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Proof

8 questions, 36 marks, every one with its mark scheme.

Or open them one at a time, as you finish each question.

Question 1

5 marks
1(a)[3 marks]

Prove by exhaustion that, for all integers n with 10 ≤ n ≤ 14, n2 + 2 is not a multiple of 5.

Mark scheme
  • Evaluates n2 + 2 for n = 10, 11, 12, 13 and 14M1
  • 102, 123, 146, 171, 198, and states none ends in 0 or 5, so none is a multiple of 5A1
  • So n2 + 2 is not a multiple of 5 for all integers 10 ≤ n ≤ 14A1

Do not accept: Testing only some of the values from 10 to 14; Listing the values with no concluding statement

1(b)[2 marks]

Prove that the statement

“n2 + 4 is not a multiple of 5 for any integer n”

is untrue.

Mark scheme
  • Uses a value of n that disproves the statement, for example n = 1M1
  • 12 + 4 = 5, which is a multiple of 5, so the statement is untrueA1

Question 2

4 marks
2(a)[4 marks]

Prove, using algebra, that (n + 3)2 − n is odd for all n ∈ ℕ.

Mark scheme
  • When n is even, n = 2k: (2k + 3)2 − 2k = 4k2 + 10k + 9M1
  • = 2(2k2 + 5k + 4) + 1, which is oddA1
  • When n is odd, n = 2k + 1: (2k + 4)2 − (2k + 1) = 4k2 + 14k + 15M1
  • = 2(2k2 + 7k + 7) + 1, which is odd; hence (n + 3)2 − n is odd for all n ∈ ℕA1cso

Do not accept: n = 2n; n ∈ ℝ; Ending on 2(2k2 + 7k + 7) + 1 with no concluding statement

Question 3

4 marks

f(x) = x2 − 6x + 14

3(a)[2 marks]

By completing the square, find the constants a and b such that f(x) ≡ (x + a)2 + b.

Mark scheme
  • f(x) = (x − 3)2 ± ...M1
  • f(x) = (x − 3)2 + 5, so a = −3, b = 5A1
3(b)[2 marks]

Hence prove that f(x) > 0 for all real values of x.

Mark scheme
  • (x − 3)2 ≥ 0 for all real xM1
  • so (x − 3)2 + 5 ≥ 5 > 0, hence f(x) > 0 for all real xA1ft

Question 4

7 marks
4(a)[4 marks]

Prove, using algebra, that n2 + n is even for all integers n.

Mark scheme
  • When n is even, n = 2k: (2k)2 + 2k = 4k2 + 2kM1
  • = 2(2k2 + k), which is evenA1
  • When n is odd, n = 2k + 1: (2k + 1)2 + 2k + 1 = 4k2 + 6k + 2M1
  • = 2(2k2 + 3k + 1), which is even; hence n2 + n is even for all integers nA1cso

Do not accept: n = 2n; Showing the result for particular values of n only

4(b)[3 marks]

Hence prove that the square of any odd integer is one more than a multiple of 8.

Mark scheme
  • Writes the odd integer as 2m + 1: (2m + 1)2 = 4m2 + 4m + 1 = 4(m2 + m) + 1M1
  • Uses part (a): m2 + m = 2j for some integer j, so (2m + 1)2 = 8j + 1dM1
  • Hence the square of any odd integer is one more than a multiple of 8A1cso

Question 5

4 marks
5(a)[4 marks]

Given that n is an integer, use algebra to prove by contradiction that if n2 + 4n + 3 is even, then n is odd.

Mark scheme
  • Assume there exists an integer n such that n2 + 4n + 3 is even and n is evenB1
  • Let n = 2p, where p is an integer: (2p)2 + 4(2p) + 3 = 4p2 + 8p + 3M1
  • = 2(2p2 + 4p + 1) + 1, which is oddA1
  • This contradicts n2 + 4n + 3 being even, so if n2 + 4n + 3 is even, then n is oddA1cso

Do not accept: Assuming n is odd at the start; Ending on 2(2p2 + 4p + 1) + 1 with no contradiction stated

Question 6

5 marks
6(a)[5 marks]

Use proof by contradiction to prove that there are no integers m and n such that

m2 − 4n = 7

Mark scheme
  • Assume there exist integers m and n such that m2 − 4n = 7B1
  • m2 = 4n + 7 = 2(2n + 3) + 1, which is odd, so m is oddB1
  • Lets m = 2k + 1, where k is an integer: 4k2 + 4k + 1 − 4n = 7M1
  • 2(k2 + k − n) = 3: the left side is even and 3 is oddA1
  • This is a contradiction, so there are no integers m and n such that m2 − 4n = 7A1cso

Question 7

4 marks
7(a)[4 marks]

Use proof by contradiction to prove that, for all real values of x,

cos2 x + 4 sin x ≤ 4

Mark scheme
  • Assume there exists a real x such that cos2 x + 4 sin x > 4B1
  • Uses cos2 x = 1 − sin2 x to reach sin2 x − 4 sin x + 3 < 0M1
  • (sin x − 1)(sin x − 3) < 0, so 1 < sin x < 3A1
  • This contradicts −1 ≤ sin x ≤ 1, so cos2 x + 4 sin x ≤ 4 for all real xA1cso

Do not accept: Checking particular values of x only

Question 8

3 marks
8(a)[3 marks]

Given that √3 is irrational, use proof by contradiction to prove that 5 − 2√3 is irrational.

Mark scheme
  • Assume that 5 − 2√3 is a rational number, so 5 − 2√3 = pq, where p and q are integers, q ≠ 0B1
  • Rearranges: √3 = 5q − p2qM1
  • 5q − p and 2q are integers, so √3 is rational. This contradiction implies that 5 − 2√3 is irrationalA1cso
All Edexcel A-level Mathematics question setsThe notes for this topic