The correct answer is Option 1: injective
Checking for Injectivity (One-to-One)
A function is injective if every distinct element in the domain maps to a distinct element in the codomain.
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The domain is $[0, \frac{\pi}{2}]$.
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When $x$ ranges from $0$ to $\frac{\pi}{2}$, the argument of the sine function, $\frac{x}{2}$, ranges from $0$ to $\frac{\pi}{4}$.
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In the interval $[0, \frac{\pi}{4}]$, the sine function is strictly increasing.
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Since the function is strictly monotonic (constantly increasing) over the given domain, it will never repeat a value. Therefore, it is injective.
Checking for Surjectivity (Onto)
A function is surjective if its Range is equal to its Codomain.
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Codomain: Given as $[0, \infty)$.
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Range Calculation: * When $x = 0$, $y = \sin(0) = 0$.
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When $x = \frac{\pi}{2}$, $y = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \approx 0.707$.
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Thus, the range is $[0, \frac{1}{\sqrt{2}}]$.
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Since the range $[0, \frac{1}{\sqrt{2}}]$ is only a small subset of the codomain $[0, \infty)$, the function is not surjective.
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