Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.
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Fullmark schemes
Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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1IGMC-042-00-01Expanding, factorising and algebraic fractionsPaper CEasy5 marks
No calculator
Answer all parts.
(a)
Expand \(4(3x - 5)\).
(1)
(b)
Expand and simplify \(3(2y + 1) - 2(y - 4)\).
(2)
(c)
Factorise completely \(6a^2 - 9ab\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
12x − 20 B1
✓ 1
Part (b)
M1 for 6y + 3 − 2y + 8 seen, with the sign of +8 correct
✓ 1
A1 4y + 11
✓ 1
Part (c)
3a(2a − 3b) B2
✓ 2
[B1 for a correct partial factorisation such as a(6a − 9b) or 3(2a2 − 3ab)]
—
Common error: In part (b), −2 × −4 = +8. Writing −8 is the most frequent slip when a negative number multiplies a bracket.
2IGMC-042-00-05Expanding, factorising and algebraic fractionsPaper CEasy4 marks
Calculator
\(E = 2x(x + 4) - x(3 - x)\)
(a)
Expand and simplify E.
(2)
(b)
Factorise your answer to the first part completely.
(1)
(c)
Find the value of E when x = 2.6.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 2x2 + 8x − 3x + x2 seen
✓ 1
A1 3x2 + 5x
✓ 1
Part (b)
B1 x(3x + 5)
✓ 1
Part (c)
B1 33.28
✓ 1
Common error: −x × −x = +x². Getting this sign wrong gives x² + 5x, and that leads to a wrong value of E in the final part.
3IGMC-042-02-02Quadratic equations and inequalitiesPaper CEasy4 marks
No calculator
(a) Solve 3x2 = 48. (b) Solve x2 − 9x = 0.
(a)
Solve 3x2 = 48.
(2)
(b)
Solve x2 − 9x = 0.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 x2 = 16
✓ 1
A1 x = 4 or x = −4
✓ 1
Part (b)
M1 x(x − 9) = 0
✓ 1
A1 x = 0 or x = 9
✓ 1
Common error: Dividing both sides of x² − 9x = 0 by x loses the solution x = 0. Factorise instead of dividing by the variable.
4IGMC-042-02-04Quadratic equations and inequalitiesPaper CEasy5 marks
No calculator
(a) Solve the inequality 4x + 3 > 2x − 9. (b) Find all the integer values of n that satisfy −3 < 2n + 1 ≤ 7.
(a)
Solve the inequality 4x + 3 > 2x − 9.
(2)
(b)
Find all the integer values of n that satisfy −3 < 2n + 1 ≤ 7.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2x > −12
✓ 1
A1 x > −6
✓ 1
Part (b)
M1 −4 < 2n ≤ 6
✓ 1
A1 −2 < n ≤ 3
✓ 1
B1 −1, 0, 1, 2, 3
✓ 1
Common error: Check the end values. −2 is excluded because the inequality is strict (<), but 3 is included because of ≤.
5IGMC-042-02-12Quadratic equations and inequalitiesPaper CEasy4 marks
Find the values of x and y. You must show all your working.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for a correct method to eliminate one variable, e.g. adding the equations to get 8x = 24, or a correct substitution
✓ 1
A1 for x = 3
✓ 1
A1 for y = 3.5 or \(\frac{7}{2}\)
✓ 1
Exam tip: The y coefficients already have opposite signs, so adding the equations removes y straight away. Then substitute x = 3 back into either equation and check the pair in the other one.
9IGMC-042-01-04Linear and simultaneous equations and rearranging formulaePaper CEasy5 marks
No calculator
An adult ticket for a museum costs $x. A child ticket costs $4 less than an adult ticket. Two adult tickets and three child tickets cost $63 in total.
(a)
Show that \(5x - 12 = 63\).
(2)
(b)
Find the cost of one child ticket.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for child ticket = x − 4
✓ 1
B1 for 2x + 3(x − 4) = 63, simplified correctly to 5x − 12 = 63
✓ 1
Part (b)
M1 for 5x = 75
✓ 1
A1 for x = 15
✓ 1
A1 for child ticket = $11
✓ 1
Common error: Read the question carefully. Solving gives x = 15, but that is the adult price. The final answer is the child price, $11.
10IGMC-042-01-05Linear and simultaneous equations and rearranging formulaePaper CEasy5 marks
No calculator
Solve these equations involving fractions.
(a)
\(\frac{x+4}{3} = 2x - 7\)
(3)
(b)
\(\frac{x}{2} + \frac{x}{5} = 14\)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for x + 4 = 3(2x − 7)
✓ 1
M1 for 6x − 21 = x + 4, leading to 5x = 25
✓ 1
A1 for x = 5
✓ 1
Part (b)
M1 for multiplying every term by 10: 5x + 2x = 140
✓ 1
A1 for x = 20
✓ 1
Common error: When you multiply both sides by 3, the whole right-hand side must be multiplied: 3(2x − 7), not 6x − 7.
11IGMC-042-03-01Sequences and graphs of functionsPaper CEasy7 marks
No calculator
The first four terms of a sequence are 7, 11, 15, 19, ...
(a)
Write down the next term.
(1)
(b)
Find an expression, in terms of n, for the nth term.
(2)
(c)
Find the 50th term.
(2)
(d)
Show that 150 is not a term of this sequence.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 23
✓ 1
Part (b)
B2 4n + 3
✓ 2
(B1 for 4n + k, where k ≠ 3)
—
Part (c)
M1 for 4 × 50 + 3
✓ 1
A1 203
✓ 1
Part (d)
M1 for 4n + 3 = 150 leading to n = 36.75 or \(\frac{147}{4}\), OR for stating that every term is odd (odd start, even common difference), OR for 150 − 7 = 143 not being a multiple of 4
✓ 1
A1 correct conclusion: n is not a whole number / 150 is even / 143 is not divisible by 4, so 150 is not a term
✓ 1
Exam tip: The number multiplying n is the common difference. Always check your expression by substituting n = 1 and n = 2.
12IGMC-042-03-02Sequences and graphs of functionsPaper CEasy4 marks
No calculator
The first four terms of a sequence are 40, 34, 28, 22, ...
(a)
Find an expression, in terms of n, for the nth term.
