Edexcel GCSE Maths coverage

Algebra

Section A
25 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

Open the printable pack
A1

Use and interpret algebraic manipulation: ab for a × b, 3y for y + y + y and 3 × y, a² for a × a, a³ for a × a × a, a²b for a × a × b, a/b for a ÷ b, coefficients as fractions, brackets

  • Algebraic notation shortens repeated operations: abab means a×ba\times b, a3a^3 means a×a×aa\times a\times a, and 3y3y means three lots of yy.
  • Write number factors before letters, collect repeated factors as powers, and use a fraction bar for division; brackets show which whole expression an operation acts on.
  • For example, a×a×b÷4a\times a\times b\div4 is written a2b4\frac{a^2b}{4}, with 14\frac14 as the coefficient of a2ba^2b.
  • Do not read a2a^2 as 2a2a, or 3y3y as 3+y3+y; powers describe repeated multiplication and coefficients describe multiplication.

Tier 1 · Easy

1 mark
ORIGINAL

Write m×m×n×n×nm\times m\times n\times n\times n using indices.

Tier 2 · Standard

2 marks
ORIGINAL

Write p×p×p×q÷5p\times p\times p\times q\div5 in conventional algebraic notation, and state its coefficient.

Tier 3 · Hard

3 marks
ORIGINAL

A rectangle has length 3x2\frac{3x}{2} and width yy. Write its area and its perimeter in conventional algebraic notation.

A2

Substitute numerical values into formulae and expressions, including scientific formulae

  • Substitution replaces every occurrence of a letter by its given numerical value while keeping the operations in the original order.
  • Put negative or fractional values in brackets, evaluate powers before multiplication, and keep extra calculator figures until the final rounding step.
  • For example, if E=12mv2E=\frac12mv^2, m=6m=6 and v=4v=4, then E=12×6×42=48E=\frac12\times6\times4^2=48.
  • A common error is to substitute v=4v=4 into v2v^2 as 2×42\times4; the square applies to the complete substituted value.

Tier 1 · Easy

2 marks
ORIGINAL

Work out 2x252x^2-5 when x=3x=-3.

Tier 2 · Standard

3 marks
ORIGINAL

The kinetic energy of an object is given by E=12mv2E=\frac12mv^2. Work out EE when m=3.2kgm=3.2\,\text{kg} and v=5m s1v=5\,\text{m s}^{-1}.

Tier 3 · Hard

3 marks
ORIGINAL

Use E=mc2E=mc^2 to calculate EE when m=4.2×108kgm=4.2\times10^{-8}\,\text{kg} and c=3×108m s1c=3\times10^8\,\text{m s}^{-1}. Give your answer in standard form.

A3

Understand and use the concepts and vocabulary of expressions, equations, formulae, identities, inequalities, terms and factors

  • An expression has no equality sign; an equation is true only for particular values, while an identity is true for every permitted value and uses \equiv.
  • A formula links quantities, an inequality compares their possible values, terms are separated by addition or subtraction, and factors are multiplied together.
  • In 6x215x=3x(2x5)6x^2-15x=3x(2x-5), the left side has two terms and the right side shows the factors 3x3x and (2x5)(2x-5).
  • Do not call every statement containing == an identity; test whether it is always true or only true when the variable has specific values.

Tier 1 · Easy

1 mark
ORIGINAL

State whether 5x+75x+7 is an expression, equation or inequality.

Tier 2 · Standard

3 marks
ORIGINAL

For 8x212x=4x(2x3)8x^2-12x=4x(2x-3), state the number of terms on the left and name the two factors on the right.

Tier 3 · Hard

3 marks
ORIGINAL

Classify each statement as an equation, an identity or an inequality: 3(2x1)=93(2x-1)=9, 3(2x1)6x33(2x-1)\equiv6x-3, and 3(2x1)<93(2x-1)<9.

A4

Simplify and manipulate algebraic expressions (incl. surds and algebraic fractions): like terms, common factors, expanding two or more binomials, factorising quadratics incl. ax² + bx + c, indices

  • Only like terms can be collected; index laws apply to matching bases, while surds are combined only after simplifying them to like surds.
  • Expand by multiplying every term required, and factorise by first removing common factors before choosing factors that reproduce every quadratic term.
  • For example, 2x2+7x+3=(2x+1)(x+3)2x^2+7x+3=(2x+1)(x+3) because the cross-terms are 6x+x=7x6x+x=7x.
  • Do not cancel terms across addition in an algebraic fraction; factorise the complete numerator and denominator, then cancel common factors and retain excluded values.

