Wavy Curve Method
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Direct answer
Solving a rational inequality such as ((x + 2)(x − 1)(x − 3))/(x(x − 2)) ≤ 0 reduces to a sign scheme: factor everything into linear terms (x − a) with positive leading coefficients, mark all critical values on a number line, then start from the rightmost region with a positive sign and let the curve wave leftward, flipping sign each time it crosses a factor of odd multiplicity and gliding through without flipping at even-multiplicity factors. The solution set is then read off as the union of negative regions, plus or minus the critical points depending on whether the inequality admits equality. Denominator roots are always excluded, whatever the inequality sign — division by zero has no sign.
What you must remember
- Preparation is compulsory: every factor must look like (x − a) with the variable's coefficient +1; an inequality like (2 − x)(x + 1) > 0 must first be rewritten as (x − 2)(x + 1) < 0, flipping the inequality when multiplying by −1.
- Rightmost region is positive: after standard form, the sign for x greater than the largest critical value is always +; the wave alternates moving left across each simple root.
- Odd versus even multiplicity: sign changes at odd-multiplicity roots (like (x − 1) or (x − 1)³) and stays the same at even-multiplicity roots (like (x − 2)²) — the single most tested distinction.
- Endpoint inclusion: strict inequality (< or >) excludes all critical points; weak inequality (≤ or ≥) includes numerator roots but never denominator roots.
- Repeated-factor shortcut: (x − a)² is non-negative everywhere, so it can be dropped from sign analysis (keeping only x = a for equality cases); (x − a)^even behaves identically.
- Irrational and quadratic factors: irreducible quadratics (negative discriminant) are always positive and vanish from the scheme; never split them into complex roots.
- Final answer as intervals: JEE numerical-answer and matrix-match formats expect interval unions, e.g. (−∞, −2] ∪ (0, 1] ∪ (2, 3] — writing only endpoints loses marks.
One inequality, start to finish
Solve ((x + 2)(x − 1)(x − 3))/(x(x − 2)) ≤ 0 completely. The factors are already in standard form, so mark the five critical values −2, 0, 1, 2, 3 on the line. For x > 3 all five factors are positive, so the expression is positive there; crossing 3 flips to negative on (2, 3); crossing 2 flips again to positive on (1, 2); crossing 1 flips to negative on (0, 1); crossing 0 flips to positive on (−2, 0); crossing −2 flips to negative on (−∞, −2). The weak inequality ≤ collects every negative region together with numerator zeros: the answer is (−∞, −2] ∪ (0, 1] ∪ (2, 3]. Note x = 0 and x = 2 are barred forever — they belong to the denominator — while −2, 1 and 3 enter because equality is permitted.
Contrast the classic even-multiplicity trap: (x − 1)(x − 2)² ≤ 0. The squared factor never makes the product negative, so the sign is decided by (x − 1) alone: negative for x < 1, zero at 1 and at 2. The solution is (−∞, 1] ∪ {2} — an interval plus an isolated point, exactly the shape that breaks answer options built only from intervals.
Signs and endpoints: exam pitfalls
The first casualty in wavy-curve questions is preparation: students wave the curve on (a − x) or (a + bx) factors without standardising, and the scheme inverts somewhere silently. The second is endpoint discipline — a weak inequality with a denominator root "included" because it was marked on the line is the most common single-mark loss in JEE Main sessions. Advanced raises the stakes by burying the inequality inside a domain condition (logarithm bases or square roots hiding beneath) so that the final answer intersects two sign schemes. A viva-style habit worth training: state the excluded values aloud before reading intervals; if any region boundary is an excluded value, it must wear a round bracket, never a square one.
Frequently asked questions
What is the first step before drawing the wavy curve?
Rewrite the inequality so every factor is (x − a) with positive leading coefficient, moving everything to one side; multiply or divide by negatives with an inequality flip.
Where does the sign change fail to alternate?
At repeated roots of even multiplicity: crossing (x − a)² or (x − a)⁴ keeps the sign unchanged, so the wave bounces instead of crossing.
Which critical points enter the solution of a weak rational inequality?
Numerator roots only; denominator roots are excluded even under ≤ or ≥.
How do you handle an irreducible quadratic factor in the scheme?
Verify its discriminant is negative, conclude it is positive for all real x, and delete it from the sign analysis.
What is the solution set of (x − 1)(x − 2)² ≤ 0?
(−∞, 1] ∪ {2}: the squared factor contributes only the isolated zero at x = 2, never a negative region.