The function f(x) = |x - 2| + |2.5 - x| + |3.6 - x|, where x is a real number, attains minimum at
Option (b)
f(x) = | x – 2| + | 2.5 – x | + | 3.6 – x | can attain minimum value when either of the terms = 0.
Case I :
When | x – 2 | = 0 => x = 2, value of f(x) = 0.5 + 1.6 = 2.1.
Case II.
When | 2.5 – x | = 0 => x = 2.5
value of f(x)
= 0.5 + 0 + 1.1 = 1.6.
Case III.
When | 3.6 – x | = 0 => x = 3.6
f(x) = 1.6 + 1.1 + 0 = 2.7. Hence the minimum value of f(x) is 1.6 at x = 2.5.