A Algebra — revision question pack

22 specification points · notes, questions, answers and worked methods

Checked against AQA 8365 section A. Review basis: the qualification registry sourced from the AQA Level 2 Certificate in Further Mathematics (8365) specification; registry verification recorded 11 July 2026.

How this checking works

Answer all questions in the spaces provided.

A1 · The basic processes of algebra, including use of the associative, commutative and distributive laws

Explanation

The commutative laws permit the order to change in addition and multiplication: $a+b=b+a$ and $ab=ba$ . The associative laws permit regrouping: $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$ .
The distributive law connects multiplication with addition or subtraction: $a(b+c)=ab+ac$ and $a(b-c)=ab-ac$ .
These laws justify rearranging, expanding, factorising and simplifying algebraic expressions.
Subtraction and division are neither commutative nor associative, so their order cannot be changed in the same way.
Examiners expect each transformation to preserve equality; a useful check is to identify the law that permits a common factor to be taken out or terms to be reordered.

Worked example

Use algebraic laws to work out $17\times48+17\times52$ without evaluating the two products separately.

1.Recognise the common factor $17$ .
2.Apply the distributive law in reverse: $17\times48+17\times52=17(48+52)$ .
3.Evaluate $17\times100$ .

Answer: $1700$ .

Common mistakes

Don't reorder a subtraction as though it were commutative, treating $a-b$ as $b-a$ .
Don't take out a common factor from only one term, for example writing $ab+ac=a(b)+c$ .
Don't change the grouping of a division and assume $(a\div b)\div c=a\div(b\div c)$ .

Exam tip

When a question asks for algebraic laws, show the factorised or regrouped line before the numerical result.

Tier 1 · Easy

1
Use a basic algebraic law to work out $19\times37+19\times63$ .
[2 marks]
2
Write $7(p+q)+3(p+q)$ as a single product.
[1 mark]

Tier 2 · Standard

1
Use algebraic laws to work out $25\times47+25\times55-25\times2$ without evaluating any product separately.
[2 marks]
2
Using algebraic laws, simplify $p(q+r)-q(p-r)$ to a single product.
[3 marks]
3
Given $p+q=17$ and $pq=60$ , work out $p^2q+pq^2$ without working out $p$ or $q$ separately.
[3 marks]

Tier 3 · Hard

1
Using algebraic laws, simplify $(a+b+c)^2-(a+b-c)^2$ to a single product.
[3 marks]
2
Expand and simplify $a(b-c)+b(a+c)-2ab$ to a single product.
[3 marks]
3
Given $x+y=5$ and $xy=-3$ , work out $x^3y+2x^2y^2+xy^3$ .
[4 marks]
4
The numbers $x$ , $y$ and $z$ satisfy $x(y+z)=26$ , $y(x+z)=17$ and $x-y=3$ . Work out $z$ using algebraic laws.
[3 marks]
5
Work out $999^2-998\times1000$ using algebraic laws rather than long multiplication.
[3 marks]

A2 · Definition of a function; notation f(x)

Explanation

A function assigns exactly one output to each permitted input. In the notation $f(x)$ , $x$ is a placeholder for the input and $f(3)$ means that $3$ replaces every occurrence of $x$ in the function rule.
A negative or algebraic input should be enclosed in brackets before simplifying.
The equation $f(x)=k$ asks for input values whose output is $k$ ; it does not mean $f$ multiplied by $x$ .
Different inputs may share one output, but one input cannot have two outputs if the relation is a function.
Examiners expect accurate substitution, complete expansion of bracketed inputs and a distinction between evaluating a function and solving an equation involving it.

Worked example

For $f(x)=2x^2-3x+1$ , work out $f(-2)$ .

1.Substitute with brackets: $f(-2)=2(-2)^2-3(-2)+1$ .
2.Evaluate the powers and products: $2(4)+6+1$ .
3.Add the terms.

Answer: $f(-2)=15$ .

Common mistakes

Don't calculate $(-2)^2$ as $-4$ because the substituted negative value is not bracketed.
Don't treat $f(-2)$ as $f\times(-2)$ instead of applying the function rule.
Don't substitute the input into only one occurrence of $x$ .

Exam tip

For a negative or algebraic input, replace every $x$ with a bracketed copy before any simplification.

Tier 1 · Easy

1
For $f(x)=4x-7$ , work out the value of $f(5)$ .
[1 mark]
2
For $h(t)=t^2+2t$ , work out the value of $h(-3)$ .
[1 mark]

Tier 2 · Standard

1
Given $f(x)=x^2-4x+1$ , work out $f(-2)$ and simplify $f(a+1)$ .
[3 marks]
2
Given $f(x)=3x^2-2$ , simplify $f(2t)-4f(t)$ .
[3 marks]
3
For $f(x)=x^2-4x+7$ , work out and simplify $f(a+2)+f(2-a)$ .
[3 marks]

Tier 3 · Hard

1
A function has the form $g(x)=ax+b$ . Given that $g(2)=7$ and $g(-1)=-2$ , work out $g(5)$ .
[4 marks]
2
A function has the form $h(x)=x^2+px+q$ . Given that $h(0)=6$ and $h(2)=0$ , work out $h(-3)$ .
[4 marks]
3
A function has the form $f(x)=ax^2+bx+c$ . For every value of $x$ , $f(x+1)-f(x)=4x+1$ , and $f(0)=3$ . Work out $f(-2)$ .
[4 marks]
4
For every value of $x$ , a function satisfies $f(3x-2)=12x+7$ . Work out an expression for $f(x)$ .
[3 marks]
5
A function $f$ has domain $\{1,2,3\}$ and range exactly $\{4,7\}$ . Given $f(1)=a+1$ , $f(2)=2a-2$ and $f(3)=7$ , work out $f(1)+f(2)+f(3)$ .
[4 marks]

A3 · Domain and range of a function

Explanation

The domain is the set of permitted input values; the range is the set of output values actually produced. For a finite domain, evaluate every input and list distinct outputs.
For a continuous restricted domain, consider both endpoints and any turning point within the interval before stating the range.
Domain restrictions can arise because a denominator cannot be zero or an even square root cannot contain a negative value.
The domain of a function and its range must be expressed with the correct variable and inequality direction.
Examiners expect endpoint inclusion to match symbols such as $\leq$ or $<$ , and a graph or algebraic argument that identifies the true maximum and minimum outputs.

A restricted quadratic domain whose range is determined by its turning point and endpoints.

Worked example

For $f(x)=(x-1)^2+2$ with domain $-1\leq x\leq4$ , work out the range.

1.The turning point occurs at $x=1$ in the domain, giving minimum $f(1)=2$ .
2.Evaluate the endpoints: $f(-1)=6$ and $f(4)=11$ .
3.The larger endpoint output is $11$ , and both endpoints are included.

Answer: $2\leq f(x)\leq11$ .

Common mistakes

Don't use only the endpoint values and miss the minimum at the turning point.
Don't state the domain again instead of listing the function’s output range.
Don't exclude $2$ or $11$ even though the domain uses inclusive inequalities.

Exam tip

For a restricted quadratic, test the turning point only if it lies inside the domain, then compare both endpoint outputs.

Tier 1 · Easy

1
The function $f(x)=x+1$ has domain $\{-2,0,3\}$ . Write down its range.
[1 mark]
2
The function $g(x)=x^2$ has domain $-2\leq x\leq3$ . Work out the range of $g$ .
[2 marks]

Tier 2 · Standard

1
For $f(x)=x^2+2$ with domain $-2\leq x\leq3$ , work out the range of $f$ .
[3 marks]
2
For $f(x)=5-2x$ with domain $-1<x\leq4$ , work out the range of $f$ .
[2 marks]
3
The function $f(x)=(x-2)^2-3$ has domain $-1<x\leq4$ . Work out the range of $f$ .
[3 marks]

Tier 3 · Hard

1
The function $h(x)=\dfrac{6}{x-2}$ has domain $3\leq x\leq8$ . Work out its range.
[3 marks]
2
The function $g(x)=3+\dfrac{2}{x+1}$ has domain $0\leq x<3$ . Work out its range.
[3 marks]
3
The function $g(x)=\dfrac{1}{x^2-4}$ has domain $x\leq-3$ or $x\geq3$ . Work out the range of $g$ .
[4 marks]
4
Work out the domain and range of $f(x)=\sqrt{12-3x}+1$ .
[3 marks]
5
The function $h(x)=x^2+1$ has domain $-4\leq x\leq-3$ or $1<x<2$ . Work out the range of $h$ .
[4 marks]

A4 · Composite functions (the result of two or more functions acting in succession)

Explanation

A composite function applies functions in succession. In AQA notation, $fg(x)$ means $f(g(x))$ : $g$ acts first and its output becomes the input to $f$ .
To form an algebraic composite, substitute the entire inner expression into every occurrence of the variable in the outer function, using brackets where needed.
In general $fg(x)\ne gf(x)$ , because reversing the order changes the intermediate value.
Domain restrictions from both stages must be respected when they matter.
Examiners expect the action order to be stated or made visible in the substitution; multiplying the two function formulae together does not form a composite function.

In AQA notation, g acts first in the composite fg(x).

Worked example

Let $f(x)=2x-3$ and $g(x)=x^2+1$ . Work out and simplify $fg(x)$ .

1.Since $g$ acts first, write $fg(x)=f(x^2+1)$ .
2.Substitute the whole inner expression into $f$ : $2(x^2+1)-3$ .
3.Simplify the result.

Answer: $fg(x)=2x^2-1$ .

Common mistakes

Don't form $f(x)g(x)$ by multiplying the two rules together.
Don't calculate $g(f(x))$ even though $fg(x)$ means $f(g(x))$ .
Don't substitute the inner function into only one occurrence of the outer variable.

Exam tip

Rewrite the notation as nested brackets, such as $fg(x)=f(g(x))$ , before substituting.

Tier 1 · Easy

1
Let $f(x)=2x+1$ and $g(x)=x^2$ . Work out $fg(3)$ .
[2 marks]
2
Let $f(x)=x+5$ and $g(x)=3x$ . Work out $gf(-2)$ .
[2 marks]

Tier 2 · Standard

1
Let $f(x)=x-4$ and $g(x)=3x^2$ . Work out and simplify $gf(x)$ .
[3 marks]
2
Let $f(x)=2x-1$ and $g(x)=x^2$ . Work out and simplify both $fg(x)$ and $gf(x)$ .
[3 marks]
3
Let $f(x)=x^2+2$ and $g(x)=3x-1$ . Work out $fg(-2)-gf(-2)$ .
[3 marks]

Tier 3 · Hard

1
Let $f(x)=3x-2$ and $g(x)=x^2+1$ . Solve $fg(x)=13$ .
[4 marks]
2
Let $f(x)=ax+2$ and $g(x)=x^2-1$ . Given that $fg(2)=17$ , work out $a$ and then solve $gf(x)=3$ .
[4 marks]
3
Let $f(x)=x^2+1$ and $g(x)=x^2-1$ . The composites $fg(x)$ and $gf(x)$ have the same value. Work out the possible values of $x$ .
[4 marks]
4
The function $f$ is linear, $g(x)=x^2+1$ , and $fg(x)=6x^2+11$ for every real value of $x$ . Work out $f(x)$ .
[3 marks]
5
Let $f(x)=2x-1$ , and let $fff(x)$ mean $f(f(f(x)))$ . Work out $x$ when $fff(x)=x+14$ .
[3 marks]

A5 · Inverse functions (domains chosen for f to make f one-one)

Explanation

An inverse function reverses the action of the original function, so $f^{-1}(f(x))=x$ for inputs in the chosen domain. To find an inverse, write $y=f(x)$ , rearrange to make $x$ the subject, then exchange the variable labels.
The domain of $f^{-1}$ is the range of $f$ , and its range is the domain of $f$ .
A function must be one-one on its domain to have an inverse function; a quadratic therefore needs a restriction to one side of its turning point.
Graphs of a function and its inverse are reflections in $y=x$ .
Examiners expect the correct square-root branch to follow from any stated domain restriction.

The graphs of a one-one function and its inverse reflect in the line y = x.

Worked example

Given $f(x)=3x+7$ , work out $f^{-1}(x)$ and verify the result by composition.

1.Write $y=3x+7$ and rearrange: $x=\frac{y-7}{3}$ .
2.Exchange labels: $f^{-1}(x)=\frac{x-7}{3}$ .
3.Check: $f^{-1}(f(x))=\frac{(3x+7)-7}{3}=x$ .

Answer: $f^{-1}(x)=\dfrac{x-7}{3}$ .

Common mistakes

Don't write $f^{-1}(x)=\frac{1}{3x+7}$ , confusing an inverse function with a reciprocal.
Don't change the sign of $7$ but forget to divide the whole numerator by $3$ .
Don't keep both square-root branches when a restricted quadratic domain permits only one.

Exam tip

After rearranging an inverse, check one composition simplifies exactly to $x$ .

Tier 1 · Easy

1
Given $f(x)=7-2x$ , work out $f^{-1}(x)$ .
[2 marks]
2
Given $f(x)=\dfrac{x-6}{4}$ , work out $f^{-1}(x)$ .
[2 marks]

Tier 2 · Standard

1
The function $f(x)=4-(x-2)^2$ has domain $x\geq2$ . Work out $f^{-1}(x)$ and state its domain.
[3 marks]
2
The function $f(x)=4x+1$ has domain $-2\leq x<3$ . Work out $f^{-1}(x)$ and state the range of $f^{-1}$ .
[3 marks]
3
The function is $f(x)=(x-1)^3+4$ . Work out $f^{-1}(x)$ .
[3 marks]

Tier 3 · Hard

1
For $f(x)=\dfrac{2x+5}{x-1}$ , work out $f^{-1}(x)$ and state the value excluded from its domain.
[4 marks]
2
Let $f(x)=2x-3$ and $g(x)=(x+1)^2$ for $x\geq-1$ . The function $h$ is defined by $h(x)=fg(x)$ . Work out $h^{-1}(x)$ and state its domain.
[4 marks]
3
The function $f(x)=2-(x+1)^2$ has domain $x\leq-1$ . A number $r$ satisfies $f^{-1}(r)=r$ . Work out $r$ .
[4 marks]
4
For the one-one function $f(x)=2x^3-5$ , $f^{-1}(3k+2)=2$ . Work out $k$ .
[3 marks]
5
A one-one function has a linear inverse $f^{-1}(x)=ax+b$ . Given $f(2)=7$ and $f^{-1}(1)=-1$ , work out $f(x)$ .
[4 marks]

A6 · Expanding brackets and collecting like terms

Explanation

Expanding brackets applies the distributive law: every term in one bracket multiplies every term in the other. Products should be written with their signs before like terms are collected.
Terms are like only when their variable parts and powers match, so $x^2$ and $x$ cannot be combined.
When an entire expanded expression is subtracted, every sign in that expression changes.
For three brackets, multiply two brackets first, simplify the result, then multiply by the remaining bracket.
Examiners expect a fully expanded and collected polynomial in descending powers unless another form is requested; omitted cross-products and mishandled negative signs are the usual sources of lost accuracy.

Worked example

Expand and simplify $(2x+1)(x-4)-(x+2)^2$ .

1. $(2x+1)(x-4)=2x^2-7x-4$ .
2. $(x+2)^2=x^2+4x+4$ .
3.Subtract the whole second expression: $2x^2-7x-4-x^2-4x-4=x^2-11x-8$ .

Answer: $x^2-11x-8$ .

Common mistakes

Don't omit one of the two cross-products when expanding a pair of binomials.
Don't write $(x+2)^2=x^2+4$ and miss the middle term $4x$ .
Don't change only the first sign after the subtraction instead of every term in $x^2+4x+4$ .

Exam tip

Keep each bracket expansion on a separate line before collecting like powers.

Tier 1 · Easy

1
Expand and simplify $(x+4)(x-3)$ .
[2 marks]
2
Expand and simplify $3(2x-5)-4(x+1)$ .
[2 marks]

Tier 2 · Standard

1
Expand and simplify $(2x-3)(x+5)-(x-1)^2$ .
[3 marks]
2
Expand and simplify $(3x+2)^2-2x(x-5)$ .
[3 marks]
3
Expand and simplify $(x-2)(x^2+2x+4)$ .
[3 marks]

Tier 3 · Hard

1
Expand and simplify $(x+2)(x-1)(2x-3)$ .
[4 marks]
2
Expand and simplify $(2x-1)(x^2+3x-4)-(x+2)(2x^2-5)$ .
[4 marks]
3
Expand and simplify $(2x+1)^3-(2x-1)^3$ .
[4 marks]
4
Expand and simplify fully $(2a-b)(a+3b)^2$ .
[4 marks]
5
The expansion of $(x+p)(x^2-2x+5)$ contains no $x^2$ term. Work out the coefficient of $x$ .
[3 marks]

A7 · Expand (a + b)^n for positive integer n; use of Pascal's triangle

Explanation

Row $n$ of Pascal’s triangle gives the coefficients in the expansion of $(a+b)^n$ .
Each interior entry is the sum of the two entries above it.
Across the expansion, the power of the first term decreases from $n$ to $0$ , while the power of the second increases from $0$ to $n$ ; the two powers in every term add to $n$ .
Numerical coefficients and negative signs inside either term must also be raised to the indicated powers.
Examiners may ask for a full expansion or one coefficient, so identifying which powers produce the requested variable term can avoid unnecessary expansion.

