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A-Level · Pure Mathematics

P1 Pure: Proof

EDEXCEL 9MA0 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question involves proof by deduction using algebraic representations of odd and even integers.

(a)Prove that the sum of any two odd integers is always even.(3)

(b)Prove that the square of any odd integer is always odd.(3)

(c)Prove that the sum of any two consecutive integers is always odd.(2)

(Total for Question 1 is 8 marks)

This question explores proof by exhaustion and its limitations, using the statement: for every positive integer n, n³+n is even.

(a)Prove by exhaustion that n³+n is even for every integer n such that 1 ≤ n ≤ 4.(4)

(b)Prove, by considering the cases n even and n odd, that n³+n is even for every positive integer n.(4)

(c)Explain why the working in part (a) alone does not prove that n³+n is even for every positive integer n.(2)

(Total for Question 2 is 10 marks)

In each part, find a counter-example to disprove the given statement.

(a)Disprove the statement: "For every real number x, x² ≥ x."(2)

(b)Disprove the statement: "For every integer n, if n is prime then n is odd."(2)

(c)Disprove the statement: "For all real numbers a and b, (a+b)² = a²+b^2."(2)

(d)Disprove the statement: "For 0 ≤ θ ≤ 360 (degrees), sin(θ) ≤ cos(θ) implies θ ≤ 45."(2)

(Total for Question 3 is 8 marks)

Priya defines f(n) = n² - n + 11 for positive integers n. She calculates f(1)=11, f(2)=13, f(3)=17, f(4)=23 and f(5)=31, all of which are prime, and concludes: "f(n) is prime for every positive integer n."

(a)Show that f(11) = 121, and explain why this disproves Priya's conclusion.(3)

(b)Explain why testing values of n from 1 to 5, as Priya did, could never have proved her conclusion, even before a counter-example was found.(3)

(Total for Question 4 is 6 marks)

This question involves proof by deduction using cases.

(a)Prove that the product of any two consecutive integers, n(n+1), is always even.(3)

(b)Hence prove that n²+n+1 is always odd, for any integer n.(3)

(Total for Question 5 is 6 marks)

This question involves an identity established using the compound angle formulae.

(a)Show that sin(θ+60) + sin(θ-60) = sin(θ), where θ is measured in degrees.(4)

(b)Hence solve, for 0 ≤ θ < 360 (degrees), the equation sin(θ+60) + sin(θ-60) = 0.5.(3)

(Total for Question 6 is 7 marks)

This question involves proof using algebraic inequalities.

(a)Prove that x²+y² ≥ 2xy for all real numbers x and y.(3)

(b)Hence, by substituting x=a and y=b into the result of part (a), prove that a/b + b/a ≥ 2 for all real a, b > 0.(3)

(Total for Question 7 is 6 marks)

f(x) = x³ + 6x² + 15x - 2 for all real x.

(a)Find f'(x).(2)

(b)Show that f'(x) can be written in the form 3[(x+2)²+1], and hence show that f'(x) > 0 for all real values of x.(3)

(c)State the conclusion that follows about the function f.(1)

(Total for Question 8 is 6 marks)

This question proves that n³-n is divisible by 6 for every positive integer n.

(a)Fully factorise n³-n.(2)

(b)Explain why (n-1)n(n+1) is always divisible by 2.(2)

(c)Explain why (n-1)n(n+1) is always divisible by 3.(2)

(d)Hence prove that n³-n is divisible by 6 for every positive integer n.(2)

(Total for Question 9 is 8 marks)

This question builds a proof that √2 is irrational.

(a)Prove that if n² is even, where n is an integer, then n is even.(2)

(b)Prove, by contradiction, that √2 is irrational.(7)

(Total for Question 10 is 9 marks)

This question proves, by contradiction, that there are infinitely many prime numbers.

