Question

The midpoint of the interval in which $x^2-2(\sqrt{-x}{)}^2-3<0$ is satisfied, is

• A. $\frac{-3}{2}$
• B. -2
• C. $\frac{1}{2}$
• D. $\frac{-3}{4}$

Answer 1

Answer: (A)

$x^2-2(\sqrt{-x}{)}^2-3<0$ is valid only if $x<0$
Then $x^2+2x-3<0\Rightarrow (x+3)(x-1)<0\Rightarrow -3<x<1$
Since $x<0$ hence $x\in (-3,0)$
The midpoint of the interval is $\frac{-3}{2}$