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+4 x-4 x^{2} \geq 0 $
$\Rightarrow 4 x^{2}-4 x-3 \leq 0$
$\Rightarrow 4 x^{2}-6 x+2 x-3 \leq 0 $
$\Rightarrow 2 x(2 x-3)+1(2 x-3) \leq 0$
$\Rightarrow (2 x-3)(2 x+1) \leq 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] $
$=\left[-\frac{1}{2}, 1\right)$