IB DP Maths AA SL1.6 :Simple deductive proof, numerical and algebraic HL Paper 1- Exam Style Questions
▶️ Answer/Explanation
(a) Arithmetic sequence: \( p = a + d \), \( q = a + 2d \).
Then: \( q – p = (a + 2d) – (a + d) = d \), so \( p = a + (q – p) \).
Rearrange: \( 2p – q = a \).
(b) Geometric sequence: \( s = ar \), \( t = ar^2 \).
Then: \( \frac{s}{a} = r \), \( \frac{t}{s} = r \), so \( \frac{s}{a} = \frac{t}{s} \).
Cross-multiply: \( s^2 = at \).
(c) Given \( q = t = 1 \).
From (a): \( 2p – 1 = a \).
From (b): \( s^2 = a \cdot 1 = a \).
Substitute: \( 2p – 1 = s^2 \).
Since \( s^2 > 0 \), \( 2p – 1 > 0 \implies p > \frac{1}{2} \).
(d) Given \( a = 9 \), \( q = t = 1 \), \( s > 0 \).
(i) Arithmetic: \( 2p – 1 = 9 \implies p = 5 \). Sequence: \( a, p, q, q + d \).
\( d = q – p = 1 – 5 = -4 \). Terms: \( 9, 5, 1, 1 + (-4) = -3 \).
(ii) Geometric: \( s^2 = 9 \cdot 1 = 9 \implies s = 3 \). Sequence: \( a, s, t, tr \).
\( r = \frac{t}{s} = \frac{1}{3} \). Terms: \( 9, 3, 1, 1 \cdot \frac{1}{3} = \frac{1}{3} \).
(e)
(i) \( u_1 = 9 + \ln 9 \), \( u_2 = 5 + \ln 3 \), \( u_3 = 1 + \ln 1 \).
Since \( \ln 1 = 0 \), \( \ln 9 = 2 \ln 3 \), compute:
\( d = u_2 – u_1 = (5 + \ln 3) – (9 + 2 \ln 3) = 5 – 9 + \ln 3 – 2 \ln 3 = -4 – \ln 3 \).
(ii) Sum: \( S_{10} = \frac{10}{2} \left( 2 u_1 + 9d \right) \).
\( u_1 = 9 + 2 \ln 3 \), \( d = -4 – \ln 3 \).
\( S_{10} = 5 \left( 2 (9 + 2 \ln 3) + 9 (-4 – \ln 3) \right) = 5 (18 + 4 \ln 3 – 36 – 9 \ln 3) = 5 (-18 – 5 \ln 3) = -90 – 25 \ln 3 \).
Markscheme
(a) \( d = p – a \), \( q – a = 2d \), \( q – p = d \), so \( p – a = q – p \implies 2p – q = a \quad \mathbf{(M1)(A1)} \).
(b) \( r = \frac{s}{a} \), \( r = \frac{t}{s} \), so \( \frac{s}{a} = \frac{t}{s} \implies s^2 = at \quad \mathbf{(M1)(A1)} \).
(c) \( 2p – 1 = a \), \( s^2 = a \cdot 1 \), so \( 2p – 1 = s^2 \), \( s^2 > 0 \implies p > \frac{1}{2} \quad \mathbf{(M1)(A1)} \).
(d)(i) \( 2p – 1 = 9 \implies p = 5 \), \( d = 1 – 5 = -4 \), terms: 9, 5, 1, -3 \quad \mathbf{A1} \).
(d)(ii) \( s^2 = 9 \implies s = 3 \), \( r = \frac{1}{3} \), terms: 9, 3, 1, \(\frac{1}{3} \quad \mathbf{A1} \).
(e)(i) \( d = (5 + \ln 3) – (9 + 2 \ln 3) = -4 – \ln 3 \quad \mathbf{(M1)(A1)} \).
(e)(ii) \( S_{10} = \frac{10}{2} (2 (9 + 2 \ln 3) + 9 (-4 – \ln 3)) = 5 (-18 – 5 \ln 3) = -90 – 25 \ln 3 \quad \mathbf{(M1)(A1)} \).
[6 marks]
▶️ Answer/Explanation
(a) Sum: \( (n – 1) + n + (n + 1) = 3n \).
Since \( 3n \) is divisible by 3 for any integer \( n \), the sum is always divisible by 3.
(b) Sum of squares: \( (n – 1)^2 + n^2 + (n + 1)^2 \).
Expand: \( (n^2 – 2n + 1) + n^2 + (n^2 + 2n + 1) = 3n^2 + 2 \).
Divide by 3: \( \frac{3n^2 + 2}{3} = n^2 + \frac{2}{3} \).