(2)
(b)
Find the value of the first term in the sequence that is negative.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B2 46 − 6n
✓ 2
(B1 for k − 6n or −6n + k, where k ≠ 46)
—
Part (b)
M1 for 46 − 6n < 0 or continuing the sequence 16, 10, 4, ...
✓ 1
A1 −2
✓ 1
Common error: In a decreasing sequence the coefficient of n is negative. Writing 6n + 46 gives an increasing sequence.
13IGMC-042-03-03Sequences and graphs of functionsPaper CEasy6 marks
Write down the equation of the line of symmetry of the graph of y = x2 − 3x − 2.
(1)
(c)
Find the minimum value of y.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 p = 2
✓ 1
B1 q = −4
✓ 1
B1 r = 2
✓ 1
Part (b)
B1 x = 1.5 (accept x = \(\frac{3}{2}\))
✓ 1
Part (c)
M1 substituting x = 1.5: 1.52 − 3 × 1.5 − 2, or completing the square to get (x − 1.5)2 − 4.25
✓ 1
A1 −4.25 (accept \(-\frac{17}{4}\))
✓ 1
Exam tip: Equal y-values in a table of a quadratic sit either side of the line of symmetry; the line passes halfway between their x-values.
14IGMC-042-03-04Sequences and graphs of functionsPaper CEasy6 marks
No calculator
A pattern of shapes is made from matchsticks. Pattern 1 uses 5 matchsticks, Pattern 2 uses 9 matchsticks and Pattern 3 uses 13 matchsticks. The number of matchsticks increases by the same amount each time.
(a)
Find the number of matchsticks in Pattern 6.
(1)
(b)
Find an expression for the number of matchsticks in Pattern n.
(2)
(c)
Rana has 150 matchsticks. She makes the largest pattern she can. Find the pattern number and the number of matchsticks she has left over.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 25
✓ 1
Part (b)
B2 4n + 1
✓ 2
(B1 for 4n + k, where k ≠ 1)
—
Part (c)
M1 for 4n + 1 ≤ 150 or (150 − 1) ÷ 4 = 37.25
✓ 1
A1 Pattern 37
✓ 1
A1 1 matchstick left over
✓ 1
Common error: Rounding 37.25 up to 38 gives a pattern that needs 153 matchsticks, which is more than Rana has. Round down when resources are limited.
15IGMC-042-01-07Linear and simultaneous equations and rearranging formulaePaper CMedium5 marks
Calculator
At a café, 3 coffees and 2 muffins cost $13.30. 5 coffees and 3 muffins cost $21.10. A coffee costs $c and a muffin costs $m.
(a)
Write down two equations in c and m.
(2)
(b)
Solve your equations to find the cost of one coffee and the cost of one muffin.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for 3c + 2m = 13.30
✓ 1
B1 for 5c + 3m = 21.10
✓ 1
Part (b)
M1 for multiplying to give equal coefficients and eliminating one variable, e.g. 10c + 6m = 42.20 minus 9c + 6m = 39.90
✓ 1
A1 for c = 2.30, so a coffee costs $2.30
✓ 1
A1 for m = 3.20, so a muffin costs $3.20
✓ 1
Exam tip: Money answers should be written to 2 decimal places, e.g. $3.20, not $3.2.
16IGMC-042-01-10Linear and simultaneous equations and rearranging formulaePaper CMedium4 marks
No calculator
The three angles of a triangle are (2x + 10)°, (3x − 5)° and (x + 25)°.
(a)
Write down an equation in x and solve it.
(3)
(b)
Find the size of the largest angle.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for (2x + 10) + (3x − 5) + (x + 25) = 180
✓ 1
A1 for 6x + 30 = 180
✓ 1
A1 for x = 25
✓ 1
Part (b)
B1 for 70°
✓ 1
Exam tip: Check your answer by working out all three angles (60°, 70°, 50°) and confirming that they add up to 180°.
17IGMC-042-01-12Linear and simultaneous equations and rearranging formulaePaper CMedium6 marks
No calculator
A straight line L passes through the points (−2, 7) and (4, −5).
(a)
Find the equation of L in the form y = mx + c.
(3)
(b)
Find the coordinates of the point where L meets the line y = x − 6.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for \(\frac{-5-7}{4-(-2)}\)
✓ 1
A1 for gradient = −2
✓ 1
A1 for y = −2x + 3
✓ 1
Part (b)
M1 for −2x + 3 = x − 6
✓ 1
A1 for x = 3
✓ 1
A1 for (3, −3)
✓ 1
Exam tip: Substitute one of the given points into your equation to check it. For example, at x = −2: −2(−2) + 3 = 7 ✓.
18IGMC-042-03-06Sequences and graphs of functionsPaper CMedium5 marks
No calculator
The first five terms of a sequence are 4, 7, 12, 19, 28, ...
(a)
Write down the next term.
(1)
(b)
Find an expression, in terms of n, for the nth term.
(2)
(c)
Which term of the sequence is equal to 403?
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 39
✓ 1
Part (b)
B2 n2 + 3
✓ 2
(B1 for n2 seen in an expression, or for second differences of 2 identified)
—
Part (c)
M1 for n2 + 3 = 403 leading to n2 = 400
✓ 1
A1 20th term
✓ 1
Exam tip: When the first differences increase by 2 each time, compare the sequence with the square numbers 1, 4, 9, 16, ... and look at what has been added.
19IGMC-042-03-09Sequences and graphs of functionsPaper CMedium8 marks
Calculator
The graphs of y = \(\frac{12}{x}\) and y = x + 1 intersect at two points, A and B.
(a)
The point P lies on the curve y = \(\frac{12}{x}\) and has x-coordinate 2.5. Find the y-coordinate of P.
(1)
(b)
Show that the x-coordinates of A and B satisfy the equation x2 + x − 12 = 0.
(2)
(c)
Find the coordinates of A and B.
(3)
(d)
Calculate the length of AB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 4.8
✓ 1
Part (b)
M1 for \(\frac{12}{x}\) = x + 1
✓ 1
A1 for 12 = x2 + x leading to x2 + x − 12 = 0 with no errors
✓ 1
Part (c)
M1 for (x + 4)(x − 3) = 0, or correct substitution into the quadratic formula, or equivalent valid method
✓ 1
A1 x = 3 and x = −4
✓ 1
A1 (3, 4) and (−4, −3), accepted with either point labelled A
✓ 1
Part (d)
M1 for \(\sqrt{(3-(-4))^2 + (4-(-3))^2}\) or \(\sqrt{7^2 + 7^2}\)
✓ 1
A1 9.90, also accepting 9.9, 9.899… or \(7\sqrt{2}\)
✓ 1
Exam tip: In a 'show that' part, write down every step. Show the multiplication of both sides by x clearly before giving the final rearranged equation.