Tier 1 · Easy

2 marks
ORIGINAL

Simplify 7a+3b2a+5b7a+3b-2a+5b.

Tier 2 · Standard

3 marks
ORIGINAL

Factorise 6x2+x26x^2+x-2.

Tier 3 · Hard

4 marks
ORIGINAL

Simplify x29x2+x6\frac{x^2-9}{x^2+x-6}, stating every value of xx excluded from the original expression.

A5

Understand and use standard mathematical formulae; rearrange formulae to change the subject

  • The subject of a formula is the variable isolated on one side of the equality; changing the subject keeps an equivalent relationship.
  • Undo operations in reverse order and perform the same operation on both sides, clearing fractions or brackets before collecting terms containing the new subject.
  • For example, from v=u+atv=u+at, subtracting uu and then dividing by aa gives t=vuat=\frac{v-u}{a}.
  • A common error is to move a term by changing its sign without applying an operation to the whole side, especially when the subject appears more than once.

Tier 1 · Easy

1 mark
ORIGINAL

Make ww the subject of A=lwA=lw.

Tier 2 · Standard

2 marks
ORIGINAL

Make tt the subject of v=u+atv=u+at.

Tier 3 · Hard

4 marks
ORIGINAL

Make aa the subject of P=a+babP=\frac{a+b}{a-b}.

A6

Know the difference between an equation and an identity; argue mathematically to show algebraic expressions are equivalent, and use algebra to support and construct arguments and proofs

  • An equation is satisfied by particular values, whereas an identity states that two expressions are equivalent for every permitted value.
  • For an algebraic proof, define an integer with a letter, translate the claim into algebra, and rearrange it into a form that guarantees the required property.
  • For example, consecutive integers nn and n+1n+1 have sum 2n+12n+1, which is odd because it is one more than a multiple of 22.
  • Checking several numerical examples is evidence but not a proof; the algebra must cover every value allowed by the claim.

Tier 1 · Easy

2 marks
ORIGINAL

Show that 4(n+2)3(n1)4(n+2)-3(n-1) is equivalent to n+11n+11.

Tier 2 · Standard

3 marks
ORIGINAL

Prove algebraically that the sum of two consecutive integers is odd.

Tier 3 · Hard

4 marks
ORIGINAL

An odd number is written as 2n+12n+1. Demonstrate algebraically that squaring it leaves remainder 11 after division by 88.

A7

Interpret simple expressions as functions with inputs and outputs; interpret the reverse as the 'inverse function' and two successive functions as a 'composite function' (formal notation expected)

  • A function maps each allowed input to one output; f(a)f(a) means substitute the complete input aa into the rule for ff.
  • The inverse f1f^{-1} reverses a one-to-one function, while the composite fg(x)=f(g(x))fg(x)=f(g(x)) applies gg first and then ff.
  • For example, if f(x)=3x+1f(x)=3x+1, then f1(x)=x13f^{-1}(x)=\frac{x-1}{3} because subtracting 11 and dividing by 33 reverses the rule.
  • Do not interpret f1(x)f^{-1}(x) as 1/f(x)1/f(x) or reverse the order of a composite; in fgfg, the right-hand function acts first.

Tier 1 · Easy

1 mark
ORIGINAL

Given f(x)=3x4f(x)=3x-4, work out f(6)f(6).

Tier 2 · Standard

3 marks
ORIGINAL

Given f(x)=5x+2f(x)=5x+2, find f1(x)f^{-1}(x) and work out f1(27)f^{-1}(27).

Tier 3 · Hard

4 marks
ORIGINAL

Let f(x)=2x+3f(x)=2x+3 and g(x)=x21g(x)=x^2-1. Solve fg(x)=19fg(x)=19, where fg(x)=f(g(x))fg(x)=f(g(x)).

