The correct option is D −4
||x−|x−2|+3|−4|=3
Case 1: x<2
⇒||x+x−2+3|−4|=3
⇒||2x+1|−4|=3
If x≥−12 or If x<−12
|2x−3|=3 or |−2x−5|=3
⇒2x−3=±3 or −2x−5=±3
⇒x=0 (3 rejected as x<2) or x=−4,−1
∴x=−4,−1,0
Case 2: x≥2
||x−x+2+3|−4|=3
⇒1=3 No solution
∴ The solutions are x=−4,−1,0