20IGMC-042-01-14Linear and simultaneous equations and rearranging formulaePaper CHard6 marks
No calculator
A fraction is \(\frac{n}{d}\). When 1 is added to both the numerator and the denominator, the fraction becomes \(\frac{2}{3}\). When 1 is subtracted from both the numerator and the denominator, the fraction becomes \(\frac{1}{2}\).
(a)
Show that \(3n - 2d = -1\).
(2)
(b)
Form a second equation and solve the two equations to find the fraction \(\frac{n}{d}\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for \(\frac{n+1}{d+1} = \frac{2}{3}\) leading to 3(n + 1) = 2(d + 1)
✓ 1
A1 for 3n + 3 = 2d + 2, rearranged correctly to 3n − 2d = −1
✓ 1
Part (b)
M1 for \(\frac{n-1}{d-1} = \frac{1}{2}\) leading to 2(n − 1) = d − 1
✓ 1
A1 for 2n − d = 1
✓ 1
M1 for a correct method to solve the two equations simultaneously, e.g. substituting d = 2n − 1 into 3n − 2d = −1
✓ 1
A1 for \(\frac{3}{5}\)
✓ 1
Exam tip: Check your fraction against both conditions: \(\frac{3+1}{5+1} = \frac{4}{6} = \frac{2}{3}\) and \(\frac{3-1}{5-1} = \frac{2}{4} = \frac{1}{2}\).
21IGMC-042-00-02Expanding, factorising and algebraic fractionsPaper EEasy6 marks
No calculator
Answer all parts.
(a)
Expand and simplify \((x + 5)(x - 3)\).
(2)
(b)
Expand and simplify \((2x - 1)^2\).
(2)
(c)
Factorise \(x^2 - 7x + 12\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for at least 3 of the 4 terms x2, −3x, +5x, −15 correct
✓ 1
A1 x2 + 2x − 15
✓ 1
Part (b)
M1 for (2x − 1)(2x − 1) expanded with at least 3 of 4 terms correct
✓ 1
A1 4x2 − 4x + 1
✓ 1
Part (c)
(x − 3)(x − 4) B2
✓ 2
[B1 for two brackets whose product gives two of the three terms correctly]
—
Common error: \((2x-1)^2\) is not \(4x^2 + 1\). Write the bracket out twice and multiply every term so the middle term −4x is not lost.
22IGMC-042-00-03Expanding, factorising and algebraic fractionsPaper EEasy5 marks
No calculator
Answer all parts.
(a)
Factorise \(x^2 - 49\).
(1)
(b)
Factorise completely \(5x^2 - 20\).
(2)
(c)
Use factorisation to work out the exact value of \(83^2 - 17^2\). Show your working.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(x − 7)(x + 7) B1
✓ 1
Part (b)
M1 for 5(x2 − 4) seen
✓ 1
A1 5(x − 2)(x + 2)
✓ 1
Part (c)
M1 for (83 − 17)(83 + 17) or 66 × 100
✓ 1
A1 6600
✓ 1
Exam tip: "Factorise completely" means take out the common factor first, then check whether what is left is a difference of two squares.
23IGMC-042-00-04Expanding, factorising and algebraic fractionsPaper EEasy4 marks
No calculator
Simplify fully.
(a)
\(\frac{12x^3y}{4xy^2}\)
(2)
(b)
\(\frac{x}{3} + \frac{x}{5}\)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 for two of: coefficient 3, x2 in numerator, y in denominator
✓ 1
B1 \(\frac{3x^2}{y}\) fully correct
✓ 1
Part (b)
M1 for \(\frac{5x}{15} + \frac{3x}{15}\) or common denominator 15 used correctly
✓ 1
A1 \(\frac{8x}{15}\)
✓ 1
Common error: Adding the numerators and the denominators to get \(\frac{2x}{8}\) is wrong. Fractions can only be added once they share a common denominator.
24IGMC-042-02-05Quadratic equations and inequalitiesPaper EEasy3 marks
No calculator
(a) Show that x = 3 is a solution of 2x2 − 5x − 3 = 0. (b) Find the other solution of the equation.
(a)
Show that x = 3 is a solution of 2x2 − 5x − 3 = 0.
(1)
(b)
Find the other solution of the equation.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 2 × 9 − 15 − 3 = 0 shown
✓ 1
Part (b)
M1 (x − 3)(2x + 1) = 0
✓ 1
A1 x = \(-\frac{1}{2}\)
✓ 1
Exam tip: If one solution is x = 3, then (x − 3) must be a factor. This gives you one bracket and makes the factorising quicker.
25IGMC-042-04-01Functions and differentiationPaper EEasy4 marks
No calculator
The functions f and g are defined by f(x) = 3x − 5 g(x) = x2 + 1
(a)
Find f(4).
(1)
(b)
Find g(−3).
(1)
(c)
Find fg(2).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 7
✓ 1
Part (b)
B1: 10
✓ 1
Part (c)
M1: g(2) = 22 + 1 = 5
✓ 1
A1: fg(2) = f(5) = 3(5) − 5 = 10
✓ 1
Exam tip: In fg(2), work from the inside out. Apply g first, then put that answer into f.
26IGMC-042-04-02Functions and differentiationPaper EEasy5 marks
No calculator
The function f is defined by f(x) = \(\frac{x+7}{2}\).
(a)
Find f(5).
(1)
(b)
Find f−1(x).
(2)
(c)
Solve the equation f(x) = f−1(x).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 6
✓ 1
Part (b)
M1: write y = (x + 7)/2 and rearrange to x = 2y − 7, or swap x and y and rearrange
✓ 1
A1: f−1(x) = 2x − 7
✓ 1
Part (c)
M1: (x + 7)/2 = 2x − 7 leading to x + 7 = 4x − 14
✓ 1
A1: x = 7
✓ 1
Common error: f⁻¹(x) is the inverse function, not the reciprocal. Do not write 2/(x + 7).
