MathematicsRelations and FunctionsJEE

Relations and Functions · Part 9 of 20

Inequalities and the Wavy Curve Method

Domains and ranges need inequalities done safely: multiplying by negatives, squaring and inverting ranges, and the wavy curve (sign scheme) method for rational inequalities.

Builds on: Part 4 · Functions: Definition, Intervals and Domain.

Video coming soon

Squaring

x² for −3 < x ≤ 4

The interval crosses 0, so x² starts at 0: x² ∈ [0, 16]. Squaring only the ends gives [9, 16], which misses everything below 9.

Parabola with an interval and its range. — Read the range off the parabola.

Hyperbola with the two pieces of the range. — −3 is open, so −1/3 is open.

Reciprocals

1/x for −3 < x ≤ 4

Same sign: $x \in [a, b] \Rightarrow \frac{1}{x} \in [\frac{1}{b}, \frac{1}{a}]$ . Crossing 0, it splits: $\frac{1}{x} \in (- \infty, - \frac{1}{3}) \cup [\frac{1}{4}, \infty)$ .

Sign scheme

The wavy curve

Mark the zeros of numerator and denominator. Start with + on the far right, change sign at odd powers, bounce at even powers, and never include denominator zeros. For the expression shown: + on (0, 1) ∪ (1, 3) ∪ (5, ∞), − on (−∞, −2) ∪ (−2, 0) ∪ (3, 5).

Wavy curve over −2, 0, 1, 3, 5. — Bounces at −2 and 1 (even powers).

Summary

Key formulas

x^{2} \leq a^{2} ⟺ - a \leq x \leq a

Squares and reciprocals of ranges

x^{2} \geq a^{2} ⟺ x \leq - a or x \geq a

Squares and reciprocals of ranges

ab > 0 : \frac{1}{x} \in [\frac{1}{b}, \frac{1}{a}]

Squares and reciprocals of ranges

a < 0 < b : \frac{1}{x} \in (- \infty, \frac{1}{a}] \cup [\frac{1}{b}, \infty)

Squares and reciprocals of ranges

Worked examples

From the video

1. If −3 < x ≤ 4, find the range of x² and of 1/x.

x² ∈ [0, 16]; 1/x ∈ (−∞, −1/3) ∪ [1/4, ∞).

2. Solve 2/x − 3 < 0.

x < 0 or x > 2/3.

3. Solve x² − x − 1 ≤ 0.

(1 − √5)/2 ≤ x ≤ (1 + √5)/2.

JEE-style question

Your turn

The solution set of (x − 1)²(x + 2)/(x − 3) ≤ 0 is:

(a)[−2, 3)

(b)[−2, 3]

(c)[−2, 1) ∪ (1, 3)

(d)(−∞, −2] ∪ (3, ∞)

Show the answer and the traps

+ right of 3, − on (1, 3), bounce at 1, − on (−2, 1), + left of −2. Include −2 and 1, exclude 3: [−2, 3). Option (a).

(b) includes 3, where the denominator is zero. (c) drops x = 1, where E = 0 satisfies ≤.

(d) is where E is positive: the wrong side.

Watch out

Common mistakes

Squaring the ends of an interval containing 0x² starts at 0 there.

Cross-multiplying by an expression of unknown signBring everything to one side and use the sign scheme.

Including zeros of the denominatorThe expression is undefined there.

Practice

Try these

1. If −2 ≤ x ≤ 3, find the ranges of x² and 1/x (x ≠ 0).

x² ∈ [0, 9]; 1/x ∈ (−∞, −1/2] ∪ [1/3, ∞).

2. Solve (x − 2)/(x + 3) ≥ 0.

x < −3 or x ≥ 2.

3. Solve (x + 1)²(x − 4)/(x − 1) < 0.

1 < x < 4.

4. Solve x² − 5x + 6 > 0.

x < 2 or x > 3.