Number and algebra

Exam analysis

Most tested topics

1 Number

When several operations appear together, their priority fixes the meaning: evaluate brackets first, then powers or roots, then multiplication and division, then addition and subtraction. Operations at the same priority are completed from left to right.

18-3(4-1)+8\div2=18-9+4=13

Work one priority level at a time and rewrite the whole expression after each step. A minus sign attached to a number is part of that number, so $(-3)^2=9$ but $-3^2=-(3^2)=-9$ .

Do not automatically do multiplication before division, or addition before subtraction: each pair has equal priority, so read from left to right. Enter brackets explicitly on a calculator when the expression contains them.

A prime number has exactly two positive factors, $1$ and itself. Writing a positive integer as a product of primes exposes every factor and makes highest common factor (HCF) and lowest common multiple (LCM) systematic.

Task	Prime-factor choice
HCF	take each common prime with the smaller exponent
LCM	take every prime present with the larger exponent

28=2^2 imes7,\quad120=2^3 imes3 imes5[2pt]\operatorname{HCF}=2^2=4,\quad\operatorname{LCM}=2^3 imes3 imes5 imes7=840

HCF must divide both numbers; LCM must be a multiple of both. The number $1$ is neither prime nor composite.

Form	Result	Condition
$a^m a^n$	$a^{m+n}$	same base
$a^m\div a^n$	$a^{m-n}$	same non-zero base
$(a^m)^n$	$a^{mn}$	multiply the indices
$a^0$	$1$	$a
e0$
$a^{-n}$	$1/a^n$	$a
e0$
$a^{p/q}$	$\sqrt[q]{a^p}$	use a real root in the stated domain

16^{3/4}=(\sqrt[4]{16})^3=2^3=8,\qquad rac{x^5}{x^{-2}}=x^7;(x
e0)

Multiplication combines repeated factors, so indices add; division cancels factors, so indices subtract. A negative index means reciprocal, not a negative value.

Index laws combine powers only when the bases match. In particular, $a^m+a^n$ cannot usually be replaced by one power.

A surd is an irrational root kept in exact form. Simplify it by extracting the largest square factor; only like surds can then be added or subtracted.

\sqrt{48}=\sqrt{16 imes3}=4\sqrt3,\qquad5\sqrt3+2\sqrt3=7\sqrt3

Products use $\sqrt a\sqrt b=\sqrt{ab}$ when the roots are defined. For example, $(2\sqrt3)(3\sqrt6)=6\sqrt{18}=18\sqrt2$ .

Do not split addition inside a root: $\sqrt{a+b}$ is not generally $\sqrt a+\sqrt b$ . Decimal approximations lose the exact value, so keep the surd until an approximation is requested.

Rationalising removes a surd from the denominator without changing the value. For a binomial denominator, multiply numerator and denominator by its conjugate: keep the terms but reverse the sign between them.

(a-b)(a+b)=a^2-b^2

rac{15}{\sqrt7-2} imesrac{\sqrt7+2}{\sqrt7+2}=rac{15(\sqrt7+2)}{7-4}=5(\sqrt7+2)

The conjugate is used because the middle surd terms cancel. Multiply the entire numerator and denominator, and simplify only after checking that the original denominator is non-zero.

Set	Meaning	Examples
$\mathbb N$	natural counting numbers	$1,2,3,\ldots$ ; follow the question's convention about $0$
$\mathbb Z$	integers	$\ldots,-2,-1,0,1,2,\ldots$
$\mathbb Q$	numbers expressible as $p/q$ , $q
e0 $\|$ -3 $,$ 2/5 $,$ 0.\overline7$
irrational	real but not rational	$\sqrt2$ , $\pi$

\mathbb N\subset\mathbb Z\subset\mathbb Q\subset\mathbb R,\qquad ext{irrationals}=\mathbb R\setminus\mathbb Q

Terminating and recurring decimals are rational. Simplify an expression before classifying it: for example, $\sqrt{20}/\sqrt5=\sqrt4=2$ is natural, integer and rational.

A root symbol does not automatically make a number irrational: $\sqrt{49}=7$ . Proofs of irrationality are not required here.

