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Question

Given the function f(x) such that 2f(x)+xf(1x)2f(2sinπ(x+14))=4cos2πx2+xcos(πx), then which one of the following is correct ?

A
f(2)+f(12)=1
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B
f(1)=1, but the values of f(2),f(12) cannot be determined
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C
f(2)+f(1)=f(12)
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D
f(2)+f(1)=0
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Solution

The correct options are
A f(2)+f(1)=f(12)
B f(2)+f(12)=1
C f(2)+f(1)=0
Given 2f(x)+xf(1x)2f(2sinπ(x+14))=4cos2πx2+xcos(πx) .....(1)

Put x=1 in (1)
2f(1)+1f(1)2f(2sin5π4)=1
3f(1)2f(1)=1
f(1)=1

Now put x=2 in (1)
2f(2)+2f(12)2f(1)=4cos2π+2cosπ2
2f(2)+2f(12)2f(1)=4
f(2)+f(12)=1 .......... (2)

Now put x=12 in (1), we get
2f(12)+12f(2)2f(2sin(3π4))=4cos2π4+12cos2π
4f(12)+f(2)=1 .......... (3)

From (2) and (3)
f(12)=0 & f(2)=1
Hence, f(2)+f(1)=0=f(12)

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