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Question

Let f(x)=∣ ∣cosxsinxcosxcos2xsin2x2cos2xcos3xsin3x3cos3x∣ ∣
Then

A
f(0)=0
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B
f(π/2)=4
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C
f(π)=0
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D
f(π)=1
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Solution

The correct options are
A f(0)=0
B f(π/2)=4
C f(π)=0
f(x)=∣ ∣cosxsinxcosxcos2xsin2x2cos2xcos3xsin3x3cos3x∣ ∣f(x)=∣ ∣sinxsinxcosx2sin2xsin2x2cos2x3sin3xsin3x3cos3x∣ ∣+∣ ∣cosxcosxcosxcos2x2cos2x2cos2xcos3x3cos3x3cos3x∣ ∣+∣ ∣cosxsinxsinxcos2xsin2x4sin2xcos3xsin3x9sin3x∣ ∣f(0)=0=f(π),f(π2)=4

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