A8

Work with coordinates in all four quadrants

  • A coordinate (x,y)(x,y) gives horizontal position first and vertical position second; the signs determine the quadrant.
  • Use differences in coordinates for movements, and average corresponding coordinates to find the midpoint of a line segment.
  • For example, the midpoint of (3,5)(-3,5) and (7,1)(7,-1) is ((3+7)/2,(51)/2)=(2,2)(({-3+7})/2,(5-1)/2)=(2,2).
  • Do not swap the coordinate order or ignore negative signs when subtracting endpoints; write each coordinate calculation separately.

Tier 1 · Easy

1 mark
ORIGINAL

State the quadrant containing the point (4,3)(-4,3).

Tier 2 · Standard

2 marks
ORIGINAL

Find the midpoint of the line segment joining (5,2)(-5,2) and (3,4)(3,-4).

Tier 3 · Hard

3 marks
ORIGINAL

The point PP divides the line segment from A(6,5)A(-6,5) to B(4,5)B(4,-5) in the ratio AP:PB=3:2AP:PB=3:2. Find the coordinates of PP.

A9

Plot graphs of straight-line equations; use y = mx + c to identify parallel and perpendicular lines; find the equation of a line through two given points, or one point with a given gradient

  • A straight line has equation y=mx+cy=mx+c, where mm is its gradient and (0,c)(0,c) is its intercept on the yy-axis.
  • Find a gradient using m=y2y1x2x1m=\frac{y_2-y_1}{x_2-x_1}, then substitute one known point to determine cc; two accurate points are enough to draw the line.
  • For example, the line of gradient 33 through (2,1)(2,1) satisfies 1=3(2)+c1=3(2)+c, so c=5c=-5 and the equation is y=3x5y=3x-5.
  • Parallel lines have equal gradients and perpendicular non-vertical lines have gradients whose product is 1-1; do not merely change the sign.

Tier 1 · Easy

1 mark
ORIGINAL

Write the equation of the line with gradient 44 and yy-intercept 7-7.

Tier 2 · Standard

3 marks
ORIGINAL

Find the equation of the line through (2,5)(-2,5) and (4,1)(4,-1).

Tier 3 · Hard

4 marks
ORIGINAL

Find the equation of the line through (3,5)(3,-5) that is perpendicular to 2x3y=62x-3y=6. Give your answer in the form y=mx+cy=mx+c.

A10

Identify and interpret gradients and intercepts of linear functions graphically and algebraically

  • The gradient measures change in the vertical quantity per unit change in the horizontal quantity; its sign shows whether the line rises or falls.
  • Read the yy-intercept where x=0x=0 and the xx-intercept where y=0y=0, using the axis scales and units before interpreting either value.
  • For example, in C=12t+35C=12t+35, the gradient 1212 is the cost added per unit of tt, while 3535 is the cost when t=0t=0.
  • Do not describe a contextual gradient or intercept as a bare number; include what it represents and the correct compound or original unit.

Tier 1 · Easy

2 marks
ORIGINAL

State the gradient and yy-intercept of y=3x+8y=-3x+8.

Tier 2 · Standard

3 marks
ORIGINAL

A straight line crosses the axes at (0,12)(0,12) and (6,0)(6,0). Find its gradient and both intercepts.

Tier 3 · Hard

4 marks
ORIGINAL

A straight-line graph of water volume VV litres against time tt minutes passes through (0,120)(0,120) and (8,72)(8,72). Find and interpret its gradient and VV-intercept, then write VV in terms of tt.

A11

Identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square

  • Roots are the xx-coordinates where a quadratic graph meets the xx-axis, while the yy-intercept is found by setting x=0x=0.
  • Factorising can reveal the roots, and completing the square into a(xh)2+ka(x-h)^2+k reveals the turning point (h,k)(h,k) and axis of symmetry x=hx=h.
  • For example, x26x+5=(x3)24x^2-6x+5=(x-3)^2-4, so its turning point is (3,4)(3,-4) and its roots are 11 and 55.
  • In completed-square form, the turning point's xx-coordinate has the opposite sign to the value inside the bracket; do not report (h,k)(-h,k) for (xh)2+k(x-h)^2+k.

Tier 1 · Easy

2 marks
ORIGINAL

Find the roots and the yy-intercept of y=x25x+6y=x^2-5x+6.

Tier 2 · Standard

3 marks
ORIGINAL

For y=(x4)29y=(x-4)^2-9, state the turning point and axis of symmetry, and find the roots.

