Pure 1: Algebra and Functions

Pure 1: Algebra and Functions — CIE A Level Mathematics 9709

1. Quadratics

A quadratic function has the form f(x) = ax² + bx + c (a ≠ 0).

Three forms to know:

| Form | Expression | Use |

|---|---|---|

| Standard | ax² + bx + c | Identifying a, b, c for formulae |

| Factored | a(x − p)(x − q) | Finding roots (p and q) |

| Vertex / Completed square | a(x − h)² + k | Finding vertex (h, k), range, axis of symmetry |

Completing the square — method:

x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4

→ Vertex at (−3, −4); axis of symmetry x = −3.

Quadratic formula: For ax² + bx + c = 0:

$$x = rac{-b pm sqrt{b^2 - 4ac}}{2a}$$

Discriminant Δ = b² − 4ac:

• Δ > 0: two distinct real roots
• Δ = 0: one repeated real root (tangent to x-axis)
• Δ < 0: no real roots (complex only)

Worked example: Show that x² − 5x + 7 = 0 has no real roots.

Δ = (−5)² − 4(1)(7) = 25 − 28 = −3 < 0 ✓ No real roots.

2. Functions

Definition: A function f: A → B maps each element of domain A to exactly one element of codomain B. The **range** is the set of actual output values.

Types:

• One-to-one (injective): distinct inputs give distinct outputs → inverse exists
• Many-to-one: multiple inputs can give the same output (e.g., f(x) = x²)

Composite functions: (f ∘ g)(x) = f(g(x)) — apply g first, then f.

Note: f ∘ g ≠ g ∘ f in general.

Inverse functions:

• Only exists when f is one-to-one (restrict domain if necessary)
• To find f⁻¹: write y = f(x), rearrange to make x the subject, replace x with f⁻¹(x)
• Graphically: y = f⁻¹(x) is the reflection of y = f(x) in the line y = x
• Key: domain of f⁻¹ = range of f; range of f⁻¹ = domain of f

Example: f(x) = 2x + 3. Find f⁻¹(x).

Let y = 2x + 3 → x = (y − 3)/2 → f⁻¹(x) = (x − 3)/2

Modulus function |f(x)|: Reflects the part of y = f(x) below the x-axis above it.

To solve |2x − 1| = 3: case 1: 2x − 1 = 3 → x = 2; case 2: 2x − 1 = −3 → x = −1.

3. Simultaneous Equations

Linear and quadratic system (A Level 9709 Paper 1 staple):

Solve: y = 2x − 1 and x² + y² = 25

Substitute y = 2x − 1 into circle equation:

x² + (2x − 1)² = 25

x² + 4x² − 4x + 1 = 25

5x² − 4x − 24 = 0

(5x + 12)(x − 2) → But check discriminant: Δ = 16 + 480 = 496 → use formula.

Key exam technique: For line y = mx + c tangent to curve, substitute and set Δ = 0.

4. Inequalities

Solving quadratic inequalities:

Find roots, sketch parabola, read off which region satisfies the inequality.

Example: x² − 5x + 6 > 0 → (x − 2)(x − 3) > 0 → x < 2 or x > 3

Modulus inequalities:

|x − 3| < 5 means −5 < x − 3 < 5 → −2 < x < 8

|x − 3| > 5 means x − 3 > 5 or x − 3 < −5 → x > 8 or x < −2

5. The Binomial Expansion

(a + b)ⁿ for positive integer n:

$$(a + b)^n = sum_{r=0}^{n} inom{n}{r} a^{n-r} b^r$$

where $inom{n}{r} = rac{n!}{r!(n-r)!}$

Pascal's Triangle gives the coefficients: 1, 4, 6, 4, 1 for n = 4.

Worked example: Expand (2 + 3x)⁴ up to x² term.

• r=0: C(4,0)(2)⁴(3x)⁰ = 16
• r=1: C(4,1)(2)³(3x)¹ = 4×8×3x = 96x
• r=2: C(4,2)(2)²(3x)² = 6×4×9x² = 216x²

→ (2 + 3x)⁴ ≈ 16 + 96x + 216x² + ...

Finding a specific term: The (r+1)th term of (a + b)ⁿ = C(n,r)aⁿ⁻ʳbʳ.

6. Polynomials: Factor and Remainder Theorem

Remainder theorem: When f(x) is divided by (x − a), the remainder = f(a).

Factor theorem: (x − a) is a factor of f(x) ⟺ f(a) = 0.

Example: Show (x − 2) is a factor of x³ − 6x² + 11x − 6.

f(2) = 8 − 24 + 22 − 6 = 0 ✓ → (x − 2) is a factor.

Long division or inspection gives: x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3).