Question

$f\left(x\right)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1<x<1$
and $g\left(x\right)=\sqrt{(3+4x-4x^2)}$.
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 $[\frac{-1}{2},1)$

$g\left(x\right)=\sqrt{(3+4x-4x^2)}=\sqrt{(1+2x)(3-2x)}\Rightarrow \frac{-1}{2}\le x\le \frac{3}{2}$
$f\left(x\right)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1<x<1$
$\therefore \text{domain of (f+g) is}[\frac{-1}{2},1)$