m^2/(1-m) > m
m^2/(1-m)-(m)>0
m((m/(1-m))-1)>0
m*(2m-1)/(1-m)>0?
So the question can be rephrased to
If m!=1, m*(2m-1)/(1-m)>0?
1) m>0
We shall consider plugging in numbers in three intervals (0,1/2), (1/2,1), and >1
m = 1/4, m*(2m-1)/(1-m) = +ve*(-ve)/(+ve) < 0
m = 3/4, m*(2m-1)/(1-m) = +ve*(+ve)/(+ve) > 0
m = 2, m*(2m-1)/(1-m) = +ve*(+ve)/(-ve) < 0.
Insufficient to answer the question
m = 1/4, m*(2m-1)/(1-m) = +ve*(-ve)/(+ve) < 02) m is not an integer
m = 3/4, m*(2m-1)/(1-m) = +ve*(+ve)/(+ve) > 0
Insufficient to answer the question
m = 1/4, m*(2m-1)/(1-m) = +ve*(-ve)/(+ve) < 0From 1 and 2
m = 3/4, m*(2m-1)/(1-m) = +ve*(+ve)/(+ve) > 0
Insufficient to answer the question
IMO E
