$\sqrt{\text{x}-1}$
$\sqrt{\text{x}}$
$\sqrt{\text{x}+1}$
$-\sqrt{\text{x}}$
Solution:
Given that
$\text{f(x)}=\sin^2\text{x}$ and the composite function $\text{g(f(x))}=|\sin\text{x}|$We will do it using trial and error method.
If we take
$\text{g(x)}=-\sqrt{\text{x}}$ and $\text{f(x)}=\sin^2\text{x}$$\text{g(f(x))}=\text{g}(\sin^2\text{x})$
$=-\sin\text{x}$
Which contradicts to the $\text{g(f(x))}=|\sin\text{x}|$
Hence, we take $\text{g(x)}=\sqrt{\text{x}}$
$\text{g(f(x))}=\text{g}(\sin^2\text{x})$
$=\sqrt{\sin^2\text{x}}=|\sin\text{x}|$
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