Since \( \frac{2}{3} \) is not an integer, \( 3n^2 + 2 \) is never divisible by 3.
Markscheme
(a) \( (n – 1) + n + (n + 1) = 3n \), which is always divisible by 3.
(b) \( (n – 1)^2 + n^2 + (n + 1)^2 = 3n^2 + 2 \quad \mathbf{M1} \).
Attempts to expand \( (n – 1)^2 \) or \( (n + 1)^2 \). Do not accept \( n^2 – 1 \) or \( n^2 + 1 \).
\( \frac{3n^2 + 2}{3} = n^2 + \frac{2}{3} \), \( \frac{2}{3} \) not integer, so never divisible by 3 \quad \mathbf{A1} \).
OR \( 3n^2 \) divisible by 3, 2 not divisible by 3, so \( 3n^2 + 2 \) never divisible by 3.
▶️ Answer/Explanation
(a) Expand RHS: \( (x + 5)(2x^2 – 3x + 1) = x(2x^2 – 3x + 1) + 5(2x^2 – 3x + 1) \)
\( = 2x^3 – 3x^2 + x + 10x^2 – 15x + 5 = 2x^3 + 7x^2 – 14x + 5 \).
Matches LHS, so identity holds.
(b) Expand RHS: \( (x + 5)(ax^2 + bx + c) = ax^3 + bx^2 + cx + 5ax^2 + 5bx + 5c \)
\( = ax^3 + (b + 5a)x^2 + (c + 5b)x + 5c \).
Compare with LHS \( 3x^3 + 13x^2 – 3x + 35 \):
\( a = 3 \), \( b + 5a = 13 \implies b + 15 = 13 \implies b = -2 \),
\( c + 5b = -3 \implies c – 10 = -3 \implies c = 7 \), \( 5c = 35 \implies c = 7 \).
Thus, \( a = 3 \), \( b = -2 \), \( c = 7 \).
(c) Expand RHS: \( (x + 5)(2x^2 + dx + 2) = x(2x^2 + dx + 2) + 5(2x^2 + dx + 2) \)
\( = 2x^3 + dx^2 + 2x + 10x^2 + 5dx + 10 = 2x^3 + (d + 10)x^2 + (2 + 5d)x + 10 \).
Compare with LHS \( ax^3 + bx^2 – 23x + c \):
\( a = 2 \), \( b = d + 10 \), \( 2 + 5d = -23 \implies 5d = -25 \implies d = -5 \),
\( b = -5 + 10 = 5 \), \( c = 10 \).
Thus, \( a = 2 \), \( b = 5 \), \( c = 10 \), \( d = -5 \).
Markscheme
(a) \( (x + 5)(2x^2 – 3x + 1) = 2x^3 – 3x^2 + x + 10x^2 – 15x + 5 = 2x^3 + 7x^2 – 14x + 5 \quad \mathbf{A1} \).
(b) \( (x + 5)(ax^2 + bx + c) = ax^3 + (b + 5a)x^2 + (c + 5b)x + 5c \).
Compare with \( 3x^3 + 13x^2 – 3x + 35 \): \( a = 3 \), \( b + 5a = 13 \implies b = -2 \), \( 5c = 35 \implies c = 7 \quad \mathbf{(M1)(A1)} \).
(c) \( (x + 5)(2x^2 + dx + 2) = 2x^3 + (d + 10)x^2 + (2 + 5d)x + 10 \).
Compare with \( ax^3 + bx^2 – 23x + c \): \( a = 2 \), \( b = d + 10 \), \( 2 + 5d = -23 \implies d = -5 \), \( b = 5 \), \( c = 10 \quad \mathbf{(M1)(A1)} \).
▶️ Answer/Explanation
Assume \( n \) is odd, so \( n = 2k + 1 \).
Compute: \( n^2 + 2n + 5 = (2k + 1)^2 + 2(2k + 1) + 5 \)
\( = 4k^2 + 4k + 1 + 4k + 2 + 5 = 4k^2 + 8k + 8 = 4(k^2 + 2k + 2) \).
This is even, contradicting the assumption that \( n^2 + 2n + 5 \) is odd.
Hence, if \( n^2 + 2n + 5 \) is odd, \( n \) must be even.
Markscheme
Assume \( n = 2k + 1 \): \( n^2 + 2n + 5 = (2k + 1)^2 + 2(2k + 1) + 5 \quad \mathbf{M1} \).
Expand: \( 4k^2 + 4k + 1 + 4k + 2 + 5 = 4k^2 + 8k + 8 \quad \mathbf{A1} \).
Recognize \( 4k^2 + 8k + 8 \) is even, contradicting odd assumption \quad \mathbf{M1} \).
Conclude \( n \) is even \quad \mathbf{A1} \).