Pascal’s triangle from row 0 to row 5 supplies binomial coefficients.

Worked example

Work out the coefficient of $x^3$ in $(2+x)^5$ .

1.The $x^3$ term uses three factors of $x$ and two factors of $2$ .
2.The Pascal coefficient is $\binom53=10$ .
3.The term is $10(2^2)x^3=40x^3$ .

Answer: The coefficient of $x^3$ is $40$ .

Common mistakes

Don't use the Pascal coefficient $10$ but omit the factor $2^2$ .
Don't select the position for $x^2$ instead of the requested $x^3$ term.
Don't treat a negative second term as positive in every power.

Exam tip

For one requested coefficient, state the two term powers and the matching Pascal coefficient before evaluating constants.

Tier 1 · Easy

1
Expand $(x+2)^3$ .
[2 marks]
2
Write down the coefficient of $a^2b$ in the expansion of $(a+b)^3$ .
[1 mark]

Tier 2 · Standard

1
Work out the coefficient of $x^2$ in the expansion of $(2x-1)^4$ .
[3 marks]
2
Expand $(1-2x)^4$ in ascending powers of $x$ .
[3 marks]
3
Using Pascal's triangle, work out the terms up to and including the $x^2$ term in the expansion of $(2+3x)^5$ .
[3 marks]

Tier 3 · Hard

1
The coefficient of $x^2$ in $(2x+k)^4$ is $216$ , where $k$ is an integer. Work out the possible values of $k$ .
[4 marks]
2
In the expansion of $(1+ax)^5$ , the coefficients of $x$ and $x^2$ are equal and non-zero. Work out $a$ .
[3 marks]
3
Work out the coefficient of $x^2$ in $(1+x)^4(1-2x)^3$ .
[4 marks]
4
Using Pascal's triangle, work out the constant term in the expansion of $\left(x+\dfrac2x\right)^4$ .
[3 marks]
5
In $(1+2x)^n$ , where $n$ is a positive integer, the coefficient of $x$ is $12$ . Using Pascal's triangle, work out the coefficient of $x^2$ .
[4 marks]

A8 · Factorising

Explanation

Factorising reverses expansion by writing an expression as a product. First remove the greatest common numerical and algebraic factor.
A quadratic can then be factorised by choosing binomial factors whose leading and constant products are correct and whose cross-terms produce the middle coefficient.
The identity $a^2-b^2=(a-b)(a+b)$ handles a difference of two squares.
Expressions such as quadratics in $x^2$ may need more than one stage. ‘Factorise fully’ means continue until no factor can be factorised further over the required number set.
Examiners expect a product, and expansion of the final answer is an efficient check that every original term and sign returns.

Worked example

Factorise fully $6x^3-24x$ .

1.Take out the greatest common factor: $6x^3-24x=6x(x^2-4)$ .
2.Recognise a difference of two squares: $x^2-4=x^2-2^2$ .
3.Factorise the difference: $x^2-4=(x-2)(x+2)$ .

Answer: $6x(x-2)(x+2)$ .

Common mistakes

Don't take out $6$ but leave the common factor $x$ inside the bracket.
Don't stop at $6x(x^2-4)$ even though the instruction says factorise fully.
Don't write $x^2-4=(x-2)^2$ instead of conjugate factors.

Exam tip

After extracting the greatest common factor, scan every remaining factor for a quadratic or difference of two squares.

Tier 1 · Easy

1
Factorise fully $6x^2-15x$ .
[2 marks]
2
Factorise fully $12a^2b+18ab^2$ .
[2 marks]

Tier 2 · Standard

1
Factorise $2x^2-7x-15$ .
[3 marks]
2
Factorise fully $x^3+4x^2-9x-36$ .
[3 marks]
3
Factorise fully $6x^2+9xy-4xz-6yz$ .
[3 marks]

Tier 3 · Hard

1
Factorise fully $x^4-5x^2+4$ .
[4 marks]
2
Factorise fully $a^4-16b^4$ .
[4 marks]
3
Factorise fully $2x^3-5x^2-8x+20$ .
[4 marks]
4
Factorise fully $4(x-1)^2-25(y+2)^2$ .
[3 marks]
5
Factorise fully $6x^2-xy-2y^2$ .
[3 marks]

A9 · Manipulation of rational expressions: use of + - x / for algebraic fractions with numeric, linear or quadratic denominators

Explanation

Rational expressions are algebraic fractions. Factor numerators and denominators before cancelling, because only common factors—not individual terms—may cancel.
Addition or subtraction requires a common denominator, with brackets protecting each adjusted numerator. Multiplication combines numerators and denominators after any valid factor cancellation.
Division means multiplying by the reciprocal of the divisor; the divisor must also be non-zero. Values that make any original denominator zero remain excluded even if a factor later cancels.
Examiners expect one fully simplified fraction and may require stated restrictions.
A reliable check is to factor first, identify exclusions from the original expression, then perform the requested operation.

Worked example

Write $\dfrac{3}{x+1}+\dfrac{2}{x-2}$ as one simplified fraction and state the excluded values.

1.Use common denominator $(x+1)(x-2)$ .
2.Combine numerators: $3(x-2)+2(x+1)=3x-6+2x+2=5x-4$ .
3.The original denominators exclude $x=-1$ and $x=2$ .

Answer: $\dfrac{5x-4}{(x+1)(x-2)}$ , where $x\ne-1$ and $x\ne2$ .

Common mistakes

Don't add denominators and write a denominator of $2x-1$ .
Don't cancel an $x$ term across a sum in the numerator.
Don't forget that $x=-1$ and $x=2$ remain excluded from the original expression.

Exam tip

State restrictions from the original denominators before any cancellation can hide them.

Tier 1 · Easy

1
Simplify $\dfrac{x^2-9}{x+3}$ , stating the excluded value.
[2 marks]
2
Simplify $\dfrac{3x}{x+2}\times\dfrac{x+2}{9}$ , stating the excluded value.
[2 marks]

Tier 2 · Standard

1
Simplify $\dfrac{2}{x-1}-\dfrac{3}{x+2}$ into one fraction.
[3 marks]
2
Simplify $\dfrac{x^2-4}{x+1}\div\dfrac{x-2}{3x+3}$ , then state the values of $x$ for which the original expression is not defined, including any value that would make a divisor zero.
[3 marks]
3
Simplify fully $\dfrac{x^2-5x+6}{x^2-9}\times\dfrac{x+3}{x-2}$ , stating every excluded value of $x$ .
[3 marks]

Tier 3 · Hard

1
Simplify fully $\dfrac{x^2+5x+6}{x^2-4}\div\dfrac{x^2+6x+9}{2x-4}$ , stating the excluded values of $x$ .
[4 marks]
2
Write $\dfrac{2}{x^2-x-6}+\dfrac{1}{x^2+5x+6}$ as one simplified fraction, stating the excluded values.
[4 marks]
3
Simplify fully $\left(\dfrac1x+\dfrac1{x+2}\right)\div\dfrac{2x+2}{x^2-4}$ , stating every excluded value of $x$ .
[4 marks]
4
For non-zero $x$ and $y$ , simplify fully $\dfrac{\frac1x-\frac1y}{\frac1x+\frac1y}$ , stating any further restriction.
[4 marks]
5
Work out $\dfrac{x^2-4}{x^2-x-2}-\dfrac{x-2}{x+1}$ as one simplified fraction, stating every excluded value of $x$ .
[4 marks]

A10 · Use and manipulation of formulae and expressions

Explanation

A formula expresses one quantity in terms of others. Substitution should use consistent units and bracket negative values.
To change the subject, perform inverse operations on both sides while preserving equality. Fractions may first need clearing by multiplication.
If the new subject appears in more than one term, collect those terms on one side, factor out the subject, then divide by the remaining factor.
Removing a square can introduce positive and negative roots unless a stated domain or physical context fixes the sign.
Examiners expect a final expression with the requested subject alone and no occurrence of it elsewhere; substituting the rearranged result back into the original formula can verify equivalence.

Worked example

Make $a$ the subject of $v^2=u^2+2as$ , where $s\ne0$ .

1.Subtract $u^2$ from both sides: $v^2-u^2=2as$ .
2.Divide both sides by $2s$ .
3.Check that $a$ appears once and is isolated.

Answer: $a=\dfrac{v^2-u^2}{2s}$ .

Common mistakes

Don't divide only $u^2$ by $2s$ instead of the whole numerator $v^2-u^2$ .
Don't change $v^2-u^2$ to $u^2-v^2$ while moving terms.
Don't leave $a$ on both sides after rearranging a formula with several $a$ -terms.

Exam tip

Use a fraction bar or brackets around an entire numerator when the last step divides more than one term.

Tier 1 · Easy

1
Rearrange $y=3x-7$ to make $x$ the subject.
[2 marks]
2
The formula is $C=2a+3b^2$ . Work out the value of $C$ when $a=-1$ and $b=4$ .
[1 mark]

Tier 2 · Standard

1
The formula $P=2\pi\sqrt{\dfrac{k}{m}}$ has positive variables. Make $k$ the subject.
[3 marks]
2
Make $x$ the subject of $P=ax+bx+c$ , where $a+b\neq0$ .
[2 marks]
3
The formula is $q=\dfrac{a+bt}{c-dt}$ . Work out an expression for $t$ , where $b+qd\neq0$ .
[3 marks]

Tier 3 · Hard

1
Given $y=\dfrac{3x-4}{x+2}$ , make $x$ the subject and hence work out $x$ when $y=-5$ .
[4 marks]
2
Given $\dfrac1f=\dfrac1u+\dfrac1v$ , make $v$ the subject. Hence work out $v$ when $f=6$ and $u=10$ .
[4 marks]
3
Positive quantities $A$ and $x$ satisfy $A=x+\dfrac9x$ , where $A\geq6$ . Work out the two possible expressions for $x$ in terms of $A$ .
[4 marks]
4
Positive quantities $S$ , $r$ and $h$ satisfy $S=2\pi r^2+2\pi rh$ . Work out $r$ in terms of $S$ and $h$ .
[4 marks]
5
Given $T=\dfrac{x^2-y^2}{x-y}+xy$ , where $x\neq y$ and $y\neq-1$ , work out $x$ in terms of $T$ and $y$ .
[4 marks]

A11 · Use of the factor theorem for rational values of the variable for polynomials

Explanation

The factor theorem states that $x-a$ is a factor of a polynomial $f(x)$ exactly when $f(a)=0$ . For a proposed factor $px-q$ , the corresponding root is $x=\frac qp$ , so rational substitutions must be handled exactly.
Substitution can verify a factor or determine an unknown coefficient.
Once one factor is established, polynomial division or coefficient comparison produces the remaining factor.
A zero remainder proves the proposed factor, but a request to factorise fully or solve the polynomial requires further work.
Examiners expect the substituted value, the evaluation to zero and an explicit conclusion invoking the factor theorem; merely stating that a factor ‘works’ is insufficient.

Worked example

Use the factor theorem to verify that $2x-1$ is a factor of $f(x)=2x^3-5x^2-4x+3$ , then factorise $f(x)$ fully.

1. $f\left(\frac12\right)=2\left(\frac18\right)-5\left(\frac14\right)-4\left(\frac12\right)+3=0$ , so $2x-1$ is a factor.
2.Dividing by $2x-1$ gives $x^2-2x-3$ .
3.Factorise the quotient: $x^2-2x-3=(x-3)(x+1)$ .

Answer: $f(x)=(2x-1)(x-3)(x+1)$ .

Common mistakes

Don't substitute $x=1$ for the factor $2x-1$ instead of $x=\frac12$ .
Don't obtain $f(a)=0$ and fail to state that $x-a$ is therefore a factor.
Don't stop after identifying one factor even though the question asks for full factorisation.

Exam tip

For a rational factor $px-q$ , solve $px-q=0$ first and substitute that exact fraction into the polynomial.

Tier 1 · Easy

1
Use the factor theorem to show that $x-2$ is a factor of $x^3-3x^2-4x+12$ .
[2 marks]
2
Let $f(x)=6x^2+x-2$ . Show that $3x+2$ is a factor of $f(x)$ .
[2 marks]

Tier 2 · Standard

1
The polynomial $p(x)=x^3+2x^2+kx-6$ has factor $x+1$ . Work out $k$ .
[3 marks]
2
The polynomial $p(x)=3x^3+kx^2-8x+4$ has factor $3x-2$ . Work out $k$ .
[3 marks]
3
For $p(x)=3x^3-5x^2-17x-5$ , use $p\left(-\dfrac13\right)$ to justify the factor $3x+1$ , then work out the remaining quadratic factor.
[3 marks]

Tier 3 · Hard

1
Use the factor theorem, with a rational value of $x$ , to verify that $2x-1$ divides $f(x)=2x^3+x^2-13x+6$ , then factorise $f(x)$ fully.
[4 marks]
2
The polynomial $p(x)=2x^3+kx^2-11x-6$ has factor $2x+1$ . Use the factor theorem to work out $k$ , then factorise $p(x)$ fully.
[4 marks]
3
The polynomial $p(x)=6x^3+ax^2+bx-6$ has factors $2x-1$ and $3x+1$ . Work out $a$ and $b$ using the factor theorem.
[4 marks]
4
Exactly one of $2x-1$ , $2x+1$ and $3x+1$ is a factor of $p(x)=6x^3+23x^2-6x-8$ . Work out which one, and hence factorise $p(x)$ fully.
[4 marks]
5
The polynomial $p(x)=2x^3+ax^2+bx-4$ has $(x-1)^2$ as a factor. Work out $a$ and $b$ using the factor theorem.
[4 marks]

A12 · Completing the square

Explanation

Completing the square rewrites a quadratic in the form $a(x-h)^2+k$ , exposing its turning point $(h,k)$ and supporting equation solving.
For $x^2+bx+c$ , halve the coefficient $b$ inside the bracket, then compensate outside: $(x+\frac b2)^2+c-\frac{b^2}{4}$ .
If the coefficient of $x^2$ is not $1$ , first factor it from the quadratic and linear terms.
Any compensation inside the bracket is multiplied by that outside factor.
Examiners expect an identity that expands back to the original expression; the signs in the bracket also reveal the opposite-signed $x$ -coordinate of the turning point.

Completed-square form shows the turning point (h,k) of a quadratic.

Worked example

Write $3x^2-12x+7$ in the form $a(x-h)^2+k$ .

1.Factor $3$ from the quadratic and linear terms: $3(x^2-4x)+7$ .
2.Complete the square inside: $x^2-4x=(x-2)^2-4$ .
3.Substitute and simplify: $3[(x-2)^2-4]+7=3(x-2)^2-5$ .

Answer: $3(x-2)^2-5$ .

Common mistakes

Don't write $(x-2)^2$ for $x^2-4x$ without subtracting the compensating $4$ .
Don't forget that the outside factor $3$ also multiplies the compensation $-4$ .
Don't read the turning-point coordinate as $(-2,-5)$ instead of $(2,-5)$ .

Exam tip

Expand the completed-square answer mentally to check the linear coefficient and constant before finalising it.

Tier 1 · Easy

1
Write $x^2+8x+3$ in the form $(x+a)^2+b$ .
[2 marks]
2
Work out $k$ if $x^2+10x+k=(x+5)^2$ for all values of $x$ .
[1 mark]

Tier 2 · Standard

1
Write $x^2-6x+11$ in completed-square form and state the minimum point of $y=x^2-6x+11$ .
[3 marks]
2
Express $3x^2-18x+20$ in completed-square form. Hence state its minimum value.
[3 marks]
3
Solve $x^2-10x+7=0$ by completing the square, giving the solutions in exact form.
[3 marks]

Tier 3 · Hard

1
Write $2x^2+12x-5$ in the form $a(x+b)^2+c$ . Hence solve $2x^2+12x-5=13$ , giving exact answers.
[4 marks]
2
The greatest value of $-2x^2+8x+c$ is $11$ . Work out $c$ and the value of $x$ at which the greatest value occurs.
[4 marks]
3
The minimum value of $3x^2+kx+12$ is $5$ . Work out the two possible values of $k$ and the value of $x$ giving the minimum in each case.
[4 marks]
4
The two solutions of $x^2-8x+m=0$ differ by $6$ . Work out $m$ and both solutions.
[4 marks]
5
For $-1\le x\le4$ , work out the range of values taken by $-3x^2+6x+5$ .
[4 marks]

A13 · Drawing and sketching of functions; interpretation of graphs (linear, quadratic, exponential y = ab^x and y = ab^-x, functions restricted to no more than 3 domains)

Explanation

A function sketch must show its defining shape and important features: intercepts, turning points, asymptotes and any restricted domains. Linear graphs are straight; quadratic graphs have a single turning point; exponential graphs $y=ab^x$ and $y=ab^{-x}$ pass through $(0,a)$ when defined.
For positive $a$ and $b>1$ , $ab^x$ increases and $ab^{-x}$ decreases, while both remain positive and approach the horizontal asymptote $y=0$ in one direction.
A function may be restricted to no more than three domains.
Included endpoints use filled points and excluded endpoints use open points.
Examiners expect labelled features and correct endpoint types, not a scale-perfect plot.

