Edexcel International A Level Maths · P4 · Pure Mathematics 4
Proof by contradiction and partial fractions
28 free, high-quality past-paper-style questions on Proof by contradiction and partial fractions for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 229 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
Figure 1 shows an incomplete proof by contradiction that √7 is irrational. You may use the fact that, for any integer n, if n2 is a multiple of 7 then n is a multiple of 7.
(a)
Write down the equation that belongs in box (i).
(1)
(b)
Write down the equation that belongs in box (ii).
(1)
(c)
Write a suitable final line for box (iii) to complete the proof.
(2)
(d)
A student copies the proof, replacing every 7 by 49, in an attempt to prove that √49 is irrational. Identify the first line of the copied proof that is false, and give a counter-example to show that it is false.
(2)
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Answer
Mark
Notes
Part (a)
a2 = 7b2
B1
Part (b)
(7c)2 = 7b2, i.e. 49c2 = 7b2
B1
Part (c)
a and b are both multiples of 7, so they have a common factor of 7
M1
this contradicts a/b being in its simplest form, so the assumption is false and √7 is irrational
A1
Part (d)
Line 3: “a2 is a multiple of 49, so a is a multiple of 49” is false (49 is not prime)
B1
e.g. a = 7: a2 = 49 is a multiple of 49, but 7 is not a multiple of 49
B1
Common error: Stopping at “so b is a multiple of 7”. The proof is complete only when you state the clash: a and b share the factor 7, but a/b was in its simplest form.
A lattice point is a point whose coordinates are both integers. Figure 1 shows lattice points on a grid. The line L1 passes through the origin O and the lattice point A. The line L2 has equation y = √3 x. You may assume that √3 is irrational.
(a)
Use Figure 1 to find the gradient of L1, and write down the coordinates of a lattice point on L1 other than O and A.
(2)
(b)
Prove by contradiction that O is the only lattice point on L2.
(3)
(c)
The lattice point B appears to lie on L2. Explain why it does not.
(1)
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Answer
Mark
Notes
Part (a)
A is (3, 2), so the gradient of L1 is 2/3
B1
e.g. (6, 4) (or (−3, −2), (9, 6), …)
B1
Part (b)
assume L2 passes through a lattice point (p, q) other than O, where p and q are integers
M1
q = √3 p; if p = 0 then q = 0, which is O, so p ≠ 0 and √3 = q/p
M1
q/p is a ratio of integers with a non-zero denominator, so √3 would be rational — a contradiction, so O is the only lattice point on L2
A1
Part (c)
B(4, 7) is a lattice point other than O, so by part (b) it is not on L2 (check: 4√3 ≈ 6.93 ≠ 7)
B1
Exam tip: If a line through the origin had a lattice point (p, q) with p ≠ 0, its gradient would be q/p, a ratio of integers. An irrational gradient therefore rules out every lattice point except O.
Figure 1 shows the graph of y = f(x), where f(x) = (ax + b)/((x + p)(x + q)) and a, b, p and q are constants with p > q. The dashed lines are the vertical asymptotes of the curve. The curve passes through the points A(0, 2) and B(3, 0).
(a)
Use Figure 1 to write down the values of p and q.
(1)
(b)
Find the values of a and b.
(2)
(c)
Express f(x) in partial fractions.
(3)
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Answer
Mark
Notes
Part (a)
the asymptotes are x = −3 and x = 1, so p = 3 and q = −1
B1
Part (b)
B(3, 0): 3a + b = 0; A(0, 2): b/(3 × (−1)) = 2
M1
b = −6, a = 2
A1
Part (c)
2x − 6 ≡ A(x − 1) + B(x + 3)
M1
x = 1: −4 = 4B, so B = −1
A1
x = −3: −12 = −4A, so A = 3: f(x) = 3/(x + 3) − 1/(x − 1)
A1
Common error: Reading the asymptote x = −3 as a factor (x − 3). The bracket must be zero on the asymptote, so x = −3 comes from (x + 3).
