Question 1
1(a)
Prove that 5(2x - 3) - 2 \equiv 10x - 17
Line 1: 5(2x - 3) - 2 \equiv 10x - 17
Line 2: [ Missing Line ]
Line 3: 10x - 17 \equiv 10x - 17
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: 10x - 15 - 2 \equiv 10x - 17
B: 7x - 8 - 2 \equiv 10x - 17
C: 10x - 8 - 2 \equiv 10x - 17
D: 10x + 15 - 2 \equiv 10x - 17
Answer: A
Workings:
Line 1: 5(2x - 3) - 2 \equiv 10x - 17
Line 2: \color{red}10x - 15 - 2 \equiv 10x - 17
Line 3: 10x - 17 \equiv 10x - 17
Marks = 2
1(b):
Prove (n-2)^2 + 3 \equiv n^2 - 4n + 7
Line 1: (n-2)^2 + 3 \equiv n^2 -4n + 7
Line 2: [ Missing Line ]
Line 3: n^2 - 4n + 7 \equiv n^2 - 4n + 7
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: n^2 - 4n + 4 + 3 \equiv n^2 - 4n + 7
B: n^2 - 2n + 2 + 3 \equiv n^2 - 4n + 7
C: n^2 - 4n + 2 + 3 \equiv n^2 - 4n + 7
D: n^2 + 4n - 4 + 3 \equiv n^2 - 4n + 7
Answer: A
Workings:
Line 1: (n-2)^2 + 3 \equiv n^2 -4n + 7
Line 2: \color{red} n^2 - 4n + 4 + 3 \equiv n^2 - 4n + 7
Line 3: n^2 - 4n + 7 \equiv n^2 - 4n + 7
Marks = 2
1(c):
Prove (x+1)^2 - x^2 \equiv 2x + 1
Line 1: (x+1)^2 - x^2 \equiv 2x + 1
Line 2: [ Missing Line ]
Line 3: 2x + 1 \equiv 2x + 1
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: x^2 + x + x + 1 - x^2
B: x^2 + 2x + x + 1 - x^2
C: x^2 + x + x + 2 - x^2
D: x^2 + x + 2x + 2 - x^2
Answer: A
Workings:
Line 1: (x+1)^2 - x^2 \equiv 2x + 1
Line 2: \color{red} x^2 + x + x + 1 - x^2
Line 3: 2x + 1 \equiv 2x + 1
Marks = 2
Question 2
2(a):
Prove 5(3x - 5) - 2(2x + 9) \equiv 11x - 43
Line 1: 5(3x - 5) - 2(2x + 9) \equiv 11x - 43
Line 2: [ Missing Line ]
Line 3: 11x - 43 \equiv 11x - 43
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: 15x - 4x - 25 - 18 \equiv 11x - 43
B: 15x - 8x - 25 - 11 \equiv 11x - 43
C: 15x - 6x - 20 - 18 \equiv 11x - 43
D: 15x - 7x - 20 - 11 \equiv 11x - 43
Answer: A
Workings:
Line 1: 5(3x - 5) - 2(2x + 9) \equiv 11x - 43
Line 2: \color{red} 15x - 4x - 25 - 18 \equiv 11x - 43
Line 3: 11x - 43 \equiv 11x - 43
Marks = 3
2(b):
Prove (n - 2)^2 - (n - 5)^2 \equiv 3(2n - 7)
Line 1: (n - 2)^2 - (n - 5)^2 \equiv 3(2n - 7)
Line 2: n^2 - 4n + 4 - n^2 + 10n - 25 \equiv 3(2n - 7)
Line 3: [ Missing Line ]
Line 4: 3(2n - 7) \equiv 3(2n - 7)
What is the missing Line 3 from the working above?