Tier 3 · Hard

4 marks
ORIGINAL

Complete the square for y=2x2+8x10y=2x^2+8x-10. Hence state the turning point and find the roots.

A12

Recognise, sketch and interpret graphs of linear, quadratic and simple cubic functions, the reciprocal y = 1/x (x ≠ 0), exponential y = k^x (k > 0), and y = sin x, cos x, tan x for angles of any size

  • Recognise each family by its invariant shape: constant-gradient linear, symmetric quadratic, S-shaped cubic, two-branch reciprocal, and constant-ratio exponential.
  • Sketch by marking intercepts, roots, turning points, asymptotes and representative values; for trigonometric graphs also use their periods and standard exact values.
  • For example, y=1xy=\frac1x has asymptotes x=0x=0 and y=0y=0, with branches in quadrants I and III because xx and yy have the same sign.
  • Do not draw a reciprocal graph touching an axis or treat exponential growth as a straight line; asymptotes may be approached without being reached.

Tier 1 · Easy

1 mark
ORIGINAL

A graph passes through (0,1)(0,1) and its yy-value doubles whenever xx increases by 11. Name the function y=2xy=2^x as linear, quadratic, cubic, reciprocal or exponential.

Tier 2 · Standard

3 marks
ORIGINAL

For the graph y=6xy=\frac6x, state both asymptotes and the two quadrants containing its branches.

Tier 3 · Hard

4 marks
ORIGINAL

For y=cosxy=\cos x on 180x360-180^\circ\leq x\leq360^\circ, list the xx-intercepts and the coordinates of every maximum and minimum needed for an accurate sketch.

A13

Sketch translations and reflections of a given function [Higher only]

  • For y=f(x)+ay=f(x)+a, translate the graph vertically by vector (0a)\begin{pmatrix}0\\a\end{pmatrix}; for y=f(xa)y=f(x-a), translate it horizontally by vector (a0)\begin{pmatrix}a\\0\end{pmatrix}.
  • The graph of y=f(x)y=-f(x) is the reflection of y=f(x)y=f(x) in the xx-axis, while y=f(x)y=f(-x) is its reflection in the yy-axis.
  • For example, a point (p,q)(p,q) on y=f(x)y=f(x) becomes (p+3,q2)(p+3,q-2) on y=f(x3)2y=f(x-3)-2.
  • Horizontal changes act inside the function with the opposite apparent sign; f(x+4)f(x+4) moves the graph 44 units left, not right.

Tier 1 · Easy

2 marks
ORIGINAL

Describe fully the transformation from y=f(x)y=f(x) to y=f(x)+4y=f(x)+4.

Tier 2 · Standard

3 marks
ORIGINAL

The point (3,2)(3,-2) lies on y=f(x)y=f(x). Find the corresponding point on y=f(x)y=f(-x) and name the transformation.

Tier 3 · Hard

4 marks
ORIGINAL

The point (1,5)(-1,5) lies on y=f(x)y=f(x). Find the corresponding point on y=f(2x)3y=f(2-x)-3, and describe the reflection and translations that produce the new graph.

A14

Plot and interpret graphs (including reciprocal and exponential graphs) and graphs of non-standard functions in real contexts, to find approximate solutions e.g. simple kinematic problems

  • A graph represents all ordered pairs that satisfy a rule; intercepts, turning points and asymptotes help describe its behaviour.
  • Reciprocal graphs such as y=k/xy=k/x have two branches and approach the axes without meeting them, while exponential graphs change by a constant multiplier for equal changes in xx.
  • To solve two equations graphically, plot both relations on the same axes and read the coordinates of their intersection points.
  • In a real context, state what an intersection or graph feature means and use sensible units; a common error is to report a coordinate with no interpretation.

Tier 1 · Easy

2 marks
ORIGINAL

The time tt hours for a fixed journey is modelled by t=24/vt=24/v, where vv is the average speed in km/h. Work out the point on this graph when v=16v=16 and interpret it.

Tier 2 · Standard

4 marks
ORIGINAL

Higher only: Plot y=2xy=2^x and y=92xy=9-2x for 1x31\le x\le3. Use the intersection to estimate the solution of 2x=92x2^x=9-2x to one decimal place.

