Algebra and functions: A-level Maths OCR revision notes

1.02

Surds and indices

Surds 1.02

Definitions

Surd: an irrational root — such as $2$ or $35$ — left in exact form rather than evaluated as a rounded decimal.

Key results

Simplifying: extract the largest perfect-square factor, e.g. $75 = 25 \times 3 = 53$ .
Multiplying: $a \times b = ab$ , e.g. $3 \times 12 = 36 = 6$ .
Adding/subtracting: surds combine only once they share the same irrational part, e.g. $75 + 12 = 53 + 23 = 73$ .
Rationalising a single surd denominator: multiply top and bottom by that surd, e.g. $\frac{1}{2} = \frac{2}{2}$ .
Rationalising a binomial denominator $a + b$ : multiply by its conjugate $a - b$ , since $(a + b) (a - b) = a^{2} - b$ removes the root entirely.

Notes

Two surds only look 'unlike' if they haven't yet been simplified — always simplify every surd in an expression before deciding whether terms combine.
Rationalising with a conjugate works because $(a + b) (a - b) = a^{2} - b$ is a difference of two squares — the same idea used to remove a root from any binomial denominator.

Indices 1.02

Key results

$a^{m} \times a^{n} = a^{m + n}$
$a^{m} \div a^{n} = a^{m - n}$
$(a^{m})^{n} = a^{mn}$
$a^{0} = 1$ for any $a \neq = 0$
$a^{- n} = \frac{1}{a ^{n}}$
$a^{\frac{1}{n}} = n a$ , and more generally $a^{\frac{m}{n}} = (n a)^{m}$

Notes

These laws apply equally to algebraic bases (e.g. $x^{2} \times x^{5} = x^{7}$ ), not only numerical ones.
A fractional index is a root and a power combined; take the root first where possible, since it keeps the numbers smaller, e.g. $8^{\frac{2}{3}} = (38)^{2} = 2^{2} = 4$ rather than $3 8^{2} = 364$ (same answer, harder arithmetic).
A negative index and a negative value are unrelated ideas — $2^{- 3} = \frac{1}{8}$ is small and positive, not negative; don't let the word 'negative' bleed from the index into the result.

Worked example

Simplify $50 + 8 - 18$ fully.

Show worked solution

Simplify each surd separately:

50 = 25 \times 2 = 52

8 = 4 \times 2 = 22

18 = 9 \times 2 = 32

All three now share the irrational part $2$ , so they combine directly:

52 + 22 - 32 = 42

1.02

Quadratics and the discriminant

The discriminant and completing the square 1.02

Definitions

Discriminant of $a x^{2} + b x + c = 0$ : $Δ = b^{2} - 4 a c$ .

Key results

The completed-square form gives the turning point directly: the vertex of $y = a (x - h)^{2} + k$ is $(h, k)$ .
Discriminant test on $a x^{2} + b x + c = 0$ : two distinct real roots when $Δ > 0$ ; one repeated real root when $Δ = 0$ ; no real roots (a complex-conjugate pair) when $Δ < 0$ .
The quadratic formula, $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , is exactly completing the square carried out once, in general, on $a x^{2} + b x + c = 0$ .

Method

Divide the whole quadratic by $a$ so the coefficient of $x^{2}$ is $1$ (skip this step if $a = 1$ ).
Halve the coefficient of $x$ and write $x^{2} + p x$ as $(x + \frac{p}{2})^{2} - \frac{p ^{2}}{4}$ .
Substitute back and simplify to reach the vertex form $a (x + \frac{b}{2 a})^{2} + (c - \frac{b ^{2}}{4 a})$ .

In practiceWrite $2 x^{2} - 12 x + 7$ in completed-square form, and state its vertex and discriminant.

$2 x^{2} - 12 x + 7 = 2 (x^{2} - 6 x) + 7$ take the 2 out of the $x$ terms
$= 2 [(x - 3)^{2} - 9] + 7$ half of $- 6$ is $- 3$
$= 2 (x - 3)^{2} - 11$ vertex $(3, - 11)$
$Δ = (- 12)^{2} - 4 (2) (7) = 88 > 0$ two distinct real roots: a minimum below the axis agrees

Notes

The discriminant is the fastest route to 'how many times does this line meet this curve' — substitute the line into the curve's equation, form the resulting quadratic, and inspect $Δ$ without solving anything further.
Completing the square and the quadratic formula always agree — if a quick check by formula gives a different answer to a completed-square method, the error is in the algebra, not a discrepancy between methods.

