MCQ
If $f(x) = \sin^2x$ and the composite function $\text{g(f(x))} = |\sin\text{x}|,$ then $g(x)$ is equal to:
- A$\sqrt{\text{x}-1}$
- ✓$\sqrt{\text{x}}$
- C$\sqrt{\text{x}+1}$
- D$-\sqrt{\text{x}}$
✓
Answer
Correct option: B.
$\sqrt{\text{x}}$
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 $\text{g(x)}=\sqrt{\text{x}}$
$\text{g(f(x))}=\text{g}(\sin^2\text{x})$
$=\sqrt{\sin^2\text{x}}=|\sin\text{x}|$
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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