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8.3 Proof

EDEXCEL 1MA1 · Calculator allowed · about 80 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

1

Prove that the sum of three consecutive integers is always a multiple of 3.

(Total for Question 1 is 3 marks)

2

Show that (n + 1)² - n² = 2n + 1

(Total for Question 2 is 2 marks)

3

Prove algebraically that the product of any two even numbers is always a multiple of 4.

(Total for Question 3 is 2 marks)

4

Priya says, 'When you square a number, the answer is always bigger than the number you started with.' Show that Priya is not correct.

(Total for Question 4 is 2 marks)

5

Prove algebraically that the sum of any two consecutive odd numbers is a multiple of 4.

(Total for Question 5 is 3 marks)

6

The diagram shows triangle ABC, where AB = AC. D is a point on BC such that angle BAD = angle CAD. Prove that BD = DC. Diagram: triangle ABC with a line from A to D on BC, marks showing AB = AC and angle BAD = angle CAD.

Diagram NOT accurately drawn

(Total for Question 6 is 4 marks)

7

Prove algebraically that (2n + 1)² - (2n - 1)² is always a multiple of 8, for any positive integer n.

(Total for Question 7 is 3 marks)

8

Prove that n² - n is an even number for every integer value of n.

(Total for Question 8 is 3 marks)

9

The diagram shows triangle PQR, where PQ = PR. Angle QPR = (4x + 20) degrees. S is a point on PR extended beyond R, so that angle QRS = (5x + 10) degrees. Prove that x = 30. Diagram: isosceles triangle PQR with PR extended to S beyond R.

Diagram NOT accurately drawn

(Total for Question 9 is 4 marks)

10

Prove that (3 + √2)(3 - √2) is a rational number.

(Total for Question 10 is 3 marks)

11

Prove algebraically that the difference between the squares of any two consecutive even numbers is always a multiple of 4.

(Total for Question 11 is 3 marks)

12

OAB is a triangle, where OA = vector a and OB = vector b. M is the midpoint of OA and N is the midpoint of OB. Prove that MN is parallel to AB, and that MN = (1/2) AB. Diagram: triangle OAB with M the midpoint of OA and N the midpoint of OB.

Diagram NOT accurately drawn

(Total for Question 12 is 5 marks)

13

Prove that the recurring decimal 0.181818... (0.18 recurring) is equal to 2/11.

(Total for Question 13 is 4 marks)

14

Kwame says, 'For every positive integer value of n, the expression n² + n + 41 gives a prime number.' Show, by finding a counter-example, that Kwame is not correct.

(Total for Question 14 is 3 marks)

15

Prove algebraically that the sum of any four consecutive integers is always even, but is never a multiple of 4.

(Total for Question 15 is 4 marks)

16

The diagram shows a circle with centre O. A, B and C are points on the circumference. Prove that angle AOC = 2 x angle ABC (the angle at the centre is twice the angle at the circumference subtended by the same arc). Diagram: circle centre O with points A, B, C on the circumference, radii OA, OB, OC drawn, and chords AB, BC, AC drawn.

Diagram NOT accurately drawn

(Total for Question 16 is 5 marks)

17

Prove algebraically that the sum of the squares of any two consecutive odd numbers is always even, but is never a multiple of 4.

(Total for Question 17 is 4 marks)

18

Prove, by exhaustion, that n² + n + 1 is an odd number for every integer n from 1 to 5 inclusive.

(Total for Question 18 is 4 marks)

19

ABCD is a trapezium with AB parallel to DC, and DC = 3 x AB. The diagonals AC and BD intersect at X. Relative to A, the position vector of B is b and the position vector of D is d. Diagram: trapezium ABCD with AB parallel to DC, DC three times the length of AB, and diagonals AC and BD meeting at X.

Diagram NOT accurately drawn

(a)Show that the position vector of C (relative to A) is d + 3b.(2)

(b)Prove that AX : XC = 1 : 3.(3)

(Total for Question 19 is 5 marks)

20

Prove, by completing the square, that x² - 6x + 10 > 0 for all real values of x.

(Total for Question 20 is 4 marks)

Mark scheme · 8.3 Proof

Question 1

M1 Consecutive integers written algebraically as n, n+1, n+2 (oe)
M1 Sum simplified to 3n + 3 (oe)
A1 Factorised to 3(n+1) with conclusion that this is a multiple of 3 (cso)
Answer: n + (n+1) + (n+2) = 3n + 3 = 3(n+1), which is a multiple of 3 for any integer n.