When two resistors with resistances R1 ohms and R2 ohms are connected in parallel, they are equivalent to a single resistor of resistance R ohms, where 1/R = 1/R1 + 1/R2. Figure 1 shows such a pair, with resistances (x + 1)/A ohms and (x + 4)/B ohms, where A and B are positive constants and x > 0, and the equivalent single resistor. An engineer needs the pair to be equivalent to R = (x + 1)(x + 4)/(5x + 11).
(a)
Show that 1/R = A/(x + 1) + B/(x + 4).
(1)
(b)
Express (5x + 11)/((x + 1)(x + 4)) in partial fractions, and hence find the values of A and B.
(3)
(c)
Find the two resistances when x = 5, and check that together they give the required value of R.
(2)
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Answer
Mark
Notes
Part (a)
1/R1 = A/(x + 1) and 1/R2 = B/(x + 4), and these add to give 1/R
B1
Part (b)
5x + 11 ≡ A(x + 4) + B(x + 1)
M1
x = −1: 6 = 3A, so A = 2
A1
x = −4: −9 = −3B, so B = 3
A1
Part (c)
(x + 1)/2 = 3 ohms and (x + 4)/3 = 3 ohms
B1
1/3 + 1/3 = 2/3, so R = 1.5 ohms, and (6 × 9)/36 = 1.5 ✓
B1
Exam tip: In context each partial fraction can mean something on its own: here each term is 1/(resistance) for one resistor of the pair.
Prove by contradiction that if a2 is even, then a is even.
(2)
(b)
Prove by contradiction that there are no integers a and b such that a2 − 4b = 2.
(4)
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Answer
Mark
Notes
Part (a)
assume a2 is even and a is odd, a = 2k + 1: a2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1
M1
this is odd, contradicting a2 even, so a is even
A1
Part (b)
assume integers a, b exist with a2 = 4b + 2
M1
4b + 2 = 2(2b + 1) is even, so a2 is even and, by (a), a is even
M1
a = 2c: 4c2 = 4b + 2, so 2c2 − 2b = 1
A1
the left side is even and 1 is odd — a contradiction, so no such integers exist
A1
Exam tip: When an equation has no integer solutions, make one side obviously even and the other obviously odd, or one a multiple of 4 and the other not.
The cubic denominator of the function below is given in expanded form, so it must be factorised before the function can be split up. h(x) = (4x2 + 5x − 21)/(x3 − 2x2 − 5x + 6)
(a)
Show that (x − 1) is a factor of x3 − 2x2 − 5x + 6, and hence factorise it completely.
(3)
(b)
Express h(x) in partial fractions.
(5)
Show mark scheme
Answer
Mark
Notes
Part (a)
substituting x = 1: 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor
f(x) = (kx2 + x + 3)/((2x − 1)2(x + 2)), where k is a constant. Given that f(x) ≡ A/(2x − 1) + B/(2x − 1)2 + C/(x + 2), where A, B and C are constants, and that C = 1,
(a)
find the value of k,
(2)
(b)
find the values of A and B,
(4)
(c)
verify your values by substituting x = 0 into your identity.
Figure 1 shows a square of side 4 units with corners at (0, 0) and (4, 4), divided into four smaller squares of side 2. Five points A, B, C, D and E with integer coordinates are marked; a point may lie on the boundary of a square.
(a)
Use Figure 1 to identify the two marked points that lie in the same small square, and find the distance between them.
(2)
(b)
Show that any two points in a square of side 2, including its boundary, are at most 2√2 units apart.
(1)
(c)
Prove by contradiction that, however five points are placed in the large square (including its boundary), at least two of them are at most 2√2 units apart.
(3)
(d)
Show that the result in part (c) is not true for four points.
