CAT 2003 Slot 1 — QA Question 17

f(x) = |x − 2| + |2.5 − x| + |3.6 − x| = g(x) + |2.5 – x|, where g(x) = |x − 2| + |3.6 − x|

When 2 ≤ x ≤ 3.6, g(x) attains a fixed value.

This happens as in this range |x – 2| = x – 2 and |3.6 – x| = 3.6 – x

∴ |x − 2| + |3.6 − x| = x – 2 + 3.6 – x = 1.6
When x < 2, x – 2 < 0, |x – 2| > 0

Also, as –x > –2, 3.6 – x > 3.6 – 2

∴ 3.6 – x > 1.6

∴ |x − 2| + |3.6 − x| > 1.6

Similarly, when x > 3.6,

|3.6 – x| > 0 and |x – 2| > 1.6

∴ |x − 2| + |3.6 − x| > 1.6
Thus we can say that g(x) has the minimum value in the range 2 ≤ x ≤ 3.6

As f(x) = g(x) + |2.5 – x|,

∴ f(x) attains the minimum value when 2 ≤ x ≤ 3.6 and |2.5 – x| is minimum.

This happens when x = 2.5

∴ f(x) attains minimum when x = 2.5

Hence, option (b).
Alternatively,

f(x) = |x − 2| + |2.5 − x| + |3.6 − x|

Substituting the value of x from the given options in the function,

when x = 2.3

f(x) = 0.3 + 0.2 + 1.3 = 1.8

when x = 2.5

f(x) = 0.5 + 0 + 1.1 = 1.6

when x = 2.7

f(x) = 0.7 + 0.2 + 0.9 = 1.8

Substituting any arbitrary real values of x in f(x),

when x = 2

f(x) = 0 + 0.5 + 1.6 = 2.1

when x = 3

f(x) = 1 + 0.5 + 0.6 = 2.1

For any other value of x, f(x) will be greater than 1.6

Hence, f(x) is minimum when x = 2.5

Hence, option (b).