Question

Consider the function $f\left(x\right)=\frac{x^2}{x^2-1}$.The interval in which f is increasing is

• A. $(-1,1)$
• B. $(-\infty ,-1)\cup (-1,0)$
• C. $(-\infty ,\infty )-(-1,1)$
• D. $(0,1)\cup (1,\infty )$

Answer 1

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

$f\left(x\right)=\frac{x^2}{x^2-1}$
$f^{\prime }\left(x\right)=\frac{(x^2-1)2x-x^2.2x}{(x^2-1{)}^2}=\frac{-2x}{(x^2-1{)}^2}$
For $f\left(x\right)$ to be increasing
$f^{\prime }\left(x\right)>0\Rightarrow \frac{-2x}{(x^2-1{)}^2}>0$
$\Rightarrow xϵ(-\infty ,-1)\cup (-1,0)$