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Questions on the domain of a function, the sign of a quadratic or the intervals where a function increases all end in the same step: solve an inequality with several factors. Testing values one at a time is slow. The wavy curve method, also called the method of intervals, does it in one pass. This post shows how it works, why, and where students go wrong.
Where does it show up in JEE?
Neither syllabus names the method, so it is a tool rather than a topic. The JEE (Main) 2026 syllabus lists quadratic equations, functions and "monotonic-increasing and decreasing functions". The JEE (Advanced) 2026 brochure lists domain and range of functions. Each needs you to find where an expression is positive or negative. The basic rules of inequalities are taught in the NCERT Class 11 chapter on Linear Inequalities, which our post on linear inequalities in Class 11 covers.
What goes wrong without a method?
Tsamir and Almog (2001) studied how secondary school students solved linear, quadratic, rational and square-root inequalities. Students who used a graph of the parabola usually got rational and quadratic inequalities right. Errors came from failing to reject excluded values and from choosing the wrong "and" or "or" between conditions. The most common source of difficulty was treating inequalities as if they were equations. The wavy curve method is a graph in compressed form, which is why it helps.
How does the method work?
Write the expression as a product and quotient of factors of the form (x − a), each with a positive coefficient of x.
Mark critical points. These are the zeros of every factor, top and bottom, on a number line.
Start on the right. For x larger than every critical point, every factor is positive, so the expression is positive. Write + there.
Move left and alternate. Each time you cross a critical point, a factor changes sign, so the expression changes sign.
Except at even powers. A factor such as (x − 2)² does not change sign, so the sign stays the same across that point.
Read off the answer. Take the intervals with the sign you need. Include zeros of the numerator for ≥ or ≤. Never include zeros of the denominator.
Worked example 1: a basic rational inequality
Solve (x − 1)(x + 2)/(x − 3) ≥ 0.
Critical points: −2, 1, 3. The point 3 makes the denominator zero, so it is excluded.
For x > 3: +.
Between 1 and 3: −. Between −2 and 1: +. Below −2: −.
We need ≥ 0, so take the + intervals and the zeros of the numerator.
Answer: −2 ≤ x ≤ 1, or x > 3.
Check: at x = 0, (−1)(2)/(−3) = 2/3, positive. At x = 2, (1)(4)/(−1) = −4, negative. Both agree.
Worked example 2: a repeated factor
Solve (x − 2)²(x + 1)/(x − 4) < 0.
Critical points: −1, 2 (squared), 4 (excluded).
For x > 4: +. Between 2 and 4: −.
Crossing 2 does not change the sign, so between −1 and 2: −.
Below −1: +.
We need < 0, strictly. At x = 2 the expression equals 0, so remove it.
Answer: −1 < x < 2, or 2 < x < 4.
Check: at x = 0, (4)(1)/(−4) = −1, negative. At x = 3, (1)(4)/(−1) = −4, negative.
Worked example 3: do not cross-multiply
Solve (2x − 1)/(x + 3) > 1.
A common wrong step is to multiply both sides by x + 3, giving x > 4. That assumes x + 3 is positive. Instead, bring everything to one side:
(2x − 1)/(x + 3) − 1 > 0, which simplifies to (x − 4)/(x + 3) > 0.
Critical points −3 and 4. Signs from the right: +, −, +.
Answer: x < −3, or x > 4. The cross-multiplying method loses the whole interval x < −3. Check x = −4: (−9)/(−1) = 9, which is greater than 1.
Which mistakes cost marks?
A negative coefficient of x. For (3 − x)(x + 1) ≥ 0, rewrite as −(x − 3)(x + 1) ≥ 0, then multiply by −1 and flip: (x − 3)(x + 1) ≤ 0, giving −1 ≤ x ≤ 3.
Factors that are always positive. x² + x + 1 has a negative discriminant, so it is positive for every x. Drop it; it does not affect the sign.
Including denominator zeros. Use open circles for them every time.
Flipping at even powers. A squared factor touches zero but does not change sign.
Cross-multiplying by an expression of unknown sign. Bring all terms to one side instead.
A practice routine
Solve six inequalities: two with only linear factors, two with a squared factor, two that need rearranging first.
For each, check one point in every interval.
Find the domain of two functions with square roots of rational expressions.
Sketch y = (x − 1)(x + 2)/(x − 3) roughly and compare it with your sign chart.
If seeing the curve helps, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Before your next maths session, solve (x + 2)(x − 5)³/(x − 1)² ≤ 0 and check one point in each interval.