Tier 3 · Hard

5 marks
ORIGINAL

For 0t50\le t\le5, two moving objects have distances from a marker modelled by d=3t2+2d=3t^2+2 and d=14td=14t, with dd in metres and tt in seconds. Draw both graphs and estimate the later time when the objects are equally far from the marker.

A15

Calculate or estimate gradients of graphs and areas under graphs (incl. quadratic and other non-linear); interpret e.g. distance-time, velocity-time and financial graphs (not calculus) [Higher only]

  • The gradient between two points is change in the vertical coordinate divided by change in the horizontal coordinate; its units come from those axes.
  • For a curve, draw a tangent at the required point and calculate the gradient using two well-separated points on that tangent.
  • Estimate an area under a curve by splitting it into strips and using trapezia; on a velocity-time graph this area represents displacement.
  • Read the scale before calculating and interpret the sign and units; a common error is to use two points on the curve instead of two points on the tangent.

Tier 1 · Easy

3 marks
ORIGINAL

A distance-time graph is a straight line from (12,150)(12,150) to (32,510)(32,510), where time is in seconds and distance is in metres. Calculate and interpret its gradient.

Tier 2 · Standard

4 marks
ORIGINAL

A velocity-time graph joins the points (0,0)(0,0), (6,15)(6,15), (14,15)(14,15) and (20,3)(20,3) with straight lines. Work out the distance travelled in the first 2020 seconds.

Tier 3 · Hard

6 marks
ORIGINAL

A curved velocity-time graph passes through the values v=4,9,18,32v=4,9,18,32 m/s at t=0,2,4,6t=0,2,4,6 seconds. Use three trapezia to estimate the distance travelled. A tangent at t=4t=4 passes through (2,11)(2,11) and (6,25)(6,25); estimate the acceleration then.

A16

Recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point [Higher only]

  • A circle with centre (0,0)(0,0) and radius rr has equation x2+y2=r2x^2+y^2=r^2.
  • A point lies on the circle when its coordinates make the left-hand side equal to r2r^2.
  • The tangent at a point is perpendicular to the radius through that point, so their gradients multiply to 1-1 when both gradients are defined.
  • Substitute the given point into the final tangent equation to check it; a common error is to use the radius gradient without taking its negative reciprocal.

Tier 1 · Easy

1 mark
ORIGINAL

Write down the equation of the circle with centre (0,0)(0,0) and radius 77.

Tier 2 · Standard

4 marks
ORIGINAL

The point P(5,12)P(5,12) lies on the circle x2+y2=169x^2+y^2=169. Determine the tangent's equation there.

Tier 3 · Hard

5 marks
ORIGINAL

The tangent to x2+y2=50x^2+y^2=50 at Q(1,7)Q(1,7) meets the positive coordinate axes. Find the exact area of the triangle enclosed by the tangent and the axes.

A17

Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph

  • Keep an equation balanced by carrying out the same operation on both sides until the unknown is isolated.
  • Expand brackets and collect unknown terms on one side before collecting constants on the other.
  • A graphical solution is the xx-coordinate where the graphs representing the two sides of the equation intersect.
  • Clear fractions using a common multiple and check by substitution; a common error is to change a sign when moving a term without applying a valid operation.

Tier 1 · Easy

2 marks
ORIGINAL

Solve 7x+5=337x+5=33.

Tier 2 · Standard

4 marks
ORIGINAL

Draw y=4.5x2y=4.5x-2 and y=9.2xy=9.2-x on the same axes. Use the intersection to solve 4.5x2=9.2x4.5x-2=9.2-x, giving an estimate to one decimal place.

Tier 3 · Hard

4 marks
ORIGINAL

Solve 3(x2)4x+13=56\frac{3(x-2)}{4}-\frac{x+1}{3}=\frac56.

A18

Solve quadratic equations (including those requiring rearrangement) algebraically by factorising, by completing the square and by using the quadratic formula; find approximate solutions using a graph

  • Rearrange a quadratic equation into ax2+bx+c=0ax^2+bx+c=0 before choosing a solution method.
  • Factorising uses the zero-product rule, while completing the square rewrites the quadratic so that a squared expression can be isolated.
  • For ax2+bx+c=0ax^2+bx+c=0, the quadratic formula is x=b±b24ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}; a graph gives roots as xx-intercepts or intersections.
  • Keep both roots unless the context excludes one; a common error is to lose the negative value when taking a square root.