ANSWER: Multiple Choice (Type 1)
A: 6n - 21 \equiv 3(2n - 7)
B: 2n^2 + 6n - 21 \equiv 3(2n - 7)
C: -14n - 21 \equiv 3(2n - 7)
D: 6n - 29 \equiv 3(2n - 7)
Answer: A
Workings:
Line 1: (n - 2)^2 - (n - 5)^2 \equiv 3(2n - 7)
Line 2: n^2 - 4n + 4 - n^2 + 10n - 25 \equiv 3(2n - 7)
Line 3: \color{red} 6n - 21 \equiv 3(2n - 7)
Line 4: 3(2n - 7) \equiv 3(2n - 7)
Marks = 3
2(c):
Prove (n + 2)^2 - 3(n + 4) \equiv (n + 4)(n - 3) + 4
Line 1: (n + 2)^2 - 3(n + 4) \equiv (n + 4)(n - 3) + 4
Line 2: n^2 + 2n + 2n + 4 - 3n - 12 \equiv (n + 4)(n - 3) + 4
Line 3: [ Missing Line ]
Line 4: (n + 4)(n - 3) + 4 \equiv (n + 4)(n - 3) + 4
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: n^2 + n - 12 + 4 \equiv (n + 4)(n - 3) + 4
B: n^2 + 4n - 8 + 4 \equiv (n + 4)(n - 3) + 4
C: n^2 + 2n + 8 \equiv (n + 4)(n - 3) + 4
D: n^2 + n - 4 + 12 \equiv (n + 4)(n - 3) + 4
Answer: A
Workings:
Line 1: (n + 2)^2 - 3(n + 4) \equiv (n + 4)(n - 3) + 4
Line 2: n^2 + 2n + 2n + 4 - 3n - 12 \equiv (n + 4)(n - 3) + 4
Line 3: \color{red} n^2 + n - 12 + 4 \equiv (n + 4)(n - 3) + 4
Line 4: (n + 4)(n - 3) + 4 \equiv (n + 4)(n - 3) + 4
Marks = 3
2(d):
Prove 3(n + 3)(n - 1) - 3(1 - n) \equiv (3n - 3)(n + 4)
Line 1: 3(n + 3)(n - 1) - 3(1 - n) \equiv (3n - 3)(n + 4)
Line 2: 3(n^2 + 2n - 3) - 3 + 3n \equiv (3n - 3)(n + 4)
Line 3: [ Missing Line ]
Line 4: (3n - 3)(n + 4) \equiv (3n - 3)(n + 4)
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: 3n^2 + 9n - 12 \equiv (3n - 3)(n + 4)
B: 3n^2 + 5n - 12 \equiv (3n - 3)(n + 4)
C: 3n^2 + 6n - 9 \equiv (3n - 3)(n + 4)
D: 3n^2 + 4n - 9 \equiv (3n - 3)(n + 4)
Answer: A
Workings:
Line 1: 3(n + 3)(n - 1) - 3(1 - n) \equiv (3n - 3)(n + 4)
Line 2: 3(n^2 + 2n - 3) - 3 + 3n \equiv (3n - 3)(n + 4)
Line 3: \color{red} 3n^2 + 9n - 12 \equiv (3n - 3)(n + 4)
Line 4: (3n - 3)(n + 4) \equiv (3n - 3)(n + 4)
Marks = 3
Question 3
3(a):
Prove (3n + 1)(n + 3) - n(3n + 7) \equiv 3(n + 1)
Line 1: (3n + 1)(n + 3) - n(3n + 7) \equiv 3(n + 1)
Line 2: 3n^2 + 9n + n + 3 - 3n^2 - 7n \equiv 3(n + 1)
Line 3: [ Missing Line ]
Line 4: 3(n + 1) \equiv 3(n + 1)
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: 3n + 3 \equiv 3(n + 1)
B: 3n - 3 \equiv 3(n + 1)
C: -3n - 3 \equiv 3(n + 1)
D: -3n + 3 \equiv 3(n + 1)
Answer: A
Workings:
Line 1: (3n + 1)(n + 3) - n(3n + 7) \equiv 3(n + 1)
Line 2: 3n^2 + 9n + n + 3 - 3n^2 - 7n \equiv 3(n + 1)
Line 3: \color{red} 3n + 3 \equiv 3(n + 1)
Line 4: 3(n + 1) \equiv 3(n + 1)
Marks = 3
3(b):