For a > 0 and b > 1, y = ab⁻ˣ is decreasing with y-intercept (0,a) and asymptote y = 0.

Worked example

Sketch $y=2\times3^{-x}$ , stating its $y$ -intercept, direction and horizontal asymptote.

1.At $x=0$ , $y=2\times3^0=2$ , so the intercept is $(0,2)$ .
2.As $x$ increases, $3^{-x}$ decreases, so the curve is decreasing and remains above the $x$ -axis.
3.As $x\to\infty$ , $3^{-x}\to0$ , giving horizontal asymptote $y=0$ .

Answer: A decreasing positive exponential through $(0,2)$ with horizontal asymptote $y=0$ .

Common mistakes

Don't draw $y=2\times3^{-x}$ as increasing because the base $3$ is greater than $1$ .
Don't label $(0,1)$ instead of using the multiplier to obtain $(0,2)$ .
Don't draw the curve crossing its asymptote $y=0$ .

Exam tip

Label the $y$ -intercept and asymptote, then make the curve’s increasing or decreasing direction unmistakable.

Tier 1 · Easy

1
For $y=3\times2^x$ , state the $y$ -intercept and the equation of the horizontal asymptote.
[2 marks]
2
For $y=5\times3^{-x}$ , state whether the graph increases or decreases, and work out $y$ when $x=1$ .
[2 marks]

Tier 2 · Standard

1
Sketch $y=2^{-x}+1$ . Label its $y$ -intercept and its horizontal asymptote, and state whether it is increasing or decreasing.
[3 marks]
2
Sketch $y=-(x-2)^2+4$ . Label its turning point and every intercept with the coordinate axes.
[3 marks]
3
The graph of $f(x)=x^2+1$ for $x\le2$ joins the graph of $f(x)=mx-3$ for $x>2$ without a jump. Work out $m$ and state the endpoint type for each domain at $x=2$ .
[3 marks]

Tier 3 · Hard

1
On one set of axes, sketch the three-domain rule $f(x)=x+4$ for $x<-1$ , $f(x)=x^2$ for $-1\le x\le2$ , and $f(x)=7-x$ for $x>2$ . Mark endpoint types and solve $f(x)=3$ .
[4 marks]
2
A function is defined by $f(x)=2^x$ for $x\le1$ and $f(x)=4-x$ for $x>1$ . Sketch both domains with correct endpoint types, then solve $f(x)=2$ .
[4 marks]
3
The graph $y=ab^x$ , where $a>0$ and $0<b<1$ , passes through $(0,6)$ and $(2,\frac32)$ . Work out $a$ , $b$ and the coordinates where the graph meets $y=3$ .
[4 marks]
4
Work out every intercept and the turning point of $y=(x+1)(x-5)$ , then use them to sketch the graph.
[4 marks]
5
Work out both $y$ -intercepts and the point of intersection of $y=3\times2^x$ and $y=12\times2^{-x}$ , then sketch both curves on one set of axes with their common horizontal asymptote.
[4 marks]

A14 · Solution of linear and quadratic equations (by factorisation, graph, completing the square or formula)

Explanation

A linear equation is solved by applying inverse operations to both sides until the unknown is isolated. A quadratic should first be rearranged into $ax^2+bx+c=0$ .
It may then be solved by factorisation, graph, completing the square or the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ .
Factorisation is efficient when integer or simple rational factors exist; the formula works generally and may produce exact surds.
A graphical solution is read from relevant intersections and is approximate unless exact points are evident.
Examiners expect all real solutions and method appropriate to the requested accuracy, with the entire formula numerator divided by $2a$ .

Worked example

Solve $3x^2+2x-5=0$ by factorisation.

1.Factorise: $3x^2+2x-5=(3x+5)(x-1)$ .
2.Set each factor equal to zero: $3x+5=0$ or $x-1=0$ .
3.Solve the two linear equations.

Answer: $x=-\frac53$ or $x=1$ .

Common mistakes

Don't start factorising before rearranging a quadratic equation to equal zero.
Don't set only one factor equal to zero and report a single root.
Don't divide the square-root term by $2a$ but not the $-b$ term in the quadratic formula.

Exam tip

After solving a quadratic, substitute each root or check the sum and product of roots against the coefficients.

Tier 1 · Easy

1
Solve $x^2-x-12=0$ .
[2 marks]
2
Solve $4(2x-3)=5x+9$ .
[2 marks]

Tier 2 · Standard

1
Solve $2x^2+5x-4=0$ , giving exact answers.
[3 marks]
2
Solve $x^2+4x-1=0$ by completing the square. Give exact answers.
[3 marks]
3
Solve $6x^2-x-2=0$ by factorisation.
[3 marks]

Tier 3 · Hard

1
The roots of $3x^2-7x-2=0$ are $p$ and $q$ , where $p>q$ . Work out the exact value of $p-q$ .
[4 marks]
2
The equation $x^2+px-8=0$ has one negative root and a positive root twice the magnitude of the negative root. Work out $p$ and both roots.
[4 marks]
3
A rectangle has side lengths $x+2$ centimetres and $2x-1$ centimetres. Its area is $42\text{ cm}^2$ . Work out both side lengths.
[4 marks]
4
Solve $2(x-3)^2=5x+7$ . Give both answers in exact form.
[4 marks]
5
One solution of $3x^2+mx-8=0$ is $x=\frac43$ . Work out $m$ and the other solution.
[4 marks]

A15 · Algebraic and graphical solution of simultaneous equations in two unknowns, where the equations could both be linear or one linear and one second order

Explanation

A simultaneous solution is an ordered pair satisfying both equations; graphically, it is an intersection point. Two linear equations can be solved by elimination or substitution and usually have one intersection unless they are parallel or identical.
When one equation is second order, substitution normally produces a quadratic, so zero, one or two real intersection points are possible.
Each resulting $x$ -value must be substituted back to find its paired $y$ -value.
Algebraic answers can be checked in both original equations.
Examiners expect complete coordinate pairs and, in a graphical method, intersections read to the stated accuracy rather than separate unpaired values.

A line and a quadratic can have two simultaneous solutions at their intersections.

Worked example

Solve simultaneously $y=2x+3$ and $y=x^2$ .

1.Equate the expressions: $x^2=2x+3$ , so $x^2-2x-3=0$ .
2.Factorise: $(x-3)(x+1)=0$ , giving $x=3$ or $x=-1$ .
3.Use $y=x^2$ : the paired values are $y=9$ and $y=1$ .

Answer: $(x,y)=(3,9)$ or $(-1,1)$ .

Common mistakes

Don't report $x=3$ and $x=-1$ without finding the corresponding $y$ -coordinates.
Don't pair $x=3$ with $y=1$ after substituting the roots in the wrong order.
Don't assume a line and a quadratic must have exactly two real intersections.

Exam tip

Write each simultaneous solution as an ordered pair and verify it satisfies both original equations.

Tier 1 · Easy

1
Solve simultaneously $2x+y=7$ and $x-y=2$ .
[2 marks]
2
Solve simultaneously $y=3x-4$ and $x+2y=13$ .
[2 marks]

Tier 2 · Standard

1
Solve simultaneously $y=x+2$ and $y=x^2-4$ .
[3 marks]
2
Solve simultaneously $y=x^2+1$ and $y=4x-3$ .
[3 marks]
3
Solve the simultaneous equations $3x+2y=7$ and $5x-3y=-1$ .
[3 marks]

Tier 3 · Hard

1
Solve simultaneously $x+y=7$ and $xy=10$ .
[4 marks]
2
Solve simultaneously $y=2x+1$ and $x^2+y^2=13$ .
[4 marks]
3
The line $y=2x+k$ touches the curve $y=x^2-4x+7$ . Work out $k$ and the coordinates of the point of contact.
[4 marks]
4
Two real numbers $x$ and $y$ satisfy $x-y=3$ and $x^2+y^2=89$ . Work out all possible ordered pairs $(x,y)$ , and the distance between the two corresponding points in the coordinate plane.
[4 marks]
5
Work out all solutions to the simultaneous equations $x+y=6$ and $x^2+2xy=27$ .
[4 marks]

A16 · Algebraic solution of linear equations in three unknowns

Explanation

A solution to three linear equations is an ordered triple $(x,y,z)$ satisfying all three equations. Eliminate the same unknown from two different pairs of equations to create two equations in two unknowns.
Solve that pair by elimination or substitution, then substitute the two known values into an original equation to find the third.
Multiplying an equation before addition or subtraction must affect every term.
The final triple should be checked in all three original equations, since an elimination sign error can still produce values satisfying one derived equation.
Examiners expect a systematic chain of labelled or clearly ordered equations rather than unexplained trial values.

Worked example

Solve $x+y+z=4$ , $2x-y+z=8$ and $x+2y-z=-3$ .

1.Subtract the first equation from the second: $x-2y=4$ .
2.Add the first and third equations: $2x+3y=1$ .
3.Use $x=4+2y$ in $2x+3y=1$ : $8+7y=1$ , so $y=-1$ , then $x=2$ and $z=3$ .

Answer: $(x,y,z)=(2,-1,3)$ .

Common mistakes

Don't eliminate different unknowns from the two equation pairs and gain no solvable two-variable system.
Don't multiply one side of an equation but not every term on the other side.
Don't stop after finding two variables and omit the third coordinate.

Exam tip

Check the final triple in all three original equations; one failed equality pinpoints an elimination error.

Tier 1 · Easy

1
Solve $x+y+z=9$ , $x-y=2$ and $z=3$ .
[2 marks]
2
Solve $x+y+z=12$ , $x+y=7$ and $y+z=9$ .
[2 marks]

Tier 2 · Standard

1
Solve $x+y+z=6$ , $2x-y+z=3$ and $x+2y-z=2$ .
[3 marks]
2
Solve $x+y=7$ , $y+z=10$ and $z+x=9$ .
[3 marks]
3
Solve the simultaneous equations $x+y+z=2$ , $x+2y+3z=1$ and $x+4y+9z=-5$ .
[3 marks]

Tier 3 · Hard

1
Solve $2x+3y-z=-2$ , $x-y+2z=9$ and $3x+2y+2z=10$ .
[4 marks]
2
Solve $2x+y-z=4$ , $x-2y+3z=9$ and $3x+y+2z=13$ .
[4 marks]
3
At a shop, $2$ pens, $1$ ruler and $1$ compass cost £13. One pen, $3$ rulers and $2$ compasses cost £19. Four pens, $2$ rulers and $3$ compasses cost £31. Work out the price of each item.
[4 marks]
4
A three-digit number has hundreds digit $h$ , tens digit $t$ and units digit $u$ . The digits satisfy $h+t+u=14$ , $h-t=2$ and $u=2t$ . Work out the number.
[3 marks]
5
The quadratic function $f(x)=ax^2+bx+c$ satisfies $f(1)=3$ , $f(2)=6$ and $f(-1)=9$ . Work out $a$ , $b$ , $c$ and $f(4)$ .
[4 marks]

A17 · Solution of linear and quadratic inequalities

Explanation

A linear inequality is manipulated like an equation, except that multiplying or dividing by a negative number reverses the inequality sign. For a quadratic inequality, first find the roots of the corresponding equation.
These roots split the number line into intervals; a sign diagram, test values or a sketch identifies where the quadratic is positive or negative. A positive-leading quadratic is negative between two distinct roots and positive outside them.
Strict inequalities exclude roots, while $\leq$ or $\geq$ includes them.
If two conditions must hold, their solution sets are intersected.
Examiners expect interval endpoints and inclusion symbols to match the original inequality exactly.

A positive-leading quadratic is non-positive between its two roots.

Worked example

Solve $x^2-x-6\leq0$ .

1.Factorise: $x^2-x-6=(x-3)(x+2)$ , so the roots are $-2$ and $3$ .
2.The leading coefficient is positive, so the quadratic is non-positive between the roots.
3.Equality is allowed, so include both endpoints.

Answer: $-2\leq x\leq3$ .

Common mistakes

Don't give $x\leq-2$ or $x\geq3$ , selecting where the positive-leading quadratic is positive.
Don't use open endpoints even though the inequality includes equality.
Don't divide a linear inequality by a negative number without reversing its sign.

Exam tip

Mark the roots on a sign diagram and shade only intervals whose sign matches the inequality.

Tier 1 · Easy

1
Solve $7-3x\leq19$ .
[2 marks]
2
Solve $1<2x+5\le11$ .
[2 marks]

Tier 2 · Standard

1
Solve $(x-4)(x+1)\le0$ .
[3 marks]
2
Solve $x^2+2x-8>0$ .
[3 marks]
3
Solve $2x^2+x\le6$ .
[3 marks]

Tier 3 · Hard

1
Solve the simultaneous inequalities $x^2-5x-6>0$ and $2x+1\le9$ .
[4 marks]
2
Solve $(x-1)(x+4)>6$ .
[3 marks]
3
Solve $x^2<3x+2$ , giving the exact boundary values.
[4 marks]
4
Work out all integer values of $x$ satisfying $2x^2-5x-3<0$ .
[3 marks]
5
Work out the values of $k$ for which $x^2-6x+k>0$ for every real value of $x$ .
[3 marks]

A18 · Index laws, including fractional and negative indices and the solution of equations

Explanation

For a common non-zero base, $a^ma^n=a^{m+n}$ , $a^m\div a^n=a^{m-n}$ and $(a^m)^n=a^{mn}$ . A negative index means a reciprocal: $a^{-n}=\frac1{a^n}$ .
A fractional index represents a root and a power: $a^{1/n}=\sqrt[n]{a}$ and $a^{m/n}=\sqrt[n]{a^m}$ where the real expression is defined.
Coefficients are handled separately from indices, and final forms should use positive indices where requested.
To solve an exponential equation, rewrite both sides as powers of the same base, then equate exponents.
Examiners expect exact index manipulation and relevant domain conditions, particularly where an even root or reciprocal is involved.

Worked example

Solve $8^{x+1}=4^{2x-1}$ .

1.Rewrite both sides with base $2$ : $2^{3x+3}=2^{4x-2}$ .
2.Equate exponents: $3x+3=4x-2$ .
3.Solve the linear equation.

Answer: $x=5$ .

Common mistakes

Don't multiply indices when multiplying equal bases instead of adding them.
Don't write $a^{-n}=-a^n$ instead of taking the reciprocal.
Don't equate exponents before rewriting both sides with the same base.

Exam tip

Write the common-base line explicitly before equating exponents in an index equation.

Tier 1 · Easy

1
Work out $16^{-1/2}$ .
[1 mark]
2
Given $p\ne0$ , simplify $\frac{(p^3)^2}{p}$ .
[2 marks]

Tier 2 · Standard

1
Solve $9^{x+1}=27^{2x-1}$ .
[3 marks]
2
Given $x>0$ and $y>0$ , simplify $\frac{(x^6y^{-3})^{1/2}}{x^{-1}y^{-2}}$ using positive indices.
[3 marks]
3
Given $x>0$ and $x^{3/2}=64$ , work out $x$ and $x^{-1/2}$ .
[3 marks]

Tier 3 · Hard

1
Given $a>0$ and $b>0$ , simplify fully $\frac{(81a^{-8}b^6)^{1/4}}{3a^{-1}b^{1/2}}$ .
[4 marks]
2
Solve $\frac{4^{x+1}8^{2-x}}{2^{3x}}=\frac18$ .
[4 marks]
3
Solve $2^{2x}-5(2^x)+4=0$ .
[4 marks]
4
Solve $2^{x}\times8^{x+1}=\left(\dfrac{1}{16}\right)^{x-2}$ .
[3 marks]
5
For all $x>0$ and $y>0$ , $\frac{(x^3y^{-2})^p}{(x^{-1}y)^q}=x^{11}y^{-8}$ . Work out $p$ and $q$ .
[4 marks]

A19 · Algebraic proof

Explanation

An algebraic proof establishes a statement for every permitted value, so it begins with a general representation rather than selected examples. An even integer can be written $2n$ and an odd integer $2n+1$ , where $n$ is an integer.
Consecutive integers may be represented by $n,n+1,n+2$ .
To prove divisibility by $k$ , rearrange the expression as $k$ multiplied by an integer.
Identities may be proved by expanding or factorising one side until it matches the other.
Examiners expect assumptions such as ‘ $n$ is an integer’ and a final sentence connecting the algebraic form to the required conclusion; numerical evidence alone is not proof.

