Question

For the expression $\sqrt{x^2+x}+\frac{1}{\sqrt{x^2+x}}>0,x\in$

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

Answer 1

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

For given expression to be valid $(x^2+x)>0$
$\Rightarrow x(x+1)>0$
$\therefore x\in (-\infty ,-1)\cup (0,\infty )$