27IGMC-042-04-03Functions and differentiationPaper EEasy4 marks
No calculator
A curve has equation y = 4x3 − 6x2 + 2x − 9.
(a)
Find \(\frac{dy}{dx}\).
(2)
(b)
Find the gradient of the curve at the point where x = 2.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: at least two of the terms 12x2, −12x, +2 correct
✓ 1
A1: dy/dx = 12x2 − 12x + 2, fully correct with the constant removed
✓ 1
Part (b)
M1: substitute x = 2 into their dy/dx: 12(4) − 12(2) + 2
✓ 1
A1: 26
✓ 1
Common error: A constant term such as −9 differentiates to 0. It does not stay in the derivative.
28IGMC-042-04-04Functions and differentiationPaper EEasy5 marks
No calculator
A curve has equation y = x2 − 8x + 3.
(a)
Find \(\frac{dy}{dx}\).
(1)
(b)
Find the coordinates of the turning point of the curve.
(3)
(c)
State whether the turning point is a maximum or a minimum. Give a reason.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 2x − 8
✓ 1
Part (b)
M1: set 2x − 8 = 0
✓ 1
A1: x = 4
✓ 1
A1: y = 16 − 32 + 3 = −13, giving the point (4, −13)
✓ 1
Part (c)
B1: minimum, with a valid reason, e.g. the coefficient of x2 is positive, or d2y/dx2 = 2 > 0
✓ 1
Exam tip: After finding x at a turning point, always substitute back into the original equation, not into dy/dx, to find y.
29IGMC-042-04-05Functions and differentiationPaper EEasy6 marks
No calculator
The function f is defined by f(x) = 5 − 2x.
(a)
Find the value of x when f(x) = 11.
(2)
(b)
Find f−1(x).
(2)
(c)
Find ff(x), giving your answer in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 5 − 2x = 11
✓ 1
A1: x = −3
✓ 1
Part (b)
M1: y = 5 − 2x rearranged to x = (5 − y)/2, or equivalent
✓ 1
A1: f−1(x) = (5 − x)/2, or equivalent
✓ 1
Part (c)
M1: 5 − 2(5 − 2x)
✓ 1
A1: 4x − 5
✓ 1
Common error: ff(x) means f applied twice, not [f(x)]². Substitute the whole expression 5 − 2x in place of x.
30IGMC-042-00-06Expanding, factorising and algebraic fractionsPaper EMedium5 marks
No calculator
Answer all parts.
(a)
Factorise \(2x^2 + 7x - 15\).
(2)
(b)
Hence solve \(2x^2 + 7x - 15 = 0\).
(1)
(c)
Factorise completely \(12x^2 - 27\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(2x − 3)(x + 5) B2
✓ 2
[B1 for (2x + 3)(x − 5), or for brackets (2x + a)(x + b) with ab = −15 or with 2b + a = 7]
—
Part (b)
x = \(\frac{3}{2}\) (or 1.5) and x = −5, both required B1
✓ 1
Part (c)
M1 for 3(4x2 − 9), or for a correct partial factorisation such as (6x − 9)(2x + 3) or (2x − 3)(6x + 9); implied by a correct final answer
✓ 1
A1 3(2x − 3)(2x + 3)
✓ 1
Exam tip: Always check a factorisation by expanding it again. For 'factorise completely', take out any common numerical factor first, then look for a difference of two squares.
31IGMC-042-00-07Expanding, factorising and algebraic fractionsPaper EMedium5 marks
No calculator
Simplify fully.
(a)
\(\frac{x^2 - 9}{x^2 + x - 6}\)
(3)
(b)
\(\frac{3x + 6}{x^2 - 4}\)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 (x − 3)(x + 3)
✓ 1
B1 (x + 3)(x − 2)
✓ 1
B1 \(\frac{x-3}{x-2}\)
✓ 1
Part (b)
M1 for \(\frac{3(x+2)}{(x-2)(x+2)}\)
✓ 1
A1 \(\frac{3}{x-2}\)
✓ 1
Common error: You can only cancel whole factors, never single terms. Crossing out the x² in the numerator and denominator earns no marks.
32IGMC-042-00-08Expanding, factorising and algebraic fractionsPaper EMedium4 marks
No calculator
Answer both parts.
(a)
Write \(\frac{3}{x+2} - \frac{2}{x-1}\) as a single fraction in its simplest form.
Exam tip: Write the formula, or your completed square, with the values substituted before using your calculator. If the final answers are wrong, the method marks can still be earned from this line.
37IGMC-042-02-07Quadratic equations and inequalitiesPaper EMedium5 marks
No calculator
This question is about the quadratic expression x2 − x − 12.
(a)
Factorise x2 − x − 12.
(2)
(b)
Hence solve the inequality x2 − x − 12 < 0.
(2)
(c)
Write down the number of integers that satisfy x2 − x − 12 < 0.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (x + a)(x + b) with ab = −12 and a + b = −1, or a pair of brackets giving the correct x2 and constant terms
✓ 1
A1 (x − 4)(x + 3)
✓ 1
Part (b)
M1 critical values −3 and 4 identified
✓ 1
A1 −3 < x < 4
✓ 1
Part (c)
B1 6 (the integers −2, −1, 0, 1, 2, 3)
✓ 1
Common error: Writing x 4 gives the region where the expression is positive. A '< 0' quadratic with a positive x² term is negative between its roots.
38IGMC-042-02-08Quadratic equations and inequalitiesPaper EMedium7 marks
No calculator
A rectangle has length (x + 4) cm and width (x − 1) cm. The area of the rectangle is 36 cm2. (a) Show that x2 + 3x − 40 = 0. (b) Solve x2 + 3x − 40 = 0 and explain why only one solution is valid. (c) Find the perimeter of the rectangle.
(a)
Show that x2 + 3x − 40 = 0.
(2)
(b)
Solve x2 + 3x − 40 = 0 and explain why only one solution is valid.
(3)
(c)
Find the perimeter of the rectangle.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (x + 4)(x − 1) = 36
✓ 1
A1 x2 + 3x − 4 = 36 leading to x2 + 3x − 40 = 0 with no errors
✓ 1
Part (b)
M1 (x + 8)(x − 5) = 0
✓ 1
A1 x = 5 or x = −8
✓ 1
B1 x = −8 rejected because the width x − 1 (and length) would be negative
✓ 1
Part (c)
M1 2 × (9 + 4)
✓ 1
A1 26 cm
✓ 1
Exam tip: In a 'show that' part, write every line of working down to the given equation. A result that appears without working earns no marks.
