Question

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

• 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 $[\frac{-1}{2},1)$

$f\left(x\right)+g\left(x\right)=\sqrt{3+4x-4x^2}+\frac{1}{2}-tan\left(\frac{\pi x}{2}\right)$
Therefore, for real values of $(f+g)$,
$3+4x-4x^2\ge 0$
$4x^2-4x-3\le 0$
$4x^2-6x+2x-3\le 0$
$2x(2x-3)+(2x-3)\le 0$
$(2x+1)(2x-3)\le 0$
$x\ge \frac{-1}{2}$ and $x\le 3$
Therefore
$xϵ[\frac{-1}{2},3]$ ...(i)
Also consider $tan\left(\frac{\pi .x}{2}\right)$. Since $tanx$ is discontinuous for all $xϵ\frac{(2n+1)\pi }{2})$ hence for $tan\left(\frac{\pi .x}{2}\right)$ the domain is $R-1,3,\mathrm{5...}(2n+1)$ ...(ii)
Hence from i and ii
$xϵ[\frac{-1}{2},1)\cup (1,3)$