Question

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

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

Answer 1

Answer: (A)

Domain of $(f+g)=$ Domain of $f\cap$ Domain of $g$
Now, domain of $f(x)=(-1,1)$
Domain of $g(x)$ is defined when
$3+4x-4x^2\ge 0$
$\Rightarrow 4x^2-4x-3\le 0$
$\Rightarrow 4x^2-6x+2x-3\le 0$
$\Rightarrow 2x(2x-3)+1(2x-3)\le 0$
$\Rightarrow (2x-3)(2x+1)\le 0$
$\Rightarrow x\in \left[\frac{-1}{2},\frac{3}{2}\right]$
$\therefore$ Domain of $(f+g)=(-1,1)\cap \left[-\frac{1}{2},\frac{3}{2}\right]$
$=\left[-\frac{1}{2},1\right)$