IB Mathematics SL 1.6 Deductive Proof Numerical and Algebraic AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) Suppose \( f(x_1) = f(x_2) \). Then \( 2x_1 + 3 = 2x_2 + 3 \). M1
Subtract 3 from both sides: \( 2x_1 = 2x_2 \Rightarrow x_1 = x_2 \). A1
[2 marks]
(b) Suppose \( g(x_1) = g(x_2) \). Then \( \frac{x_1 + 3}{2x_1 – 4} = \frac{x_2 + 3}{2x_2 – 4} \). M1
Cross-multiply: \( (x_1 + 3)(2x_2 – 4) = (x_2 + 3)(2x_1 – 4) \). M1
Expand: \( 2x_1 x_2 – 4x_1 + 6x_2 – 12 = 2x_1 x_2 – 4x_2 + 6x_1 – 12 \).
Simplify: \( -4x_1 + 6x_2 = 6x_1 – 4x_2 \Rightarrow 10x_2 = 10x_1 \Rightarrow x_1 = x_2 \). A1
[3 marks]
(c) This is the contrapositive of (b): If \( g(x_1) = g(x_2) \), then \( x_1 = x_2 \). Thus, if \( x_1 \neq x_2 \), then \( g(x_1) \neq g(x_2) \). M1
Alternatively, suppose \( x_1 \neq x_2 \) and \( g(x_1) = g(x_2) \). From (b), this implies \( x_1 = x_2 \), a contradiction. R1
[2 marks]
(d) Counterexample: Let \( x_1 = 2 \), \( x_2 = -2 \). Then \( x_1 \neq x_2 \), but \( h(x_1) = 2^2 + 1 = 5 \), \( h(x_2) = (-2)^2 + 1 = 5 \), so \( h(x_1) = h(x_2) \). A1
[1 mark]
Total [8 marks]
▶️ Answer/Explanation
Part (a)(i)
Assume \( a, b, c \) are even integers.
Let \( a = 2 \times n \), \( b = 2 \times m \), \( c = 2 \times k \), where \( n, m, k \in \mathbb{Z} \) (M1).
Then: \( d = a + b + c = (2 \times n) + (2 \times m) + (2 \times k) = 2 \times (n + m + k) \).
Since \( n + m + k \in \mathbb{Z} \), \( d \) is divisible by 2, hence even (A1 N2).
[2 marks]
Part (a)(ii)
Assume \( a, b, c \) are odd integers.
Let \( a = 2 \times n + 1 \), \( b = 2 \times m + 1 \), \( c = 2 \times k + 1 \), where \( n, m, k \in \mathbb{Z} \) (M1).
Then: \( d = a + b + c = (2 \times n + 1) + (2 \times m + 1) + (2 \times k + 1) = 2 \times (n + m + k + 1) + 3 – 2 = 2 \times (n + m + k + 1) + 1 \).
Since \( n + m + k + 1 \in \mathbb{Z} \), \( d \) is of the form \( 2 \times p + 1 \), hence odd (A1 N2).
[1 mark]
Part (b)
Counterexample: Let \( a = 2 \), \( b = 4 \), \( c = 5 \).
Then: \( d = 2 + 4 + 5 = 11 \), which is odd.
But \( a = 2 \), \( b = 4 \) are even, so \( a, b, c \) are not all odd (A1 N1).
[1 mark]
Part (c)
Statement: “If \( d \) is even then \( a, b, c \) are all even” is false.
Counterexample: Let \( a = 2 \), \( b = 3 \), \( c = 5 \) (M1).
Then: \( d = 2 + 3 + 5 = 10 \), which is even.
But \( b = 3 \), \( c = 5 \) are odd, so \( a, b, c \) are not all even (A1 N2).
[2 marks]
Part (d)
Assume \( d \) is even and suppose, for contradiction, that \( a, b, c \) are all odd (M1).
By Part (a)(ii), if \( a, b, c \) are odd, then \( d = a + b + c \) is odd.
This contradicts the assumption that \( d \) is even.
Hence, at least one of \( a, b, c \) must be even (A1 N2).
[2 marks]
Total [8 marks]