Question 2

M1 (n+1)² expanded correctly to n² + 2n + 1
A1 n² + 2n + 1 - n² simplified to 2n + 1 (cso, answer given)
Answer: (n+1)² - n² = n² + 2n + 1 - n² = 2n + 1, as required.

Question 3

M1 Even numbers written as 2n and 2m, product formed as 2n x 2m = 4nm (oe)
A1 Conclusion that 4nm is a multiple of 4 for integers n and m (cso)
Answer: 2n x 2m = 4nm, which is a multiple of 4 for all integers n and m.

Question 4

B1 A valid counter-example chosen, e.g. n = 0.2 (any value with 0 < n < 1, or n = 0, or n = 1), with the square correctly evaluated
B1 Conclusion that the square is smaller than (or equal to) the original number, so Priya's statement is not correct (cao, any valid counter-example)
Answer: n = 0.2 gives 0.2² = 0.04, and 0.04 < 0.2, so Priya's statement is false.

Question 5

M1 Consecutive odd numbers written as 2n + 1 and 2n + 3 (oe, e.g. 2n - 1 and 2n + 1)
M1 Sum simplified to 4n + 4 (oe)
A1 Factorised to 4(n + 1) with conclusion that this is a multiple of 4 (cso)
Answer: (2n+1) + (2n+3) = 4n + 4 = 4(n+1), which is a multiple of 4 for any integer n.

Question 6

B1 AB = AC (given) and angle BAD = angle CAD (given) stated with reasons
B1 AD = AD (common side to both triangles) stated with reason
dM1 Triangles ABD and ACD are congruent, reason SAS (two sides and the included angle equal)
A1 BD = DC, since corresponding sides in congruent triangles are equal (cso)
Answer: Triangles ABD and ACD are congruent (SAS), so BD = DC.

Question 7

M1 Both squares expanded correctly: 4n² + 4n + 1 and 4n² - 4n + 1
M1 Subtraction simplified to 8n (oe)
A1 Conclusion that 8n is a multiple of 8 for any positive integer n (cso)
Answer: (2n+1)² - (2n-1)² = (4n²+4n+1) - (4n²-4n+1) = 8n, a multiple of 8.

Question 8

M1 n² - n factorised to n(n - 1)
M1 n and n - 1 identified as two consecutive integers
A1 Conclusion that one of any two consecutive integers must be even, so their product n(n-1) is always even (cso)
Answer: n² - n = n(n-1), the product of two consecutive integers. Exactly one of n and n-1 is even, so the product is always even.

Question 9

M1 Base angles PQR = PRQ = (180 - (4x+20))/2 = 80 - 2x (oe), using the angle sum of a triangle and isosceles property
M1 Exterior angle QRS = 180 - (80 - 2x) = 100 + 2x (oe), using angles on a straight line
M1 Equation 100 + 2x = 5x + 10 formed and rearranged
A1 x = 30 (cso, answer given)
Answer: x = 30

Question 10

M1 Expansion using difference of two squares: 3² - (√2)²
A1 Simplified to 9 - 2 = 7 (cao)
B1 Conclusion that 7 is a rational number (an integer, so rational), cso
Answer: (3+√2)(3-√2) = 9 - 2 = 7, which is rational.

Question 11

M1 Consecutive even numbers written as 2n and 2n + 2 (oe)
M1 (2n+2)² - (2n)² expanded and simplified to 8n + 4 (oe)
A1 Factorised to 4(2n+1) with conclusion that this is a multiple of 4 (cso)
Answer: (2n+2)² - (2n)² = 8n + 4 = 4(2n+1), a multiple of 4.

Question 12

M1 OM = (1/2)a stated
M1 ON = (1/2)b stated
M1 MN = ON - OM = (1/2)b - (1/2)a (oe)
A1 AB = OB - OA = b - a, so MN = (1/2)(b - a) = (1/2)AB (oe)
B1 Conclusion: since MN is a scalar multiple of AB, MN is parallel to AB, and MN = (1/2)AB (cso)
Answer: MN = (1/2)(b - a) = (1/2)AB, so MN is parallel to AB and half its length.