Tier 1 · Easy

2 marks
ORIGINAL

Solve x2+2x35=0x^2+2x-35=0 by factorising.

Tier 2 · Standard

4 marks
ORIGINAL

Higher only: Solve x28x+3=0x^2-8x+3=0 by completing the square. Give exact answers.

Tier 3 · Hard

5 marks
ORIGINAL

Higher only: The curves y=x2y=x^2 and y=5x+1y=5x+1 intersect twice. Use the quadratic formula to find the exact xx-coordinates, then give the values a graph should show to two decimal places.

A19

Solve two simultaneous equations in two variables (linear/linear or linear/quadratic) algebraically; find approximate solutions using a graph

  • A simultaneous solution is an ordered pair that satisfies both equations and appears graphically as an intersection point.
  • For two linear equations, eliminate one variable by adding or subtracting suitable multiples of the equations.
  • When one equation is quadratic, substitute the linear expression into it, solve the resulting quadratic and find the paired value for every root.
  • Check every ordered pair in both original equations; a common error is to give two xx-values without their corresponding yy-values.

Tier 1 · Easy

3 marks
ORIGINAL

Solve simultaneously 2x+y=112x+y=11 and xy=1x-y=1.

Tier 2 · Standard

4 marks
ORIGINAL

On the same axes draw y=0.6x+1y=0.6x+1 and y=50.8xy=5-0.8x. Use the graph to estimate their point of intersection to one decimal place.

Tier 3 · Hard

5 marks
ORIGINAL

Higher only: Solve simultaneously y=x+1y=x+1 and x2+y2=25x^2+y^2=25.

A20

Find approximate solutions to equations numerically using iteration [Higher only]

  • Iteration rewrites an equation as x=g(x)x=g(x) and repeatedly applies xn+1=g(xn)x_{n+1}=g(x_n) from a stated starting value.
  • Keep extra calculator digits during the repeated substitutions and round only the reported value.
  • A stable decimal answer is supported when successive iterates agree to the requested accuracy, provided the iteration converges.
  • Substitute the approximation into the original equation as a check; a common error is to reuse x0x_0 instead of the latest iterate.

Tier 1 · Easy

2 marks
ORIGINAL

The iteration xn+1=10xnx_{n+1}=\sqrt{10-x_n} starts with x0=3x_0=3. Work out x1x_1 and x2x_2, giving each to three decimal places.

Tier 2 · Standard

4 marks
ORIGINAL

Use xn+1=(18xn)/2x_{n+1}=\sqrt{(18-x_n)/2} with x0=3x_0=3 to find x4x_4. Give the result to four decimal places.

Tier 3 · Hard

6 marks
ORIGINAL

Let f(x)=x3+x12f(x)=x^3+x-12. Show that f(x)=0f(x)=0 has a root between 22 and 33. Starting with x0=2.2x_0=2.2, use xn+1=12xn3x_{n+1}=\sqrt[3]{12-x_n} to find this root to four decimal places.

A21

Translate simple situations or procedures into algebraic expressions or formulae; derive an equation (or two simultaneous equations), solve the equation(s) and interpret the solution

  • Choose a variable and state what it represents before translating each quantity into an algebraic expression.
  • Use the relationship in the situation to form an equation or a pair of simultaneous equations, including consistent units.
  • Solve the equation, then translate the mathematical result back into the quantities asked for.
  • Reject values that are impossible in context, such as negative lengths or ages; a common error is to stop at an unlabelled value of the variable.

Tier 1 · Easy

3 marks
ORIGINAL

A rectangle has width xx cm and length (x+3)(x+3) cm. Its perimeter is 3434 cm. Form and solve an equation to find both dimensions.

Tier 2 · Standard

4 marks
ORIGINAL

A club sells 3838 tickets. Adult tickets cost £7 and junior tickets cost £4. The total received is £203. Form two equations and find how many tickets of each type were sold.

Tier 3 · Hard

5 marks
ORIGINAL

Mira is 44 years older than Theo. In 33 years, the product of their ages will be 192192. Form an equation and find their current ages.

