2f(x)+xf(1x)−2f(|2sin(π(x+14)|)=4cos2πx2+xcosπx , ∀x∈R,x≠0  for some real function f , then

replacing x by 2, we get 2f(2)+2f(12)−2+|2sin9π4| =4cos2π+2cosπ2 ⇒2f(2)+2f(12)−2f(1)=4 f(2)+f(12)=2+f(1)                         ……(1)Now replacing x by 1, we get 2f(1)+f(1)−2f(1)=4.0−1 ⇒f(1)=−1                                                   ……(2)Now replace x by 12 ,2f(12)+12f(2)−2f(1)=2+12 ⇒          2f(12)+12f(2)+2=52                        ….(3)⇒  f(2)  and  f(12)  can be obtained from (1) and (3)