If f(x) is a real-valued function defined as f(x)=ln⁡(1−sin⁡x), then the graph of f(x) is

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symmetric about the line x=π

symmetric about the y-axis

symmetric about the line x=π2

symmetric about the origin

answer is C.

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Detailed Solution

fπ2−x=ln⁡(1−cos⁡x) and fπ2+x=ln⁡(1−cos⁡x) Thus, fπ2+x=fπ2−x Thus, f(x) is symmetrical about line x=π2.

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