Q92 / 150 · MHT CET 2007 · Mathematics

The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

  • Correct answer
    A
    $(0, \infty)$
  • B
    $(-\infty, 0)$
  • C
    $(-\infty, \infty)$
  • D
    None of these

Solution

Given, $f(x)=\log (1+x)-\frac{2 x}{2+x}$ $$ \begin{aligned} \therefore \quad f^{\prime}(x) &=\frac{1}{1+x}-\frac{(2+x) \cdot 2-2 x}{(2+x)^{2}} \\ &=\frac{x^{2}}{(1+x)(x+2)^{2}} \end{aligned} $$ Clearly, $f^{\prime}(x)>0$ for all $x>0$. Hence, $f(x)$ is increasing on $(0, \infty)$.

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

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