Choose units that match the quantity, write the given conversion as an equality, and multiply by a conversion factor equal to $1$ . For area or volume, the linear conversion factor must be squared or cubed.

Quantity	Useful metric/SI link
length	$1 ext{ m}=100 ext{ cm}$
area	$1 ext{ m}^2=10\,000 ext{ cm}^2$
volume/capacity	$1 ext{ m}^3=1000 ext{ L}$ ; $1 ext{ L}=1000 ext{ cm}^3$
mass	$1 ext{ kg}=1000 ext{ g}$
time	$1 ext{ h}=3600 ext{ s}$

ext{average speed}=rac{ ext{total distance}}{ ext{total time}},\qquad2.4 ext{ m}^2=24,000 ext{ cm}^2

For currency, use the rate exactly as stated. If $1 ext{ GBP}=1.18 ext{ EUR}$ , multiply pounds by $1.18$ to obtain euros and divide euros by $1.18$ to obtain pounds. Round money only at the requested stage.

Average speed is based on total distance and total time; it is not normally the mean of separate speeds. Two incomplete zero-mark Question Bank fragments were excluded from the evidence.

Form	Conversion/use
fraction $a/b$	exact part of a whole; use common denominators for addition/subtraction
decimal	divide numerator by denominator; convenient for calculation
percentage	multiply a proportion by $100\%$ ; useful for comparison
ratio	compares parts in the same units; simplify by a common factor

rac ab\divrac cd=rac ab imesrac dc\quad(c,d
e0),\qquad ext{percentage multiplier}=1\pmrac r{100}

To divide $£420$ in the ratio $2:3:5$ , total the shares: $2+3+5=10$ . One share is $£42$ , so the amounts are $£84$ , $£126$ and $£210$ .

A ratio gives relative shares, not the actual total. For repeated percentage change, apply the multiplier each time; adding the percentages ignores the changed starting amount. Ratio and proportion questions use at most three parts in this syllabus.

Accuracy	Where counting starts	Example
decimal places (dp)	first digit after the decimal point	$18.376$ to $2$ dp is $18.38$
significant figures (sf)	first non-zero digit	$0.004786$ to $3$ sf is $0.00479$

Locate the final digit to keep, inspect the next digit, then increase the kept digit by $1$ if the next digit is $5$ or more. Replace omitted whole-number digits with zeros when needed to preserve place value.

Leading zeros are not significant, but zeros between non-zero digits are. Keep extra calculator digits during working and round the final answer unless the question directs otherwise.

x ext{ rounded to step }u ext{ as }r\quad\Longrightarrow\quad r-rac u2\le x<r+rac u2

Choose the combination that makes the required result largest or smallest. For positive quantities: an upper sum uses upper bounds; an upper difference uses the first upper and second lower; an upper product uses upper bounds; an upper quotient uses an upper numerator and lower denominator.

A_{\max}=(610.5)(155.5-75.5)=(610.5)(80)=48,840 ext{ cm}^2

Here $610$ , $155$ and $76$ are each correct to the nearest centimetre. The net height is a subtraction, so its upper bound uses $155.5-75.5$ , not $155.5-76.5$ .

A rounded value is not itself an upper or lower bound. State the interval first, and reconsider the extremum rule if a quantity can be negative.

a imes10^n,\qquad1\le a<10,\quad n\in\mathbb Z

Move the decimal point to make a coefficient between $1$ and $10$ ; the number of places moved gives the power of $10$ . Positive indices represent large place values and negative indices represent small positive place values.

Operation	Method
multiply	multiply coefficients and add indices
divide	divide coefficients and subtract indices
add/subtract	first rewrite with the same power of $10$

(3.6 imes10^7)(2 imes10^{-3})=7.2 imes10^4,\qquad 4 imes10^{100}+3.6 imes10^{101}=4.0 imes10^{101}

Normalise the final coefficient: $36 imes10^5=3.6 imes10^6$ is standard form, but $36 imes10^5$ is not. Preserve units and round only when requested.

2 Sets

A set is a well-defined collection of distinct objects, called elements. 'Well-defined' means that membership can be decided unambiguously.

