Mini-Lesson

Algebra and functions

OCR section 1.02 is the biggest in the specification. It runs from indices and surds through quadratics, polynomials and inequalities to the A-level headliners: partial fractions, the modulus function, and composite and inverse functions.

Get fluent here and the rest of the course gets easier. Work through each screen, answer the questions as you go and collect ⭐ stars. Press Start when you're ready.

Algebra · indices and surds

Indices and surds

a^m × aⁿ = a^m+n · a^m ÷ aⁿ = a^m−n · (a^m)ⁿ = a^mna⁰ = 1 · a⁻ⁿ = 1/aⁿ · a^1/n = ⁿ√a · a^m/n = (ⁿ√a)^m

Surds. Simplify by pulling out square factors, and rationalise the denominator by multiplying top and bottom by the conjugate.

Worked examples

√50 − √18 = 5√2 − 3√2 = 2√2.

8^2/3 = (∛8)² = 2² = 4.

1/(√5 − 2) = (√5 + 2)/((√5 − 2)(√5 + 2)) = (√5 + 2)/(5 − 4) = √5 + 2.

Misconception: √(a + b) is not √a + √b. Check: √(9 + 16) = 5, but √9 + √16 = 7.

Algebra · quadratics

Quadratics and the discriminant

x = (−b ± √(b² − 4ac)) / 2a · Δ = b² − 4acΔ > 0: two distinct real roots · Δ = 0: one repeated root · Δ < 0: no real roots

Completing the square gives the turning point instantly: a(x + p)² + q has vertex at (−p, q).

Worked example — x² − 8x + 21

Half of −8 is −4, so x² − 8x = (x − 4)² − 16.

∴ x² − 8x + 21 = (x − 4)² + 5. Minimum value 5, at x = 4. Since 5 > 0 the curve never crosses the x-axis — consistent with Δ = 64 − 84 = −20 < 0.

Calculate

Your turn — the discriminant

1Find the value of the discriminant Δ = b² − 4ac for 2x² − 5x + 4 = 0, and hence decide how many real roots it has.

Δ =

Hint: a = 2, b = −5, c = 4. Δ = (−5)² − 4(2)(4) = 25 − 32. A negative discriminant means no real roots.

Algebra · polynomials

Polynomials and the factor theorem

If f(a) = 0 then (x − a) is a factor of f(x)and the remainder when f(x) is divided by (x − a) is f(a) — the remainder theorem

Worked example — factorise f(x) = 2x³ − 5x² − 4x + 3

Try x = 3: f(3) = 54 − 45 − 12 + 3 = 0, so (x − 3) is a factor.

Divide: 2x³ − 5x² − 4x + 3 = (x − 3)(2x² + x − 1) = (x − 3)(2x − 1)(x + 1).

Roots: x = 3, x = ½, x = −1.

Check your roots: for ax³ + bx² + cx + d the roots sum to −b/a. Here −(−5)/2 = 2.5, and 3 + 0.5 − 1 = 2.5. ✓

Calculate

Your turn — sum of the roots

2The cubic 2x³ − 5x² − 4x + 3 = 0 has roots 3, ½ and −1. Find the sum of the three roots.

Hint: 3 + 0.5 + (−1). You can check it against −b/a = −(−5)/2.

Algebra · partial fractions

Partial fractions

Splitting one algebraic fraction into simpler ones is the key that unlocks integration and binomial expansions of algebraic fractions later in the course.

(px + q) / ((x − a)(x − b)) ≡ A/(x − a) + B/(x − b)a squared linear factor (x − a)² needs THREE terms: A/(x − a) + B/(x − a)² + C/(x − b)

Worked example — the cover-up method

(5x + 1) / ((x − 1)(x + 2)) ≡ A/(x − 1) + B/(x + 2).

Multiply through: 5x + 1 ≡ A(x + 2) + B(x − 1).

Let x = 1: 6 = 3A ⇒ A = 2. Let x = −2: −9 = −3B ⇒ B = 3.

Check: 2(x + 2) + 3(x − 1) = 2x + 4 + 3x − 3 = 5x + 1. ✓

OCR limit: denominators are no more complicated than squared linear terms, with no more than 3 terms, and numerators are constant or linear.

