Question

Let $f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1<x<1$ and $g(x)=$ $\sqrt{3+4x-4x^2}$ then the domain $(f+g)$ is

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

Answer 1

Answer: (C)

$f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right),-1<x<1$
$g(x)=\sqrt{3+4x-4x^2}$
$=\sqrt{4-(2x-1{)}^2}$
Domain of $f(x)\Rightarrow \frac{\pi x}{2}\ne (2n+1)\frac{\pi }{2}$
$x\ne 2n+1$
Domain $\Rightarrow x\in (-1,1)$
Domain of $g(x)\Rightarrow 4-(2x-1{)}^2\ge 0$
$\Rightarrow (2x-1{)}^2\le 4$
$\Rightarrow -2\le 2x-1\le 2$
$\Rightarrow -\frac{1}{2}\le x\le \frac{3}{2}$
Domain of $g(x)\Rightarrow x\in \left[-\frac{1}{2},\frac{3}{2}\right]$
Domain of $f(x)+g(x)$ will be
$x\in (-1,1)\cap x\in \left[-\frac{1}{2},\frac{3}{2}\right]$
$\Rightarrow x\in \left[-\frac{1}{2},1\right)$