The vowels in the word MATHEMATICS form the set $\{A,E,I\}$ : repeated letters are written once. The collection 'interesting numbers' is not well-defined unless a precise rule for interesting is supplied.

The order used to list elements does not change the set, so $\{2,4,6\}=\{6,2,4\}$ . Braces describe the collection; they do not mean an ordered list.

Form	Example	Meaning
words	$A$ is the set of positive even integers below $10$	states the membership rule
roster	$A=\{2,4,6,8\}$	lists every element once
set-builder	$A=\{x:x ext{ is a positive even integer and }x<10\}$	gives a rule after the colon

In a practical situation, define what one element represents before forming the set. In an abstract set, test each candidate against every condition in the rule.

Dots such as $\{2,4,6,\ldots\}$ are safe only when the continuation rule is unambiguous. Do not confuse an element such as $2$ with the one-element set $\{2\}$ .

Operation	Membership test	Venn region
$A\cap B$	in both $A$ and $B$	overlap only
$A\cup B$	in $A$ or $B$ or both	every region inside either set

A={1,2,3,4},\quad B={3,4,5}\ A\cap B={3,4},\qquad A\cup B={1,2,3,4,5}

For algebraically defined sets, translate each rule into a membership condition. Intersection means satisfying both conditions simultaneously; union means satisfying at least one.

The mathematical word 'or' in a union is inclusive: elements in the overlap are included. List each element only once.

The notation $n(A)$ means the number of distinct elements in set $A$ , not the sum of their numerical values.

A={2,4,6,8}\quad\Longrightarrow\quad n(A)=4

n(A\cup B)=n(A)+n(B)-n(A\cap B)

The intersection is subtracted once because it was counted in both $n(A)$ and $n(B)$ . On a Venn diagram, add every disjoint region belonging to the required set exactly once.

An empty region contributes $0$ elements. A written zero inside a region is a count, not an element that must be counted as one.

The complement $A'$ contains every element of the universal set that is not in $A$ . Its meaning therefore depends on the chosen universal set.

\mathcal E={1,2,3,4,5,6},\quad A={2,4,6}\quad\Longrightarrow\quad A'={1,3,5}

Expression	Equivalent description
$(A\cup B)'$	outside both $A$ and $B$ ; $A'\cap B'$
$(A\cap B)'$	not in both together; $A'\cup B'$

Complement does not mean 'negative' or 'opposite'. It means outside the named set but still inside the universal set.

Set $A$ is a subset of set $B$ when every element of $A$ is also an element of $B$ . The empty set and the whole set are subsets of every set and of itself respectively.

ext{A set with }n ext{ elements has }2^n ext{ subsets.}

For $P=\{a,b\}$ , the complete subset list is $arnothing$ , $\{a\}$ , $\{b\}$ and $\{a,b\}$ . There are $2^2=4$ subsets.

Membership and subset are different: $a$ is an element of $P$ , while $\{a\}$ is a subset of $P$ . Reordering the same elements does not create another subset.

Set	Meaning	Key consequence
universal set $\mathcal E$	all elements currently under consideration	every named set is contained in it
empty set $arnothing$	a set with no elements	$n( arnothing)=0$

If $\mathcal E=\{1,2,3,4,5,6\}$ , $A$ is the even elements and $B$ is the odd elements, then $A\cap B= arnothing$ and $A\cup B=\mathcal E$ .

The empty set is not the same as $\{0\}$ . The first has no elements; the second has one element, the number zero. The universal set changes with the stated context.

A Venn diagram divides the universal set into mutually exclusive regions. For three sets, enter information from the most specific region outward.

Place the three-set intersection. 2. Fill each 'exactly two' region after subtracting the triple intersection when necessary. 3. Fill each single-set-only region. 4. Put elements in none of the sets outside all circles but inside the rectangle. 5. Check every region against the totals.

n(\mathcal E)= ext{sum of all disjoint regions, including the outside region}

A value in an overlap belongs to every circle that contains it, but it is counted once in the universal total. Row 199715 supplied a diagram but no answer and was excluded as representative evidence.

