Question

|x²– 2x| +|x – 4| >|x²– 3x +4|.
solve for x belongs to R​

Answer 1

Answer:

(x2−2x)(x−4)x(x−2)(x−4)0<x<2orx>0>0>4

Step-by-step explanation:

Note that |x2−2x|+|x−4|=|x2−2x|+|−x+4|≥|x2−3x+4| with the equality holds if and only if x2−2x and −x+4 are of the same sign.

Therefore, |x2−2x|+|x−4|>|x2−3x+4| if and only if x2−2x and x−4 are of the same sign. So we have

(x2−2x)(x−4)x(x−2)(x−4)0<x<2orx>0>0>4