Question

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

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

Answer 1

The correct option is A $(-1,0)\cup (0,1)$

We have,

$f\left(x\right)=x+\frac{1}{x}\left(x\ne 0\right)$ is incresing if

$f\left(x\right)=⩾0i.e1-\frac{1}{x^2}⩾0$

So,$f\left(x\right)$ will be non-incresing if

$f^{\prime }\left(x\right)<0i.e1-\frac{1}{x^2}<0$

$x\in \left(-1,1\right)$ but at $x=0$

$f\left(x\right)$ is not detined.

So,Required solution is $x\in \left(-1,0\right)\cup \left(0,1\right)$

Then,

Option $A$ is correct answer.