Symbol	Read as
$x\in A$ / $x
otin A $\|$ x $is / is not an element of$ A$
$A\subseteq B$	$A$ is a subset of $B$
$A\cup B$ / $A\cap B$	union / intersection
$A'$	complement of $A$
$arnothing$	empty set
$n(A)$	number of elements in $A$
$\mathcal E$	universal set

Read a compound expression from its grouping outward. For example, $n(B'\cap C)$ asks for the number of elements that are in $C$ and not in $B$ .

Symbols describe different relationships: $3\in A$ can be true while $3\subseteq A$ is ill-formed because $3$ is an element, not a set. Preserve brackets when taking a complement of a combined set.

3 Algebra

Algebraic manipulation preserves value by applying arithmetic and index laws to terms with compatible structure. Like terms have the same variable part, including the same powers.

Move	Valid example
collect like terms	$3x^2-5x+2x^2+x=5x^2-4x$
expand	$3(2x-5)=6x-15$
multiply powers	$x^{3/2}x^{1/2}=x^2$
divide powers	$x^5/x^2=x^3$ for $x
e0$

(2x+1)^2-(2x-2)(2x+1)=(2x+1)igl[(2x+1)-(2x-2)igr]=6x+3

Terms such as $x$ and $x^2$ are not like terms. A negative or fractional power follows the same index laws, but division and negative powers require a non-zero base.

A formula states how quantities are related. Define every symbol and its units, substitute with brackets, and rearrange by performing inverse operations on both sides.

To change the subject when it appears more than once: remove fractions, expand, collect every term containing the new subject on one side, factor it out, then divide by its coefficient.

u=rac{5-4t}{3+6t}\Rightarrow u(3+6t)=5-4t\Rightarrow t(6u+4)=5-3u\Rightarrow t=rac{5-3u}{6u+4}

The rearranged formula inherits restrictions from the original: here $3+6t
e0 $, and the final division also requires$ 6u+4
e0$. Do not change a sign merely because a term crosses the equals sign; apply the same operation to both sides.

Factorising rewrites a sum or difference as a product. It is the reverse of expansion, so multiplying the factors back is a direct check.

Structure	Factorising move
common factor	take the greatest common numerical and algebraic factor outside brackets
four terms	group pairs to create a repeated bracket
difference of squares	$a^2-b^2=(a-b)(a+b)$
quadratic	find two terms whose product and sum match the quadratic

2x^3-6xz+x^2z-3z^2=2x(x^2-3z)+z(x^2-3z)=(2x+z)(x^2-3z)

A factor must divide every term in the part from which it is extracted. 'Completely factorised' means no remaining factor can be factorised further over the required number system.

(x-r) ext{ is a factor of }f(x)\iff f(r)=0

For a factor $ax-b$ , set it equal to zero first: its corresponding root is $x=b/a$ . Substitution gives the remainder without carrying out full division.

f(x)=x^3-6x^2-7x+60,\quad f(5)=125-150-35+60=0\ f(x)=(x-5)(x^2-x-12)=(x-5)(x-4)(x+3)

Testing $f(r)$ concerns the factor $(x-r)$ , not $(x+r)$ . A zero remainder proves a factor; a non-zero value is the remainder and disproves that proposed factor.

Polynomial division mirrors numerical long division: order terms by descending powers, include zero coefficients for missing powers, divide the leading terms, multiply back, subtract, and repeat.

rac{x^3-4x^2+x+6}{x+1}=x^2-5x+6\quad ext{because}\quad(x+1)(x^2-5x+6)=x^3-4x^2+x+6

When the divisor is a known factor, the remainder is zero and the quadratic quotient can often be factorised further: $x^2-5x+6=(x-2)(x-3)$ .

Keep place value aligned: if the cubic has no $x^2$ or $x$ term, write a zero placeholder. If a remainder remains, report quotient plus remainder divided by the divisor.

