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Questions  

Let f(x)=2x(sinx+tanx)2x+21ππ41,xnπ, then f is (where [ . ] represents greatest integer function)

a
an odd function
b
an even function
c
both odd and even
d
neither odd nor even

detailed solution

Correct option is A

The denominator is=2x+21ππ−41=2xπ+21−41=221+xπ−41=2xπ+1∴ f(x)=x(sinx+tanx)xπ+12⇒f(−x)=−xsin(−x)+tan(−x)−xπ+12=x(sinx+tanx)−1−xπ+12=−x(sinx+tanx)xπ+12=−f(x)

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