solving the operands, by making common denominators multiplying the denominators of alternate denominators, and we get:
$ \Rightarrow f'\left( x \right) = \dfrac{{{{\left( {2 + x} \right)}^2} - 4\left( {1 + x} \right)}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}$
Opening the numerator brackets using the expansion formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$:
$ \Rightarrow f'\left( x \right) = \dfrac{{{2^2} + {x^2} + 2.2.x - \left( {4 + 4x} \right)}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{4 + {x^2} + 4x - 4 - 4x}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{{x^2}}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}$
Now, it cannot be further simplified, so we stop here.
Since, it is given that the function is increasing so the first derivative of the function will be greater than zero and can be written as:
$ \Rightarrow f'\left( x \right) > 0$
$ \Rightarrow f'\left( x \right) = \dfrac{{{x^2}}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}} > 0$
We can see that the above equation can be greater than zero if the value of $x$ is zero or greater than zero, so that it’s positive.
Therefore, the possible range is zero to infinity, numerically written as $\left( {0,\infty } \right)$.

Hence, Option (A) $\left( {0,\infty } \right)$ is correct.

Note:
1. Similarly, if we were given to find the range for which the function is decreasing then we would have the derivative less than zero, or $f'\left( x \right) < 0$.
2. We can also check for the increasing or decreasing function by taking out the first derivative, if it is greater than zero then it's increasing and if it is less than zero, then it’s decreasing.