JEE Corner — Inequalities for JEE Main and Advanced

Why JEE Cares About This Chapter

JEE rarely asks "solve $3x + 5 < 11$" — but inequalities power half the paper from behind: domains of functions, modulus conditions, sign analysis of rational expressions, and range restrictions inside bigger problems. This corner builds the three tools the rationalised textbook stops short of: modulus inequalities, rational inequalities, and the wavy-curve method.

Key Point: Solution sets combine exactly like sets from Chapter 1 — "and" means intersection ($\cap$), "or" means union ($\cup$). JEE answers are intervals or unions of intervals; fluency in $(a, b) \cup [c, \infty)$ notation is assumed.

[JEE Tip] The single most dangerous move in this territory: multiplying an inequality by an expression whose sign you don't know (like $x - 2$). Either split into cases, or — far better — bring everything to one side and analyse signs.

Modulus Inequalities: Think Distance

$|x|$ is the distance of $x$ from 0, and $|x - c|$ the distance from $c$. Every modulus inequality is a distance statement:

For $a > 0$:

• $|x| < a \Leftrightarrow -a < x < a$ — a band around 0. Similarly $|x| \le a \Leftrightarrow -a \le x \le a$.
• $|x| > a \Leftrightarrow x < -a$ or $x > a$ — two rays, written $(-\infty, -a) \cup (a, \infty)$.
• $|x - c| < r \Leftrightarrow c - r < x < c + r$ — the band of radius $r$ around $c$.
• $|x - c| > r$ — everything outside that band.

Two degenerate cases JEE loves: $|x| < a$ has no solution when $a \le 0$ (a modulus is never negative), and $|x| > a$ holds for all $x$ when $a < 0$.

Key Point: "Less than" moduli give single intervals (bands); "greater than" moduli give unions of two rays. If your answer's shape disagrees, recheck.

[JEE Tip] For $a \le |x - c| \le b$, run both conversions at once: the solution is a band of radius $b$ minus the open band of radius $a$ — two symmetric closed intervals around $c$.

Rational Inequalities and the Wavy Curve

To solve $\frac{x - 1}{x - 2} > 0$ you may NOT multiply across by $x - 2$ — its sign is unknown. Instead:

1. Bring everything to one side as a single fraction $\frac{P(x)}{Q(x)}$ with factored $P, Q$.
2. Mark the critical points — zeros of numerator and denominator — on the number line.
3. Determine the sign in each region: for one test value, or note that the sign alternates at each simple (odd-power) factor.
4. Select regions matching the required sign. Zeros of the numerator join the answer only for slack signs ($\ge, \le$); zeros of the denominator are always excluded.

Worked instance: $\frac{(x-1)(x-3)}{x-2} > 0$. Critical points 1, 2, 3. Rightmost region positive, alternating leftwards: $+$ on $(3, \infty)$, $-$ on $(2, 3)$, $+$ on $(1, 2)$, $-$ on $(-\infty, 1)$. Answer: $(1, 2) \cup (3, \infty)$.

Quadratic inequalities are the same machine: $x^2 - 5x + 6 > 0$ factors to $(x - 2)(x - 3) > 0$, giving $(-\infty, 2) \cup (3, \infty)$.

[JEE Tip] For an even-power factor like $(x - 4)^2$, the sign does NOT alternate there — the curve touches and returns. Alternation is only for odd powers.

Solved Examples

Example 1: The four basic modulus forms

Solve (i) $|x| < 3$ (ii) $|x| \ge 2$.

Solution:

Step 1 — (i) Read as distance. Distance from 0 below 3: the BAND $-3 < x < 3$, i.e. $(-3, 3)$.

Step 2 — (ii) Read as distance. Distance from 0 at least 2: two RAYS, $x \le -2$ or $x \ge 2$.

Step 3 — Write the union. $(-\infty, -2] \cup [2, \infty)$ — square brackets from the slack sign.

Takeaway: "Less than" moduli give bands; "greater than" moduli give unions of rays.

Example 2: A shifted band

Solve $|x - 2| < 5$.