(a)State the assumption made at the start of a proof by contradiction that there are infinitely many primes.(1)

(b)Let N = p1 x p2 x ... x pk + 1. Explain why none of p1, p2, ..., pk divides N.(2)

(c)Explain why this leads to a contradiction.(2)

(d)State the conclusion that follows from this contradiction.(1)

(e)Illustrate the argument using the first three primes, 2, 3 and 5. Calculate N = 2x3x5+1, verify that N is not divisible by 2, 3 or 5, and identify the new prime this reveals.(3)

(f)Calculate N for the first six primes, 2, 3, 5, 7, 11 and 13. Given that N = 59 x 509, explain why this shows the argument in parts (a) to (d) still works even when N itself is not prime.(2)

(Total for Question 11 is 11 marks)

Mark scheme · P1 Pure: Proof

Question 1

(a) B1 represents two odd integers as 2m+1 and 2n+1, where m and n are integers, oe
(a) M1 forms the sum (2m+1)+(2n+1) and simplifies to 2m+2n+2
(a) A1 writes the sum as 2(m+n+1), a multiple of 2, hence even, cso
(a) Answer: Proof: sum = 2(m+n+1), which is even
(b) B1 represents an odd integer as 2k+1, where k is an integer, oe
(b) M1 expands (2k+1)² to 4k²+4k+1
(b) A1 writes 4k²+4k+1 as 2(2k²+2k)+1, a multiple of 2 plus 1, hence odd, cso
(b) Answer: Proof: (2k+1)² = 2(2k²+2k)+1, which is odd
(c) M1 lets the consecutive integers be n and n+1 and forms the sum n+(n+1)=2n+1, oe
(c) A1 states 2n is a multiple of 2, so 2n+1 is odd for every integer n, cso
(c) Answer: Proof: sum = 2n+1, which is odd

Question 2

(a) B1 n=1: 1³+1=2, even
(a) B1 n=2: 2³+2=10, even
(a) B1 n=3: 3³+3=30, even
(a) B1 n=4: 4³+4=68, even, and all four cases 1≤n≤4 have been checked, cao
(a) Answer: n=1,2,3,4 give 2, 10, 30, 68, all even
(b) M1 case n=2k: substitutes to get n³+n=8k³+2k
(b) A1 writes 8k³+2k=2(4k³+k), a multiple of 2, hence even
(b) M1 case n=2k+1: substitutes and factorises n³+n=(2k+1)[(2k+1)²+1]=(2k+1)(4k²+4k+2)
(b) A1 writes this as 2(2k+1)(2k²+2k+1), a multiple of 2, hence even; since every integer is even or odd, this proves the result for all positive integers, cso
(b) Answer: Proof: n³+n is even for all positive integers n
(c) B1 states that part (a) only checks a finite number of cases (n=1 to 4), but the claim is about all positive integers, an infinite set
(c) B1 states that a proof by exhaustion must check every case in the set, which is impossible for an infinite set, so a general argument (as in part (b)) is needed instead
(c) Answer: Checking 4 cases cannot establish a result for infinitely many integers

Question 3

(a) B1 chooses a valid value of x with 0<x<1, e.g. x=0.5, oe
(a) A1 shows x²<x for this value, e.g. 0.5²=0.25<0.5, so the statement is false, cao
(a) Answer: x=0.5 gives 0.25 < 0.5, so the statement is false
(b) B1 states n=2
(b) A1 justifies that 2 is prime and even, contradicting the statement, cao
(b) Answer: n=2 is prime and even, so the statement is false
(c) B1 chooses valid values with a and b both non-zero, e.g. a=1, b=1, oe
(c) A1 shows (1+1)²=4 but 1²+1²=2, and 4 is not equal to 2, so the statement is false, cao
(c) Answer: a=1, b=1 gives 4 not equal to 2, so the statement is false
(d) B1 chooses a valid value of θ with θ>45 in the given range, e.g. θ=350, oe
(d) A1 shows sin(350) is awrt -0.17 and cos(350) is awrt 0.98, so sin(θ)≤cos(θ) holds, yet θ=350 is not ≤45, so the statement is false, cao
(d) Answer: θ=350 gives sin(θ) awrt -0.17 ≤ cos(θ) awrt 0.98, but θ>45, so the statement is false

