Question

The function $\frac{x-2}{x+1},(x\ne -1)$is increasing on the interval

• A. $(–\infty ,0]$
• B. $[0,\infty )$
• C. $R$
• D. None of these

Answer 1

The correct option is C

$R$

The explanation for the correct answer:

Solve for the required interval where the function is increasing.

$f\left(x\right)=\frac{x-2}{x+1},(x\ne -1)$

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left(x+1\right)-\left(x-2\right)}{{\left(x+1\right)}^2},(x\ne -1) \\ f\text{'}\left(x\right)=\frac{3}{{(x+1)}^2} \end{gathered}$$ $\left[\because \frac{d\left({\displaystyle \frac{u}{v}}\right)}{\mathrm{dx}}=\frac{v.{\displaystyle \frac{\mathrm{du}}{\mathrm{dx}}}-u.{\displaystyle \frac{\mathrm{dv}}{\mathrm{dx}}}}{v^2}\right]$

Since $f\text{'}\left(x\right)=\frac{3}{{(x+1)}^2}$ is greater than $0$ in interval $(-\infty ,\infty )$

$\therefore$ $f\left(x\right)=\frac{x-2}{x+1},(x\ne -1)$ is increasing in interval $(-\infty ,\infty )$or $R$

Hence, option(C) is the correct answer.