Prove (n + 3)^2 - (3n + 5) \equiv (n + 1)(n + 2) + 3
Line 1: (n + 3)^2 - (3n + 5) \equiv (n + 1)(n + 2) + 3
Line 2: n^2 + 6n + 9 - 3n - 4\equiv (n + 1)(n + 2) + 3
Line 3: [ Missing Line ]
Line 4: (n + 1)(n + 2) + 3\equiv (n + 1)(n + 2) + 3
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: n^2 + 3n + 5\equiv (n + 1)(n + 2) + 3
B: n^2 - 3n + 5\equiv (n + 1)(n + 2) + 3
C: -n^2 + 3n - 5\equiv (n + 1)(n + 2) + 3
D: -n^2 - 3n - 5\equiv (n + 1)(n + 2) + 3
Answer: A
Workings:
Line 1: (n + 3)^2 - (3n + 5) \equiv (n + 1)(n + 2) + 3
Line 2: n^2 + 6n + 9 - 3n - 4\equiv (n + 1)(n + 2) + 3
Line 3: \color{red} n^2 + 3n + 5\equiv (n + 1)(n + 2) + 3
Line 4: (n + 1)(n + 2) + 3\equiv (n + 1)(n + 2) + 3
Marks = 3
3(c):
Prove (n - 3)^2 - (2n + 1) \equiv (n - 4)^2 - 8
Line 1: (n - 3)^2 - (2n + 1) \equiv (n - 4)^2 - 8
Line 2: n^2 - 6n + 9 - 2n - 1 \equiv (n - 4)^2 - 8
Line 3: [ Missing Line ]
Line 4: (n - 4)^2 - 8\equiv (n - 4)^2 - 8
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: n^2 - 8n + 8 \equiv (n - 4)^2 - 8
B: n^2 - 4n + 4 \equiv (n - 4)^2 - 8
C: n^2 - 2n + 10 \equiv (n - 4)^2 - 8
D: n^2 - 4n + 8 \equiv (n - 4)^2 - 8
Answer: A
Workings:
Line 1: (n - 3)^2 - (2n + 1) \equiv (n - 4)^2 - 8
Line 2: n^2 - 6n + 9 - 2n - 1 \equiv (n - 4)^2 - 8
Line 3: \color{red} n^2 - 8n + 8 \equiv (n - 4)^2 - 8
Line 4: (n - 4)^2 - 8\equiv (n - 4)^2 - 8
Marks = 3
3(d):
Prove \dfrac{1}{8}(4n + 1)(n + 8) - \dfrac{1}{8} n(4n + 1) \equiv 4n + 1
Line 1: \dfrac{1}{8}(4n + 1)(n + 8) - \dfrac{1}{8} n(4n + 1) \equiv 4n + 1
Line 2: \dfrac{1}{8}(4n^2 + n + 32n + 8) - \dfrac{1}{8}(4n^2 + n) \equiv 4n + 1
Line 3: [ Missing Line ]
Line 4: 4n + 1 \equiv 4n + 1
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: (\dfrac{1}{2}n^2 + \dfrac{33}{8}n + 1) - (\dfrac{1}{2}n^2 + \dfrac{n}{8}) \equiv 4n + 1
B: (\dfrac{1}{4}n^2 + \dfrac{33}{4}n + 1) - (\dfrac{1}{4}n^2 + \dfrac{n}{4}) \equiv 4n + 1
C: (\dfrac{1}{4}n^2 + \dfrac{31}{8}n + 1) - (\dfrac{1}{4}n^2 + \dfrac{3n}{8}) \equiv 4n + 1
D: (\dfrac{1}{2}n^2 + \dfrac{33}{4}n + 1) - (\dfrac{1}{2}n^2 + \dfrac{n}{4}) \equiv 4n + 1
Answer: A
Workings:
Line 1: \dfrac{1}{8}(4n + 1)(n + 8) - \dfrac{1}{8} n(4n + 1) \equiv 4n + 1
Line 2: \dfrac{1}{8}(4n^2 + n + 32n + 8) - \dfrac{1}{8}(4n^2 + n) \equiv 4n + 1
Line 3: \color{red} (\dfrac{1}{2}n^2 + \dfrac{33}{8}n + 1) - (\dfrac{1}{2}n^2 + \dfrac{n}{8}) \equiv 4n + 1
Line 4: 4n + 1 \equiv 4n + 1
Marks = 3
Question 4
Prove the product of two even numbers is always even.