(1)
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Answer
Mark
Notes
Part (a)
C(3, 0) and E(3, 1) both lie in the bottom right-hand small square
B1
CE = 1 unit
B1
Part (b)
their x-coordinates differ by at most 2 and their y-coordinates by at most 2, so the distance is at most √(22 + 22) = 2√2
B1
Part (c)
assume five points can be placed with every pair more than 2√2 units apart
M1
each point lies in one of the four small squares (a point on a dividing line is given to just one of them); five points and four squares, so two points lie in the same small square
M1
by part (b) these two are at most 2√2 apart, contradicting the assumption, so some pair is at most 2√2 apart
A1
Part (d)
e.g. the four corners (0, 0), (4, 0), (0, 4), (4, 4): every pair is 4 or 4√2 apart, more than 2√2
B1
Exam tip: With more objects than boxes, two objects must share a box. Choose the boxes so that “sharing a box” gives exactly the bound you need.
Figure 1 shows the value of a2 + b2 for each pair of integers a and b with 0 ≤ a ≤ 7 and 0 ≤ b ≤ 7. Each cell is shaded according to the remainder when its value is divided by 4.
(a)
Use Figure 1 to write down the value of a2 + b2 when a = 5 and b = 6, and state which remainder on division by 4 does not occur in the table.
(2)
(b)
Prove that, for every integer n, n2 leaves a remainder of 0 or 1 when it is divided by 4.
(2)
(c)
Prove by contradiction that there are no integers a, b and k for which a2 + b2 = 4k + 3.
(3)
(d)
Hence show that 2027 cannot be written as the sum of two square numbers.
(1)
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Answer
Mark
Notes
Part (a)
52 + 62 = 61
B1
remainder 3 does not occur
B1
Part (b)
n even, n = 2k: n2 = 4k2, remainder 0
M1
n odd, n = 2k + 1: n2 = 4(k2 + k) + 1, remainder 1
A1
Part (c)
assume there are integers a, b and k with a2 + b2 = 4k + 3
M1
by part (b), a2 + b2 leaves remainder 0 + 0, 0 + 1 or 1 + 1, that is 0, 1 or 2, on division by 4
M1
but 4k + 3 leaves remainder 3 — a contradiction, so there are no such integers
A1
Part (d)
2027 = 4 × 506 + 3, so by part (c) it is not equal to a2 + b2 for any integers a and b
B1
Common error: Treating the table as the proof. The table only covers 0 ≤ a, b ≤ 7; the cases n = 2k and n = 2k + 1 cover every integer.
Figure 1 shows the graph of y = f(x), where f(x) = (ax2 + bx + c)/((x + 1)(x − 2)), x ≠ −1, x ≠ 2, and a, b and c are constants. The dashed lines x = −1, x = 2 and y = 2 are asymptotes of the curve. The curve crosses the x-axis at (−4, 0) and (1, 0), and crosses the y-axis at (0, 4).
(a)
Explain why a = 2.
(1)
(b)
Find the values of b and c.
(2)
(c)
Express f(x) in the form A + B/(x + 1) + C/(x − 2), where A, B and C are constants.
(4)
(d)
The curve crosses its horizontal asymptote at the point K, shown in Figure 1. Use your answer to part (c) to find the coordinates of K.
(2)
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Answer
Mark
Notes
Part (a)
for large |x|, f(x) gets close to ax2/x2 = a; the horizontal asymptote is y = 2, so a = 2
B1
Part (b)
the numerator is zero at x = −4 and x = 1, so it is 2(x + 4)(x − 1)
M1
= 2x2 + 6x − 8, so b = 6, c = −8 (check: f(0) = −8/(−2) = 4 ✓)
A1
Part (c)
(x + 1)(x − 2) = x2 − x − 2 and 2x2 + 6x − 8 = 2(x2 − x − 2) + 8x − 4, so A = 2
M1
8x − 4 ≡ B(x − 2) + C(x + 1)
M1
x = −1: −12 = −3B, so B = 4
A1
x = 2: 12 = 3C, so C = 4: f(x) = 2 + 4/(x + 1) + 4/(x − 2)
Exam tip: When the top and bottom have the same degree, the constant A in the partial fractions is the horizontal asymptote — a quick check on your division.
