If $\log (1+x)-\frac{2 x}{2+x}$ is increasing, then
If $\log (1+x)-\frac{2 x}{2+x}$ is increasing, then
- $0 < x < \infty$
- $-\infty < x < 0$
- $-\infty < x < \infty$
- $-1 < x < 2$
Solution
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 $\infty$
Asked in: AP EAMCET 2002
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