MathematicsRelations and FunctionsJEE

Relations and Functions · Part 11 of 20

Modulus Function

Read |x| as a distance and most modulus problems become one picture: V-shaped graphs, the properties of modulus, and ranges of expressions with modulus.

Builds on: Part 9 · Inequalities and the Wavy Curve Method.

Video coming soon

Graphs

V-shapes

y = |x| has its vertex at O; y = |x − a| moves it to (a, 0) and passes through (0, a). |f(x)| flips the part below the axis up, as for |x² − 4|.

Three modulus graphs. — Purple |x|, orange |x − 2|, green |x² − 4|.

Properties

Rules of modulus

$∣ ab ∣ = ∣ a ∣∣ b ∣$ , $\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}$ , $x^{2} = ∣ x ∣^{2} = ∣ x^{2} ∣$ (the book's ≠ is a misprint), $∣ x ∣ = a > 0 \Rightarrow x = \pm a$ , and $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ .

Ranges

Read the V, not the ends

For −3 ≤ x < 1 the interval crosses the vertex, so |x| ∈ [0, 3]; the ends alone give only 1 and 3.

V graph with an interval and its range. — Range marked on the y-axis.

Graph of x plus |x|. — Flat for x < 0, slope 2 after.

Example

x + |x|

x + |x| is 2x for x ≥ 0 and 0 for x < 0: range [0, ∞). 1/√(x + |x|) needs x + |x| > 0, so its domain is (0, ∞).

Summary

Key formulas

∣ x ∣ = x (x \geq 0), - x (x < 0)

Definition and graph

∣ f (x) ∣ = f (x) (f \geq 0), - f (x) (f < 0)

Definition and graph

∣ ab ∣ = ∣ a ∣ ∣ b ∣

Properties of modulus

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

Properties of modulus

x^{2} = ∣ x ∣^{2} = ∣ x^{2} ∣

Properties of modulus

∣ x ∣ = a (a > 0) \Rightarrow x = \pm a

Properties of modulus

∣ a \pm b ∣^{2} = a^{2} + b^{2} \pm 2 ab

Properties of modulus

(∣ a ∣ \pm ∣ b ∣)^{2} = a^{2} + b^{2} \pm 2∣ a ∣∣ b ∣

Properties of modulus

Worked examples

From the video

1. Values of |x| for 2 < x < 4, −3 ≤ x ≤ −1, −3 ≤ x < 1; |x − 2| for −5 < x < 7; |2x − 7| for 1 ≤ x ≤ 5.

(2, 4); [1, 3]; [0, 3]; [0, 7); [0, 5].

2. Range of x + |x| and domain of 1/√(x + |x|).

[0, ∞); (0, ∞).

JEE-style question

Your turn

The domain of f(x) = 1/√(|x| − x) is:

(a)(−∞, 0)

(b)(0, ∞)

(c)(−∞, 0]

(d)R

Show the answer and the traps

|x| − x = 0 for x ≥ 0 and −2x > 0 for x < 0. So the domain is (−∞, 0). Option (a).

(b) is the domain of 1/√(x + |x|), the mirror case. (c) includes 0, where the denominator is 0.

(d) forgets that |x| − x is 0 for every x ≥ 0.

Watch out

Common mistakes

Writing √(x²) = x√(x²) = |x|.

Taking |·| of the interval ends onlyIf the interval crosses the vertex, the range starts at 0.

Believing x² ≠ |x²| (book misprint)x² = |x|² = |x²| for every real x.

Practice

Try these

1. Solve |x − 3| = 5.

x = 8 or x = −2.

2. Solve |2x − 1| < 3.

−1 < x < 2.

3. Range of |x − 1| for −2 ≤ x < 4.

[0, 3].

4. Range of f(x) = |x| − x.

[0, ∞).