Q109 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics

If $x>0$, then $\frac{x}{1+x}-\log (1+x)$

  • Correct answer
    A
    is less than zero
  • B
    is greater than zero
  • C
    is equal to zero
  • D
    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, $\quad f(x) < 0$

Hence, the correct answer is option: 'A'.

Relevant Questions