Solution:

Step 1 — Read as distance from 2. The distance is below 5: $2 - 5 < x < 2 + 5$.

Step 2 — Interval. $(-3, 7)$ — the open band of radius 5 around 2.

Takeaway: $|x - c| < r$ is always the band $(c - r, c + r)$.

Example 3: A shifted pair of rays

Solve $|x + 3| \ge 4$.

Solution:

Step 1 — Rewrite the centre. $|x + 3| = |x - (-3)|$: distance from $-3$ at least 4.

Step 2 — Write the rays. $x \le -3 - 4 = -7$ or $x \ge -3 + 4 = 1$.

Step 3 — Union. $(-\infty, -7] \cup [1, \infty)$.

Takeaway: $x + 3$ hides the centre $-3$ — rewrite before placing the rays.

Example 4: An annulus on the line (JEE classic)

Solve $1 \le |x - 1| \le 3$.

Solution:

Step 1 — Outer bound. $|x - 1| \le 3$ gives the band $-2 \le x \le 4$.

Step 2 — Inner bound. $|x - 1| \ge 1$ gives the rays $x \le 0$ or $x \ge 2$.

Step 3 — Intersect. $[-2, 0] \cup [2, 4]$ — two symmetric closed intervals around 1.

Takeaway: A two-sided modulus condition is band-minus-band: run both conversions and intersect.

Example 5: Modulus of a linear expression

Solve $|3x - 2| \le 4$.

Solution:

Step 1 — Convert to a chain. $-4 \le 3x - 2 \le 4$.

Step 2 — Add 2 throughout. $-2 \le 3x \le 6$.

Step 3 — Divide by 3. $-\frac{2}{3} \le x \le 2$: interval $\left[-\frac{2}{3}, 2\right]$.

Takeaway: Inside the modulus may be any linear expression — convert, then run the usual chain algebra.

Example 6: Degenerate moduli

Solve (i) $|x - 1| < -2$ (ii) $|x + 5| > -3$.

Solution:

Step 1 — (i) Compare against zero. A modulus is never negative, so it can never be below $-2$: solution $\varnothing$.

Step 2 — (ii) Same comparison. A modulus is always $\ge 0 > -3$: every real works — solution $\mathbb{R}$.

Takeaway: No algebra needed — only the definition. JEE plants these to reward definition-level thinking.

Example 7: First rational inequality

Solve $\frac{x - 1}{x - 2} > 0$.

Solution:

Step 1 — Mark the critical points. Numerator zero: 1; denominator zero: 2.

Step 2 — Sign-test each region. $x < 1$: $\frac{(-)}{(-)} = +$ ✓; $1 < x < 2$: $\frac{(+)}{(-)} = -$ ✗; $x > 2$: $\frac{(+)}{(+)} = +$ ✓.

Step 3 — Collect the positive regions. $(-\infty, 1) \cup (2, \infty)$.

Takeaway: A fraction is positive when numerator and denominator share a sign — never multiply across by the unknown-sign denominator.

Example 8: Slack sign — endpoints diverge

Solve $\frac{x + 2}{x - 3} \le 0$.

Solution:

Step 1 — Critical points and sign. Points $-2$ and 3; the fraction is negative between them (opposite signs).

Step 2 — Decide each endpoint separately. $x = -2$ makes the fraction 0 — allowed by $\le$: INCLUDE. $x = 3$ kills the denominator: ALWAYS exclude.

Step 3 — Interval. $[-2, 3)$.

Takeaway: Numerator zeros join under slack signs; denominator zeros never join.

Example 9: Full wavy curve

Solve $\frac{(x - 1)(x - 3)}{x - 2} \ge 0$.

Solution:

Step 1 — Mark the critical points. 1, 2, 3 — all simple (odd-power) factors, so the sign alternates at each.

Step 2 — Read the wavy curve. Rightmost region positive; alternating leftwards: $-, +, -, +$ left to right.

Step 3 — Collect non-negative regions with endpoint care. $[1, 2) \cup [3, \infty)$ — numerator zeros 1, 3 enter (slack sign); denominator zero 2 stays out.