Question 4

(a) M1 substitutes n=11 into f(n): 11²-11+11
(a) A1 simplifies to 121, cso
(a) B1 states 121=11x11, which is not prime, so n=11 is a counter-example that disproves the conclusion, cao
(a) Answer: f(11)=121=11x11, not prime, so the conclusion is false
(b) B1 states that Priya's conclusion is a claim about every positive integer n, an infinite set, but she only checked 5 cases
(b) B1 states that finitely many successful checks (proof by exhaustion of only some cases) cannot establish a universal statement over an infinite set
(b) B1 states that a single valid counter-example, such as n=11, is sufficient to disprove a universal claim, whereas no finite number of confirming cases could ever prove it, oe
(b) Answer: Finitely many confirming cases can never prove a claim about all positive integers, but one counter-example disproves it

Question 5

(a) M1 considers the case n=2k (n even) and writes n(n+1)=2k(2k+1), a multiple of 2
(a) A1 considers the case n=2k+1 (n odd) and writes n(n+1)=(2k+1)(2k+2)=2(2k+1)(k+1), a multiple of 2
(a) A1 since every integer n is even or odd, and the product is a multiple of 2 in both cases, n(n+1) is always even, cso
(a) Answer: Proof: n(n+1) is always even
(b) M1 recognises n²+n=n(n+1), oe
(b) A1 uses the result from part (a) that n(n+1) is even, ft
(b) A1 states that even + 1 is odd, so n²+n+1 is always odd, cso
(b) Answer: Proof: n²+n+1 = n(n+1)+1 = even+1 = odd

Question 6

(a) M1 expands sin(θ+60) = sin(θ)cos60 + cos(θ)sin60
(a) M1 expands sin(θ-60) = sin(θ)cos60 - cos(θ)sin60
(a) A1 adds the two expansions so the cos(θ)sin60 terms cancel, leaving 2sin(θ)cos60
(a) A1 substitutes the exact value cos60=1/2 to obtain 2sin(θ)x(1/2)=sin(θ), cso
(a) Answer: Proof: sin(θ+60)+sin(θ-60) = sin(θ)
(b) M1 uses the identity from part (a) to rewrite the equation as sin(θ)=0.5, ft
(b) A1 θ=30
(b) A1 θ=150, cao, both values and no extras in range
(b) Answer: θ = 30 or θ = 150

Question 7

(a) B1 states (x-y)² ≥ 0 for all real x and y, since a square cannot be negative
(a) M1 expands (x-y)² to x²-2xy+y²
(a) A1 rearranges x²-2xy+y²≥0 to x²+y²≥2xy, cso
(a) Answer: Proof: since (x-y)² ≥ 0, x²+y² ≥ 2xy
(b) M1 substitutes x=a, y=b into part (a) to obtain a²+b²≥2ab, ft
(b) B1 states ab>0 since a>0 and b>0, so dividing both sides of the inequality by ab does not change its direction
(b) A1 divides both sides by ab to obtain a/b+b/a≥2, cso
(b) Answer: Proof: dividing a²+b²≥2ab by ab>0 gives a/b+b/a≥2

Question 8

(a) M1 differentiates term by term
(a) A1 f'(x)=3x²+12x+15, cao
(a) Answer: f'(x) = 3x² + 12x + 15
(b) M1 takes out the factor of 3: f'(x)=3(x²+4x+5)
(b) A1 completes the square: x²+4x+5=(x+2)²+1, giving f'(x)=3[(x+2)²+1]
(b) A1 states (x+2)²≥0 for all real x, so (x+2)²+1≥1, hence f'(x)≥3>0 for all real x, cso
(b) Answer: f'(x) = 3[(x+2)²+1] ≥ 3 > 0 for all real x
(c) B1 states that since f'(x)>0 for all real x, f is an increasing function for all real values of x, cso, ft
(c) Answer: f(x) is an increasing function for all real x

