IB DP Maths AA SL1.6 :Simple deductive proof, numerical and algebraic HL Paper 1- Exam Style Questions
▶️ Answer/Explanation
(a)
Factorize the two quadratic expressions.
\(x^2-8x+7=(x-1)(x-7)\)
\(x^2-1=(x-1)(x+1)\)
Substitute these factorizations into the left-hand side.
\(\dfrac{x}{(x-1)(x-7)}\times\dfrac{(x-1)(x+1)}{x+1}\)
Cancel the common factors \((x-1)\) and \((x+1)\).
\(\dfrac{x}{(x-1)(x-7)}\times\dfrac{(x-1)(x+1)}{x+1}=\dfrac{x}{x-7}\)
Since \(x>7\), none of the cancelled factors or denominators is zero.
✅ Hence, \(\dfrac{x}{x^2-8x+7}\times\dfrac{x^2-1}{x+1}=\dfrac{x}{x-7}\).
(b)
Move the logarithm on the right-hand side to the left.
\(\log_2\left[x(x^2-1)\right]-\log_2\left[(x^2-8x+7)(x+1)\right]=1\)
Using the quotient law of logarithms:
\(\log_2\left[\dfrac{x(x^2-1)}{(x^2-8x+7)(x+1)}\right]=1\)
From part (a), the expression inside the logarithm simplifies to \(\dfrac{x}{x-7}\).
\(\log_2\left(\dfrac{x}{x-7}\right)=1\)
Since \(1=\log_2 2\), the arguments must be equal.
\(\dfrac{x}{x-7}=2\)
\(x=2(x-7)\)
\(x=2x-14\)
\(x=14\)
The value \(x=14\) satisfies \(x>7\), so all logarithms are defined.
✅ Answer: \(x=14\)
▶️ 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]