Figure 1 shows two cuboids, P and Q, with edge lengths in centimetres, where x > 0. For each cuboid, S cm2 is its total surface area and V cm3 is its volume.
(a)
Show that, for cuboid P, S/V = (6x2 + 16x + 6)/(x(x + 1)(x + 3)).
(2)
(b)
Express S/V for cuboid P in partial fractions.
(3)
(c)
Show that, for any cuboid with edges of length p, q and r, S/V = 2/p + 2/q + 2/r.
(1)
(d)
Hence write down S/V for cuboid Q in partial fractions, and find the value of x for which S/V = 5/3 for cuboid Q.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
S = 2[x(x + 1) + x(x + 3) + (x + 1)(x + 3)]
M1
= 2(3x2 + 8x + 3) = 6x2 + 16x + 6, and V = x(x + 1)(x + 3), giving the result
Exam tip: If you start cuboid Q from the single fraction over (x + 1)²(x + 4), the term over (x + 1)² comes out as 0. Part (c) lets you write the answer down without that work.
Figure 1 shows the graph of y = q(x), where q(x) is a cubic polynomial in which the coefficient of x3 is 1. The curve crosses the x-axis at (−2, 0), touches it at (1, 0) and crosses the y-axis at (0, 2). f(x) = (3x2 + 6)/q(x), x ≠ −2, x ≠ 1.
(a)
Explain how Figure 1 shows that q(x) = (x + 2)(x − 1)2, and use the point (0, 2) to check this.
(2)
(b)
Write down the form of the partial fractions of f(x).
(1)
(c)
Find the values of the constants in your answer to part (b).
(4)
(d)
Verify your answer by finding f(0) from both forms of f(x).
(1)
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Answer
Mark
Notes
Part (a)
the curve crosses the axis at x = −2, giving a factor (x + 2), and touches it at x = 1, giving a repeated factor (x − 1)2; the coefficient of x3 is 1, so q(x) = (x + 2)(x − 1)2
Exam tip: A graph that touches the x-axis without crossing it marks a repeated root. That repeated factor needs two partial fractions, one over (x − 1) and one over (x − 1)².
Every odd integer can be written in the form 4k + 1 or 4k + 3, where k is an integer. This question leads to a proof that there are infinitely many primes of the form 4k + 3.
(a)
Show that the product of two integers of the form 4k + 1 is also of the form 4k + 1.
(2)
(b)
The primes 3, 7 and 11 are of the form 4k + 3. Find the value of N = 4 × 3 × 7 × 11 − 1, and show that N has a prime factor of the form 4k + 3 that is not 3, 7 or 11.
(3)
(c)
Prove by contradiction that there are infinitely many primes of the form 4k + 3.
(6)
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Answer
Mark
Notes
Part (a)
(4m + 1)(4n + 1) = 16mn + 4m + 4n + 1
M1
= 4(4mn + m + n) + 1, which is of the form 4k + 1
A1
Part (b)
N = 923
B1
923 = 13 × 71, with 13 and 71 both prime
M1
71 = 4 × 17 + 3 is of the form 4k + 3 and is not 3, 7 or 11
A1
Part (c)
assume there are finitely many, and that p1, p2, …, pn is the complete list of primes of the form 4k + 3
B1
let N = 4p1p2…pn − 1 = 4(p1…pn − 1) + 3, which is odd, greater than 1 and of the form 4k + 3
M1
N is odd, so each of its prime factors is odd, of the form 4k + 1 or 4k + 3
M1
if every prime factor were of the form 4k + 1, then by (a) N would be of the form 4k + 1; so N has a prime factor q of the form 4k + 3
A1
q is in the list, so q divides 4p1…pn; it also divides N, so it divides 4p1…pn − N = 1, which is impossible
M1
this contradiction shows there are infinitely many primes of the form 4k + 3
A1
Exam tip: Part (b) is a rehearsal for part (c): build the same kind of number, then argue in general that one of its prime factors must be missing from the list.
Exam tip: Differentiating a fraction in partial-fraction form is much quicker than using the quotient rule on the original: each term is just k(ax + b)ⁿ.
