Given inequality is ∣x−1∣+∣x−3∣≤8Case Ix<1⇒−(x−1)−(x−3)≤8⇒−x+1−x+3≤8⇒−2x+4≤8⇒−2x≤4⇒x≥−2 Interval [-2,1) Case II1≤x<3⇒x−1−x+3≤8⇒2≤8 Interval [1,3] Case IIIx≥3⇒x−1+x−3≤8⇒2x−4≤8⇒2x≤12⇒x≤6 Interval [3,6] From all the interval
values of x lie in the interval [-2,6] Hence, (b) is the correct option.