A22

Solve linear inequalities in one or two variable(s), and quadratic inequalities in one variable; represent the solution set on a number line, using set notation and on a graph

  • Solve a linear inequality like an equation, but reverse the inequality sign when multiplying or dividing by a negative number.
  • On a number line use a filled endpoint for \le or \ge and an open endpoint for << or >>.
  • For two-variable inequalities, draw each boundary and test a point to identify the required region; strict inequalities use dashed boundaries.
  • For a quadratic inequality, find the roots and test the intervals they create; a common error is to assume the answer always lies between the roots.

Tier 1 · Easy

3 marks
ORIGINAL

Higher only: Solve 3x7113x-7\le11. Give the answer in set notation and describe its number-line representation.

Tier 2 · Standard

5 marks
ORIGINAL

Higher only: On coordinate axes, show the region satisfying both y2x1y\ge2x-1 and x+y<5x+y<5. State the intersection of the boundary lines and identify which boundaries are included.

Tier 3 · Hard

4 marks
ORIGINAL

Higher only: Solve (2x+3)(x2)>0(2x+3)(x-2)>0. Give the solution in set notation and describe it on a number line.

A23

Generate terms of a sequence from either a term-to-term or a position-to-term rule

  • A term-to-term rule produces each new term from the preceding term or terms, so begin with every stated starting value.
  • A position-to-term rule gives the term directly from its position nn; substitute n=1,2,3,n=1,2,3,\ldots in order.
  • Write down intermediate terms when a repeated procedure is used so that alternating or multi-step rules remain in the correct order.
  • Check whether the first term corresponds to n=1n=1; a common error is to substitute n=0n=0 unless the sequence explicitly starts there.

Tier 1 · Easy

2 marks
ORIGINAL

A sequence has position-to-term rule 6n26n-2. Write down its first four terms.

Tier 2 · Standard

3 marks
ORIGINAL

A sequence starts 2,52,5. Each later term is one more than the sum of the previous two terms. Write down the next three terms.

Tier 3 · Hard

4 marks
ORIGINAL

The nnth term of a sequence is n23n+4n^2-3n+4. Generate the first six terms.

A24

Recognise and use triangular, square and cube numbers, arithmetic progressions, Fibonacci type sequences, quadratic sequences, simple geometric progressions (r^n, r rational > 0 or a surd) and others

  • Square and cube numbers have forms n2n^2 and n3n^3, while triangular numbers have form n(n+1)/2n(n+1)/2.
  • Arithmetic progressions have a constant first difference; quadratic sequences have constant second differences.
  • In a Fibonacci-type sequence, later terms are formed from preceding terms, while a geometric progression has a constant multiplier between consecutive terms.
  • Check more than one step before naming a pattern; a common error is to assume a sequence is arithmetic from a single pair of terms.

Tier 1 · Easy

2 marks
ORIGINAL

The sequence 1,4,9,16,1,4,9,16,\ldots is made from a named type of number. Name the type and write down the next two terms.

Tier 2 · Standard

3 marks
ORIGINAL

A Fibonacci-type sequence begins 3,7,10,17,27,3,7,10,17,27,\ldots, with each term after the second equal to the sum of the previous two. Find the eighth term.

Tier 3 · Hard

4 marks
ORIGINAL

Higher only: A geometric progression begins 2,23,6,63,2,2\sqrt3,6,6\sqrt3,\ldots. State the common ratio and find the eighth term in exact form.

A25

Deduce expressions to calculate the nth term of linear and quadratic sequences

  • For a linear sequence with common difference dd, begin with dndn and adjust the constant so that the expression gives the first term.
  • For a quadratic sequence an2+bn+can^2+bn+c, the constant second difference is 2a2a.
  • After finding aa, subtract the values of an2an^2 from the sequence; the remaining linear sequence determines bb and cc.
  • Verify the expression against several given terms; a common error is to use the first term itself as the constant in the nnth-term expression.

Tier 1 · Easy

2 marks
ORIGINAL

Find an expression for the nnth term of 7,11,15,19,7,11,15,19,\ldots.

Tier 2 · Standard

4 marks
ORIGINAL

Higher only: Find an expression for the nnth term of 2,7,14,23,34,2,7,14,23,34,\ldots.

Tier 3 · Hard

6 marks
ORIGINAL

Higher only: A quadratic sequence starts 4,15,32,554,15,32,55. Deduce its nnth term and determine the position of the term equal to 207207.