Question

The function $x+\left(\frac{1}{x}\right),(x\ne 0)$ is a non-increasing function in the interval

• A. $[–1,1]$
• B. $[0,1]$
• C. $[–1,0)\cup (0,1]$
• D. $[–1,2]$

Answer 1

The correct option is C

$[–1,0)\cup (0,1]$

The explanation for the correct answer:

Solve for the interval where the function $x+\left(\frac{1}{x}\right),(x\ne 0)$ is non-increasing

$$\begin{gathered} f\left(x\right)=x+\left(\frac{1}{x}\right) \\ f’\left(x\right)=1–\left(\frac{1}{x^2}\right)<0 \\ \Rightarrow \frac{x^2–1}{x^2}<0 \\ \Rightarrow \frac{\left(x+1\right)\left(x-1\right)}{x^2}<0 \\ \Rightarrow \left(x+1\right)\left(x-1\right)<0 \end{gathered}$$ $\left[\therefore \frac{d}{\mathrm{dx}}\left(\frac{u}{v}\right)=\frac{v\times {\displaystyle \frac{\mathrm{du}}{\mathrm{dx}}}-u\times {\displaystyle \frac{\mathrm{dv}}{\mathrm{dx}}}}{v^2}\right]$

So, $x\in \left[-1,1\right]$ but $x\ne 0$

$\Rightarrow x\in [-1,0)\cup (0,1]$

Hence, option(C) is the correct answer.