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Q.

If f(x)=sinx-cosx is written as f1(x)+f2(x) where f1(x) is even and f2(x) is odd then

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a

f1(x)=cosx

b

f1(x)=-cosx

c

f2(x)=-sinx+cosx

d

f2(x)=sin(2π-x)

answer is B.

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Detailed Solution

f1(x)=f(x)+f(-x)2=12Sinx-cosx-sinx-cosx=-cosx f2(x)=f(x)-f(-x)2=12sinx-cosx+sinx+cosx=sinx
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