Operation	Reliable move
simplify	factor numerator and denominator, then cancel common factors
add/subtract	use a common denominator, combine numerators, then factor
multiply	factor first, cancel, then multiply
divide	multiply by the reciprocal of the second fraction

rac6{x-2}+rac4{x+3}=rac{6(x+3)+4(x-2)}{(x-2)(x+3)}=rac{10(x+1)}{(x-2)(x+3)}

State excluded values from every original denominator. In the worked expression, $x
e2,-3$, even if later simplification were to cancel one of those factors.

Only factors cancel; terms joined by addition or subtraction do not. For example, $x$ cannot be cancelled from $(x+3)/x$ .

Degree/form	Useful solution route
linear	collect the unknown terms and isolate the unknown
quadratic	factorise, complete the square, use the formula, or read graph intersections
cubic	find a linear factor/root, divide to a quadratic, then solve the quadratic

ax^2+bx+c=0\Rightarrow x=rac{-b\pm\sqrt{b^2-4ac}}{2a}\quad(a
e0)

2x^2-17x-33=0\Rightarrow x=rac{17\pm\sqrt{553}}4\Rightarrow xpprox10.1 ext{ or }-1.63;(3 ext{ s.f.})

Substitute each solution into the original equation, not only a rearranged form. Clear fractions only after recording denominator restrictions, and reject a root that makes an original denominator zero.

A simultaneous solution is one ordered pair $(x,y)$ that satisfies both linear equations. Algebraically, eliminate one unknown or substitute an expression for it; graphically, it is the intersection of the two lines.

2x+3y=2.5,\quad4x+2y=7\ 4x+6y=5\Rightarrow4y=-2\Rightarrow y=-0.5,\quad x=2

For elimination, first scale one or both equations so one pair of coefficients is equal or opposite. Add or subtract the complete equations, solve the remaining one-variable equation, then substitute back.

Parallel distinct lines have no solution; the same line written twice has infinitely many solutions. Always check the pair in both original equations.

Use the linear equation to express one variable in terms of the other, substitute into the quadratic equation, solve the resulting quadratic, then find the matching second coordinate for every valid root.

x^2+y^2=26,\quad2x+y=9\Rightarrow y=9-2x\ x^2+(9-2x)^2=26\Rightarrow5x^2-36x+55=0\ (x,y)=(5,-1) ext{ or }\left(rac{11}{5},rac{23}{5}
ight)

The pairs are the intersection points of a line and a quadratic curve. There may be zero, one or two real pairs, so do not stop after finding the first root.

Keep each $x$ value paired with the $y$ obtained from it. Substitution must replace every occurrence of the chosen variable, including squared occurrences.

Situation	Representation rule
multiply/divide by a negative	reverse the inequality sign
$<$ or $>$ on a number line	open endpoint
$\le$ or $\ge$ on a number line	closed endpoint
strict boundary in 2D	dashed line
inclusive boundary in 2D	solid line

-13\le5x-3<12\Rightarrow-10\le5x<15\Rightarrow-2\le x<3

For a two-dimensional inequality, draw its boundary line, test a point not on the line, and shade the half-plane whose test point makes the inequality true. For simultaneous inequalities, keep only the overlap.

Do not reverse the sign when adding or subtracting. This syllabus may ask for simultaneous graphical inequalities, but not linear programming or optimisation.

Move every term to one side, factorise or find the roots, place the critical values on a number line, then determine where the quadratic has the required sign.

2x^2+9x+7<0\Rightarrow(2x+7)(x+1)<0\Rightarrow-rac72<x<-1

The roots split the number line into intervals on which the sign cannot change without passing through a root. A positive-leading quadratic is negative between two distinct real roots and positive outside them.

Include a root only for $\le$ or $\ge$ , never for $<$ or $>$ . Do not solve the inequality by treating the inequality sign as an equals sign and reporting only the roots.

Pattern	Recognition	Continuation example
arithmetic	constant first difference	$4,9,14,19,\ldots$ adds $5$
geometric	constant multiplier	$-729,243,-81,27,\ldots$ multiplies by $-1/3$
square numbers	$1^2,2^2,3^2,\ldots$	$1,4,9,16,25,\ldots$
triangular numbers	add $2,3,4,\ldots$	$1,3,6,10,15,\ldots$

Compare consecutive terms first. If differences are constant, continue by adding that common difference. If ratios are constant and defined, continue by multiplying. Otherwise test a familiar integer pattern or alternating rule.

