this ques can be solved by using a neat “trick”.
basically, |a| is less than equal to 1 if and only if a^2 is less than equal to 1. this can be proved simply by noting that a^2 – 1= (|a| – 1)(|a|+1).
so we can square both sides in the given inequality.
we get (x^2 – 5x+4)^2/(x – 2)^2(x+2)^2 is less than equal to 1.
or (x^2 – 5x+4)^2/(x – 2)^2(x+2)^2 – 1 is less than equal to 0
or [(x^2 – 5x+4)^2 – (x^2 – 4)^2]/(x – 2)^2(x+2)^2 is less than equal to 0
factorise the numerator.
x(x – 5/2)(x – 8/5)/(x – 2)^2(x+2)^2 is greater than equal to 0.
by wavy curve method, we obtain the final ans as
x belongs to [0, 8/5] U [5/2, infinity).
kindly approve :)