KEAM 2016 Math Paper

Given inequality is
|x−1|+|x−3|≤8
Case I x<1
⇒‌‌−(x−1)−(x−3)≤8
⇒‌‌−x+1−x+3≤8
⇒‌‌−2x+4≤8
⇒‌‌−2x≤4⇒x≥−2
Interval [-2,1)
Case II 1≤x<3
⇒ x−1−x+3≤8
⇒‌‌2≤8
Interval [1,3]
Case III x≥3
⇒‌‌x−1+x−3≤8
⇒‌‌2x−4≤8
⇒‌‌2x≤12
⇒‌‌x≤6
Interval [3,6]
From all the intervalvalues of x lie in the interval [-2,6]
Hence, (b) is the correct option.