If $x>0$, then $\frac{x}{1+x}-\log (1+x)$
If $x>0$, then $\frac{x}{1+x}-\log (1+x)$
- is less than zero
- is greater than zero
- is equal to zero
- takes all the real values
Solution
Let a function $f(x)=\frac{x}{1+x}-\log (1+x)$, for $x>0$
$
\begin{aligned}
\because \quad f^{\prime}(x) & =\frac{(x+1)-x}{(1+x)^2}-\frac{1}{(1+x)} \\
& =\frac{1}{(1+x)^2}-\frac{1}{1+x}=\frac{1-(1+x)}{(1+x)^2} \\
& =\frac{-x}{(1+x)^2} < 0, \text { for } x>0
\end{aligned}
$
$\because \quad f^{\prime}(x) < 0, \forall x>0$
$\therefore f(x)$ is a strictly decreasing function for $x < 0$.
Now, $f\left(0^{+}\right)=0$
So, $\quad f(x) < 0$
Asked in: AP EAMCET 2018 (22 Apr Shift 2)
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