Question 13

M1 Let x = 0.181818... and form 100x = 18.181818... (oe)
M1 Subtraction 100x - x = 18.181818... - 0.181818... to give 99x = 18
A1 x = 18/99 (oe)
A1 Simplified fully to x = 2/11 (cao)
Answer: x = 18/99 = 2/11

Question 14

B1 A valid counter-example value chosen, e.g. n = 40 (or n = 41)
M1 Expression correctly evaluated, e.g. 40² + 40 + 41 = 1681
A1 1681 shown to be composite (e.g. 1681 = 41 x 41), so not prime, hence Kwame's statement is false (cao)
Answer: n = 40 gives 1681 = 41 x 41, which is not prime, so Kwame is not correct.

Question 15

M1 Four consecutive integers n, n+1, n+2, n+3 summed and simplified to 4n + 6
A1 Factorised to 2(2n + 3), showing the sum is even (has factor 2)
M1 2n + 3 identified as odd (2n is even, plus 3 is odd)
A1 Conclusion: 2(2n+3) is even but, since 2n+3 is odd, it cannot be a multiple of 4 (cso)
Answer: n + (n+1) + (n+2) + (n+3) = 4n + 6 = 2(2n+3). This is even, but since 2n+3 is odd, the sum is never a multiple of 4.

Question 16

B1 OA = OB = OC (radii of the circle), so triangles OAB and OBC are isosceles
B1 Base angles labelled: angle OAB = angle OBA = x, and angle OCB = angle OBC = y
M1 Angle AOB = 180 - 2x and angle BOC = 180 - 2y (angle sum of a triangle = 180 degrees)
dM1 Angle AOC = 360 - (180 - 2x) - (180 - 2y) = 2x + 2y (angles around a point sum to 360 degrees)
A1 Angle ABC = x + y, so angle AOC = 2(x+y) = 2 x angle ABC (cso)
Answer: Angle AOC = 2x + 2y = 2(x + y) = 2 x angle ABC.

Question 17

M1 Consecutive odd numbers written as 2n + 1 and 2n + 3 (oe)
M1 Squares expanded and summed correctly to 8n² + 16n + 10 (oe)
A1 Factorised to 2(4n² + 8n + 5), showing the sum is even (has factor 2)
A1 Explanation that 4n² + 8n + 5 is odd (since 4n² and 8n are even, plus 5 is odd), so the expression is 2 x (an odd number) and cannot be a multiple of 4 (cso)
Answer: (2n+1)² + (2n+3)² = 8n² + 16n + 10 = 2(4n²+8n+5). This is even, but since 4n²+8n+5 is odd, it is never a multiple of 4.

Question 18

M1 n² + n + 1 evaluated correctly for at least three values of n in the range 1 to 5
A1 All five values correct: n=1 gives 3, n=2 gives 7, n=3 gives 13, n=4 gives 21, n=5 gives 31
B1 Each of the five values identified as odd
B1 Conclusion: since every integer from 1 to 5 has been checked (exhausted) and each gives an odd result, the statement is proved true for this range (cso)
Answer: Values are 3, 7, 13, 21, 31, all odd, so n² + n + 1 is odd for every integer n from 1 to 5.

Question 19

(a) M1 DC identified as parallel to AB and 3 times as long, so DC = 3b (vector)
(a) A1 AC = AD + DC = d + 3b (cso, answer given)
(a) Answer: AC = d + 3b
(b) M1 X on AC written as AX = t(d + 3b), and X on BD written as AX = b + s(d - b), for some scalars t and s
(b) M1 Coefficients of the independent vectors d and b equated: t = s (coefficient of d) and 3t = 1 - s (coefficient of b), and solved simultaneously to give t = 1/4
(b) A1 Conclusion that AX = (1/4)AC, so AX : XC = 1 : 3 (cso)
(b) Answer: AX : XC = 1 : 3

Question 20

M1 x² - 6x + 10 written as (x - 3)² - 9 + 10 (oe)
A1 Simplified correctly to (x - 3)² + 1
M1 Statement that (x - 3)² ≥ 0 for all real values of x
A1 Conclusion that (x-3)² + 1 ≥ 1 > 0 for all real x, so x² - 6x + 10 > 0 for all real x (cso)
Answer: x² - 6x + 10 = (x-3)² + 1 ≥ 1 > 0 for all real x.

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