In this question you rewrite an improper fraction in a simpler form, then use that form to prove a property of the function and to solve an equation. g(x) = (2x3 − 2x2 + 5x + 1)/(x(x − 1)2), x ≠ 0, x ≠ 1.
(a)
Express g(x) in the form A + B/x + C/(x − 1) + D/(x − 1)2, where A, B, C and D are constants.
(5)
(b)
Prove by contradiction that there is no value of x > 1 for which g(x) ≤ 2.
(3)
(c)
Solve the equation g(x) = 2.
(4)
(d)
Explain why your answers to part (c) do not contradict part (b).
part (b) is only about x > 1; both solutions are negative, so they lie outside that range
B1
Exam tip: Once a function is in partial fractions, the sign of each term for a given range of x is easy to see — that is what makes the contradiction in part (b) so short.
A lattice point is a point whose coordinates are both integers. Figure 1 shows the triangle OAB, whose vertices O, A and B are lattice points, inside a dashed square. You may assume that √3 is irrational, and that the triangle with vertices (0, 0), (p, q) and (r, s) has area ½|ps − qr|.
(a)
Show that triangle OAB is isosceles but not equilateral.
(2)
(b)
By subtracting three right-angled triangles from the dashed square, find the area of triangle OAB. Check your answer using the area formula given.
(3)
(c)
Show that an equilateral triangle with sides of length d has area (√3/4)d2.
(2)
(d)
Prove by contradiction that there is no equilateral triangle whose three vertices are all lattice points.
(5)
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Answer
Mark
Notes
Part (a)
OA2 = 42 + 12 = 17 and OB2 = 12 + 42 = 17
B1
AB2 = 32 + 32 = 18 ≠ 17, so OA = OB ≠ AB: isosceles but not equilateral
B1
Part (b)
the three right-angled triangles have areas ½ × 4 × 1 = 2, ½ × 1 × 4 = 2 and ½ × 3 × 3 = 9/2
M1
area of OAB = 16 − 2 − 2 − 9/2 = 15/2
A1
formula: ½|4 × 4 − 1 × 1| = 15/2 ✓
B1
Part (c)
area = ½ × d × d × sin 60°
M1
= ½d2 × √3/2 = (√3/4)d2
A1
Part (d)
assume there is one; translate it so that one vertex is at O — the other vertices move to lattice points (p, q) and (r, s)
M1
the side length d satisfies d2 = p2 + q2, a positive integer
A1
the area ½|ps − qr| is an integer divided by 2, so it is rational
A1
by part (c), √3 = 4 × area ÷ d2, a quotient of rational numbers, so √3 would be rational
M1
this contradicts √3 being irrational, so no such triangle exists
A1
Exam tip: Work out one quantity in two ways — here the area, from the side length and from the coordinates. One way gives an irrational number and the other a rational one, and that is the contradiction.
Figure 1 shows the graph of y = x3 − 3x + 1, with five points on the curve labelled.
(a)
Use the labelled points to show that the equation x3 − 3x + 1 = 0 has a root in each of the intervals −2 < x < −1, 0 < x < 1 and 1 < x < 2.
(2)
(b)
Suppose that a root can be written as p/q, where p and q are integers with no common factor and q > 0. Show that p3 − 3pq2 + q3 = 0.
(1)
(c)
By considering whether p and q are odd or even, prove by contradiction that every root of x3 − 3x + 1 = 0 is irrational.
