Question

$f\left(x\right)=\frac{1}{2}-tan\left(\frac{\pi x}{2}\right)[-1<x<1]$

and g(x) =$\sqrt{(3+4x-4x^2)}$

Find domain of (f + g).

• A. $[\frac{-1}{2},1)$
• B. $(\frac{-1}{2},1]$
• C. $[-\frac{1}{2},\frac{3}{2}]$
• D. $(-1,1)$

Answer 1

The correct option is A

$[\frac{-1}{2},1)$

$4x^2-4x-3\le 0$

$(2x+1)(2x-3)\le 0$

$\Rightarrow x\in \left[\begin{array}{c}\frac{-1}{2},\frac{3}{2}\end{array}\right],x\in (-1,1)$

$\therefore x\in \left[\begin{array}{c}\frac{-1}{2},1\end{array}\right)$