Worked examples

Worked example

Solve:

2 x^{2} - 8 x + 3 = 0

by completing the square, giving your answer in exact surd form.

Show worked solution

Divide by 2:

x^{2} - 4 x + \frac{3}{2} = 0

Complete the square on $x^{2} - 4 x$ :

(x - 2)^{2} - 4 + \frac{3}{2} = 0

so:

(x - 2)^{2} = \frac{5}{2}

Taking square roots:

x - 2 = \pm \frac{5}{2} = \pm \frac{10}{2}

So:

x = 2 \pm \frac{10}{2} = \frac{4 \pm 10}{2}

As a check, the quadratic formula gives:

x = \frac{8 \pm 64 - 24}{4} = \frac{8 \pm 40}{4} = \frac{8 \pm 2 10}{4} = 2 \pm \frac{10}{2}

confirming the same result.

Worked example

Find the values of $k$ for which the line $y = k x + 1$ is a tangent to the curve $y = x^{2} + 3$ .

Show worked solution

At an intersection, $x^{2} + 3 = k x + 1$ , so $x^{2} - k x + 2 = 0$ .

A tangent meets the curve at exactly one point, so this quadratic must have a repeated root: $Δ = 0$ .

Here:

Δ = (- k)^{2} - 4 (1) (2) = k^{2} - 8

so $k^{2} - 8 = 0$ , giving:

k = \pm 8 = \pm 22

1.02

Simultaneous equations and inequalities

Simultaneous equations (linear and quadratic) 1.02

Key results

Example: $y = x + 1$ and $x^{2} + y^{2} = 13$ give $x^{2} + (x + 1)^{2} = 13$ , so $2 x^{2} + 2 x - 12 = 0$ , i.e. $x^{2} + x - 6 = 0$ and $(x + 3) (x - 2) = 0$ . The solutions are $(2, 3)$ and $(- 3, - 2)$ .
Tangency: $y = 2 x + k$ meets $y = x^{2}$ where $x^{2} - 2 x - k = 0$ , with $Δ = 4 + 4 k$ . The line is a tangent when $k = - 1$ , meets the curve twice when $k > - 1$ , and misses it when $k < - 1$ .

Method

Rearrange the linear equation to make one variable the subject.
Substitute that expression into the quadratic equation.
Solve the resulting one-variable quadratic (factorising, completing the square, or the formula).
Substitute each root back into the linear equation to find its paired value of the other variable.

In practiceSolve $y = 2 x - 1$ and $x^{2} + y^{2} = 10$ simultaneously.

$x^{2} + (2 x - 1)^{2} = 10$ substitute the linear equation
$5 x^{2} - 4 x - 9 = 0 ⟹ (5 x - 9) (x + 1) = 0$
$x = \frac{9}{5}, y = \frac{13}{5} or x = - 1, y = - 3$ each $x$ paired with its $y$ from $y = 2 x - 1$

Notes

A quadratic typically has two roots, so expect (and check for) two pairs of solutions — stopping after the first pair is a very common error.
The number of solution pairs matches the discriminant of the resulting quadratic: two pairs if $Δ > 0$ , one repeated pair if $Δ = 0$ , none if $Δ < 0$ (the line does not meet the curve).
Substituting the linear equation into the quadratic (rather than the reverse) keeps the algebra manageable — substituting a rearranged quadratic into the linear equation instead usually creates unnecessary square roots.

Quadratic inequalities 1.02

Key results

$x^{2} - 5 x + 6 > 0 ⟺ (x - 2) (x - 3) > 0 ⟺ x < 2$ or $x > 3$ — outside the roots, since $x^{2} - 5 x + 6$ is an upward-opening parabola.

Method

Solve the corresponding equation to find the critical values (the roots).
Sketch the parabola, or draw a sign diagram, across those critical values.
Read off the region(s) satisfying the inequality from the sign of the expression in each region.

In practiceSolve $2 x^{2} + x - 6 > 0$ .

$2 x^{2} + x - 6 = (2 x - 3) (x + 2) = 0 ⟹ x = - 2, \frac{3}{2}$ critical values
The parabola opens upwards, so it lies above the axis outside the roots.
$x < - 2 or x > \frac{3}{2}$ strict: the roots themselves are excluded

Notes

Never divide or multiply a quadratic inequality by an expression whose sign is unknown; always work from the critical values and a sign diagram instead of manipulating the inequality algebraically as if it were an equation.
A common slip is stating the solution the wrong way round — for an upward-opening parabola, ' $> 0$ ' is satisfied outside the roots and ' $< 0$ ' between them; sketching the parabola removes any doubt.