39IGMC-042-02-10Quadratic equations and inequalitiesPaper EMedium5 marks
No calculator
(a) Show that the equation \(\frac{3}{x+1} + \frac{2}{x} = 2\) can be written as 2x2 − 3x − 2 = 0. (b) Solve 2x2 − 3x − 2 = 0.
(a)
Show that \(\frac{3}{x+1} + \frac{2}{x} = 2\) can be written as 2x2 − 3x − 2 = 0.
(3)
(b)
Solve 2x2 − 3x − 2 = 0.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 multiplying every term by x(x + 1): 3x + 2(x + 1) = 2x(x + 1)
✓ 1
M1 5x + 2 = 2x2 + 2x
✓ 1
A1 rearranged to 2x2 − 3x − 2 = 0 with no errors
✓ 1
Part (b)
M1 (2x + 1)(x − 2) = 0
✓ 1
A1 x = 2 or x = \(-\frac{1}{2}\)
✓ 1
Common error: When clearing fractions, the 2 on the right-hand side must also be multiplied by x(x + 1). Leaving it as 2 is a frequent slip.
40IGMC-042-01-11Linear and simultaneous equations and rearranging formulaePaper EMedium5 marks
No calculator
The lens formula is \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\).
(a)
Make v the subject of the formula. Write your answer as a single fraction.
(3)
(b)
Find v when f = 12 and u = 20.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
✓ 1
M1 for combining into a single fraction: \(\frac{1}{v} = \frac{u-f}{uf}\)
✓ 1
A1 for \(v = \frac{uf}{u-f}\)
✓ 1
Part (b)
M1 for \(\frac{20 \times 12}{20 - 12}\)
✓ 1
A1 for 30
✓ 1
Common error: \(\frac{1}{f} - \frac{1}{u}\) is not \(\frac{1}{f-u}\). Use a common denominator uf before you take the reciprocal.
41IGMC-042-03-07Sequences and graphs of functionsPaper EMedium6 marks
No calculator
The first five terms of a sequence are 2, 9, 20, 35, 54, ...
(a)
Find an expression, in terms of n, for the nth term.
(3)
(b)
Factorise your expression from part (a) and hence explain why no term after the first is a prime number.
(2)
(c)
Find the 30th term.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for second differences of 4 identified
✓ 1
M1 for 2n2 seen in an expression
✓ 1
A1 2n2 + n − 1
✓ 1
Part (b)
B1 (2n − 1)(n + 1)
✓ 1
B1 for n ≥ 2 both factors are at least 3, so every term is a product of two whole numbers greater than 1
✓ 1
Part (c)
B1 1829
✓ 1
Exam tip: Halve the second difference to get the coefficient of n², then subtract that part from each term to find the linear part.
42IGMC-042-03-08Sequences and graphs of functionsPaper EMedium8 marks
Calculator
A curve has equation y = x3 − 6x2 + 9x + 1.
(a)
Find \(\frac{dy}{dx}\).
(2)
(b)
Find the coordinates of the two turning points.
(4)
(c)
Determine whether each turning point is a maximum or a minimum. Show how you decide.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B2 3x2 − 12x + 9
✓ 2
(B1 for two of the three terms correct)
—
Part (b)
M1 for 3x2 − 12x + 9 = 0
✓ 1
M1 for 3(x − 1)(x − 3) = 0 or x = 1 and x = 3 found
✓ 1
A1 (1, 5)
✓ 1
A1 (3, 1)
✓ 1
Part (c)
M1 for a valid method, e.g. \(\frac{d^2y}{dx^2}\) = 6x − 12 evaluated at both points, or gradient values either side
✓ 1
A1 (1, 5) is a maximum and (3, 1) is a minimum
✓ 1
Common error: Finding the x-values of the turning points is not enough: substitute back into the equation of the curve, not into \(\frac{dy}{dx}\), to find the y-coordinates.
43IGMC-042-03-10Sequences and graphs of functionsPaper EMedium7 marks
No calculator
A curve has equation y = 2x2 − 5x + 3.
(a)
Find \(\frac{dy}{dx}\).
(1)
(b)
Find the gradient of the curve at the point where x = 2.
(1)
(c)
The tangent to the curve at point P is parallel to the line y = 7x − 4. Find the coordinates of P.
(3)
(d)
Find the equation of the tangent at P, giving your answer in the form y = mx + c.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 4x − 5
✓ 1
Part (b)
B1 3
✓ 1
Part (c)
M1 for 4x − 5 = 7
✓ 1
A1 x = 3
✓ 1
A1 (3, 6)
✓ 1
Part (d)
M1 for 6 = 7 × 3 + c or y − 6 = 7(x − 3)
✓ 1
A1 y = 7x − 15
✓ 1
Common error: Parallel lines have equal gradients. Setting the curve's equation equal to 7x − 4 finds where the curve meets the line, not where its gradient is 7.
44IGMC-042-03-11Sequences and graphs of functionsPaper EMedium6 marks
Calculator
The number of bacteria, y, in a sample after x hours is modelled by y = 500 × 1.2x.
(a)
Write down the number of bacteria at the start.
(1)
(b)
Find the number of bacteria after 6 hours.
(2)
(c)
Find the value of x when the number of bacteria is 2000. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 500
✓ 1
Part (b)
M1 for 500 × 1.26
✓ 1
A1 1493 or 1490 (from 1492.99…)
✓ 1
Part (c)
M1 for 1.2x = 4 or 500 × 1.2x = 2000
✓ 1
M1 for a correct trial between 7 and 8, e.g. 500 × 1.27.6 = 1998.7, or for x = log 4 ÷ log 1.2
✓ 1
A1 7.60
✓ 1
Exam tip: In trial and improvement, test the value halfway between your two closest trials (here 7.605) to decide which way to round.
45IGMC-042-04-06Functions and differentiationPaper EMedium7 marks
No calculator
f(x) = 2x + 1 g(x) = x2 − 3
(a)
Find gf(x), giving your answer in the form ax2 + bx + c.
