IB Mathematics SL 2.9 The function ax and its graph. AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) 9% (A1, N1).
Working:
Car value: \( C = 40,000(0.91)^t \).
Depreciation factor: 0.91.
Depreciation rate: \( 1 – 0.91 = 0.09 = 9\% \).
[1 mark]
(b) \( \$24,960 \) (A1A1, N2).
Working:
Years from 2023 to 2028: \( t = 5 \).
Car value: \( C = 40,000(0.91)^5 \).
Calculate: \( (0.91)^5 \approx 0.6240 \).
\( C \approx 40,000 \times 0.6240 = 24,960 \).
[2 marks]
(c) \( M \approx \$7,570.50 \) (A1A1A1, N3).
Working:
Alvie’s investment: \( FV = 15,000(1.03)^t \).
For \( t = 5 \): \( FV = 15,000(1.03)^5 \).
Calculate: \( (1.03)^5 \approx 1.1593 \).
\( FV \approx 15,000 \times 1.1593 \approx 17,389.50 \).
Car value from (b): \( 24,960 \).
Extra amount: \( M = 24,960 – 17,389.50 \approx 7,570.50 \).
[3 marks]
Total [6 marks]
▶️ Answer/Explanation
(a) \( A = 10 \) (M1A1, AG, N0).
Working:
Substituting (0, 13) into function (M1)
e.g. \( 13 = A e^0 + 3 \).
\( 13 = A + 3 \) (A1)
\( A = 10 \).
[2 marks]
(b) \( k = -0.201 \) (A1M1A1, N2).
Working:
Substituting into \( f(15) = 3.49 \) (A1)
e.g. \( 3.49 = 10 e^{15k} + 3 \), \( 0.49 = 10 e^{15k} \), \( 0.049 = e^{15k} \).
Evidence of solving equation (M1)
e.g. using \( \ln \), \( 15k = \ln(0.049) \).
\( k = \frac{\ln(0.049)}{15} \approx -0.201 \) (A1).
[3 marks]
(c) (i) \( f'(x) = -2.01 e^{-0.201x} \) (A1A1A1, N3).
Working:
\( f(x) = 10 e^{-0.201x} + 3 \).
Differentiate: \( f'(x) = 10 \times (-0.201) e^{-0.201x} \) (A1 for \( 10 e^{-0.201x} \), A1 for \( \times -0.201 \), A1 for derivative of 3 is zero).
\( f'(x) = -2.01 e^{-0.201x} \).
[1 mark]
(c) (ii) \( f \) is a decreasing function (R1, N1).
Working:
Valid reason with reference to derivative (R1)
e.g. \( f'(x) < 0 \), derivative always negative.
[1 mark]
(c) (iii) \( y = 3 \) (A1, N1).
Working:
Equation of the horizontal asymptote.
[1 mark]
(d) Area = 19.5 (A1A1M1A1A2, N4).
Working:
Finding limits \( 3.8953\ldots \), \( 8.6940\ldots \) (seen anywhere) (A1A1).
Evidence of integrating and subtracting functions (M1).
Correct expression (A1)
e.g. \( \int_{3.90}^{8.69} [g(x) – f(x)] \, dx \), \( \int_{3.90}^{8.69} [(-x^2 + 12x – 24) – (10 e^{-0.201x} + 3)] \, dx \).
Area \( = 19.5 \) (A2).
[6 marks]
Total [16 marks]