Worked example

Prove that the difference between the squares of two consecutive integers is odd.

1.Let the consecutive integers be $n$ and $n+1$ , where $n$ is an integer.
2. $(n+1)^2-n^2=n^2+2n+1-n^2=2n+1$ .
3.Since $n$ is an integer, $2n+1$ is odd.

Answer: The difference is of the form $2n+1$ , so it is odd for all consecutive integers.

Common mistakes

Don't check several numerical pairs but never introduce a general integer $n$ .
Don't obtain $2n+1$ and fail to state why this form is odd.
Don't use $n$ and $n+2$ for consecutive integers.

Exam tip

Finish an algebraic proof by naming the required form—such as $2k$ , $2k+1$ or a multiple of $k$ —and stating that its parameter is an integer.

Tier 1 · Easy

1
Prove algebraically that the sum of two odd integers is even.
[2 marks]
2
Prove algebraically that the square of any even integer is divisible by $4$ .
[2 marks]

Tier 2 · Standard

1
Prove that the product of three consecutive integers is divisible by $6$ .
[3 marks]
2
Prove that the square of every odd integer is one more than a multiple of $8$ .
[3 marks]
3
Prove that the difference between the squares of any two odd integers is divisible by $8$ .
[3 marks]

Tier 3 · Hard

1
Given $x\ne0$ , prove algebraically that $x^2+\frac1{x^2}\ge2$ .
[4 marks]
2
Prove that the product of four consecutive integers, increased by $1$ , is a perfect square.
[4 marks]
3
Prove that there are no integers $a$ and $b$ for which $a^2-b^2=2026$ .
[4 marks]
4
Show that $n^4+4=(n^2+2)^2-(2n)^2$ . Hence prove that $n^4+4$ is never prime for any integer $n>1$ .
[4 marks]
5
Prove that $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4x}{x^2-1}$ for $x\ne1$ and $x\ne-1$ .
[3 marks]

A20 · Using nth terms of sequences; limiting value of a sequence as n approaches infinity

Explanation

An nth-term formula gives a sequence value directly by substituting a positive integer for $n$ .
For a rational expression in $n$ , divide numerator and denominator by the highest power of $n$ present.
Terms such as $\frac1n$ and $\frac1{n^2}$ approach $0$ as $n\to\infty$ , leaving the ratio of leading coefficients when numerator and denominator have the same degree.
The limiting value is the number the terms approach; it need not be reached by any term.
Examiners expect a clear limiting argument rather than substitution of an arbitrarily large number, and any requested finite term must still be found by direct substitution.

Sequence terms can approach a limiting value without reaching it.

Worked example

Work out the limiting value of $u_n=\dfrac{7n+2}{2n-1}$ as $n\to\infty$ .

1.Divide numerator and denominator by $n$ : $u_n=\frac{7+2/n}{2-1/n}$ .
2.As $n\to\infty$ , both $2/n$ and $1/n$ approach $0$ .
3.Evaluate the remaining ratio $\frac72$ .

Answer: The limiting value is $\frac72$ .

Common mistakes

Don't substitute $n=\infty$ as though infinity were an ordinary number.
Don't use the ratio of constant terms $2\div(-1)$ instead of the leading coefficients.
Don't claim the sequence must contain a term exactly equal to $\frac72$ .

Exam tip

For a rational nth term, divide by the highest power of $n$ before taking the limit.

Tier 1 · Easy

1
State the limiting value of $u_n=\frac{5n-2}{n}$ as $n$ approaches infinity.
[1 mark]
2
For $u_n=\frac{3n+4}{n+2}$ , work out $u_5$ .
[1 mark]

Tier 2 · Standard

1
For $v_n=\frac{3n^2+2}{n^2+4n}$ , work out $v_2$ and the limiting value as $n$ approaches infinity.
[2 marks]
2
The sequence $u_n=\frac{kn+3}{2n-5}$ has limiting value $4$ . Work out $k$ .
[2 marks]
3
The sequence $u_n=\frac{kn^2-5}{3n^2+n}$ has limiting value $4$ . Work out $k$ and $u_1$ .
[3 marks]

Tier 3 · Hard

1
The sequence $w_n=\frac{2n+1}{n+3}$ has limiting value $L$ . Work out $L$ and the least positive integer $n$ for which $w_n$ differs from $L$ by less than $0.01$ .
[4 marks]
2
The sequence $u_n=\frac{an+2}{n+b}$ has limiting value $3$ and $u_1=1$ . Work out $a$ , $b$ and $u_{10}$ .
[4 marks]
3
For positive integers $n$ , let $u_n=\frac{2n-5}{n+1}$ and $v_n=\frac{3n+4}{2n}$ . Work out the least value of $n$ for which $u_n>v_n$ .
[4 marks]
4
For positive integers $n$ , let $u_n=\frac{4n+1}{n+2}$ and $v_n=\frac{kn-3}{2n+1}$ . The two sequences have the same limiting value. Work out $k$ , then state which term is greater for each possible value of $n$ .
[4 marks]
5
For positive integers $n$ , $u_n=\frac{(n+1)(3n-2)}{n(2n+5)}$ . Work out the limiting value of the sequence and the value of $n$ for which $u_n=\frac{25}{26}$ .
[4 marks]

A21 · nth terms of linear sequences

Explanation

A linear sequence has a constant first difference and nth term of the form $an+b$ . The coefficient $a$ is the common difference.
The constant $b$ can be found by substituting a known term, or the equivalent form $u_n=u_1+(n-1)d$ can be used with first term $u_1$ and difference $d$ .
A negative difference produces a decreasing sequence.
To decide whether a value occurs, set the nth term equal to that value and check that the solution for $n$ is a positive integer.
Examiners expect indexing from $n=1$ , so checking the formula against the first two terms catches a common constant-term error.

Worked example

Work out the nth term of $13,9,5,1,\ldots$ .

1.The common difference is $-4$ , so begin with $-4n$ .
2.At $n=1$ , $-4n=-4$ , which needs $17$ added to give the first term $13$ .
3.Check $n=2$ : $-8+17=9$ .

Answer: $u_n=17-4n$ .

Common mistakes

Don't use the first term $13$ as the coefficient of $n$ instead of the common difference $-4$ .
Don't write $13-4n$ , which gives $9$ rather than $13$ when $n=1$ .
Don't accept a non-integer or non-positive value of $n$ when deciding whether a term occurs.

Exam tip

Substitute $n=1$ and $n=2$ into an nth-term formula before presenting it.

Tier 1 · Easy

1
Work out the nth term of $7,11,15,19,\ldots$ .
[2 marks]
2
A sequence has nth term $5n-8$ . Work out its 20th term.
[1 mark]

Tier 2 · Standard

1
A linear sequence has 12th term $-17$ and 30th term $-71$ . Work out its nth term.
[3 marks]
2
Show that $200$ is not a term of the linear sequence $11,17,23,29,\ldots$ .
[2 marks]
3
Two linear sequences have nth terms $7n-4$ and $3n+40$ . Work out the term number at which their values are equal and the common value.
[3 marks]

Tier 3 · Hard

1
A linear sequence has 5th term $18$ . The sum of its 8th and 12th terms is $76$ . Work out the nth term. Work out which term is equal to $202$ .
[4 marks]
2
A linear sequence has 7th term $31$ and 20th term $83$ . Work out its nth term and the first term greater than $250$ .
[4 marks]
3
The 5th, 6th and 7th terms of an increasing linear sequence have sum $42$ and product $2520$ . Work out the nth term of the sequence.
[4 marks]
4
A linear sequence has nth term $u_n=an+b$ and $u_4=15$ . For every positive integer $n$ , $u_{2n}=2u_n-3$ . Work out the nth term and the 25th term.
[4 marks]
5
Two linear sequences have nth terms $8n-3$ and $6n+1$ . Work out the smallest number greater than $50$ that occurs in both sequences.
[3 marks]

A22 · nth terms of quadratic sequences

Explanation

A quadratic sequence has constant second differences and nth term $an^2+bn+c$ . If the constant second difference is $D$ , then $a=\frac D2$ .
Subtract the sequence $an^2$ from the original terms; the remainders form a linear sequence whose nth term supplies $bn+c$ .
Alternatively, substitute three term values to form three simultaneous equations for $a,b,c$ .
The completed formula should be checked against several original terms.
Examiners expect the second-difference table or an equivalent algebraic method; using the second difference itself as the coefficient of $n^2$ doubles the leading term.

Worked example

Work out the nth term of $5,12,23,38,\ldots$ .

1.First differences are $7,11,15$ , so the constant second difference is $4$ and $a=2$ .
2.Subtract $2n^2$ : the remainders are $3,4,5,6$ , with nth term $n+2$ .
3.Combine the parts.

Answer: $u_n=2n^2+n+2$ .

Common mistakes

Don't use $4n^2$ because the second difference is $4$ , instead of halving it.
Don't find the $2n^2$ part but omit the remaining linear sequence.
Don't build first differences with inconsistent subtraction direction.

Exam tip

After finding $a=\frac12$ of the second difference, subtract $an^2$ term by term and solve the simpler linear remainder.

Tier 1 · Easy

1
Work out the nth term of $4,13,28,49,\ldots$ .
[2 marks]
2
A quadratic sequence has nth term $n^2-2n+6$ . Work out its 7th term.
[1 mark]

Tier 2 · Standard

1
Work out the nth term of $2,9,20,35,54,\ldots$ .
[3 marks]
2
Is $200$ a term of the quadratic sequence with nth term $3n^2-2$ ? Justify your answer.
[2 marks]
3
A quadratic sequence has nth term $u_n=n^2+kn+5$ and $u_4-u_3=11$ . Work out $k$ and $u_{10}$ .
[3 marks]

Tier 3 · Hard

1
A quadratic sequence has nth term $an^2+bn+c$ . Its 1st, 2nd and 4th terms are $6$ , $15$ and $45$ . Work out its 10th term.
[4 marks]
2
A quadratic sequence satisfies $u_{n+1}-u_n=6n+5$ and $u_1=4$ . Work out its nth term and the least $n$ for which $u_n>500$ .
[4 marks]
3
The quadratic sequence $u_n=2n^2+n+4$ and the linear sequence $v_n=23n-44$ have equal values for two positive term numbers. Work out both term numbers and their corresponding common values.
[4 marks]
4
For positive integers $n$ , a quadratic sequence has nth term $u_n=n^2-11n+30$ . Work out every term number for which $u_n<3$ .
[4 marks]
5
Pattern $n$ contains a rectangular block with $n+2$ rows and $2n-1$ tiles in each row, together with $3$ extra tiles. Work out the nth term for the number of tiles and the pattern number that contains $120$ tiles.
[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

A1 · The basic processes of algebra, including use of the associative, commutative and distributive laws

Tier 1 · Easy

Mark scheme for A1 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$1900$	2	Use the distributive law to take out the common factor: $19\times37+19\times63=19(37+63)=19\times100=1900$ .
2	$10(p+q)$	1	The repeated factor is $p+q$ . Apply the distributive law in reverse: $7(p+q)+3(p+q)=(7+3)(p+q)=10(p+q)$ .

Tier 2 · Standard

Mark scheme for A1 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$2500$	2	All three terms share the factor $25$ , so distribute in reverse across both the addition and the subtraction: $25\times47+25\times55-25\times2=25(47+55-2)=25\times100=2500$ .
2	$r(p+q)$	3	Distribute across both brackets: $p(q+r)-q(p-r)=pq+pr-pq+qr$ . The $pq$ terms cancel, leaving $pr+qr$ . Factor out $r$ to obtain $r(p+q)$ .
3	$1020$	3	Use the distributive law twice: $p^2q+pq^2=pq(p+q)$ . Substitute the given values to obtain $60\times17=1020$ .

Tier 3 · Hard

Mark scheme for A1 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$4c(a+b)$	3	Use the difference of two squares with $U=a+b+c$ and $V=a+b-c$ . Then $U^2-V^2=(U-V)(U+V)=(2c)(2a+2b)=4c(a+b)$ .
2	$c(b-a)$ (or $-c(a-b)$ )	3	Expand and collect like terms: $a(b-c)+b(a+c)-2ab=ab-ac+ab+bc-2ab=bc-ac$ . Factor out the common factor $c$ to obtain $c(b-a)$ .
3	$-75$	4	Factor out $xy$ : $x^3y+2x^2y^2+xy^3=xy(x^2+2xy+y^2)$ . The bracket is $(x+y)^2$ , so the expression is $xy(x+y)^2$ . Substitution gives $-3\times5^2=-75$ .
4	$z=3$	3	Subtract the second expression from the first: $x(y+z)-y(x+z)=26-17=9$ . Distributing and using $xy=yx$ gives $xy+xz-yx-yz=z(x-y)$ . Hence $3z=9$ , so $z=3$ .
5	$1$	3	Let $n=999$ , so $998=n-1$ and $1000=n+1$ . Distributing gives $(n-1)(n+1)=n^2-1$ . Therefore $999^2-998\times1000=n^2-(n^2-1)=1$ .

A2 · Definition of a function; notation f(x)

Tier 1 · Easy

Mark scheme for A2 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$f(5)=13$	1	Substitute $x=5$ : $f(5)=4(5)-7=20-7=13$ .
2	$h(-3)=3$	1	Substitute $t=-3$ using brackets: $h(-3)=(-3)^2+2(-3)=9-6=3$ .

Tier 2 · Standard

Mark scheme for A2 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$f(-2)=13$ $f(a+1)=a^2-2a-2$	3	$f(-2)=(-2)^2-4(-2)+1=4+8+1=13$ . Also, $f(a+1)=(a+1)^2-4(a+1)+1=a^2+2a+1-4a-4+1=a^2-2a-2$ .
2	$6$	3	$f(2t)=3(2t)^2-2=12t^2-2$ , while $4f(t)=4(3t^2-2)=12t^2-8$ . Therefore $f(2t)-4f(t)=12t^2-2-(12t^2-8)=6$ .
3	$2a^2+6$	3	$f(a+2)=(a+2)^2-4(a+2)+7=a^2+3$ . Also, $f(2-a)=(2-a)^2-4(2-a)+7=a^2+3$ . Their sum is $2a^2+6$ .

Tier 3 · Hard

Mark scheme for A2 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$g(5)=16$	4	The information gives $2a+b=7$ and $-a+b=-2$ . Subtracting the second equation from the first gives $3a=9$ , so $a=3$ . Then $-3+b=-2$ , so $b=1$ . Therefore $g(5)=3(5)+1=16$ .
2	$h(-3)=30$	4	From $h(0)=6$ , $q=6$ . Then $h(2)=0$ gives $4+2p+6=0$ , so $2p=-10$ and $p=-5$ . Hence $h(-3)=(-3)^2-5(-3)+6=9+15+6=30$ .
3	$f(-2)=13$	4	Expanding the difference gives $f(x+1)-f(x)=2ax+a+b$ . Comparing this with $4x+1$ gives $2a=4$ and $a+b=1$ , so $a=2$ and $b=-1$ . Since $f(0)=3$ , $c=3$ . Therefore $f(-2)=2(-2)^2-(-2)+3=13$ .
4	$f(x)=4x+15$	3	Set $u=3x-2$ , so $x=(u+2)/3$ . Then $f(u)=12(u+2)/3+7=4u+15$ . Replacing the placeholder $u$ by $x$ gives $f(x)=4x+15$ .
5	$15$	4	Every output must be $4$ or $7$ . From $a+1\in\{4,7\}$ , $a=3$ or $a=6$ . From $2a-2\in\{4,7\}$ , $a=3$ or $a=9/2$ . The only common value is $a=3$ , which gives outputs $4$ , $4$ and $7$ . Their sum is $15$ .

A3 · Domain and range of a function

Tier 1 · Easy

Mark scheme for A3 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\{-1,1,4\}$	1	Apply $f(x)=x+1$ to each domain value: $f(-2)=-1$ , $f(0)=1$ and $f(3)=4$ . Hence the range is $\{-1,1,4\}$ .
2	$0\leq g(x)\leq9$	2	The minimum of $x^2$ is $0$ , attained at $x=0$ , which is inside the domain. The maximum occurs at the endpoint with greatest magnitude: $g(3)=9$ . Therefore $0\leq g(x)\leq9$ .

