given in question f (1)=1
f(2n)=2f(n)-1{equation one}
f(2n+1)=2f(n)+1 {equation second }
lets start with finding
putting n=1 in equation 1
f(2)=2f(1)-1=1
f(3)=3 using equation second by putting n=1
using equation alternate
f(4)=1
f(5)=3
f(6)=5
f(7)=7
f(8)=1
f(9)=3
f(10)=5
going option number A
put n=2
f(3)=3 valid so A is correct
going option B put n=2
f(4)=1 valid option B also correct
going option c put n=1
f(10)= 5 valid so option c is correct
going option D put n=2
f(5)=5 invalid so D is false
so overall option A ,B,C is correct