Calculate

Your turn — partial fractions

3Write (5x + 1) / ((x − 1)(x + 2)) as A/(x − 1) + B/(x + 2). Find the value of A.

A =

Hint: 5x + 1 ≡ A(x + 2) + B(x − 1). Substitute x = 1 to make the B term vanish: 6 = 3A.

Quick check

The right form

?Which is the correct form of the partial fractions for (3x + 5) / ((x − 1)(x + 4)²)?

Algebra · functions

Composite and inverse functions

A function maps each input in its domain to exactly one output; the set of outputs is the range.

Composite fg(x) = f(g(x)) — apply g first, then f. Read it right to left.
Inverse f⁻¹ undoes f. It exists only if f is one-to-one (often you must restrict the domain first).
Domain of f⁻¹ = range of f; range of f⁻¹ = domain of f. The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.

Worked examples — f(x) = 3x − 2, g(x) = x² + 1

fg(4): g(4) = 16 + 1 = 17, then f(17) = 3(17) − 2 = 49.

gf(4): f(4) = 10, then g(10) = 101. So fg ≠ gf — order matters.

Inverse of h(x) = 3x³ + 5: let y = 3x³ + 5 ⇒ x³ = (y − 5)/3 ⇒ h⁻¹(x) = ∛((x − 5)/3).

Calculate

Your turn — composite function

4f(x) = 3x − 2 and g(x) = x² + 1. Find fg(4).

Hint: fg means g first. g(4) = 4² + 1 = 17, then f(17) = 3 × 17 − 2.

Quick check

Which order?

?For functions f and g, what does the composite fg(x) mean?

Algebra · modulus

The modulus function

|x| is the size of x, ignoring sign: |x| = x if x ≥ 0, and |x| = −x if x < 0. The graph of y = |f(x)| takes the graph of f and reflects everything below the x-axis up above it.

Worked example — solve |2x − 3| = 7

Two cases: 2x − 3 = 7 ⇒ 2x = 10 ⇒ x = 5.

Or 2x − 3 = −7 ⇒ 2x = −4 ⇒ x = −2.

Check: |2(5) − 3| = |7| = 7 ✓ and |2(−2) − 3| = |−7| = 7 ✓

Always check your solutions. When you square both sides of a modulus equation you can create extra roots that do not satisfy the original. Substituting back is not optional.

Calculate

Your turn — modulus equation

5Solve |2x − 3| = 7. Give the larger of the two solutions.

x =

Hint: either 2x − 3 = 7 (giving x = 5) or 2x − 3 = −7 (giving x = −2). The larger is 5.

Algebra · graphs

Graph transformations

Four transformations, and one rule that decides them all: a change inside f( ) affects x and does the opposite of what it looks like; a change outside affects y and does exactly what it looks like.

y = f(x) + a — translation up by a.
y = f(x + a) — translation left by a (inside ⇒ opposite).
y = k f(x) — vertical stretch, scale factor k.
y = f(kx) — horizontal stretch, scale factor 1/k.
y = −f(x) — reflection in the x-axis. y = f(−x) — reflection in the y-axis.

Quick check

Which way does it move?

?The graph of y = f(x) is transformed to y = f(x + 3). What happens?

Sort it

Classify the transformation

Tap a transformation, then tap the type it belongs to.

↔️ Translation

↕️ Stretch

🪞 Reflection

Match it

Simplify it

Tap an expression on the left, then its simplified form on the right.

Expression

Simplified

Quick check

Two distinct roots

?For what condition on the discriminant does ax² + bx + c = 0 have two distinct real roots?

Recap

The big ideas to know

Indices & surds: a^m/n = (∛a)^m; rationalise with the conjugate

Quadratics: Δ = b² − 4ac decides the roots; completing the square gives the vertex

Polynomials: f(a) = 0 ⇒ (x − a) is a factor; roots sum to −b/a

Partial fractions: cover-up method; a squared linear factor needs two terms

Functions: fg(x) = f(g(x)) — g first; f⁻¹ needs f one-to-one and reflects in y = x

Modulus: |2x − 3| = 7 gives two cases — and always check for extra roots

Transformations: inside f( ) ⇒ affects x and does the opposite; outside ⇒ affects y

That is OCR 1.02 — the algebra the whole A-level rests on. Press Finish to see your score.

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