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(C) Given that $f: A \rightarrow B$ is an onto function, the range of $f(x)$ must be equal to the codomain $B$.Since $g: B \rightarrow C$ is defined a

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$[-\frac{1}{2}, \frac{3}{2}]$

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(C) Given that $f: A \rightarrow B$ is an onto function, the range of $f(x)$ must be equal to the codomain $B$.Since $g: B \rightarrow C$ is defined as $g(x) = \sqrt{3 + 4x - 4x^2}$, the domain of $g$ is $B$.For $g(x)$ to be defined, the expression inside the square root must be non-negative:$3 + 4x - 4x^2 \geq 0$Multiplying by $-1$ reverses the inequality:$4x^2 - 4x - 3 \leq 0$Factoring the quadratic expression:$(2x - 3)(2x + 1) \leq 0$The roots are $x = -\frac{1}{2}$ and $x = \frac{3}{2}$.Testing the intervals, the inequality holds for $x \in [-\frac{1}{2}, \frac{3}{2}]$.Thus, the domain of $g$ is $[-\frac{1}{2}, \frac{3}{2}]$, which is the range of $f$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

Let $f: A \rightarrow B$ be defined as $f(x) = \frac{1}{2} - \tan \left(\frac{\pi x}{2}\right)$ and $g: B \rightarrow C$ be defined as $g(x) = \sqrt{3 + 4x - 4x^2}$. If $A, B, C$ are subsets of $\mathbb{R}$ and $f$ is an onto function, then the range of the function $f(x)$ is:

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