Question 9

(a) M1 takes out a factor of n: n³-n=n(n²-1)
(a) A1 factorises the difference of squares to give n(n-1)(n+1), cao
(a) Answer: n³-n = (n-1)n(n+1)
(b) B1 states that n and n+1 are consecutive integers, so exactly one of them is even, oe
(b) B1 states that since the product (n-1)n(n+1) contains an even factor, the whole product is divisible by 2, cso
(b) Answer: Divisible by 2 because n and n+1 are consecutive, so one is even
(c) B1 states that n-1, n and n+1 are three consecutive integers, and among any three consecutive integers exactly one is a multiple of 3, oe
(c) B1 states that since the product contains a multiple of 3, the whole product is divisible by 3, cso
(c) Answer: Divisible by 3 because three consecutive integers include a multiple of 3
(d) M1 states that 2 and 3 share no common factor other than 1, ft from (b) and (c)
(d) A1 since (n-1)n(n+1) is divisible by both 2 and 3, it is divisible by 2x3=6 for every positive integer n, cso
(d) Answer: Proof complete: n³-n is divisible by 6 for every positive integer n

Question 10

(a) M1 assumes n is odd, writes n=2k+1 for integer k, and expands n²=(2k+1)²=4k²+4k+1
(a) A1 writes n²=2(2k²+2k)+1, which is odd; since "n odd implies n² odd" has been shown, its contrapositive "n² even implies n even" must also be true, cso
(a) Answer: Proof: if n² is even then n is even
(b) B1 assumes, for contradiction, that √2 is rational, so √2=p/q where p and q are integers with no common factor (q not equal 0), i.e. p/q is in its lowest terms
(b) M1 squares both sides to obtain p²=2q²
(b) A1 deduces p² is even, so by part (a), p is even; writes p=2m for integer m, ft
(b) dM1 substitutes p=2m into p²=2q² to get 4m²=2q², so q²=2m², dependent on the previous M1
(b) A1 deduces q² is even, so by part (a), q is even, ft
(b) B1 states that p and q are both even, so they share a common factor of 2, contradicting the assumption that p/q has no common factor
(b) B1 concludes that the original assumption is false, hence √2 is irrational, cso
(b) Answer: Proof by contradiction: √2 is irrational

Question 11

(a) B1 states, for contradiction, that there are only finitely many primes, p1, p2, ..., pk, and that this is a complete list of every prime, oe
(a) Answer: Assume there is a complete finite list of primes p1, p2, ..., pk
(b) B1 states that each π divides the product p1xp2x...xpk exactly, so N leaves remainder 1 when divided by any π, oe
(b) B1 states that this holds for every i from 1 to k, so no prime in the list divides N, cso
(b) Answer: N leaves remainder 1 on division by each π, so no π divides N
(c) B1 states that N>1, so N must have at least one prime factor, since every integer greater than 1 has a prime factor
(c) B1 states that this prime factor cannot be any of p1, ..., pk (by part (b)), so it is a prime missing from the original list, contradicting the assumption that the list was complete, cso, ft
(c) Answer: N has a prime factor not in the original list, contradicting completeness of the list
(d) B1 states that the assumption of finitely many primes is false, so there are infinitely many primes, cso, ft
(d) Answer: There are infinitely many primes
(e) M1 calculates N=2x3x5+1=31
(e) A1 verifies 31 is not divisible by 2, 3 or 5 (e.g. remainder 1 in each case)
(e) A1 identifies that 31 is itself prime, a prime not in the original list {2, 3, 5}, cao
(e) Answer: N=31, which is prime and not in {2, 3, 5}
(f) B1 calculates N=2x3x5x7x11x13+1=30031, and notes 30031=59x509 is composite
(f) B1 states that neither 59 nor 509 is in the original list {2,3,5,7,11,13}, so N still has a prime factor missing from the list, confirming the argument holds whether or not N is prime, cso
(f) Answer: N=30031=59x509; both 59 and 509 are new primes not in the original list

Pure: Proof - Worksheets, Questions and Revision