Tier 2 · Standard

Mark scheme for A3 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$2\leq f(x)\leq11$	3	The minimum of $x^2$ occurs at $x=0$ , which is in the domain, so the minimum output is $2$ . At the endpoints, $f(-2)=6$ and $f(3)=11$ , so the maximum is $11$ . Therefore $2\leq f(x)\leq11$ .
2	$-3\leq f(x)<7$	2	The function decreases as $x$ increases. At the included endpoint $x=4$ , $f(4)=-3$ , so $-3$ is included. As $x$ approaches $-1$ from above, $f(x)$ approaches $7$ , but $x=-1$ is excluded. Therefore $-3\leq f(x)<7$ .
3	$-3\leq f(x)<6$	3	The turning point $x=2$ is in the domain, so the minimum is $f(2)=-3$ . At the included endpoint, $f(4)=1$ . As $x$ approaches $-1$ from above, $f(x)$ approaches $6$ , but $x=-1$ is excluded. Hence $-3\leq f(x)<6$ .

Tier 3 · Hard

Mark scheme for A3 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$1\leq h(x)\leq6$	3	On this domain, $x-2$ is positive and increases from $1$ to $6$ , so $6/(x-2)$ decreases. The endpoint outputs are $h(3)=6$ and $h(8)=1$ . Every intermediate output occurs, so $1\leq h(x)\leq6$ .
2	$\dfrac{7}{2}<g(x)\leq5$ (or $3.5<g(x)\leq5$ )	3	On the stated domain, $2/(x+1)$ decreases as $x$ increases. The included endpoint gives $g(0)=5$ . As $x$ approaches $3$ from below, $g(x)$ approaches $3+2/4=7/2$ , but $x=3$ is excluded. Hence $7/2<g(x)\leq5$ .
3	$0<g(x)\leq\dfrac15$	4	On the stated domain, $x^2\geq9$ , so $x^2-4\geq5$ and $g(x)$ is positive. The largest value is $1/5$ , attained when $x=\pm3$ . As $\|x\|$ increases without bound, $g(x)$ approaches $0$ but never reaches it. Therefore $0<g(x)\leq1/5$ .
4	Domain: $x\leq4$ Range: $f(x)\geq1$	3	The square-root input must be non-negative, so $12-3x\geq0$ and hence $x\leq4$ . Also, $\sqrt{12-3x}\geq0$ , with equality at $x=4$ , so the least output is $1$ . The square root grows without bound as $x$ decreases, giving the range $f(x)\geq1$ .
5	$2<h(x)<5$ or $10\leq h(x)\leq17$	4	For $-4\leq x\leq-3$ , squaring gives $9\leq x^2\leq16$ , so $10\leq h(x)\leq17$ . For $1<x<2$ , $1<x^2<4$ , so $2<h(x)<5$ . The intervals do not join, giving $2<h(x)<5$ or $10\leq h(x)\leq17$ .

A4 · Composite functions (the result of two or more functions acting in succession)

Tier 1 · Easy

Mark scheme for A4 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$fg(3)=19$	2	Apply $g$ first: $g(3)=3^2=9$ . Then apply $f$ : $f(9)=2(9)+1=19$ .
2	$gf(-2)=9$	2	Apply $f$ first: $f(-2)=-2+5=3$ . Then apply $g$ : $g(3)=3(3)=9$ .

Tier 2 · Standard

Mark scheme for A4 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$gf(x)=3x^2-24x+48$	3	In $gf(x)$ , $f$ acts first: $gf(x)=g(x-4)=3(x-4)^2$ . Expanding gives $3(x^2-8x+16)=3x^2-24x+48$ .
2	$fg(x)=2x^2-1$ $gf(x)=4x^2-4x+1$	3	$fg(x)=f(x^2)=2x^2-1$ . In the other order, $gf(x)=g(2x-1)=(2x-1)^2=4x^2-4x+1$ .
3	$34$	3	$g(-2)=-7$ , so $fg(-2)=f(-7)=51$ . Also, $f(-2)=6$ , so $gf(-2)=g(6)=17$ . Therefore $fg(-2)-gf(-2)=51-17=34$ .

Tier 3 · Hard

Mark scheme for A4 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x=-2$ or $x=2$	4	First form the composite: $fg(x)=3(x^2+1)-2=3x^2+1$ . Hence $3x^2+1=13$ , so $3x^2=12$ and $x^2=4$ . Therefore $x=-2$ or $x=2$ .
2	$a=5$ $x=0$ or $x=-\dfrac45$	4	Since $g(2)=3$ , $fg(2)=f(3)=3a+2$ . Thus $3a+2=17$ , so $a=5$ . Then $gf(x)=g(5x+2)=(5x+2)^2-1$ . Setting this equal to $3$ gives $(5x+2)^2=4$ , so $5x+2=2$ or $5x+2=-2$ . Hence $x=0$ or $x=-4/5$ .
3	$x=\pm\dfrac{\sqrt2}{2}$ (or $x=\pm\dfrac1{\sqrt2}$ )	4	$fg(x)=(x^2-1)^2+1=x^4-2x^2+2$ , while $gf(x)=(x^2+1)^2-1=x^4+2x^2$ . Equating these gives $4x^2=2$ , so $x^2=1/2$ . Hence $x=\pm\sqrt2/2$ .
4	$f(x)=6x+5$	3	Write $f(x)=ax+b$ . Then $fg(x)=f(x^2+1)=a(x^2+1)+b=ax^2+a+b$ . Comparing coefficients with $6x^2+11$ gives $a=6$ and $a+b=11$ , so $b=5$ and $f(x)=6x+5$ .
5	$x=3$	3	First, $ff(x)=f(2x-1)=2(2x-1)-1=4x-3$ . Applying $f$ again gives $fff(x)=2(4x-3)-1=8x-7$ . Hence $8x-7=x+14$ , so $7x=21$ and $x=3$ .

A5 · Inverse functions (domains chosen for f to make f one-one)

Tier 1 · Easy

Mark scheme for A5 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$f^{-1}(x)=\dfrac{7-x}{2}$	2	Write $y=7-2x$ . Then $2x=7-y$ , so $x=(7-y)/2$ . Interchanging the labels gives $f^{-1}(x)=(7-x)/2$ .
2	$f^{-1}(x)=4x+6$	2	Write $y=(x-6)/4$ . Then $4y=x-6$ , so $x=4y+6$ . Interchanging the labels gives $f^{-1}(x)=4x+6$ .

Tier 2 · Standard

Mark scheme for A5 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$f^{-1}(x)=2+\sqrt{4-x}$ Domain: $x\leq4$	3	Write $y=4-(x-2)^2$ , so $(x-2)^2=4-y$ . The restriction $x\geq2$ requires the positive square root, giving $x=2+\sqrt{4-y}$ . Hence $f^{-1}(x)=2+\sqrt{4-x}$ . The range of $f$ is $y\leq4$ , so the inverse has domain $x\leq4$ .
2	$f^{-1}(x)=\dfrac{x-1}{4}$ Range: $-2\leq f^{-1}(x)<3$	3	From $y=4x+1$ , rearrange to $x=(y-1)/4$ , so $f^{-1}(x)=(x-1)/4$ . The range of an inverse is the domain of the original function. Therefore the range is $-2\leq f^{-1}(x)<3$ .
3	$f^{-1}(x)=1+\sqrt[3]{x-4}$	3	Write $y=(x-1)^3+4$ . Then $(x-1)^3=y-4$ , so $x=1+\sqrt[3]{y-4}$ . Interchanging the labels gives $f^{-1}(x)=1+\sqrt[3]{x-4}$ .

Tier 3 · Hard

Mark scheme for A5 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$f^{-1}(x)=\dfrac{x+5}{x-2}$ Excluded value: $x=2$	4	Let $y=(2x+5)/(x-1)$ . Then $yx-y=2x+5$ , so $x(y-2)=y+5$ and $x=(y+5)/(y-2)$ . Interchange the labels to obtain $f^{-1}(x)=(x+5)/(x-2)$ . Its denominator is zero at $x=2$ , so that value is excluded.
2	$h^{-1}(x)=-1+\sqrt{\dfrac{x+3}{2}}$ (or $h^{-1}(x)=\dfrac{\sqrt{2x+6}}{2}-1$ ) Domain: $x\geq-3$	4	First, $h(x)=2(x+1)^2-3$ . Write $y=2(x+1)^2-3$ , so $(x+1)^2=(y+3)/2$ . Since the domain has $x+1\geq0$ , take the positive square root: $x=-1+\sqrt{(y+3)/2}$ . Hence $h^{-1}(x)=-1+\sqrt{(x+3)/2}$ . The minimum value of $h$ is $-3$ , so the inverse has domain $x\geq-3$ .
3	$r=\dfrac{-3-\sqrt{13}}{2}$	4	Applying $f$ to $f^{-1}(r)=r$ gives $r=f(r)$ . Hence $r=2-(r+1)^2$ , which rearranges to $r^2+3r-1=0$ . The quadratic formula gives $r=(-3\pm\sqrt{13})/2$ . Since the range of $f^{-1}$ is the domain of $f$ , $r\leq-1$ , so only $r=(-3-\sqrt{13})/2$ is valid.
4	$k=3$	3	Apply $f$ to both sides of $f^{-1}(3k+2)=2$ . This gives $3k+2=f(2)=2(2^3)-5=11$ . Therefore $3k=9$ and $k=3$ .
5	$f(x)=2x+3$	4	Since $f(2)=7$ , the inverse relation gives $f^{-1}(7)=2$ , so $7a+b=2$ . Also, $a+b=-1$ . Subtracting gives $6a=3$ , so $a=1/2$ and $b=-3/2$ . Thus $f^{-1}(x)=(x-3)/2$ ; reversing this rule gives $f(x)=2x+3$ .

A6 · Expanding brackets and collecting like terms

Tier 1 · Easy

Mark scheme for A6 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x^2+x-12$	2	Multiply each pair of terms: $x^2-3x+4x-12$ . Collecting the linear terms gives $x^2+x-12$ .
2	$2x-19$	2	Expand both brackets: $3(2x-5)-4(x+1)=6x-15-4x-4$ . Collect like terms to get $2x-19$ .

Tier 2 · Standard

Mark scheme for A6 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x^2+9x-16$	3	Expand $(2x-3)(x+5)=2x^2+7x-15$ and $(x-1)^2=x^2-2x+1$ . Subtract the entire second expression: $2x^2+7x-15-x^2+2x-1=x^2+9x-16$ .
2	$7x^2+22x+4$	3	$(3x+2)^2=9x^2+12x+4$ and $2x(x-5)=2x^2-10x$ . Subtracting the second expression gives $9x^2+12x+4-2x^2+10x=7x^2+22x+4$ .
3	$x^3-8$	3	Expanding gives $x^3+2x^2+4x-2x^2-4x-8$ . The quadratic and linear terms cancel, leaving $x^3-8$ .

Tier 3 · Hard

Mark scheme for A6 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$2x^3-x^2-7x+6$	4	First, $(x+2)(x-1)=x^2+x-2$ . Then $(x^2+x-2)(2x-3)=2x^3-3x^2+2x^2-3x-4x+6=2x^3-x^2-7x+6$ .
2	$x^2-6x+14$	4	$(2x-1)(x^2+3x-4)=2x^3+5x^2-11x+4$ . Also, $(x+2)(2x^2-5)=2x^3+4x^2-5x-10$ . Subtracting the whole second expansion gives $2x^3+5x^2-11x+4-(2x^3+4x^2-5x-10)=x^2-6x+14$ .
3	$24x^2+2$	4	$(2x+1)^3=8x^3+12x^2+6x+1$ and $(2x-1)^3=8x^3-12x^2+6x-1$ . Subtracting the whole second expansion gives $8x^3+12x^2+6x+1-8x^3+12x^2-6x+1=24x^2+2$ .
4	$2a^3+11a^2b+12ab^2-9b^3$	4	First, $(a+3b)^2=a^2+6ab+9b^2$ . Multiplying this by $2a$ gives $2a^3+12a^2b+18ab^2$ , while multiplying by $-b$ gives $-a^2b-6ab^2-9b^3$ . Collecting like terms gives $2a^3+11a^2b+12ab^2-9b^3$ .
5	$1$	3	Expanding gives $x^3+(p-2)x^2+(5-2p)x+5p$ . The missing $x^2$ term requires $p-2=0$ , so $p=2$ . The coefficient of $x$ is therefore $5-2p=5-4=1$ .

A7 · Expand (a + b)^n for positive integer n; use of Pascal's triangle

Tier 1 · Easy

Mark scheme for A7 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x^3+6x^2+12x+8$	2	Use Pascal coefficients $1,3,3,1$ : $x^3+3x^2(2)+3x(2^2)+2^3=x^3+6x^2+12x+8$ .
2	$3$	1	The Pascal coefficients for power $3$ are $1,3,3,1$ . The $a^2b$ term is $3a^2b$ , so its coefficient is $3$ .

Tier 2 · Standard

Mark scheme for A7 Tier 2 · Standard
Q	Answer	Mark	Comments
1	Coefficient of $x^2$ : $24$	3	The $x^2$ term uses two factors of $2x$ and two factors of $-1$ . Its Pascal coefficient is $\binom{4}{2}=6$ , so the term is $6(2x)^2(-1)^2=24x^2$ . The required coefficient is $24$ .
2	$1-8x+24x^2-32x^3+16x^4$	3	Use Pascal coefficients $1,4,6,4,1$ : $1^4+4(1^3)(-2x)+6(1^2)(-2x)^2+4(1)(-2x)^3+(-2x)^4=1-8x+24x^2-32x^3+16x^4$ .
3	$32+240x+720x^2$	3	The first three row-5 Pascal coefficients are $1,5,10$ . The required terms are $2^5+5(2^4)(3x)+10(2^3)(3x)^2=32+240x+720x^2$ .

Tier 3 · Hard

Mark scheme for A7 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$k=-3$ or $k=3$	4	The $x^2$ term uses two factors of $2x$ and two factors of $k$ . It is $\binom{4}{2}(2x)^2k^2=24k^2x^2$ . Hence $24k^2=216$ , so $k^2=9$ and $k=-3$ or $k=3$ .
2	$a=\dfrac12$ (or $a=0.5$ )	3	The coefficient of $x$ is $5a$ . The coefficient of $x^2$ uses the row-5 Pascal's triangle entry $10$ , so it is $10a^2$ . Hence $5a=10a^2$ . Since the coefficients are non-zero, $a\ne0$ , so division by $5a$ gives $1=2a$ and therefore $a=1/2$ .
3	$-6$	4	Up to $x^2$ , Pascal's triangle gives $(1+x)^4=1+4x+6x^2+\cdots$ and $(1-2x)^3=1-6x+12x^2+\cdots$ . The $x^2$ coefficient in the product is $1(12)+4(-6)+6(1)=-6$ .
4	$24$	3	The powers of $x$ in successive terms are $4,2,0,-2,-4$ , so the constant comes from the middle entry of row $4$ of Pascal's triangle. That entry is $6$ , and the term is $6x^2(2/x)^2=6\times4=24$ .
5	$60$	4	The second entry in row $n$ of Pascal's triangle is $n$ , so the coefficient of $x$ is $2n$ . Hence $2n=12$ and $n=6$ . The next entry in row $6$ is $15$ , so the coefficient of $x^2$ is $15\times2^2=60$ .

A8 · Factorising

Tier 1 · Easy

Mark scheme for A8 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$3x(2x-5)$	2	The greatest common factor of $6x^2$ and $-15x$ is $3x$ . Taking it out gives $6x^2-15x=3x(2x-5)$ .
2	$6ab(2a+3b)$	2	The greatest common factor is $6ab$ . Dividing each term by $6ab$ gives $2a$ and $3b$ , so the full factorisation is $6ab(2a+3b)$ .

Tier 2 · Standard

Mark scheme for A8 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$(2x+3)(x-5)$	3	The product $2\times(-15)=-30$ and the required sum is $-7$ , so split the middle term using $3$ and $-10$ : $2x^2+3x-10x-15=x(2x+3)-5(2x+3)=(2x+3)(x-5)$ .
2	$(x+4)(x-3)(x+3)$	3	Let $p(x)=x^3+4x^2-9x-36$ . Since $p(-4)=-64+64+36-36=0$ , the factor theorem shows that $x+4$ is a factor. Dividing by $x+4$ gives $x^2-9$ , and $x^2-9=(x-3)(x+3)$ . Therefore $p(x)=(x+4)(x-3)(x+3)$ .
3	$(3x-2z)(2x+3y)$	3	Group the terms: $6x^2+9xy-4xz-6yz=3x(2x+3y)-2z(2x+3y)$ . Taking out the repeated factor gives $(3x-2z)(2x+3y)$ .