Line 1: Let m and n be any integers so that 2m and 2n are both even numbers
Line 2: 2m \times 2n = 4mn
Line 3: [ Missing Line ]
Line 4: So 4mn is even.
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: 4mn = 2(2mn); any integer multiplied by an even number is even.
B: m\times n is always even.
C: 4mn = (2mn)^2; square numbers are always even
D: 4mn = \sqrt{16m^2n^2}; square roots are always even.
Answer: A
Workings:
Line 1: Let m and n be any integers so that 2m and 2n are both even numbers
Line 2: 2m \times 2n = 4mn
Line 3: \color{red} 4mn = 2(2mn); any integer multiplied by an even number is even.
Line 4: So 4mn is even.
Marks = 2
Question 5
Prove that the product of two odd numbers is always odd.
Line 1: Let m and n be any integers so that 2m + 1 and 2n + 1 are both odd numbers.
Line 2: (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1
Line 3: 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1
Line 4: [ Missing Line ]
What is the missing Line 4 from the above working?
ANSWER: Multiple Choice (Type 1)
A: 2(2mn + m + n) is even, so 2(2mn + m + n) + 1 must be odd.
B: 2(2mn + m + n) is even, so 2(2mn + m + n) + 1 must be even.
C: 2(2mn + m + n) is odd, so 2(2mn + m + n) + 1 must be even.
D: 2(2mn + m + n) is odd, so 2(2mn + m + n) + 1 must be odd.
Answer: A
Workings:
Line 1: Let m and n be any integers so that 2m + 1 and 2n + 1 are both odd numbers.
Line 2: (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1
Line 3: 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1
Line 4: \color{red} 2(2mn + m + n) is even, so \color{red}2(2mn + m + n) + 1 must be odd.
Marks = 2
Question 6
Prove algebraically that the sum of two consecutive numbers is odd.
Line 1: If a number is n, then the next number is n + 1.
Line 2: The sum is therefore n + n + 1 = 2n + 1
Line 3: [ Missing Line ]
What is the missing Line 3 from the above working?
ANSWER: Multiple Choice (Type 1)
A: 2n is even, so 2n + 1 must be odd
B: 2n is odd, so 2n + 1 must be odd
C: 2n = 0, so 2n + 1 must be odd
D: 2n is negative, so 2n + 1 must be odd
Answer: A
Workings:
Line 1: If a number is n, then the next number is n + 1.
Line 2: The sum is therefore n + n + 1 = 2n + 1
Line 3: \color{red} 2n is even, so \color{red} 2n + 1 must be odd
Marks = 3
Question 7
Prove 6x - (2x + 3)(x + 2) is always negative.
Line 1: [ Missing Line ]
Line 2: 6x - (2x + 3)(x + 2) \equiv 6x - 2x^2 - 7x - 6
Line 3: 6x - (2x + 3)(x + 2) \equiv -2x^2 - x - 6
Line 4: 6x - (2x + 3)(x + 2) \equiv -(2x^2 + x + 6)
Line 5: (2x^2 + x + 6) is always positive.
Line 6: So multiplying (2x^2 + x + 6) by -1 means answer is always negative.
What is the missing Line 1 from the above working?
ANSWER: Multiple Choice (Type 1)
A: 6x - (2x + 3)(x + 2) \equiv 6x - (2x^2 + 4x + 3x + 6)
B: 6x - (2x + 3)(x + 2) \equiv 6x - (x^2 + 4x + 3x + 9)
C: 6x - (2x + 3)(x + 2) \equiv 6x - (x^2 + 4x + 4x + 12)
D: 6x - (2x + 3)(x + 2) \equiv 6x - (x^2 + 3x + 3x + 6)
Answer: A
Workings:
Line 1: \color{red} 6x - (2x + 3)(x + 2) \equiv 6x - (2x^2 + 4x + 3x + 6)
Line 2: 6x - (2x + 3)(x + 2) \equiv 6x - 2x^2 - 7x - 6
Line 3: 6x - (2x + 3)(x + 2) \equiv -2x^2 - x - 6
Line 4: 6x - (2x + 3)(x + 2) \equiv -(2x^2 + x + 6)
Line 5: (2x^2 + x + 6) is always positive.
Line 6: So multiplying (2x^2 + x + 6) by -1 means answer is always negative.
Marks = 3