A pattern must explain every supplied transition, not only the last two terms. More than one rule can fit a short list, so use the simplest rule consistent with the stated context.

4 Functions

A function assigns exactly one output to each allowed input. The same input cannot produce two different outputs, although different inputs may share one output.

Part	Meaning
input	the chosen value from the domain
rule	the operation or correspondence applied
output	the single resulting value

x\mapsto x^2:\quad -3\mapsto9,;2\mapsto4

A relation is not a function if one allowed input points to more than one output. A function need not use every possible number as an input or produce every possible number as an output.

A mapping diagram shows inputs in one set and outputs in another. Every input must have one outgoing arrow for the correspondence to define a function.

Diagram feature	Function?
every input has one arrow	yes
two inputs point to one output	yes
one input points to two outputs	no
an output receives no arrow	still may be a function

A={-2,0,2},\quad f(x)=x^2\quad\Rightarrow\quad -2\mapsto4,;0\mapsto0,;2\mapsto4

The function condition is checked from inputs to outputs. Many-to-one is allowed; one-to-many is not.

The statements $f(x)=2x^2-3$ and $f:x\mapsto2x^2-3$ name the same rule. In $f(a)$ , replace every $x$ in the rule by $a$ .

f(x)=2x^2-3\quad\Rightarrow\quad f(-2)=2(-2)^2-3=5

The input may be an expression: $f(2p)=2(2p)^2-3=8p^2-3$ . Use brackets before simplifying.

$f(x)$ is the value produced by function $f$ ; it does not mean $f\times x$ . Also, $f(a+b)$ is generally not $f(a)+f(b)$ .

The domain is the set of allowed inputs; the range is the set of outputs actually produced. Restrictions come from the rule and from any stated context.

Feature in the rule	Domain consequence
denominator $x-a$	exclude $x=a$
square root $\sqrt{x-a}$	require $x\ge a$
stated finite input set	use only those listed inputs

h(x)=\frac{3}{x-2}\quad\Rightarrow\quad \text{domain: }x\ne2,\qquad \text{range: }h(x)\ne0

An excluded input is not an output restriction unless the rule implies it. This syllabus does not require continuity arguments; determine exclusions algebraically or from the stated domain.

For this course, $fg$ means do $g$ first and then $f$ : $(fg)(x)=f(g(x))$ . The function written nearest the input acts first.

f(x)=2x-1,\quad g(x)=x^2+3\quad\Rightarrow\quad fg(x)=f(x^2+3)=2x^2+5

gf(x)=g(2x-1)=(2x-1)^2+3=4x^2-4x+4

Composition is usually order-sensitive: $fg\ne gf$ . An input is valid only if it is in the domain of $g$ and the intermediate value $g(x)$ is in the domain of $f$ .

An inverse function undoes the original function. On compatible domains, $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ .

Write $y=f(x)$ , interchange $x$ and $y$ , then rearrange to make $y$ the subject. Finally write the result as $f^{-1}(x)$ and carry over necessary restrictions.

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

$f^{-1}(x)$ is not $1/f(x)$ . A many-to-one rule needs a restricted domain before it can have an inverse function.

Statement	Equation
$y\propto x^n$	$y=kx^n$
$y\propto1/x^n$	$y=k/x^n$
$y\propto\sqrt{x}$	$y=k\sqrt{x}$
$y\propto1/\sqrt{x}$	$y=k/\sqrt{x}$

Translate the proportionality into an equation with constant $k$ , substitute one known pair to find $k$ , then use the complete equation for the new value.

y\propto\frac1{x^3},\quad y=\frac{32}{27}\text{ when }x=\frac32\Rightarrow y=\frac{k}{x^3},;k=4\Rightarrow y=\frac4{x^3}

The frozen syllabus allows only powers $1,2,3$ and square-root forms, directly or inversely. 'Inversely proportional to $x^3$ ' means $k/x^3$ , not $k/x$ cubed after substitution errors.

A point $(x,y)$ is located by moving horizontally to $x$ and vertically to $y$ . The $x$ -coordinate is always written first.