Tier 3 · Hard

Mark scheme for A8 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$(x-2)(x-1)(x+1)(x+2)$	4	Treat the expression as a quadratic in $x^2$ : $x^4-5x^2+4=(x^2-1)(x^2-4)$ . Each factor is a difference of two squares, so $(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)$ .
2	$(a-2b)(a+2b)(a^2+4b^2)$	4	First use a difference of two squares: $a^4-16b^4=(a^2-4b^2)(a^2+4b^2)$ . Factorise $a^2-4b^2$ again as $(a-2b)(a+2b)$ . The sum $a^2+4b^2$ does not factorise further over the real numbers, giving $(a-2b)(a+2b)(a^2+4b^2)$ .
3	$(2x-5)(x-2)(x+2)$	4	Group the expression as $x^2(2x-5)-4(2x-5)$ . Taking out the repeated factor gives $(2x-5)(x^2-4)$ . The difference of two squares then gives $(2x-5)(x-2)(x+2)$ .
4	$(2x-5y-12)(2x+5y+8)$	3	This is a difference of two squares: $[2(x-1)]^2-[5(y+2)]^2$ . Its factors are $[2(x-1)-5(y+2)][2(x-1)+5(y+2)]$ . Simplifying the two brackets gives $(2x-5y-12)(2x+5y+8)$ .
5	$(3x-2y)(2x+y)$	3	Split the middle term as $3xy-4xy$ : $6x^2-xy-2y^2=6x^2+3xy-4xy-2y^2$ . Grouping gives $3x(2x+y)-2y(2x+y)$ . Taking out the repeated factor gives $(3x-2y)(2x+y)$ .

A9 · Manipulation of rational expressions: use of + - x / for algebraic fractions with numeric, linear or quadratic denominators

Tier 1 · Easy

Mark scheme for A9 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x-3$ , where $x\neq-3$	2	Factor the numerator: $x^2-9=(x-3)(x+3)$ . Cancel the common factor $x+3$ to get $x-3$ . The original denominator is zero at $x=-3$ , so this value remains excluded.
2	$\dfrac{x}{3}$ , where $x\neq-2$	2	Cancel the common factor $x+2$ and reduce $3/9$ to $1/3$ , giving $x/3$ . The original denominator $x+2$ is zero when $x=-2$ , so this value remains excluded.

Tier 2 · Standard

Mark scheme for A9 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\dfrac{7-x}{(x-1)(x+2)}$	3	Use the common denominator $(x-1)(x+2)$ . The numerator is $2(x+2)-3(x-1)=2x+4-3x+3=7-x$ . Therefore the result is $(7-x)/((x-1)(x+2))$ .
2	$3(x+2)$ , where $x\neq-1$ and $x\neq2$	3	Factorise and multiply by the reciprocal: $\dfrac{(x-2)(x+2)}{x+1}\times\dfrac{3(x+1)}{x-2}=3(x+2)$ . The original denominators exclude $x=-1$ , and the divisor equals zero at $x=2$ , so both values are excluded.
3	$1$ , where $x\neq-3$ , $x\neq2$ and $x\neq3$	3	Factorising gives $\dfrac{(x-2)(x-3)}{(x-3)(x+3)}\times\dfrac{x+3}{x-2}$ . All displayed factors cancel, giving $1$ . The original denominators exclude $x=-3$ , $x=3$ and $x=2$ .

Tier 3 · Hard

Mark scheme for A9 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\dfrac{2}{x+3}$ , where $x\neq2$ , $x\neq-2$ and $x\neq-3$	4	Factorise everything: $\dfrac{(x+2)(x+3)}{(x-2)(x+2)}\div\dfrac{(x+3)^2}{2(x-2)}$ . Division is multiplication by the reciprocal: $\dfrac{x+3}{x-2}\times\dfrac{2(x-2)}{(x+3)^2}$ . Cancel the common factors $x-2$ and $x+3$ to get $\dfrac{2}{x+3}$ . The original denominators exclude $x=2$ and $x=-2$ , and the divisor must be non-zero, excluding $x=-3$ .
2	$\dfrac{3(x+1)}{(x+2)(x-3)(x+3)}$ (or $\dfrac{3(x+1)}{x^3+2x^2-9x-18}$ ), where $x\neq-3$ , $x\neq-2$ and $x\neq3$	4	Factor the denominators: $x^2-x-6=(x-3)(x+2)$ and $x^2+5x+6=(x+2)(x+3)$ . Using common denominator $(x+2)(x-3)(x+3)$ gives numerator $2(x+3)+(x-3)=3x+3=3(x+1)$ . No factor cancels. The original denominators exclude $x=3,-2,-3$ .
3	$\dfrac{x-2}{x}$ , where $x\neq-2$ , $x\neq-1$ , $x\neq0$ and $x\neq2$	4	First, $1/x+1/(x+2)=2(x+1)/(x(x+2))$ . Dividing by the second fraction gives $\dfrac{2(x+1)}{x(x+2)}\times\dfrac{(x-2)(x+2)}{2(x+1)}=\dfrac{x-2}{x}$ . The original denominators exclude $x=-2,0,2$ , and the divisor is zero when $x=-1$ , so all four values are excluded.
4	$\dfrac{y-x}{x+y}$ , where $x\neq0$ , $y\neq0$ and $x+y\neq0$	4	The numerator is $(y-x)/(xy)$ and the denominator is $(x+y)/(xy)$ . Dividing these gives $\dfrac{y-x}{xy}\times\dfrac{xy}{x+y}=\dfrac{y-x}{x+y}$ . Besides $x\neq0$ and $y\neq0$ , the original divisor must not be zero, so $x+y\neq0$ .
5	$\dfrac4{x+1}$ , where $x\neq-1$ and $x\neq2$	4	Factorising the first fraction gives $\dfrac{(x-2)(x+2)}{(x-2)(x+1)}=\dfrac{x+2}{x+1}$ , while retaining the exclusion $x\neq2$ . The subtraction is then $\dfrac{x+2-(x-2)}{x+1}=\dfrac4{x+1}$ . The original denominators exclude $x=-1$ and $x=2$ .

A10 · Use and manipulation of formulae and expressions

Tier 1 · Easy

Mark scheme for A10 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x=\dfrac{y+7}{3}$	2	Add $7$ to both sides to obtain $y+7=3x$ . Divide both sides by $3$ , giving $x=(y+7)/3$ .
2	$C=46$	1	Substitute using brackets: $C=2(-1)+3(4^2)=-2+3(16)=-2+48=46$ .

Tier 2 · Standard

Mark scheme for A10 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$k=\dfrac{mP^2}{4\pi^2}$	3	Divide by $2\pi$ : $P/(2\pi)=\sqrt{k/m}$ . Square both sides to get $P^2/(4\pi^2)=k/m$ . Multiplying by $m$ gives $k=mP^2/(4\pi^2)$ .
2	$x=\dfrac{P-c}{a+b}$	2	Subtract $c$ : $P-c=ax+bx$ . Factor out $x$ on the right: $P-c=x(a+b)$ . Divide by $a+b$ to obtain $x=(P-c)/(a+b)$ .
3	$t=\dfrac{qc-a}{b+qd}$	3	Multiply by $c-dt$ : $qc-qdt=a+bt$ . Collect the terms containing $t$ : $qc-a=bt+qdt=t(b+qd)$ . Dividing by $b+qd$ gives $t=(qc-a)/(b+qd)$ .

Tier 3 · Hard

Mark scheme for A10 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x=\dfrac{2y+4}{3-y}$ When $y=-5$ , $x=-\dfrac{3}{4}$	4	Multiply by $x+2$ : $yx+2y=3x-4$ . Collect the $x$ -terms: $x(y-3)=-4-2y$ . Thus $x=(-4-2y)/(y-3)=(2y+4)/(3-y)$ . For $y=-5$ , $x=(-10+4)/(3+5)=-6/8=-3/4$ .
2	$v=\dfrac{fu}{u-f}$ When $f=6$ and $u=10$ , $v=15$	4	Subtract $1/u$ from both sides: $1/v=1/f-1/u=(u-f)/(fu)$ . Taking reciprocals gives $v=fu/(u-f)$ . Substituting $f=6$ and $u=10$ gives $v=60/(10-6)=60/4=15$ .
3	$x=\dfrac{A+\sqrt{A^2-36}}{2}$ or $x=\dfrac{A-\sqrt{A^2-36}}{2}$	4	Multiplying by $x$ gives $Ax=x^2+9$ , so $x^2-Ax+9=0$ . Applying the quadratic formula gives $x=(A\pm\sqrt{A^2-36})/2$ . The condition $A\geq6$ makes the roots real, and both roots are positive.
4	$r=\dfrac{-h+\sqrt{h^2+\frac{2S}{\pi}}}{2}$ (or the equivalent $r=-\dfrac{h}{2}+\sqrt{\dfrac{h^2}{4}+\dfrac{S}{2\pi}}$ )	4	Divide by $2\pi$ and collect terms to obtain $r^2+hr-S/(2\pi)=0$ . The quadratic formula gives $r=(-h\pm\sqrt{h^2+2S/\pi})/2$ . Since $r$ is positive, only the plus-square-root value is valid.
5	$x=\dfrac{T-y}{1+y}$	4	Factorising the numerator gives $(x-y)(x+y)$ , so the fraction simplifies to $x+y$ . Hence $T=x+y+xy=x(1+y)+y$ . Subtracting $y$ and dividing by $1+y$ gives $x=(T-y)/(1+y)$ .

A11 · Use of the factor theorem for rational values of the variable for polynomials

Tier 1 · Easy

Mark scheme for A11 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$f(2)=0$ , so $x-2$ is a factor	2	Let $f(x)=x^3-3x^2-4x+12$ . Then $f(2)=2^3-3(2^2)-4(2)+12=8-12-8+12=0$ . By the factor theorem, $x-2$ is a factor.
2	$f(-\tfrac23)=\tfrac{8}{3}-\tfrac{2}{3}-2=0$ , so $3x+2$ is a factor	2	The factor $3x+2$ corresponds to the root $x=-2/3$ . For $f(x)=6x^2+x-2$ , $f(-2/3)=6(4/9)+(-2/3)-2=8/3-2/3-2=0$ . Therefore $3x+2$ is a factor by the factor theorem.

Tier 2 · Standard

Mark scheme for A11 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$k=-5$	3	Since $x+1=x-(-1)$ is a factor, the factor theorem gives $p(-1)=0$ . Thus $-1+2-k-6=0$ , so $-5-k=0$ and $k=-5$ .
2	$k=1$	3	The factor $3x-2$ gives the root $x=2/3$ . By the factor theorem, $p(2/3)=0$ . Thus $3(8/27)+k(4/9)-8(2/3)+4=0$ , so $8/9+4k/9-16/3+4=0$ . Multiplying by $9$ gives $8+4k-48+36=0$ , hence $4k=4$ and $k=1$ .
3	$p(-\tfrac13)=0$ , so $3x+1$ is a factor $p(x)=(3x+1)(x^2-2x-5)$	3	$p(-1/3)=-1/9-5/9+17/3-5=0$ , so the factor theorem justifies $3x+1$ as a factor. Dividing $p(x)$ by $3x+1$ gives $x^2-2x-5$ , so $p(x)=(3x+1)(x^2-2x-5)$ .

Tier 3 · Hard

Mark scheme for A11 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$f(\tfrac12)=0$ , so $2x-1$ is a factor; $f(x)=(2x-1)(x+3)(x-2)$	4	For the factor $2x-1$ , substitute the rational root $x=\tfrac12$ : $f(\tfrac12)=2(\tfrac18)+\tfrac14-\tfrac{13}{2}+6=\tfrac14+\tfrac14-\tfrac{13}{2}+6=0$ , so $2x-1$ is a factor. Dividing $f(x)$ by $2x-1$ gives $x^2+x-6$ with zero remainder, and $x^2+x-6=(x+3)(x-2)$ . Therefore $f(x)=(2x-1)(x+3)(x-2)$ .
2	$k=3$ $p(x)=(2x+1)(x-2)(x+3)$	4	The factor $2x+1$ gives the root $x=-1/2$ . Hence $p(-1/2)=0$ : $-1/4+k/4+11/2-6=0$ . Multiplying by $4$ gives $-1+k+22-24=0$ , so $k=3$ . Dividing $2x^3+3x^2-11x-6$ by $2x+1$ gives $x^2+x-6=(x-2)(x+3)$ . Therefore $p(x)=(2x+1)(x-2)(x+3)$ .
3	$a=35$ and $b=-7$	4	From the factor $2x-1$ , $p(1/2)=0$ , giving $a+2b=21$ . From the factor $3x+1$ , $p(-1/3)=0$ , giving $a-3b=56$ . Subtracting the equations gives $-5b=35$ , so $b=-7$ and then $a=35$ .
4	$2x+1$ is the factor $p(x)=(2x+1)(3x-2)(x+4)$	4	The corresponding test values are $1/2$ , $-1/2$ and $-1/3$ . Substitution gives $p(1/2)=-9/2$ , $p(-1/2)=0$ and $p(-1/3)=-11/3$ , so the factor theorem selects $2x+1$ . Division gives $3x^2+10x-8=(3x-2)(x+4)$ , hence $p(x)=(2x+1)(3x-2)(x+4)$ .
5	$a=-8$ and $b=10$	4	First, $p(1)=0$ gives $2+a+b-4=0$ , so $a+b=2$ . Dividing by $x-1$ gives the quotient $2x^2+(a+2)x+4$ . Since a second factor $x-1$ remains, the quotient is also zero at $x=1$ : $2+a+2+4=0$ . Thus $a=-8$ , and then $b=10$ .

A12 · Completing the square

Tier 1 · Easy

Mark scheme for A12 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$(x+4)^2-13$	2	$x^2+8x+3=(x+4)^2-16+3=(x+4)^2-13$ .
2	$k=25$	1	Expanding the right-hand side gives $(x+5)^2=x^2+10x+25$ , so $k=25$ .

Tier 2 · Standard

Mark scheme for A12 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x^2-6x+11=(x-3)^2+2$ Minimum point: $(3,2)$	3	$x^2-6x+11=(x-3)^2-9+11=(x-3)^2+2$ . Since $(x-3)^2\ge 0$ , the minimum occurs when $x=3$ , giving $y=2$ .
2	$3(x-3)^2-7$ Minimum value: $-7$	3	$3x^2-18x+20=3(x^2-6x)+20=3[(x-3)^2-9]+20=3(x-3)^2-7$ . Since $3(x-3)^2\ge0$ , the minimum value is $-7$ .
3	$x=5\pm3\sqrt2$	3	$x^2-10x+7=(x-5)^2-18$ . Therefore $(x-5)^2=18$ , so $x-5=\pm3\sqrt2$ and $x=5\pm3\sqrt2$ .

Tier 3 · Hard

Mark scheme for A12 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$2x^2+12x-5=2(x+3)^2-23$ $x=-3\pm3\sqrt2$	4	$2x^2+12x-5=2(x^2+6x)-5=2[(x+3)^2-9]-5=2(x+3)^2-23$ . Setting this equal to $13$ gives $2(x+3)^2=36$ , so $(x+3)^2=18$ and $x=-3\pm3\sqrt2$ .
2	$c=3$ The greatest value occurs at $x=2$	4	$-2x^2+8x+c=-2[(x-2)^2-4]+c=-2(x-2)^2+8+c$ . Its greatest value is therefore $8+c$ , attained when $x=2$ . Hence $8+c=11$ , so $c=3$ .
3	$k=2\sqrt{21}$ with $x=-\frac{\sqrt{21}}3$ , or $k=-2\sqrt{21}$ with $x=\frac{\sqrt{21}}3$	4	$3x^2+kx+12=3\left(x+\frac{k}{6}\right)^2+12-\frac{k^2}{12}$ . Hence $12-\frac{k^2}{12}=5$ , so $k^2=84$ and $k=\pm2\sqrt{21}$ . The minimum occurs at $x=-k/6$ , giving the stated paired values.
4	$m=7$ $x=1$ or $x=7$	4	$x^2-8x+m=(x-4)^2+m-16$ . Hence the solutions are $x=4\pm\sqrt{16-m}$ , so their difference is $2\sqrt{16-m}$ . Therefore $2\sqrt{16-m}=6$ , giving $16-m=9$ and $m=7$ . The solutions are then $x=4\pm3$ , so $x=1$ or $x=7$ .
5	$-19\le -3x^2+6x+5\le8$	4	$-3x^2+6x+5=-3(x-1)^2+8$ , so its greatest value is $8$ at $x=1$ . At the endpoints, the values are $-4$ when $x=-1$ and $-19$ when $x=4$ . The least value on the interval is therefore $-19$ , giving $-19\le -3x^2+6x+5\le8$ .

A13 · Drawing and sketching of functions; interpretation of graphs (linear, quadratic, exponential y = ab^x and y = ab^-x, functions restricted to no more than 3 domains)

Tier 1 · Easy

Mark scheme for A13 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$y$ -intercept: $(0,3)$ Horizontal asymptote: $y=0$	2	At the $y$ -intercept, $x=0$ , so $y=3\times2^0=3$ . As $x$ becomes increasingly negative, $2^x$ approaches $0$ , so the horizontal asymptote is $y=0$ .
2	The graph decreases $y=\frac53$	2	As $x$ increases, $3^{-x}$ decreases. At $x=1$ , $y=5\times3^{-1}=5/3$ .

