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Basic Maths for JEE · The Algebra Toolkit

The algebra every JEE topic quietly assumes: intervals, inequalities and the wavy curve method, modulus, indices and surds, logarithms, and the AM-GM inequality, built from the ground up.

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In short

Most JEE questions that look unfamiliar are ordinary once the algebra underneath is solid. Intervals and the wavy curve method solve any polynomial inequality by sign, modulus is a distance so it splits into cases, logarithms turn products into sums, and AM-GM settles a whole class of minimum and maximum questions in one line.

Contents

1Numbers, Intervals and the Number Line
2Inequalities and the Wavy Curve Method
3The Modulus (Absolute Value)
4Indices and Surds
5Logarithms Laws, evaluation, domain trap
6Basic Coordinate Geometry and AM-GM
·Basic Maths: One-Page Formula Sheet

Numbers, Intervals and the Number Line

Every JEE problem is set somewhere on the real number line, so fluency with intervals and inequalities is the true foundation of the subject. We start there.

SETS OF NUMBERS

• Natural numbers N: 1, 2, 3, ...
• Integers Z: ..., −2, −1, 0, 1, 2, ...
• Rational numbers Q: any p/q with q ≠ 0 (terminating or recurring decimals).
• Irrational numbers: non-terminating, non-recurring (√2, π, e).
• Real numbers R: rationals and irrationals together, the whole number line.

★ INTERVAL NOTATION

• Closed [a, b]: endpoints included, a ≤ x ≤ b.
• Open (a, b): endpoints excluded, a < x < b.
• Half-open [a, b) or (a, b]: one endpoint only.
A square bracket means the point is in; a round bracket means it is out. Infinity always takes a round bracket.

Inequalities and the Wavy Curve Method

Solving polynomial and rational inequalities is a core JEE skill that appears in every chapter from functions to calculus. The wavy curve (sign-scheme) method solves them fast.

★ THE WAVY CURVE METHOD, STEP BY STEP

To solve an inequality like (x − 1)(x − 2)/(x + 3) ≥ 0:

1. Find the critical points (where each factor is zero): here x = 1, 2, and −3.
2. Mark them on the number line. Points from the numerator are filled (allowed if the inequality is ≥ or ≤); points from the denominator are always hollow (never allowed, division by zero).
3. Start from the far right, where the expression is positive, and alternate signs as you cross each simple point, moving left.
4. Pick the intervals matching the sign you want.

The sign scheme for (x−1)(x−2)/(x+3) ≥ 0. The expression is positive on the shaded stretches. x = −3 is hollow (it makes the denominator zero) so it is excluded.

EXAMPLE · A RATIONAL INEQUALITY

Solve (x − 1)(x − 2)/(x + 3) ≥ 0. Critical points: −3 (hollow), 1, 2. Testing from the right the sign pattern is +, −, +, −. The expression is ≥ 0 on (−3, 1] ∪ [2, ∞). Note −3 is open and 1, 2 are closed.

EXAMPLE · A QUADRATIC INEQUALITY

Solve x² − 5x + 6 < 0. Factor: (x − 2)(x − 3) < 0. The product of two factors is negative only between the roots, so the answer is (2, 3). A parabola opening upward is below the axis between its roots.

The Modulus (Absolute Value)

The modulus measures distance from zero and appears throughout JEE. Handling it means splitting into cases by sign.

|x| = x if x ≥ 0, and |x| = −x if x < 0
|x| = a ⇒ x = ±a · |x| < a ⇒ −a < x < a · |x| > a ⇒ x < −a or x > a

The graph of y = |x| is a V: two straight rays meeting at the origin. It is never negative, and it measures the distance of x from 0.

EXAMPLE · SOLVING A MODULUS EQUATION

Solve |x − 3| = 5. The quantity inside is 5 away from zero, so x − 3 = 5 or x − 3 = −5, giving x = 8 or x = −2. Always split into the two sign cases.

KEY MODULUS PROPERTIES

• |ab| = |a||b| and |a/b| = |a|/|b|.
• Triangle inequality: |a + b| ≤ |a| + |b|.
• |x|² = x², useful for squaring away a modulus safely.

Indices and Surds

Powers and roots are the grammar of algebra. The laws of indices must be automatic.

a^m · aⁿ = a^m+n · a^m/aⁿ = a^m−n · (a^m)ⁿ = a^mn
a⁰ = 1 · a⁻ⁿ = 1/aⁿ · a^1/n = ⁿ√a

RATIONALISING SURDS

To clear a surd from a denominator, multiply top and bottom by the conjugate. For 1/(√a + √b), multiply by (√a − √b) to get (√a − √b)/(a − b). This uses the identity (p + q)(p − q) = p² − q² to remove the roots.

Logarithms

Logarithms turn multiplication into addition and powers into products. They are essential for JEE and the natural inverse of the exponential.

log_a N = x ⇔ a^x = N (a > 0, a ≠ 1, N > 0)

★ THE LAWS OF LOGARITHMS

• log(mn) = log m + log n
• log(m/n) = log m − log n
• log(m^p) = p log m
• Change of base: log_a N = log_b N / log_b a
• log_a a = 1 and log_a 1 = 0.

EXAMPLE · EVALUATING LOGARITHMS

log₂ 8 asks: 2 to what power is 8? Since 2³ = 8, the answer is 3. Similarly log₃ 81 = 4 because 3⁴ = 81. Read a logarithm as the question that undoes a power.

⚠ DOMAIN TRAP

A logarithm is only defined for a positive argument and a positive base not equal to 1. When solving a log equation, always check that every solution keeps each argument positive; extra roots that violate this must be rejected.

Basic Coordinate Geometry and AM-GM

Two quick foundations that recur everywhere: the distance and section formulas, and a favourite inequality.

distance = √[(x₂−x₁)² + (y₂−y₁)²]
midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)
slope m = (y₂−y₁)/(x₂−x₁)

★ THE AM-GM INEQUALITY

For positive numbers, the arithmetic mean is at least the geometric mean: (a + b)/2 ≥ √(ab), with equality only when a = b. A famous consequence: for any x > 0, x + 1/x ≥ 2, and the minimum value 2 occurs exactly at x = 1. This one-line result solves many JEE minimum-value problems instantly.

EXAMPLE · MINIMUM BY AM-GM

Find the least value of x + 1/x for x > 0. By AM-GM, x + 1/x ≥ 2√(x · 1/x) = 2√1 = 2, reached when x = 1/x, that is x = 1. No calculus needed.

★

Basic Maths: One-Page Formula Sheet

Everything for revision day.

NUMBER SETS

N ⊂ Z ⊂ Q ⊂ R

irrational: √2, π, e

[ ] included, ( ) excluded

∞ always (

WAVY CURVE

mark critical points

numerator: filled

denominator: hollow

start + at far right

alternate signs leftward

MODULUS

|x|=a ⇒ x=±a

|x|<a ⇒ −a<x<a

|x|>a ⇒ x<−a or x>a

|a+b| ≤ |a|+|b|

INDICES

a^m · a^n = a^(m+n)

(a^m)^n = a^(mn)

a^−n = 1/a^n

a^(1/n) = n√a

LOGARITHMS

log(mn)=log m+log n

log(m^p)=p log m

change base: logN/loga

arg > 0 always

AM-GM

(a+b)/2 ≥ √(ab)

x + 1/x ≥ 2

equality at a=b

x=1 gives min 2

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