Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Relations and Functions

Question:

The range of the function f(x) = |x − 1| + |x − 2|, for −1 ≤ x ≤ 3, is

Options:

[2, 5]

[1, 5]

[3, 5]

none of these

Correct Answer:

[1, 5]

Explanation:

f(x) = |x −1| + |x − 2| , −1 ≤ x ≤ 3

If x < 1, f(x) = (1 − x) + (2 − x) = 3 − 2x

In this interval, f(x) is decreasing.

If 1 ≤ x < 2, f(x) = x − 1 + 2 − x = 1

In this interval, f(x) is constant.

If 2 ≤ x ≤ 3, f(x) = x − 1 + x − 2 = 2x − 3 

In this interval, f(x) is increasing

∴ Max. f(x) = the greatest among f(−1) and f(3) = 5

Min. f(x) = f(1) = 1

Hence range = [1, 5].