case 1:
for
2x-1 >=0
=> x >=1/2
f(2x-1)< f(1/3)
=> 2x-1 < 1/3
x < 2/3
solution for case 1
x >=1/2 and x x > 1/3
solution for case 2
x 1/3
ie 1/3< x < 1/2
solution for f(2x-1)
case 1:
for
2x-1 >=0
=> x >=1/2
f(2x-1)< f(1/3)
=> 2x-1 < 1/3
x < 2/3
solution for case 1
x >=1/2 and x x > 1/3
solution for case 2
x 1/3
ie 1/3< x < 1/2
solution for f(2x-1)