Q97 / 180 · KCET 2022 · Mathematics

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

  • A
    $(-\infty, \infty)$
  • B
    $(\infty,-1)$
  • Correct answer
    C
    $(-1, \infty)$
  • D
    $(-\infty, 0)$

Solution

Given, $f(x)=\log (1+x)-\frac{2 x}{2+x}$ Differentiating the function w.r.t. $x$, we get $$ \begin{aligned} f^{\prime}(x) & =\frac{1}{(1+x)}(0+1)-2\left[\frac{(2+x) \times 1-x(0+1)}{(2+x)^2}\right] \\ & =\frac{1}{1+x}-2\left[\frac{2+x-x}{(2+x)^2}\right]=\frac{1}{1+x}-\frac{4}{(2+x)^2} \\ & =\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2}=\frac{4+x^2+4 x-4-4 x}{(x+1)(x+2)^2} \\ & =\frac{x^2}{(x+1)(x+2)^2}=\left(\frac{x}{x+2}\right)^2 \cdot \frac{1}{(x+1)} \end{aligned} $$ Sign of $f^{\prime}(x)$ depends on the sign of $\frac{1}{(x+1)}$ $$ \begin{aligned} & f^{\prime}(x)>0 \text { when } x>-1 \Rightarrow \text { increasing } \\ & f^{\prime}(x) < 0 \text { when } x < -1 \Rightarrow \text { decreasing } \end{aligned} $$ Hence, $f(x)$ is increasing on $(-1, \infty)$.

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

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