Worked example

Solve the inequality $2 x^{2} - 3 x - 2 \leq 0$ .

Show worked solution

Factorise: the discriminant is:

(- 3)^{2} - 4 (2) (- 2) = 9 + 16 = 25

and $25 = 5$ , so the roots are:

x = \frac{3 \pm 5}{4}

giving $x = 2$ or:

x = - \frac{1}{2}

So:

2 x^{2} - 3 x - 2 = (x - 2) (2 x + 1)

This is an upward-opening parabola, so it is $\leq 0$ between its roots: $- \frac{1}{2} \leq x \leq 2$ .

1.02

Polynomials and the factor theorem

The factor and remainder theorems 1.02

Key results

Remainder theorem: when a polynomial $f (x)$ is divided by $(x - a)$ , the remainder equals $f (a)$ .
Factor theorem: $(x - a)$ is a factor of $f (x)$ if and only if $f (a) = 0$ — the special case of the remainder theorem where the remainder is zero.

Method

Shortlist candidate roots using the factor pairs of the constant term (for integer coefficients).
Test each candidate $a$ by evaluating $f (a)$ until $f (a) = 0$ is found.
Divide $f (x)$ by $(x - a)$ (by algebraic long division or comparing coefficients) to reduce a cubic to a quadratic.
Solve the resulting quadratic by the usual methods to complete the factorisation.

In practiceFactorise $f (x) = x^{3} - 2 x^{2} - 5 x + 6$ fully.

$candidates: \pm 1, \pm 2, \pm 3, \pm 6$ factors of the constant term
$f (1) = 1 - 2 - 5 + 6 = 0$ so $(x - 1)$ is a factor
$f (x) = (x - 1) (x^{2} - x - 6)$ compare coefficients
$= (x - 1) (x - 3) (x + 2)$

Notes

Try $a = \pm 1$ first when shortlisting candidate roots — they cost almost nothing to check and often work for exam-style polynomials.
The remainder theorem also works for a divisor $(a x - b)$ : dividing by it evaluates $f$ at $x = \frac{b}{a}$ , not at $x = b$ — a common source of error when the leading coefficient isn't $1$ .

1.02

Algebraic and partial fractions

Algebraic and partial fractions 1.02

Key results

Algebraic fractions combine and simplify by the same rules as numerical fractions: a common denominator for addition and subtraction, and cancelling shared factors (never shared terms) for simplification.
Partial fractions reverse the process of combining fractions over a common denominator: $\frac{3 x + 5}{( x + 1 ) ( x + 2 )} \equiv \frac{A}{x + 1} + \frac{B}{x + 2}$ .
A repeated linear factor $(x - p)^{2}$ requires two separate terms: $\frac{A}{x - p} + \frac{B}{( x - p ) ^{2}}$ .
An irreducible quadratic factor requires a linear numerator, not a constant: $\frac{A x + B}{x ^{2} + b x + c}$ .

Method

Write the fraction as a sum with unknown constants over each factor, using the repeated-factor and quadratic-factor rules above where relevant.
Multiply both sides by the original denominator to clear all fractions.
Find each constant by substituting values of $x$ that eliminate all but one unknown (usually the roots of each linear factor), then compare remaining coefficients for any unknowns not yet found.

In practiceExpress $\frac{3 x + 1}{( x - 1 ) ( x + 1 ) ^{2}}$ in partial fractions.

$\frac{3 x + 1}{( x - 1 ) ( x + 1 ) ^{2}} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{( x + 1 ) ^{2}}$ a repeated factor takes two terms
$3 x + 1 = A (x + 1)^{2} + B (x - 1) (x + 1) + C (x - 1)$ multiply through
$x = 1 : 4 = 4 A ⟹ A = 1; x = - 1 : - 2 = - 2 C ⟹ C = 1$
$x^{2} : 0 = A + B ⟹ B = - 1$ check $x = 0$ : $A - B - C = 1$

Notes

A common error is cancelling an $x$ that appears in a sum rather than as a factor of the entire numerator and denominator — e.g. wrongly cancelling the $x$ 's in $\frac{x + 2}{x}$ , which is not valid.
Before splitting into partial fractions, check the numerator's degree is strictly less than the denominator's — an improper fraction needs polynomial division first to extract a whole-number part.

Worked example

Express:

\frac{x + 5}{( x - 1 ) ^{2} ( x + 1 )}

in partial fractions.