(2)
(b)
Solve gf(x) = 6.
(3)
(c)
Find f−1(g(2)).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: (2x + 1)2 − 3
✓ 1
A1: 4x2 + 4x − 2
✓ 1
Part (b)
M1: 4x2 + 4x − 8 = 0, or x2 + x − 2 = 0
✓ 1
M1: correct factorisation (x + 2)(x − 1) = 0, or a correct method for solving
✓ 1
A1: x = 1 and x = −2
✓ 1
Part (c)
B1: g(2) = 1, or f−1(x) = (x − 1)/2
✓ 1
B1: 0
✓ 1
Exam tip: When you expand (2x + 1)², write it as (2x + 1)(2x + 1). Squaring each term alone loses the middle term 4x.
46IGMC-042-04-07Functions and differentiationPaper EMedium8 marks
No calculator
A curve has equation y = x3 − 3x2 − 9x + 5.
(a)
Find \(\frac{dy}{dx}\).
(2)
(b)
Find the coordinates of the two stationary points on the curve.
(4)
(c)
Determine the nature of each stationary point. Show your working.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: at least two terms correct
✓ 1
A1: 3x2 − 6x − 9
✓ 1
Part (b)
M1: 3x2 − 6x − 9 = 0
✓ 1
M1: 3(x − 3)(x + 1) = 0, or equivalent
✓ 1
A1: x = 3 and x = −1
✓ 1
A1: (3, −22) and (−1, 10)
✓ 1
Part (c)
M1: d2y/dx2 = 6x − 6 evaluated at both points, or a valid gradient-sign test either side of each point
✓ 1
A1: (−1, 10) is a maximum and (3, −22) is a minimum
✓ 1
Exam tip: Take out the common factor 3 before factorising. The quadratic x² − 2x − 3 is much easier to handle.
47IGMC-042-04-08Functions and differentiationPaper EMedium7 marks
No calculator
A curve has equation y = 2x2 − \(\frac{8}{x}\).
(a)
Find \(\frac{dy}{dx}\).
(2)
(b)
Find the gradient of the curve at the point where x = 2.
(1)
(c)
Find the equation of the tangent to the curve at the point where x = 2. Give your answer in the form y = mx + c.
(3)
(d)
Find the x-coordinate of the point where this tangent crosses the x-axis.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 4x
✓ 1
B1: + 8/x2, or + 8x−2
✓ 1
Part (b)
B1: 10
✓ 1
Part (c)
B1: y = 2(4) − 8/2 = 4 when x = 2
✓ 1
M1: y − 4 = 10(x − 2), or 4 = 10(2) + c
✓ 1
A1: y = 10x − 16
✓ 1
Part (d)
B1: 1.6
✓ 1
Common error: Rewrite −8/x as −8x⁻¹ before differentiating. Its derivative is +8x⁻². The sign changes because the power is negative.
48IGMC-042-04-09Functions and differentiationPaper EMedium6 marks
No calculator
The function f is defined by f(x) = \(\frac{3x-2}{x+4}\), x ≠ −4.
(a)
Find f(1).
(1)
(b)
Find f−1(x).
(3)
(c)
Solve f−1(x) = 2.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 1/5
✓ 1
Part (b)
M1: y(x + 4) = 3x − 2, or x(y + 4) = 3y − 2 after swapping
✓ 1
M1: collect the terms in the subject on one side and factorise, e.g. x(y − 3) = −4y − 2
✓ 1
A1: f−1(x) = (4x + 2)/(3 − x), or equivalent
✓ 1
Part (c)
M1: (4x + 2)/(3 − x) = 2 leading to 4x + 2 = 6 − 2x, or use f−1(x) = 2 ⇔ x = f(2) = 4/6
✓ 1
A1: x = 2/3
✓ 1
Exam tip: To make x the subject when it appears twice, gather every x term on one side and then factorise x out.
49IGMC-042-04-10Functions and differentiationPaper EMedium7 marks
No calculator
A small robot moves along a straight track. Its distance, s metres, from a fixed point after t seconds is given by s = t3 − 6t2 + 15t. The velocity v m/s is the rate of change of s, so v = \(\frac{ds}{dt}\).
(a)
Find an expression for v in terms of t.
(2)
(b)
Find the velocity when t = 4.
(1)
(c)
Show that the robot is never at rest.
(2)
(d)
The acceleration is \(\frac{dv}{dt}\). Find the velocity of the robot when its acceleration is zero.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: at least two terms correct
✓ 1
A1: v = 3t2 − 12t + 15
✓ 1
Part (b)
B1: 15 m/s
✓ 1
Part (c)
M1: a valid method, e.g. discriminant (−12)2 − 4(3)(15) = −36, or 3[(t − 2)2 + 1]
✓ 1
A1: correct conclusion that v is never 0 (in fact v > 0 for all t), so the robot is never at rest
✓ 1
Part (d)
M1: 6t − 12 = 0 so t = 2
✓ 1
A1: v = 3 m/s
✓ 1
Exam tip: "At rest" means velocity = 0. A "show that" question needs a clear concluding statement.
50IGMC-042-04-11Functions and differentiationPaper EMedium7 marks
No calculator
A curve has equation y = ax2 + bx, where a and b are constants. The gradient of the curve is 4 when x = 1. The gradient of the curve is 16 when x = 3.
(a)
Find the value of a and the value of b.
(4)
(b)
Find the coordinates of the point on the curve where the gradient is zero.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: dy/dx = 2ax + b
✓ 1
M1: 2a + b = 4 and 6a + b = 16
✓ 1
A1: a = 3
✓ 1
A1: b = −2
✓ 1
Part (b)
M1: 6x − 2 = 0
✓ 1
A1: x = 1/3
✓ 1
A1: y = 3(1/9) − 2(1/3) = −1/3, giving the point (1/3, −1/3)
✓ 1
Common error: The gradient conditions go into dy/dx, not into the equation of the curve. Substituting x = 1 and y = 4 into y = ax² + bx is wrong.
51IGMC-042-00-12Expanding, factorising and algebraic fractionsPaper EHard5 marks
No calculator
\(\frac{2}{x-1} + \frac{3}{x+2} = 2\)
(a)
Show that this equation simplifies to \(2x^2 - 3x - 5 = 0\).
