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Question

Given that
f(α)=∣ ∣cos(x+α)sin(x+α)cos2αsinxcosxsinαcosxsinxcosα∣ ∣=A+Bcos2α
then

A
A=0, B=2
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B
A=1, B=1
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C
A=1, B=0
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D
none of these
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Solution

The correct option is A A=0, B=2
f(α)=∣ ∣cos(x+α)sin(x+α)cos2αsinxcosxsinαcosxsinxcosα∣ ∣


=∣ ∣ ∣cosxcosαsinxsinαsinxcosαcosxsinαcos2αsin2αsinxcosxsinαcosxsinxcosα∣ ∣ ∣


=∣ ∣cosxcosαsinxcosαcos2αsinxcosxsinαcosxsinxcosα∣ ∣∣ ∣ ∣sinxsinαcosxsinαsin2α000000∣ ∣ ∣


=cosα∣ ∣cosxsinxcosαsinxcosxsinαcosxsinxcosα∣ ∣

Applying R3R3+R1

f(α)=cosα∣ ∣cosxsinxcosαsinxcosxsinα002cosα∣ ∣=2cos2α

By comparing with given A=0 and B=2

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