IB Mathematics SL 1.6 Deductive Proof Numerical and Algebraic AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a) [2 marks]
Method: Compute the sum of the three consecutive integers: \( (n – 1) + n + (n + 1) \).
Simplify: \( (n – 1) + n + (n + 1) = n + n + n – 1 + 1 = 3n \).
Since \( 3n = 3 \cdot n \), the sum is a multiple of 3 and thus always divisible by 3 for any integer \( n \).
Examples: If \( n = 1 \), sum = \( 0 + 1 + 2 = 3 \), divisible by 3. If \( n = 0 \), sum = \( -1 + 0 + 1 = 0 \), divisible by 3.
Answer: The sum \( 3n \) is always divisible by 3.
Part (b) [4 marks]
Method 1: Compute the sum of the squares: \( (n – 1)^2 + n^2 + (n + 1)^2 \).
Expand: \( (n – 1)^2 = n^2 – 2n + 1 \), \( n^2 = n^2 \), \( (n + 1)^2 = n^2 + 2n + 1 \).
Add: \( (n^2 – 2n + 1) + n^2 + (n^2 + 2n + 1) = n^2 + n^2 + n^2 – 2n + 2n + 1 + 1 = 3n^2 + 2 \).
Check divisibility: \( 3n^2 + 2 \div 3 = n^2 + \frac{2}{3} \). Since \( n^2 \) is an integer and \( \frac{2}{3} \) is not, the result is not an integer, so not divisible by 3.
Method 2: Use modulo 3: \( 3n^2 \equiv 0 \pmod{3} \), \( 2 \equiv 2 \pmod{3} \). Thus, \( 3n^2 + 2 \equiv 0 + 2 \equiv 2 \pmod{3} \), leaving a remainder of 2.
Examples: If \( n = 1 \), \( 0^2 + 1^2 + 2^2 = 0 + 1 + 4 = 5 \), \( 5 \div 3 = 1 + \frac{2}{3} \). If \( n = 2 \), \( 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \), \( 14 \div 3 = 4 + \frac{2}{3} \).
Answer: The sum \( 3n^2 + 2 \) is never divisible by 3.
▶️ Answer/Explanation
Part (a) [Proof of Identities]
(i) Prove \( (x – 1)^2 \equiv x^2 – 2x + 1 \):
LHS: \( (x – 1)^2 = (x – 1) \times (x – 1) \).
Using the fact \( a^2 = a \times a \), expand:
\[ (x – 1)(x – 1) = x \cdot x + x \cdot (-1) + (-1) \cdot x + (-1) \cdot (-1) \]
\[ = x^2 – x – x + 1 = x^2 – 2x + 1 \]
RHS: \( x^2 – 2x + 1 \).
Since LHS = RHS, the identity is proven.
(ii) Prove \( (x – 1)^3 \equiv x^3 – 3x^2 + 3x – 1 \):
LHS: \( (x – 1)^3 = (x – 1)^2 \times (x – 1) \).
Using the fact \( a^3 = a^2 \times a \), and from (i), \( (x – 1)^2 = x^2 – 2x + 1 \):
\[ (x – 1)^3 = (x^2 – 2x + 1) \times (x – 1) \]
Expand: \( (x^2 – 2x + 1)(x – 1) = x^2 \cdot x + x^2 \cdot (-1) + (-2x) \cdot x + (-2x) \cdot (-1) + 1 \cdot x + 1 \cdot (-1) \)
\[ = x^3 – x^2 – 2x^2 + 2x + x – 1 \]
\[ = x^3 + (-x^2 – 2x^2) + (2x + x) – 1 = x^3 – 3x^2 + 3x – 1 \]
RHS: \( x^3 – 3x^2 + 3x – 1 \).
Since LHS = RHS, the identity is proven.
Part (b) [Verification of (a)(ii)]
(i) For \( x = 2 \):
LHS: \( (x – 1)^3 = (2 – 1)^3 = 1^3 = 1 \).
RHS: \( x^3 – 3x^2 + 3x – 1 = 2^3 – 3 \cdot 2^2 + 3 \cdot 2 – 1 = 8 – 12 + 6 – 1 = 1 \).
LHS = RHS = 1, so the identity holds.
(ii) For \( x = 3 \):
LHS: \( (x – 1)^3 = (3 – 1)^3 = 2^3 = 8 \).
RHS: \( x^3 – 3x^2 + 3x – 1 = 3^3 – 3 \cdot 3^2 + 3 \cdot 3 – 1 = 27 – 27 + 9 – 1 = 8 \).
LHS = RHS = 8, so the identity holds.
Answer: (i) LHS = 1, RHS = 1; (ii) LHS = 8, RHS = 8.