Quadrant	Sign of $(x,y)$
I	$(+,+)$
II	$(-,+)$
III	$(-,-)$
IV	$(+,-)$

A=(-3,2),\quad B=(4,-1)\quad\Rightarrow\quad \overrightarrow{AB}=(4-(-3),-1-2)=(7,-3)

A point on an axis is not in a quadrant. Plotting $(2,-5)$ as $(-5,2)$ reverses the coordinate order and gives a different point.

y=mx+c,\qquad m=\frac{y_2-y_1}{x_2-x_1},\qquad (0,c)\text{ is the }y\text{-intercept}

To find a line through two points, calculate the gradient, substitute one point into $y=mx+c$ to find $c$ , then check the other point.

(2,-10),(-3,5):\quad m=\frac{5-(-10)}{-3-2}=-3,\quad -10=-3(2)+c\Rightarrow y=-3x-4

A vertical line has equation $x=a$ and an undefined gradient, so it cannot be written as $y=mx+c$ . Parallel non-vertical lines have equal gradients.

The required family is $y=Ax^3+Bx^2+Cx+D+E/x+F/x^2$ , with numerical constants and at least three constants zero. Tables, key features and smooth curves connect an equation to its graph.

Step	Action
1	identify excluded $x$ -values and a sensible scale
2	calculate missing table values accurately
3	plot points and join with a smooth curve
4	read roots, inequalities, turning points or intersections

f(x)=g(x)\quad\Longleftrightarrow\quad \text{the graphs }y=f(x)\text{ and }y=g(x)\text{ intersect}

Do not join a nonlinear graph with straight segments or draw across an excluded value. Graphical solutions are estimates and should be quoted to precision supported by the scale.

The gradient at a point on a curve is the gradient of the tangent there. A tangent follows the curve's direction locally without being a chord through two separated curve points.

Draw a reasonable tangent at the named point. Mark two well-separated points on the tangent, construct a large right-angled triangle, and calculate vertical change divided by horizontal change.

\text{gradient}\approx\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}

Read both points from the tangent, not from the curve unless they lie on the tangent. Preserve the sign: a tangent falling from left to right has negative gradient.

\frac{d}{dx}(ax^n)=anx^{n-1}\qquad\text{for integer }n

Rewrite reciprocals as negative powers, differentiate term by term, and write a constant's derivative as zero. Use $dy/dx$ for the gradient function.

y=2x^4+\frac6x-5=2x^4+6x^{-1}-5\Rightarrow\frac{dy}{dx}=8x^3-6x^{-2}=8x^3-\frac6{x^2}

Multiply by the old power before reducing it by one. The derivative of $x^{-2}$ is $-2x^{-3}$ , not $-2x^{-1}$ .

$dy/dx$ measures instantaneous rate of change. A stationary point occurs where $dy/dx=0$ ; it may be a maximum, a minimum or a stationary point of inflection.

Gradient change through the point	Classification
positive to negative	local maximum
negative to positive	local minimum
no sign change	stationary inflection possible

y=x^3-3x^2+3\Rightarrow\frac{dy}{dx}=3x(x-2)=0\Rightarrow x=0,2\Rightarrow(0,3),(2,-1)

Solving $dy/dx=0$ finds stationary $x$ -values; substitute into the original function for coordinates and use the graph or gradient sign to classify them.

v=\frac{ds}{dt},\qquad a=\frac{dv}{dt}

Graph	Gradient means	Area means
distance–time	speed	not used here
displacement–time	velocity	not used here
speed–time	acceleration in signed-gradient contexts	distance travelled
velocity–time	acceleration	displacement

s=t^3-9t^2+15t\Rightarrow v=3t^2-18t+15,\quad a=6t-18

Distance and speed are non-negative scalars; displacement, velocity and acceleration are signed. On a distance–time graph, a steeper tangent means greater instantaneous speed; on a speed–time graph, distance is area under the graph.

5 Matrices

A matrix is a rectangular array whose rows and columns give positions to data. Its order is rows by columns.

