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