Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

The minimum value of $f(x)=|2 x-1|$ is

Options:

$-\infty$

0

$\frac{1}{2}$

1

Correct Answer:

0

Explanation:

f(x) = |2x - 1| → since it is a modules function O well be minimum value

looking from its graph

f(x) is zero at $x = \frac{1}{2}$

Min value = 0