Given inequality is |x−1|+|x−3|≤8 Case 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.