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If $\log (1+x)-\frac{2 x}{2+x}$ is increasing, then
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Verified Answer
The correct answer is:
$-\infty < x < \infty$
Let $f(x)=\log (1+x)-\frac{2 x}{2+x^2}$
$$
f^{\prime}(x)=\frac{1}{(1+x)}-\frac{2}{2+x}+\frac{2 x}{(2+x)^2}
$$
For increasing function, $f^{\prime}(x)>0$
$$
\begin{aligned}
\Rightarrow \quad(2+x)^2-2(2+x)(1+x)+2 x(1+x) & >0 \\
\Rightarrow 4+x^2+4 x-4-6 x-2 x^2+2 x+2 x^2 & >0 \\
x^2 & >0
\end{aligned}
$$
This shows $x$ lies between $-\infty$ to $\infty$.
$$
f^{\prime}(x)=\frac{1}{(1+x)}-\frac{2}{2+x}+\frac{2 x}{(2+x)^2}
$$
For increasing function, $f^{\prime}(x)>0$
$$
\begin{aligned}
\Rightarrow \quad(2+x)^2-2(2+x)(1+x)+2 x(1+x) & >0 \\
\Rightarrow 4+x^2+4 x-4-6 x-2 x^2+2 x+2 x^2 & >0 \\
x^2 & >0
\end{aligned}
$$
This shows $x$ lies between $-\infty$ to $\infty$.
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