Show worked solution

Since $(x - 1)$ is a repeated factor, write:

\frac{x + 5}{( x - 1 ) ^{2} ( x + 1 )} \equiv \frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C}{x + 1}

Multiplying through by $(x - 1)^{2} (x + 1)$ :

x + 5 \equiv A (x - 1) (x + 1) + B (x + 1) + C (x - 1)^{2}

Substituting $x = 1$ : $6 = 2 B$ , so $B = 3$ .

Substituting $x = - 1$ : $4 = 4 C$ , so $C = 1$ .

Comparing $x^{2}$ coefficients (the left side has none): $0 = A + C$ , so $A = - 1$ .

Therefore:

\frac{x + 5}{( x - 1 ) ^{2} ( x + 1 )} = \frac{- 1}{x - 1} + \frac{3}{( x - 1 ) ^{2}} + \frac{1}{x + 1}

1.02

Functions and transformations of graphs

Domain, range and composite functions 1.02

Definitions

Domain: the set of inputs for which a function is defined.
Range: the resulting set of outputs — both domain and range must be stated alongside the function itself to describe it fully.

Key results

Division by zero and square roots of negative numbers are the two most common sources of domain restriction: $f (x) = \frac{1}{x - 3}$ is undefined at $x = 3$ ; $f (x) = x - 2$ requires $x \geq 2$ .
A composite function $f g (x)$ means apply $g$ first, then $f$ to the result: $f g (x) = f (g (x))$ .

Notes

The domain of a composite $f g (x)$ must respect both the domain of $g$ and the domain of $f$ applied to $g$ 's output — a value of $x$ can be excluded even if it lies in the domain of $g$ alone, if $g (x)$ then falls outside the domain of $f$ .
Finding the range often needs a sketch or knowledge of the parent function's shape rather than pure algebra — e.g. the range of $f (x) = x^{2} + 3$ is $f (x) \geq 3$ , read directly from the graph's minimum.

Inverse functions and graph transformations 1.02

Definitions

Inverse function $f^{- 1}$ : the function that undoes $f$ , satisfying $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Key results

An inverse exists only where $f$ is one-to-one on its domain (each output comes from exactly one input).
The graph of $y = f^{- 1} (x)$ is the reflection of $y = f (x)$ in the line $y = x$ .

Method

Write $y = f (x)$ .
Swap $x$ and $y$ .
Rearrange the new equation to make $y$ the subject; this rearranged expression is $f^{- 1} (x)$ .

In practiceFind $f^{- 1} (x)$ for $f (x) = \frac{2 x + 3}{x - 1}$ , $x \neq = 1$ .

$y = \frac{2 x + 3}{x - 1}$
$x = \frac{2 y + 3}{y - 1}$ swap $x$ and $y$
$x y - x = 2 y + 3 ⟹ y (x - 2) = x + 3$ collect the $y$ terms
$f^{- 1} (x) = \frac{x + 3}{x - 2}, x \neq = 2$ check: $f (0) = - 3$ and $f^{- 1} (- 3) = 0$

Notes

Graph transformations follow consistent rules: $f (x) + a$ shifts vertically by $a$ ; $f (x + a)$ shifts horizontally by $- a$ (opposite to the sign of $a$ ); $a f (x)$ stretches vertically by factor $a$ ; $f (a x)$ stretches horizontally by factor $\frac{1}{a}$ .
When combining several transformations, apply them in the order the function is built up algebraically — check the result by tracking what happens to one known point on the original graph.
$f^{- 1}$ only exists on the whole domain of $f$ if $f$ is one-to-one there; where it isn't, the domain must first be restricted (e.g. to $x \geq 0$ ) before an inverse can be defined.

Worked example

Find $f^{- 1} (x)$ for:

f (x) = \frac{2 x + 1}{x - 3}

$x \neq = 3$ , and state its domain.

Show worked solution

Write:

y = \frac{2 x + 1}{x - 3}

and swap $x$ and $y$ :

x = \frac{2 y + 1}{y - 3}

Multiply out:

x (y - 3) = 2 y + 1

so:

x y - 3 x = 2 y + 1

Collect $y$ -terms:

x y - 2 y = 1 + 3 x

so:

y (x - 2) = 1 + 3 x

giving:

y = \frac{3 x + 1}{x - 2}

So:

f^{- 1} (x) = \frac{3 x + 1}{x - 2}

with domain $x \neq = 2$ (the value excluded from $f$ 's range).

As a check, composing $f (f^{- 1} (x))$ simplifies back to $x$ for all $x \neq = 2$ .