A=\begin{pmatrix}12&15&9\8&11&14\end{pmatrix}\quad\text{has order }2\times3

State what each row and column represents before interpreting an entry. Here, for example, rows might be two shops and columns three products.

The order $2 imes3$ means two rows and three columns, not the reverse. Entries in different positions need not represent the same quantity.

Operation	Condition	Result order
$A+B$	same order	same order
$AB$	columns of $A$ = rows of $B$	rows of $A$ by columns of $B$

\begin{pmatrix}1&-2\3&4\end{pmatrix}+\begin{pmatrix}5&1\-3&2\end{pmatrix}=\begin{pmatrix}6&-1\0&6\end{pmatrix}

\begin{pmatrix}1&2\3&4\end{pmatrix}\begin{pmatrix}5\-1\end{pmatrix}=\begin{pmatrix}1(5)+2(-1)\3(5)+4(-1)\end{pmatrix}=\begin{pmatrix}3\11\end{pmatrix}

Matrix multiplication is not entry-by-entry and is generally not commutative: $AB$ and $BA$ may differ or one may be undefined.

A scalar is an ordinary number. Scalar multiplication multiplies every entry of the matrix and leaves its order unchanged.

-3\begin{pmatrix}2&-1\0&4\end{pmatrix}=\begin{pmatrix}-6&3\0&-12\end{pmatrix}

3A-2B\quad\text{means calculate }3A\text{ and }2B\text{ entry by entry, then subtract corresponding entries}

The scalar applies to every entry, including zeros and negative entries. Do not multiply only a row, column or diagonal unless explicitly stated.

I_2=\begin{pmatrix}1&0\0&1\end{pmatrix},\quad O_{2\times2}=\begin{pmatrix}0&0\0&0\end{pmatrix}

Matrix	Addition role	Multiplication role
zero matrix	$A+O=A$	compatible products give a zero matrix
identity matrix	not an additive identity	$AI=IA=A$ when orders fit

For order $3 imes3$ , the identity has ones on the main diagonal and zeros elsewhere. A zero matrix may be rectangular.

The identity matrix is not a matrix filled with ones. Always use an identity matrix of compatible order.

A=\begin{pmatrix}a&b\c&d\end{pmatrix}\quad\Rightarrow\quad\det A=ad-bc

A^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\-c&a\end{pmatrix}\qquad(ad-bc\ne0)

\begin{pmatrix}4&-2\8&6\end{pmatrix}^{-1}=\frac1{40}\begin{pmatrix}6&2\-8&4\end{pmatrix}

Swap the two main-diagonal entries and change the signs of the off-diagonal entries. The reciprocal belongs to the determinant, not to each original entry; singular-matrix theory is outside this syllabus.

\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}=\begin{pmatrix}ax+by\cx+dy\end{pmatrix}

Transformation	Matrix
reflect in $x=0$	$\begin{pmatrix}-1&0\\0&1\end{pmatrix}$
reflect in $y=0$	$\begin{pmatrix}1&0\\0&-1\end{pmatrix}$
reflect in $y=x$	$\begin{pmatrix}0&1\\1&0\end{pmatrix}$
rotate $90^\circ$ anticlockwise	$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$
enlarge by factor $k$	$\begin{pmatrix}k&0\\0&k\end{pmatrix}$

Transform every vertex as a column vector, plot the image coordinates and join them in the original order. These transformations keep the origin fixed.

Translations cannot be represented by a $2 imes2$ matrix. Keep the coordinate vector ordered as $(x,y)^T$ .

If transformation $B$ happens first and transformation $A$ happens second, the combined matrix is $AB$ . The matrix nearest the point vector acts first.

\mathbf x\xmapsto{B}B\mathbf x\xmapsto{A}A(B\mathbf x)=(AB)\mathbf x

A=\begin{pmatrix}0&-1\1&0\end{pmatrix},\ B=\begin{pmatrix}-1&0\0&1\end{pmatrix}\Rightarrow AB=\begin{pmatrix}0&-1\-1&0\end{pmatrix}

Do not multiply matrices in the chronological left-to-right order. Because $AB\ne BA$ in general, reversing the product usually gives a different transformation.