Question 1 [3 marks]
If n is even, n² and 3n are even, even+even+5 is odd. If n is odd, n² and 3n are odd, odd+odd+5 is odd.
Mark scheme
- A [3] — cao
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If n is even, n² and 3n are even, even+even+5 is odd. If n is odd, n² and 3n are odd, odd+odd+5 is odd.
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n² + n = n(n + 1); consecutive integers, so one factor is even.
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b = 10 − a; ab = 10a − a² = 25 − (a − 5)² ≤ 25.
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Let n = 2k+1; n² − 1 = 4k² + 4k = 4k(k+1).
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Expand (x + 3)(x + 3) = x² + 3x + 3x + 9 = x² + 6x + 9.
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a² − 2ab + b² = (a − b)² ≥ 0.
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(3x − 2)(3x − 2) = 9x² − 6x − 6x + 4 = 9x² − 12x + 4.
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Multiply out: x³ + x² + x − x² − x − 1 = x³ − 1.
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Use 12[(a−b)² + (b−c)² + (c−a)²] ≥ 0, which expands to the inequality.
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Expand RHS: 4x² + 10x − 10x − 25 = 4x² − 25.
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n³ − n = n(n − 1)(n + 1), product of three consecutive integers → divisible by 2 and by 3.
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Let odds be 2a+1 and 2b+1; sum = 2a + 2b + 2 = 2(a + b + 1), a multiple of 2.
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x² − 2x + 1 = (x − 1)² ≥ 0.
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x² + 2x + 2 = (x + 1)² + 1 ≥ 1 > 0.
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p² + 2pq + q² − 4pq = (p − q)² ≥ 0.
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Expand: n² + 5n + 6 − n² − n = 4n + 6.
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Expand: 2x² − 8x + x − 4 = 2x² − 7x − 4.
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n + (n+1) + (n+2) = 3n + 3 = 3(n + 1).
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Odd² ≡ 1 (mod 4); sum of two such squares ≡ 2 (mod 4) — even but not a multiple of 4.
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Expand RHS: x² + 4x + 3x + 12 = x² + 7x + 12.
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Multiply by x: x² + 1 ≥ 2x ⇔ (x − 1)² ≥ 0 (x > 0).
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Rearrange: x² − 4x + 4 = (x − 2)² ≥ 0.
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x² − 2 + 1x² = (x − 1x)² ≥ 0, so x² + 1x² ≥ 2.
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3(2n) = 9n ≡ 1n (mod 8) since 9 ≡ 1 (mod 8), hence 9n − 1 ≡ 0 (mod 8).
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Rearrange to 12[(x−y)² + (x−1)² + (y−1)²] ≥ 0.
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Of n and n+1, one is even, so the product has a factor 2.
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2y² − 2y + 5y − 5 = 2y² + 3y − 5.
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Expand: a² − ab + ab − b² = a² − b².
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Odd integer 2k+1; (2k+1)² = 4k² + 4k + 1 = 4k(k+1) + 1; k(k+1) is even so 4×(even) is divisible by 8.
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Expand RHS: x² + 3x − 3x − 9 = x² − 9 (difference of two squares).
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