Takeaway: The wavy curve reads all three regions in one sweep — one test value fixes the rightmost sign, alternation does the rest.

Example 10: Quadratic inequality by factoring

Solve $x^2 - 5x + 6 > 0$.

Solution:

Step 1 — Factor. $(x - 2)(x - 3) > 0$.

Step 2 — Sign-analyse. A product of two factors is positive OUTSIDE the roots: $x < 2$ or $x > 3$ (between them, the parabola dips negative).

Step 3 — Union. $(-\infty, 2) \cup (3, \infty)$.

Takeaway: Quadratic inequalities are the wavy curve with no denominator.

Example 11: Never cross-multiply

Solve $\frac{1}{x - 1} \le 2$.

Solution:

Step 1 — Refuse the illegal move. Multiplying by $x - 1$ (unknown sign) is ambiguous. Instead move 2 across: $\frac{1}{x - 1} - 2 \le 0$.

Step 2 — Combine into one fraction. $\frac{1 - 2(x - 1)}{x - 1} = \frac{3 - 2x}{x - 1} \le 0$.

Step 3 — Sign-analyse. Critical points $\frac{3}{2}$ (numerator) and 1 (denominator): $x < 1$: $\frac{(+)}{(-)} = -$ ✓; $1 < x < \frac{3}{2}$: $\frac{(+)}{(+)} = +$ ✗; $x > \frac{3}{2}$: $\frac{(-)}{(+)} = -$ ✓.

Step 4 — Endpoints. Include $\frac{3}{2}$ (numerator zero, slack sign); exclude 1 (denominator). Solution: $(-\infty, 1) \cup \left[\frac{3}{2}, \infty\right)$.

Takeaway: Bring everything to one side and sign-analyse — the answer's two-piece shape could never come from naive cross-multiplication.

Example 12: Rational inequality with a constant on the right

Solve $\frac{2x - 1}{x + 2} \ge 3$.

Solution:

Step 1 — Move 3 across and combine. $\frac{2x - 1 - 3(x + 2)}{x + 2} \ge 0$, i.e. $\frac{-x - 7}{x + 2} \ge 0$.

Step 2 — Clear the leading minus. Multiply numerator by $-1$ and flip: $\frac{x + 7}{x + 2} \le 0$.

Step 3 — Sign-analyse. Negative between the critical points $-7$ and $-2$; include $-7$ (numerator zero), exclude $-2$ (denominator).

Step 4 — Interval. $[-7, -2)$.

Takeaway: Factor out $-1$ early to keep the leading coefficient positive — the flip is easier to track on a whole inequality than inside a sign chart.

Example 13: Inequalities meet domains

Find the domain of $f(x) = \sqrt{x - 2} + \sqrt{8 - x}$.

Solution:

Step 1 — Write the conditions. Square roots need non-negative inputs: $x - 2 \ge 0$ AND $8 - x \ge 0$.

Step 2 — Solve each. $x \ge 2$ and $x \le 8$.

Step 3 — Intersect. Domain $[2, 8]$.

Takeaway: Domains are systems of inequalities in disguise.

Example 14: Modulus plus a side condition

Solve $|2x - 3| < 5$ together with $x > 0$.

Solution:

Step 1 — Convert the modulus. $-5 < 2x - 3 < 5$ gives $-1 < x < 4$.

Step 2 — Intersect with the side condition. Clip the band $(-1, 4)$ by the ray $(0, \infty)$.

Step 3 — Result. $(0, 4)$.

Takeaway: Extra conditions clip the modulus band — draw both on one number line.

Example 15: Union vs intersection in one problem

Given the conditions $2 < x < 5$ and $1 < x < 4$, find the set of $x$ satisfying (i) both (ii) at least one.

Solution:

Step 1 — (i) "Both" = intersection. Overlap of $(2, 5)$ and $(1, 4)$: $(2, 4)$.

Step 2 — (ii) "At least one" = union. $(2, 5) \cup (1, 4) = (1, 5)$.

Step 3 — Note the phrasing rule. "and/simultaneously" intersects; "or/at least one" unites.

Takeaway: JEE phrasing decides the operation — read for the conjunction before computing.