(3)
(b)
Solve \(2x^2 - 3x - 5 = 0\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for 2(x + 2) + 3(x − 1) = 2(x − 1)(x + 2)
✓ 1
M1 for 5x + 1 = 2x2 + 2x − 4
✓ 1
A1 correct rearrangement to 2x2 − 3x − 5 = 0 with no errors
✓ 1
Part (b)
M1 for (2x − 5)(x + 1) = 0
✓ 1
A1 x = 2.5 and x = −1
✓ 1
Common error: When you multiply through by (x − 1)(x + 2), the 2 on the right-hand side must be multiplied as well. Leaving it as 2 is a common slip.
52IGMC-042-00-13Expanding, factorising and algebraic fractionsPaper EHard8 marks
Show that \(\frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}\).
(2)
(c)
Hence find the exact value of \(\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{9 \times 10}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 2x2 + 5x − 3 = (2x − 1)(x + 3)
✓ 1
B1 4x2 − 1 = (2x − 1)(2x + 1) and 2x2 + x = x(2x + 1)
✓ 1
M1 for inverting the second fraction and multiplying
✓ 1
A1 x
✓ 1
Part (b)
M1 for \(\frac{(n+1) - n}{n(n+1)}\)
✓ 1
A1 correct simplification to \(\frac{1}{n(n+1)}\)
✓ 1
Part (c)
M1 for writing the sum as \((1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{9} - \frac{1}{10})\)
✓ 1
A1 \(\frac{9}{10}\)
✓ 1
Exam tip: In part (c), write out the first two brackets and the last one. The middle terms cancel in pairs, leaving only the first term and the last term.
53IGMC-042-00-14Expanding, factorising and algebraic fractionsPaper EHard6 marks
No calculator
Answer both parts.
(a)
\((3x + p)(x - 2) + q \equiv 3x^2 - x + 5\) Find the value of p and the value of q.
(3)
(b)
Two consecutive odd numbers can be written as 2n − 1 and 2n + 1, where n is an integer. Prove that the difference between their squares is always a multiple of 8.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for expanding to 3x2 + (p − 6)x − 2p + q, or for a correct equation from comparing coefficients, or for substituting x = 2 to give q = 12 − 2 + 5
✓ 1
A1 p = 5
✓ 1
A1 q = 15
✓ 1
Part (b)
M1 for (2n + 1)2 − (2n − 1)2
✓ 1
M1 for correct expansion, 4n2 + 4n + 1 − (4n2 − 4n + 1), or for use of the difference of two squares giving (4n)(2)
✓ 1
A1 for 8n, with the conclusion that 8n is 8 × an integer and so is a multiple of 8
✓ 1
Exam tip: A proof needs a final statement. After reaching 8n, say that it is 8 × an integer and so is a multiple of 8.
54IGMC-042-02-11Quadratic equations and inequalitiesPaper EHard6 marks
Calculator
This question is about the quadratic expression 2x2 + 3x − 7.
(a)
Solve 2x2 + 3x − 7 = 0, giving your answers correct to 2 decimal places.
(3)
(b)
Hence solve the inequality 2x2 + 3x − 7 ≥ 0.
(2)
(c)
Find how many integers do not satisfy 2x2 + 3x − 7 ≥ 0.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Correct substitution into the quadratic formula, \(\frac{-3 \pm \sqrt{3^2 - 4(2)(-7)}}{2(2)}\), or an equivalent completing-the-square method
✓ 1
√65 (or 8.062…) seen
✓ 1
x = 1.27 and x = −2.77
✓ 1
Part (b)
Recognises that the solution lies outside the roots, because the coefficient of x2 is positive. This may be implied by a correct final answer
✓ 1
x ≤ −2.77 or x ≥ 1.27. Accept exact forms, unrounded values, or values following through from the roots found in (a)
✓ 1
Part (c)
4 (the integers −2, −1, 0 and 1)
✓ 1
Exam tip: Sketch y = 2x² + 3x − 7 as a U-shape crossing the x-axis at the two roots. The curve is on or above the axis (≥ 0) in the two outer regions, so the integers that fail the inequality lie strictly between the roots.
55IGMC-042-02-13Quadratic equations and inequalitiesPaper EHard6 marks
No calculator
A car travels 240 km at an average speed of v km/h. If the average speed had been 20 km/h faster, the journey would have taken 1 hour less. (a) Show that v2 + 20v − 4800 = 0. (b) Solve the equation and find the time the journey takes at the original speed.
(a)
Show that v2 + 20v − 4800 = 0.
(3)
(b)
Solve the equation and find the time the journey takes at the original speed.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 times written as \(\frac{240}{v}\) and \(\frac{240}{v+20}\)
✓ 1
M1 \(\frac{240}{v} - \frac{240}{v+20} = 1\)
✓ 1
A1 240(v + 20) − 240v = v(v + 20) leading to v2 + 20v − 4800 = 0 with no errors
✓ 1
Part (b)
M1 (v + 80)(v − 60) = 0
✓ 1
A1 v = 60, with v = −80 rejected
✓ 1
A1 4 hours
✓ 1
Common error: The slower speed gives the longer time, so the equation is \(\frac{240}{v} - \frac{240}{v+20} = 1\). Reversing the subtraction gives the wrong sign.
56IGMC-042-02-14Quadratic equations and inequalitiesPaper EHard8 marks
Calculator
A right-angled triangle has shorter sides of length x cm and (x + 7) cm. The hypotenuse has length (2x + 1) cm. (a) Show that x2 − 5x − 24 = 0. (b) Solve the equation and hence find the area of the triangle. (c) Calculate the smallest angle of the triangle.
(a)
Show that x2 − 5x − 24 = 0.
(3)
(b)
Solve the equation and hence find the area of the triangle.
(3)
(c)
Calculate the smallest angle of the triangle.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 x2 + (x + 7)2 = (2x + 1)2
✓ 1
M1 2x2 + 14x + 49 = 4x2 + 4x + 1
✓ 1
A1 2x2 − 10x − 48 = 0 divided by 2 to give x2 − 5x − 24 = 0 with no errors
✓ 1
Part (b)
M1 (x − 8)(x + 3) = 0
✓ 1
A1 x = 8, with x = −3 rejected
✓ 1
A1 area = ½ × 8 × 15 = 60 cm2
✓ 1
Part (c)
M1 tan θ = \(\frac{8}{15}\) or sin θ = \(\frac{8}{17}\)
✓ 1
A1 28.1°
✓ 1
Exam tip: The smallest angle is always opposite the shortest side. Here that side is 8 cm, so use it as the opposite side in your trigonometry.