Tier 2 · Standard

Mark scheme for A13 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$y$ -intercept: $(0,2)$ Horizontal asymptote: $y=1$ The function is decreasing	3	When $x=0$ , $y=2^0+1=2$ . The term $2^{-x}$ approaches $0$ as $x$ increases, so the graph approaches $y=1$ from above. Since $2^{-x}$ gets smaller as $x$ increases, the curve is decreasing.
2	Turning point: $(2,4)$ $x$ -intercepts: $(0,0)$ and $(4,0)$ $y$ -intercept: $(0,0)$	3	The completed-square form gives turning point $(2,4)$ and a downward-opening parabola. Setting $y=0$ gives $(x-2)^2=4$ , so $x=0$ or $x=4$ . Substituting $x=0$ also gives the $y$ -intercept $(0,0)$ .
3	$m=4$ Closed point at $(2,5)$ on the quadratic domain; open point at $(2,5)$ on the linear domain	3	The included quadratic value is $f(2)=2^2+1=5$ . For the linear branch to approach the same point, $2m-3=5$ , so $m=4$ . The condition $x\le2$ gives a closed point, while $x>2$ gives an open point.

Tier 3 · Hard

Mark scheme for A13 Tier 3 · Hard
Q	Answer	Mark	Comments
1	Open points at $(-1,3)$ and $(2,5)$ ; closed points at $(-1,1)$ and $(2,4)$ $x=\sqrt3$ or $x=4$	4	Draw $y=x+4$ only for $x<-1$ , ending with an open point at $(-1,3)$ . Draw the section of $y=x^2$ from the closed point $(-1,1)$ through $(0,0)$ to the closed point $(2,4)$ . Draw $y=7-x$ for $x>2$ , beginning open at $(2,5)$ . The first domain would give $x=-1$ , which is excluded. In the middle domain, $x^2=3$ gives $x=\sqrt3$ because $-\sqrt3<-1$ . In the last domain, $7-x=3$ gives $x=4$ .
2	Closed point at $(1,2)$ on the exponential domain Open point at $(1,3)$ on the linear domain $x=1$ or $x=2$	4	Draw the increasing exponential $y=2^x$ up to and including the closed point $(1,2)$ . Draw the line $y=4-x$ only for $x>1$ , starting with an open point at $(1,3)$ . The exponential equation $2^x=2$ gives $x=1$ , which is included. The linear equation $4-x=2$ gives $x=2$ , which is in its domain.
3	$a=6$ $b=\frac12$ Intersection: $(1,3)$	4	At $x=0$ , $a=6$ . Using $(2,3/2)$ gives $6b^2=3/2$ , so $b^2=1/4$ and the stated restriction gives $b=1/2$ . Then $6(1/2)^x=3$ , so $(1/2)^x=1/2$ and $x=1$ .
4	$x$ -intercepts: $(-1,0)$ and $(5,0)$ $y$ -intercept: $(0,-5)$ Turning point: $(2,-9)$ ; upward-opening parabola	4	The factors give $x$ -intercepts $(-1,0)$ and $(5,0)$ . Substituting $x=0$ gives the $y$ -intercept $(0,-5)$ . The roots are equally spaced about $x=2$ , so the turning point has $x$ -coordinate $2$ ; substituting gives $y=(3)(-3)=-9$ . The positive $x^2$ coefficient gives an upward-opening parabola.
5	$y$ -intercepts: $(0,3)$ and $(0,12)$ Horizontal asymptote: $y=0$ Intersection: $(1,6)$ ; $3\times2^x$ increases and $12\times2^{-x}$ decreases	4	At $x=0$ , the two values are $3$ and $12$ . Both curves remain positive and have horizontal asymptote $y=0$ ; the first increases and the second decreases. At their intersection, $3\times2^x=12\times2^{-x}$ , so $2^{2x}=4=2^2$ . Hence $x=1$ and $y=3\times2=6$ .

A14 · Solution of linear and quadratic equations (by factorisation, graph, completing the square or formula)

Tier 1 · Easy

Mark scheme for A14 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x=-3$ or $x=4$	2	$x^2-x-12=(x-4)(x+3)$ . Therefore $(x-4)(x+3)=0$ , so $x=4$ or $x=-3$ .
2	$x=7$	2	Expanding gives $8x-12=5x+9$ . Hence $3x=21$ , so $x=7$ .

Tier 2 · Standard

Mark scheme for A14 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x=\frac{-5\pm\sqrt{57}}{4}$	3	Here $a=2$ , $b=5$ and $c=-4$ . The quadratic formula gives $x=\frac{-5\pm\sqrt{5^2-4(2)(-4)}}{2(2)}=\frac{-5\pm\sqrt{57}}{4}$ .
2	$x=-2\pm\sqrt5$	3	$x^2+4x-1=(x+2)^2-5$ . Therefore $(x+2)^2=5$ , so $x+2=\pm\sqrt5$ and $x=-2\pm\sqrt5$ .
3	$x=-\frac12$ or $x=\frac23$	3	$6x^2-x-2=(3x-2)(2x+1)$ . Setting each factor equal to zero gives $3x-2=0$ or $2x+1=0$ , so $x=2/3$ or $x=-1/2$ .

Tier 3 · Hard

Mark scheme for A14 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$p-q=\frac{\sqrt{73}}{3}$	4	The quadratic formula gives $x=\frac{7\pm\sqrt{(-7)^2-4(3)(-2)}}{6}=\frac{7\pm\sqrt{73}}{6}$ . Thus $p=\frac{7+\sqrt{73}}6$ and $q=\frac{7-\sqrt{73}}6$ , so $p-q=\frac{2\sqrt{73}}6=\frac{\sqrt{73}}3$ .
2	$p=-2$ The roots are $-2$ and $4$	4	Write the roots as $-r$ and $2r$ , where $r>0$ . Then $x^2+px-8=(x+r)(x-2r)=x^2-rx-2r^2$ . Comparing constant terms, $-2r^2=-8$ , so $r=2$ , and the roots are $-2$ and $4$ . Comparing the coefficients of $x$ , $p=-r=-2$ .
3	$6\text{ cm}$ and $7\text{ cm}$	4	$(x+2)(2x-1)=42$ , so $2x^2+3x-44=0$ . Factorising gives $(2x+11)(x-4)=0$ , hence $x=-11/2$ or $x=4$ . The negative value would give invalid side lengths, so $x=4$ and the sides are $6\text{ cm}$ and $7\text{ cm}$ .
4	$x=\frac{17\pm\sqrt{201}}4$	4	Expanding and rearranging gives $2x^2-12x+18=5x+7$ , so $2x^2-17x+11=0$ . The quadratic formula gives $x=\frac{17\pm\sqrt{(-17)^2-4(2)(11)}}{4}=\frac{17\pm\sqrt{201}}4$ .
5	$m=2$ The other solution is $x=-2$	4	Substituting $x=4/3$ gives $3(16/9)+4m/3-8=0$ . Multiplying by $3$ gives $16+4m-24=0$ , so $m=2$ . The equation is then $3x^2+2x-8=0=(3x-4)(x+2)$ , giving $x=4/3$ or $x=-2$ .

A15 · Algebraic and graphical solution of simultaneous equations in two unknowns, where the equations could both be linear or one linear and one second order

Tier 1 · Easy

Mark scheme for A15 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x=3$ , $y=1$	2	Add the equations to eliminate $y$ : $3x=9$ , so $x=3$ . Substituting into $x-y=2$ gives $3-y=2$ , so $y=1$ .
2	$(x,y)=(3,5)$	2	Substitute $y=3x-4$ into the second equation: $x+2(3x-4)=13$ . Thus $7x=21$ , so $x=3$ and $y=5$ .

Tier 2 · Standard

Mark scheme for A15 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$(x,y)=(-2,0)$ or $(3,5)$	3	Equate the two expressions for $y$ : $x^2-4=x+2$ , so $x^2-x-6=0$ . Factorising gives $(x-3)(x+2)=0$ , hence $x=3$ or $x=-2$ . Using $y=x+2$ gives the pairs $(3,5)$ and $(-2,0)$ .
2	$(x,y)=(2,5)$	3	Equating the expressions for $y$ gives $x^2+1=4x-3$ , so $x^2-4x+4=0$ . Hence $(x-2)^2=0$ and $x=2$ . Substitution gives $y=5$ , so the single solution is $(2,5)$ .
3	$(x,y)=(1,2)$	3	Multiply the first equation by $3$ and the second by $2$ : $9x+6y=21$ and $10x-6y=-2$ . Adding gives $19x=19$ , so $x=1$ . Substitution into $3x+2y=7$ gives $y=2$ .

Tier 3 · Hard

Mark scheme for A15 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$(x,y)=(2,5)$ or $(5,2)$	4	From $x+y=7$ , write $y=7-x$ . Substitute into $xy=10$ : $x(7-x)=10$ , so $x^2-7x+10=0$ . Since $(x-2)(x-5)=0$ , $x=2$ or $x=5$ . Then $y=7-x$ , giving $(2,5)$ or $(5,2)$ .
2	$(x,y)=(-2,-3)$ or $(\frac65,\frac{17}5)$	4	Substitution gives $x^2+(2x+1)^2=13$ , so $5x^2+4x-12=0$ . Factorising gives $(5x-6)(x+2)=0$ , so $x=6/5$ or $x=-2$ . Using $y=2x+1$ gives the pairs $(6/5,17/5)$ and $(-2,-3)$ .
3	$k=-2$ Point of contact: $(3,4)$	4	At an intersection, $x^2-4x+7=2x+k$ , so $x^2-6x+7-k=0$ . Complete the square: $(x-3)^2=2+k$ . Touching means exactly one repeated solution, which requires $2+k=0$ (a positive right-hand side gives two solutions and a negative one gives none), so $k=-2$ . Then $(x-3)^2=0$ gives the repeated solution $x=3$ , and $y=2(3)-2=4$ .
4	$(x,y)=(8,5)$ or $(-5,-8)$ Distance $=13\sqrt2$	4	From $x-y=3$ , write $x=y+3$ . Substitution gives $(y+3)^2+y^2=89$ , so $2y^2+6y-80=0$ . Dividing by $2$ and factorising gives $(y+8)(y-5)=0$ , so $y=-8$ or $y=5$ . Since $x=y+3$ , the ordered pairs are $(-5,-8)$ and $(8,5)$ . The two points differ by $13$ in each coordinate, so the distance between them is $\sqrt{13^2+13^2}=13\sqrt2$ .
5	$(x,y)=(3,3)$ or $(9,-3)$	4	Use $x^2+2xy=(x+y)^2-y^2$ . Since $x+y=6$ , the second equation becomes $36-y^2=27$ , so $y^2=9$ and $y=3$ or $y=-3$ . From $x=6-y$ , these give $(x,y)=(3,3)$ or $(9,-3)$ . Both pairs satisfy the two original equations.

A16 · Algebraic solution of linear equations in three unknowns

Tier 1 · Easy

Mark scheme for A16 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x=4$ , $y=2$ , $z=3$	2	Using $z=3$ in the first equation gives $x+y=6$ . Together with $x-y=2$ , addition gives $2x=8$ , so $x=4$ . Then $y=2$ and $z=3$ .
2	$x=3$ , $y=4$ , $z=5$	2	Subtract $x+y=7$ from $x+y+z=12$ to get $z=5$ . Then $y+z=9$ gives $y=4$ , and $x+y=7$ gives $x=3$ .

Tier 2 · Standard

Mark scheme for A16 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x=1$ , $y=2$ , $z=3$	3	Subtract the first equation from the second: $x-2y=-3$ , so $x=2y-3$ . From the first equation, $z=6-x-y=9-3y$ . Substitute both into the third: $(2y-3)+2y-(9-3y)=2$ , so $7y=14$ and $y=2$ . Hence $x=1$ and $z=3$ .
2	$x=3$ , $y=4$ , $z=6$	3	Adding all three equations gives $2x+2y+2z=26$ , so $x+y+z=13$ . Subtracting $y+z=10$ gives $x=3$ ; subtracting $z+x=9$ gives $y=4$ ; and subtracting $x+y=7$ gives $z=6$ .
3	$x=1$ , $y=3$ , $z=-2$	3	Subtract the first equation from the second to get $y+2z=-1$ . Subtract the second equation from the third to get $2y+6z=-6$ , or $y+3z=-3$ . Subtraction gives $z=-2$ , then $y=3$ and $x=1$ .

Tier 3 · Hard

Mark scheme for A16 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x=2$ , $y=-1$ , $z=3$	4	Add twice the first equation to the second to eliminate $z$ : $5x+5y=5$ , so $x+y=1$ . Subtract the second equation from the third to get $2x+3y=1$ . Subtracting $2(x+y)=2$ gives $y=-1$ , so $x=2$ . Substitution into $x-y+2z=9$ gives $3+2z=9$ , hence $z=3$ .
2	$x=3$ , $y=0$ , $z=2$	4	From the first equation, $y=4-2x+z$ . Substitution into the second gives $5x+z=17$ , while substitution into the third gives $x+3z=9$ . Using $z=17-5x$ in the latter gives $x+3(17-5x)=9$ , so $x=3$ . Hence $z=2$ and $y=0$ .
3	Pen: £3 Ruler: £2 Compass: £5	4	Let the prices in pounds be $p$ , $r$ and $c$ . The equations are $2p+r+c=13$ , $p+3r+2c=19$ and $4p+2r+3c=31$ . Subtracting twice the first from the third gives $c=5$ . The first two equations then reduce to $2p+r=8$ and $p+3r=9$ , giving $p=3$ and $r=2$ .
4	$536$	3	From $h-t=2$ and $u=2t$ , use $h=t+2$ and $u=2t$ . Substitution into the digit sum gives $(t+2)+t+2t=14$ , so $4t=12$ and $t=3$ . Hence $h=5$ and $u=6$ , giving the number $536$ .
5	$a=2$ , $b=-3$ , $c=4$ $f(4)=24$	4	The conditions give $a+b+c=3$ , $4a+2b+c=6$ and $a-b+c=9$ . Subtracting the first equation from the third gives $-2b=6$ , so $b=-3$ . Subtracting the first from the second gives $3a+b=3$ , hence $a=2$ , and then $c=4$ . Therefore $f(4)=2(4)^2-3(4)+4=24$ .

A17 · Solution of linear and quadratic inequalities

Tier 1 · Easy

Mark scheme for A17 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x\geq-4$	2	Subtract $7$ to get $-3x\leq12$ . Dividing by $-3$ reverses the inequality, giving $x\geq-4$ .
2	$-2<x\le3$	2	Subtracting $5$ throughout gives $-4<2x\le6$ . Dividing throughout by $2$ gives $-2<x\le3$ .

Tier 2 · Standard

Mark scheme for A17 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$-1\le x\le4$	3	The roots are $x=-1$ and $x=4$ . The quadratic has a positive $x^2$ coefficient, so it is non-positive between the roots. Equality is allowed, hence $-1\le x\le4$ .
2	$x<-4$ or $x>2$	3	$x^2+2x-8=(x+4)(x-2)$ , so the roots are $-4$ and $2$ . The positive-leading quadratic is positive outside the roots. Since the inequality is strict, $x<-4$ or $x>2$ .
3	$-2\le x\le\frac32$	3	Rearrange to $2x^2+x-6\le0$ and factorise: $(2x-3)(x+2)\le0$ . The roots are $-2$ and $3/2$ . The positive-leading quadratic is non-positive between them, including both endpoints.

Tier 3 · Hard

Mark scheme for A17 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x<-1$	4	$x^2-5x-6=(x-6)(x+1)$ , so the first inequality gives $x<-1$ or $x>6$ . The second inequality gives $2x\le8$ , so $x\le4$ . Intersecting these sets leaves $x<-1$ .
2	$x<-5$ or $x>2$ (accept equivalent interval notation)	3	Expand: $x^2+3x-4>6$ , so $x^2+3x-10>0$ . Factorise: $(x+5)(x-2)>0$ . The graph of $y=x^2+3x-10$ is an upward parabola crossing the $x$ -axis at $x=-5$ and $x=2$ , so the expression is positive outside the roots. Therefore $x<-5$ or $x>2$ .
3	$\frac{3-\sqrt{17}}2<x<\frac{3+\sqrt{17}}2$	4	Rearrange to $x^2-3x-2<0$ . The corresponding equation has roots $x=\frac{3\pm\sqrt{9+8}}2=\frac{3\pm\sqrt{17}}2$ . Since the quadratic has positive leading coefficient, it is negative strictly between these roots.
4	$x=0,1,2$	3	$2x^2-5x-3=(2x+1)(x-3)$ , so the boundary values are $-1/2$ and $3$ . The positive-leading quadratic is negative for $-1/2<x<3$ . The integers in this interval are $0$ , $1$ and $2$ .
5	$k>9$	3	$x^2-6x+k=(x-3)^2+k-9$ . Its least value is $k-9$ , attained at $x=3$ . For the expression to be strictly positive for every real $x$ , this least value must satisfy $k-9>0$ , so $k>9$ .