(6)
(d)
Hence explain why the decimal expansion of the root between 0 and 1 does not terminate.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the curve is continuous, and y changes sign from −1 to 3 on [−2, −1], from 1 to −1 on [0, 1] and from −1 to 3 on [1, 2]
M1
so there is a root in each of the three intervals
A1
Part (b)
substituting x = p/q and multiplying through by q3 gives p3 − 3pq2 + q3 = 0
B1
Part (c)
assume a root is rational, equal to p/q as in part (b)
M1
p and q have no common factor, so they are not both even
B1
p and q both odd: p3, 3pq2 and q3 are all odd, so the left side is odd − odd + odd, which is odd
A1
p even, q odd: even − even + odd, which is odd
A1
p odd, q even: odd − even + even, which is odd
A1
in every case the left side is odd, so it cannot equal 0 — a contradiction, so every root is irrational
A1
Part (d)
a terminating decimal can be written as a fraction with a power of 10 as denominator, so it is rational; the root is irrational by part (c)
B1
Common error: Missing a parity case. With p/q in its simplest form there are exactly three cases — odd/odd, even/odd and odd/even — and each must give an odd total.
Figure 1 shows the curve C with equation y = f(x), where f(x) = (7x2 − 17x + 12)/q(x) and q(x) is a cubic polynomial in which the coefficient of x3 is 1. The dashed lines are the only vertical asymptotes of C. The curve crosses the y-axis at (0, 3) and has a turning point at T(1, 1).
(a)
Explain how the behaviour of C near its asymptotes shows that q(x) = (x + 1)(x − 2)2.
(2)
(b)
Express f(x) in partial fractions.
(4)
(c)
Hence find f′(x).
(2)
(d)
Show that T(1, 1) lies on C and that the gradient of C at T is zero.
(2)
(e)
Find f″(x), and use it to show that T is a minimum point.
(2)
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Answer
Mark
Notes
Part (a)
near x = −1, y → −∞ on one side and y → +∞ on the other, so f(x) changes sign: a single factor (x + 1)
B1
near x = 2, y → +∞ on both sides, so there is no sign change: a repeated factor (x − 2)2; with leading coefficient 1, q(x) = (x + 1)(x − 2)2
Figure 1 shows the curve with equation y = f(x), where f(x) = (x3 + kx2 + 6x + 2)/(x(x + 2)), x ≠ −2, x ≠ 0, and k is a constant. The dashed line L, which passes through the marked points (−3, −2) and (3, 4), is an asymptote of the curve. The curve crosses L at the point P.
(a)
Use algebraic division to show that f(x) = x + k − 2 + ((10 − 2k)x + 2)/(x(x + 2)).
(3)
(b)
Find an equation of L, and hence find the value of k.
(2)
(c)
Express f(x) in the form x + D + A/x + B/(x + 2), where A, B and D are constants.
(3)
(d)
Find the exact coordinates of P.
(2)
(e)
Prove by contradiction that there is no point on the curve at which the tangent is parallel to L.
(3)
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Answer
Mark
Notes
Part (a)
x(x + 2) = x2 + 2x; first term of the quotient x, leaving (k − 2)x2 + 6x + 2
M1
second term k − 2, subtracting (k − 2)(x2 + 2x)
M1
remainder (6 − 2k + 4)x + 2 = (10 − 2k)x + 2, giving the result
A1
Part (b)
gradient = (4 − (−2))/(3 − (−3)) = 1, so L is y = x + 1
B1
as |x| → ∞ the fraction → 0, so the asymptote is y = x + k − 2; k − 2 = 1, so k = 3
A1
Part (c)
with k = 3 the remainder is 4x + 2 ≡ A(x + 2) + Bx
M1
x = 0: 2 = 2A, so A = 1; x = −2: −6 = −2B, so B = 3
A1
f(x) = x + 1 + 1/x + 3/(x + 2)
A1
Part (d)
on L, 1/x + 3/(x + 2) = 0, so (x + 2) + 3x = 0
M1
x = −½, y = ½: P is (−½, ½)
A1
Part (e)
f′(x) = 1 − 1/x2 − 3/(x + 2)2
B1
assume that at some point the tangent is parallel to L, so f′(x) = 1, giving 1/x2 + 3/(x + 2)2 = 0
M1
for every x in the domain both terms are positive, so the sum is positive and cannot be 0 — a contradiction, so no tangent is parallel to L
A1
Exam tip: For an improper fraction the quotient from the division gives the sloping asymptote, and the proper part shows where the curve lies relative to it.
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