Let f(x) = frac{1}{2} - tan left(frac{pi x}{2 | vedclass

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(C) Given,$f(x) = \frac{1}{2} - 	an \left(\frac{\pi x}{2}ight)$ with domain $(-1, 1)$ and $g(x) = \sqrt{3 + 4x - 4x^2}$.To find the domain of $(f +

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$\left[-\frac{1}{2}, 1\right)$

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(C) Given,$f(x) = \frac{1}{2} - 	an \left(\frac{\pi x}{2}ight)$ with domain $(-1, 1)$ and $g(x) = \sqrt{3 + 4x - 4x^2}$.To find the domain of $(f + g)$,we need the intersection of the domains of $f(x)$ and $g(x)$.For $g(x)$ to be defined,$3 + 4x - 4x^2 \geq 0$.Multiplying by $-1$,we get $4x^2 - 4x - 3 \leq 0$.Factoring the quadratic,$(2x + 1)(2x - 3) \leq 0$.This inequality holds for $x \in \left[-\frac{1}{2}, \frac{3}{2}ight]$.The domain of $(f + g)$ is the intersection of $(-1, 1)$ and $\left[-\frac{1}{2}, \frac{3}{2}ight]$.Intersection $= (-1, 1) \cap \left[-\frac{1}{2}, \frac{3}{2}ight] = \left[-\frac{1}{2}, 1ight)$.

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 of $(f + g)$ is

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