Re: |x^2 - 2x|

Hi,

The best approach is to put different value of x and validate the inquality.
For ex, x=0, LHS=RHS (thus, inequality doesn't hold true.)
So, answer choice that has got 0 in the range will be eliminated.
a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3

Put x= 1, LHS=RHS
a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3

Put x=4, LHS = RHS
a. 1<x<3
b. -1<x<3
c. 0<x<4
d. x>3
Thus, the only choice which remains is (A)

Check, put x=2, LHS < RHS
Therefore, Answer (A).