57IGMC-042-01-06Linear and simultaneous equations and rearranging formulaePaper EHard5 marks
No calculator
\(y = \frac{2x+3}{x-1}\)
(a)
Make x the subject of the formula.
(4)
(b)
Find the value of x when y = 5. Give your answer as a fraction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for multiplying out: y(x − 1) = 2x + 3, i.e. xy − y = 2x + 3
✓ 1
M1 for collecting the x terms on one side, e.g. xy − 2x = y + 3 or 2x − xy = −y − 3
✓ 1
M1 for factorising out x, e.g. x(y − 2) = y + 3 or x(2 − y) = −y − 3
✓ 1
A1 for \(x = \frac{y+3}{y-2}\), or an equivalent such as \(x = \frac{-y-3}{2-y}\)
✓ 1
Part (b)
B1 for \(\frac{8}{3}\) (accept \(2\frac{2}{3}\)), from substituting into the answer to (a) or solving 5 = (2x + 3)/(x − 1) directly
✓ 1
Exam tip: When the new subject appears twice, collect every term containing x on one side, then take x out as a common factor.
58IGMC-042-01-13Linear and simultaneous equations and rearranging formulaePaper EHard6 marks
Find all the solutions. You must show all your working.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 for substituting to get an equation in one variable: x2 + (2x − 1)2 = 13, or equivalent such as \(\left(\frac{y+1}{2}\right)^2 + y^2 = 13\)
✓ 1
M1 for correctly expanding the bracket, e.g. (2x − 1)2 = 4x2 − 4x + 1
✓ 1
A1 for 5x2 − 4x − 12 = 0, or equivalent 5y2 + 2y − 51 = 0
✓ 1
M1 for a correct method to solve their quadratic, e.g. (5x + 6)(x − 2) = 0 or (5y + 17)(y − 3) = 0 or the quadratic formula
✓ 1
A1 for x = 2 and x = \(-\frac{6}{5}\)
✓ 1
A1 for the correctly paired values: x = 2, y = 3 and x = \(-\frac{6}{5}\), y = \(-\frac{17}{5}\)
✓ 1
Common error: (2x − 1)² is not 4x² + 1. Write it out as (2x − 1)(2x − 1) so that the middle term −4x is not lost.
59IGMC-042-03-12Sequences and graphs of functionsPaper EHard8 marks
No calculator
Sequence A has nth term 5n − 2. Sequence B has nth term n2 + 2.
(a)
For some values of n, the nth term of sequence A is equal to the nth term of sequence B. Find these values of n and the value of each of these terms.
(4)
(b)
Sequence C has first five terms 2, 10, 30, 68, 130, ... Find an expression, in terms of n, for the nth term of sequence C.
(3)
(c)
Find the 10th term of sequence C.
(1)
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Marking point
Mark
Notes
Part (a)
M1 for n2 + 2 = 5n − 2
✓ 1
M1 for n2 − 5n + 4 = 0 and (n − 1)(n − 4) = 0
✓ 1
A1 n = 1 and n = 4
✓ 1
A1 terms 3 and 18
✓ 1
Part (b)
M1 for third differences of 6 identified
✓ 1
M1 for n3 seen in an expression
✓ 1
A1 n3 + n
✓ 1
Part (c)
B1 1010
✓ 1
Exam tip: A constant third difference of 6 means the leading term is n³. Subtract the cubes 1, 8, 27, 64, 125 to reveal what remains.
60IGMC-042-03-13Sequences and graphs of functionsPaper EHard8 marks
No calculator
A curve has equation y = x3 − 3x2 − 9x + 5.
(a)
Find the range of values of x for which y is decreasing.
(4)
(b)
The tangent to the curve at the point where x = 0 meets the curve again at the point Q. Find the coordinates of Q.
(4)
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Marking point
Mark
Notes
Part (a)
B1 \(\frac{dy}{dx}\) = 3x2 − 6x − 9
✓ 1
M1 for 3x2 − 6x − 9 < 0 or = 0
✓ 1
M1 for 3(x − 3)(x + 1) or x = −1 and 3 found
✓ 1
A1 −1 < x < 3
✓ 1
Part (b)
B1 tangent is y = 5 − 9x
✓ 1
M1 for x3 − 3x2 − 9x + 5 = 5 − 9x
✓ 1
M1 for x2(x − 3) = 0 leading to x = 3
✓ 1
A1 (3, −22)
✓ 1
Common error: A curve is decreasing where the gradient is negative, so the answer is the interval between the roots of \(\frac{dy}{dx}\) = 0, not the roots themselves.
61IGMC-042-03-14Sequences and graphs of functionsPaper EHard10 marks
Calculator
The curve y = ax2 + bx + c passes through the points (0, −3), (1, 2) and (−2, 5).
(a)
Find the values of a, b and c.
(4)
(b)
Find the exact coordinates of the turning point of the curve.
(3)
(c)
Solve ax2 + bx + c = 0, giving your answers correct to 3 significant figures.
(3)
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Marking point
Mark
Notes
Part (a)
B1 c = −3
✓ 1
M1 for a + b − 3 = 2 and 4a − 2b − 3 = 5
✓ 1
M1 for correct method to eliminate one variable
✓ 1
A1 a = 3 and b = 2
✓ 1
Part (b)
M1 for \(\frac{dy}{dx}\) = 6x + 2 = 0 or completing the square
Exam tip: Use the point with x = 0 first: it gives c immediately and reduces the problem to two simultaneous equations.
62IGMC-042-04-12Functions and differentiationPaper EHard9 marks
Calculator
A square sheet of card has side 30 cm. A square of side x cm is cut from each corner. The sides are then folded up to make an open box with a square base.
(a)
Show that the volume, V cm3, of the box is given by V = 4x3 − 120x2 + 900x.
(2)
(b)
Find \(\frac{dV}{dx}\).
(2)
(c)
Find the value of x that gives the maximum volume, and find this maximum volume.
(4)
(d)
Explain why the other solution of \(\frac{dV}{dx}\) = 0 is not suitable.
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