A18 · Index laws, including fractional and negative indices and the solution of equations

Tier 1 · Easy

Mark scheme for A18 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\frac14$	1	$16^{1/2}=4$ , and the negative index takes the reciprocal, so $16^{-1/2}=1/4$ .
2	$p^5$	2	$(p^3)^2=p^6$ , and $p^6\div p=p^{6-1}=p^5$ .

Tier 2 · Standard

Mark scheme for A18 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x=\frac54$	3	Write both sides with base $3$ : $3^{2x+2}=3^{6x-3}$ . Therefore $2x+2=6x-3$ , so $5=4x$ and $x=5/4$ .
2	$x^4y^{1/2}$ $x^4\sqrt y$	3	The numerator is $x^3y^{-3/2}$ . Dividing subtracts indices, giving $x^{3-(-1)}y^{-3/2-(-2)}=x^4y^{1/2}$ .
3	$x=16$ $x^{-1/2}=\frac14$	3	Raise both sides of $x^{3/2}=64$ to the power $2/3$ : $x=64^{2/3}=4^2=16$ . Therefore $x^{-1/2}=1/\sqrt{16}=1/4$ .

Tier 3 · Hard

Mark scheme for A18 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\frac ba$	4	The numerator is $81^{1/4}a^{-8/4}b^{6/4}=3a^{-2}b^{3/2}$ . Dividing by $3a^{-1}b^{1/2}$ gives $a^{-1}b=b/a$ .
2	$x=\frac{11}{4}$	4	Writing every term with base $2$ , the left side is $2^{2x+2+6-3x-3x}=2^{8-4x}$ . Since $1/8=2^{-3}$ , equate exponents: $8-4x=-3$ . Thus $4x=11$ and $x=11/4$ .
3	$x=0$ or $x=2$	4	Let $u=2^x$ , so $u>0$ . The equation becomes $u^2-5u+4=0$ , hence $(u-1)(u-4)=0$ . Thus $2^x=1$ or $2^x=4$ , giving $x=0$ or $x=2$ .
4	$x=\dfrac{5}{8}$	3	Write every term as a power of $2$ : $2^{x}\times8^{x+1}=2^{x}\times2^{3x+3}=2^{4x+3}$ , and $\left(\dfrac{1}{16}\right)^{x-2}=2^{-4(x-2)}=2^{-4x+8}$ . Equating the indices gives $4x+3=-4x+8$ , so $8x=5$ and $x=\dfrac{5}{8}$ .
5	$p=3$ , $q=2$	4	The left side simplifies to $x^{3p+q}y^{-2p-q}$ . Equating the indices of $x$ and $y$ gives $3p+q=11$ and $-2p-q=-8$ . Adding the equations gives $p=3$ , and then $3(3)+q=11$ gives $q=2$ .

A19 · Algebraic proof

Tier 1 · Easy

Mark scheme for A19 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$(2m+1)+(2n+1)=2(m+n+1)$ , so the sum is even	2	Let the two odd integers be $2m+1$ and $2n+1$ , where $m$ and $n$ are integers. Their sum is $2m+2n+2=2(m+n+1)$ . Since $m+n+1$ is an integer, the sum is even.
2	$(2n)^2=4n^2$ , which is divisible by $4$ (any complete valid argument is credited)	2	Let the even integer be $2n$ , where $n$ is an integer. Its square is $(2n)^2=4n^2$ . Since $n^2$ is an integer, the square is divisible by $4$ .

Tier 2 · Standard

Mark scheme for A19 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$n(n+1)(n+2)$ contains a factor of $2$ and a factor of $3$ , so it is divisible by $6$	3	Represent the integers as $n$ , $n+1$ and $n+2$ . Among any three consecutive integers, one is a multiple of $3$ . At least one is even, so the product also has a factor of $2$ . Since $2$ and $3$ are coprime, $n(n+1)(n+2)$ is divisible by $6$ .
2	$(2n+1)^2=8k+1$ for an integer $k$ (any complete valid argument is credited)	3	Let the odd integer be $2n+1$ . Then $(2n+1)^2=4n(n+1)+1$ . Since one of the consecutive integers $n$ and $n+1$ is even, write $n(n+1)=2k$ for an integer $k$ . The square is therefore $8k+1$ .
3	$(2m+1)^2-(2n+1)^2=4(m-n)(m+n+1)$ , and one of $m-n$ and $m+n+1$ is even, so the difference is a multiple of $8$	3	Let the odd integers be $2m+1$ and $2n+1$ . Their squared difference is $4(m-n)(m+n+1)$ . The two factors $m-n$ and $m+n+1$ have odd sum $2m+1$ , so one is even. Their product is therefore even, making the whole expression divisible by $8$ .

Tier 3 · Hard

Mark scheme for A19 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x^2+\frac1{x^2}-2=(x-\frac1x)^2\ge0$ , so $x^2+\frac1{x^2}\ge2$	4	For real $x\ne0$ , the square $(x-1/x)^2$ is non-negative. Expanding gives $x^2-2+1/x^2\ge0$ . Adding $2$ to both sides gives $x^2+1/x^2\ge2$ , as required.
2	$n(n+1)(n+2)(n+3)+1=(n^2+3n+1)^2$ (any complete valid argument is credited)	4	Let the integers be $n,n+1,n+2,n+3$ . Put $t=n^2+3n$ . Then $n(n+3)=t$ and $(n+1)(n+2)=t+2$ . Hence their product plus $1$ is $t(t+2)+1=t^2+2t+1=(t+1)^2=(n^2+3n+1)^2$ , a perfect square.
3	$a^2-b^2=(a-b)(a+b)$ , whose factors are both even or both odd, so it is odd or a multiple of $4$ ; $2026$ is neither, so no such integers exist	4	Factorise: $a^2-b^2=(a-b)(a+b)$ . The integers $a-b$ and $a+b$ differ by $2b$ , an even number, so they are both even or both odd. If both are odd, the product is odd. If both are even, each has a factor of $2$ , so the product is a multiple of $4$ . Now $2026$ is even, and $2026=4\times506+2$ is not a multiple of $4$ . Since $2026$ is neither odd nor a multiple of $4$ , no integers $a$ and $b$ satisfy the equation.
4	$(n^2+2)^2-(2n)^2=n^4+4n^2+4-4n^2=n^4+4$ $n^4+4=(n^2-2n+2)(n^2+2n+2)$ , and for $n>1$ both factors are integers greater than $1$ , so $n^4+4$ is never prime	4	Expanding, $(n^2+2)^2-(2n)^2=n^4+4n^2+4-4n^2=n^4+4$ . The difference of two squares factorises: $n^4+4=(n^2-2n+2)(n^2+2n+2)$ . The first factor is $(n-1)^2+1$ , which is greater than $1$ when $n>1$ , and the second factor is $(n+1)^2+1$ , also greater than $1$ . Therefore $n^4+4$ has a factorisation into two integers greater than $1$ , so it is never prime.
5	$\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}=\frac{4x}{x^2-1}$	3	Using the common denominator $(x-1)(x+1)$ , the left side is $\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}$ . The numerator simplifies to $x^2+2x+1-(x^2-2x+1)=4x$ , while the denominator is $x^2-1$ . This gives the right side.

A20 · Using nth terms of sequences; limiting value of a sequence as n approaches infinity

Tier 1 · Easy

Mark scheme for A20 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$5$	1	$u_n=5-2/n$ . As $n$ approaches infinity, $2/n$ approaches $0$ , so $u_n$ approaches $5$ .
2	$u_5=\frac{19}{7}$	1	$u_5=\frac{3(5)+4}{5+2}=\frac{19}{7}$ .

Tier 2 · Standard

Mark scheme for A20 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$v_2=\frac76$ Limiting value: $3$	2	Substitution gives $v_2=\frac{3(2)^2+2}{(2)^2+4(2)}=\frac{14}{12}=\frac76$ . For the limit, divide top and bottom by $n^2$ : $v_n=\frac{3+2/n^2}{1+4/n}$ . Both $2/n^2$ and $4/n$ approach $0$ , so the limiting value is $3$ .
2	$k=8$	2	The limiting value is the ratio of the coefficients of $n$ , so $k/2=4$ . Therefore $k=8$ .
3	$k=12$ $u_1=\frac74$	3	The limiting value is the ratio of the coefficients of $n^2$ , so $k/3=4$ and $k=12$ . Then $u_1=\frac{12-5}{3+1}=7/4$ .

Tier 3 · Hard

Mark scheme for A20 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$L=2$ Least $n=498$	4	Dividing by $n$ shows $L=2$ . Also $2-w_n=\frac{2n+6-(2n+1)}{n+3}=\frac5{n+3}$ , which is positive. Differing by less than $0.01$ therefore means $5/(n+3)<1/100$ . Hence $n+3>500$ , so $n>497$ and the least integer is $498$ .
2	$a=3$ $b=4$ $u_{10}=\frac{16}{7}$	4	The limiting value is $a$ , so $a=3$ . Using $u_1=1$ gives $\frac{3+2}{1+b}=1$ , hence $b=4$ . Therefore $u_{10}=\frac{3(10)+2}{10+4}=\frac{32}{14}=\frac{16}{7}$ .
3	$n=18$	4	Both denominators are positive, so $2n(2n-5)>(n+1)(3n+4)$ . This simplifies to $n^2-17n-4>0$ . Its roots are $\frac{17\pm\sqrt{305}}2$ , and the positive root lies between $17$ and $18$ . The quadratic is positive above that root, so the least positive integer is $18$ .
4	$k=8$ $u_1=v_1$ ; for every $n\ge2$ , $v_n>u_n$	4	The limiting values are $4$ and $k/2$ , so $k=8$ . Now $u_n-v_n=\frac{(4n+1)(2n+1)-(8n-3)(n+2)}{(n+2)(2n+1)}=\frac{7(1-n)}{(n+2)(2n+1)}$ . The denominator is positive. The difference is $0$ at $n=1$ and negative for every $n\ge2$ , so the terms are equal at $n=1$ and $v_n>u_n$ thereafter.
5	Limiting value: $\frac32$ $n=4$	4	Expanding gives $u_n=\frac{3n^2+n-2}{2n^2+5n}$ , so the limiting value is $3/2$ . Setting this equal to $25/26$ gives $26(3n^2+n-2)=25(2n^2+5n)$ . Hence $28n^2-99n-52=0=(n-4)(28n+13)$ . Thus $n=4$ or $n=-13/28$ . Since $n$ is a positive integer, $n=4$ .

A21 · nth terms of linear sequences

Tier 1 · Easy

Mark scheme for A21 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$4n+3$	2	The common difference is $4$ , so start with $4n$ . This gives $4$ when $n=1$ , which is $3$ below the first term $7$ . Therefore the nth term is $4n+3$ .
2	$92$	1	Substitute $n=20$ : $5(20)-8=100-8=92$ .

Tier 2 · Standard

Mark scheme for A21 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$19-3n$	3	There are $18$ steps from the 12th to the 30th term, and the total change is $-71-(-17)=-54$ . Thus the common difference is $-54/18=-3$ . Writing the term as $-3n+b$ and using $n=12$ gives $-36+b=-17$ , so $b=19$ .
2	$6n+5=200$ gives $n=32.5$ , which is not a positive integer, so $200$ is not a term	2	The nth term is $6n+5$ . If $6n+5=200$ , then $n=195/6=32.5$ . Since a term number must be a positive integer, $200$ is not in the sequence.
3	$n=11$ Common value: $73$	3	Set the nth terms equal: $7n-4=3n+40$ . Hence $4n=44$ and $n=11$ . Substitution gives $7(11)-4=73$ .

Tier 3 · Hard

Mark scheme for A21 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$u_n=4n-2$ $202$ is the 51st term	4	Let $u_n=a+(n-1)d$ . Then $a+4d=18$ . Also $(a+7d)+(a+11d)=76$ , so $2a+18d=76$ . Doubling the first equation gives $2a+8d=36$ ; subtraction gives $10d=40$ , hence $d=4$ and $a=2$ . Thus $u_n=2+4(n-1)=4n-2$ . Solving $4n-2=202$ gives $n=51$ .
2	$u_n=4n+3$ The first term greater than $250$ is the 62nd term, equal to $251$	4	The common difference is $(83-31)/(20-7)=52/13=4$ . Write $u_n=4n+b$ ; using $u_7=31$ gives $28+b=31$ , so $b=3$ . For $4n+3>250$ , $n>247/4=61.75$ . Thus the first possible integer is $n=62$ , giving $u_{62}=251$ .
3	$u_n=4n-10$	4	Write the three terms as $14-d$ , $14$ and $14+d$ , since their sum is $42$ . Their product gives $14(14-d)(14+d)=2520$ , so $196-d^2=180$ and $d^2=16$ . The sequence is increasing, so $d=4$ . Since $u_6=14$ , $u_n=14+4(n-6)=4n-10$ .
4	$u_n=3n+3$ $u_{25}=78$	4	Using $u_n=an+b$ , the identity gives $2an+b=2(an+b)-3$ . Hence $b=3$ . Since $u_4=15$ , $4a+3=15$ , so $a=3$ . Therefore $u_n=3n+3$ and $u_{25}=3(25)+3=78$ .
5	$61$ (the 8th term of the first sequence and the 10th term of the second)	3	For the first sequence to exceed $50$ , $8n-3>50$ , so $n\ge7$ . Its 7th term is $53$ , but $(53-1)/6$ is not an integer, so $53$ is not in the second sequence. Its 8th term is $61$ , and $61=6(10)+1$ . Therefore $61$ is the smallest common value above $50$ .

A22 · nth terms of quadratic sequences

Tier 1 · Easy

Mark scheme for A22 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$3n^2+1$	2	The first differences are $9,15,21$ , so the second difference is $6$ and the $n^2$ coefficient is $3$ . Subtracting $3n^2$ from the terms leaves $1$ each time, so the nth term is $3n^2+1$ .
2	$41$	1	Substitute $n=7$ : $7^2-2(7)+6=49-14+6=41$ .

Tier 2 · Standard

Mark scheme for A22 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$2n^2+n-1$	3	The first differences are $7,11,15,19$ , so the second difference is $4$ and the leading term is $2n^2$ . Subtracting $2n^2$ from the sequence gives $0,1,2,3,4$ , whose nth term is $n-1$ . Therefore the original nth term is $2n^2+n-1$ .
2	No, $200$ is not a term	2	The 8th term is $3(8)^2-2=190$ , while the 9th term is $3(9)^2-2=241$ . The sequence is increasing for positive integer $n$ , so no term equals $200$ .
3	$k=4$ $u_{10}=145$	3	$u_4-u_3=(16+4k+5)-(9+3k+5)=7+k$ . Hence $7+k=11$ and $k=4$ . Therefore $u_{10}=10^2+4(10)+5=145$ .

Tier 3 · Hard

Mark scheme for A22 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$u_n=2n^2+3n+1$ $u_{10}=231$	4	The data give $a+b+c=6$ , $4a+2b+c=15$ and $16a+4b+c=45$ . Subtracting consecutive equations gives $3a+b=9$ and $12a+2b=30$ , so $6a+b=15$ . Hence $3a=6$ and $a=2$ ; then $b=3$ and $c=1$ . Therefore $u_{10}=2(10)^2+3(10)+1=231$ .
2	$u_n=3n^2+2n-1$ The least value is $n=13$	4	Let $u_n=an^2+bn+c$ . Then $u_{n+1}-u_n=2an+a+b$ , so comparison with $6n+5$ gives $a=3$ and $a+b=5$ , hence $b=2$ . Since $u_1=4$ , $3+2+c=4$ and $c=-1$ . Thus $u_n=3n^2+2n-1$ . Now $u_{12}=455$ and $u_{13}=532$ , so the least $n$ giving more than $500$ is $13$ .
3	$n=3$ gives $25$ $n=8$ gives $140$	4	Set the nth terms equal: $2n^2+n+4=23n-44$ , so $2n^2-22n+48=0$ . Dividing by $2$ and factorising gives $(n-3)(n-8)=0$ , so $n=3$ or $n=8$ . Substitution gives $u_3=25$ and $u_8=140$ .
4	$n=4,5,6,7$	4	$u_n<3$ gives $n^2-11n+27<0$ . The boundary values are $n=\frac{11\pm\sqrt{13}}2$ , which lie between $3$ and $4$ , and between $7$ and $8$ , respectively. The positive-leading quadratic is negative between these boundaries, so the positive integer values are $n=4,5,6,7$ .
5	$u_n=2n^2+3n+1$ Pattern $7$	4	The number of tiles is $(n+2)(2n-1)+3=2n^2+3n+1$ . Setting this equal to $120$ gives $2n^2+3n-119=0=(2n+17)(n-7)$ . Thus $n=7$ or $n=-17/2$ . A pattern number is a positive integer, so the required pattern is $7$ .