IBDP Maths SL 1.6 Deductive Proof Numerical and Algebraic AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Difference of squares: \( (n + 1)^2 – n^2 \). M1
Use difference of squares formula: \( (n + 1)^2 – n^2 = (n + 1 – n)(n + 1 + n) \). A1
\( = 1 \cdot (2n + 1) = 2n + 1 \). A1
Sum of the two integers: \( n + (n + 1) = 2n + 1 \). A1
Since \( (n + 1)^2 – n^2 = 2n + 1 = n + (n + 1) \), the difference of their squares equals the sum of the two integers. AG
[3 marks]
▶️ Answer/Explanation
(a) (i) Let \( a = 2n \), \( b = 2m \), \( c = 2k \), where \( n, m, k \in \mathbb{Z} \). M1
Then \( d = a + b + c = 2n + 2m + 2k = 2(n + m + k) \), which is even. A1
[2 marks]
(a) (ii) Let \( a = 2n + 1 \), \( b = 2m + 1 \), \( c = 2k + 1 \), where \( n, m, k \in \mathbb{Z} \). M1
Then \( d = a + b + c = (2n + 1) + (2m + 1) + (2k + 1) = 2(n + m + k + 1) + 1 \), which is odd. A1
[1 mark]
(b) Counterexample: Let \( a = 2 \), \( b = 4 \), \( c = 5 \). Then \( d = 2 + 4 + 5 = 11 \), which is odd, but \( a, b, c \) are not all odd. A1
[1 mark]
(c) The statement is false. Counterexample: Let \( a = 2 \), \( b = 3 \), \( c = 5 \). Then \( d = 2 + 3 + 5 = 10 \), which is even, but \( a, b, c \) are not all even. M1 A1
[2 marks]
(d) Suppose \( d \) is even and \( a, b, c \) are all odd (contradiction hypothesis). M1
By (a)(ii), if \( a, b, c \) are all odd, then \( d \) is odd, contradicting the assumption that \( d \) is even. R1
Thus, at least one of \( a, b, c \) must be even. AG
[2 marks]
Total [8 marks]