IB Mathematics SL 2.5 Exponential growth and decay models AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Professor Chen observed that students have difficulty remembering information presented in their lectures. They modeled the percentage of information retained, \( R \), by the function \[ R(t) = 100e^{-pt}, \quad t \geq 0 \] where \( t \) is the number of days after the lecture. They found that 1 day after a lecture, students had forgotten 50% of the information presented.
(a) Determine the value of \( p \). [2]
(b) Use this model to determine the percentage of information retained by their students 36 hours after Professor Chen’s lecture. [2]
(c) State a mathematical reason why Professor Chen believes that their students will always retain some information from the lecture. [1]
(d) State one possible limitation of the domain of the model. [1]
▶️ Answer/Explanation
Markscheme
(a)
Function: \( R(t) = 100e^{-pt} \). At \( t = 1 \), \( R(1) = 50 \). M1
Substitute: \( 50 = 100e^{-p \cdot 1} \), so \( \dfrac{50}{100} = e^{-p} \), \( 0.5 = e^{-p} \).
Solve: \( \ln(0.5) = -p \), \( p = -\ln(0.5) = \ln(2) \approx 0.693147 \).
Value: \( p = 0.693 \). A1
[2 marks]
Function: \( R(t) = 100e^{-pt} \). At \( t = 1 \), \( R(1) = 50 \). M1
Substitute: \( 50 = 100e^{-p \cdot 1} \), so \( \dfrac{50}{100} = e^{-p} \), \( 0.5 = e^{-p} \).
Solve: \( \ln(0.5) = -p \), \( p = -\ln(0.5) = \ln(2) \approx 0.693147 \).
Value: \( p = 0.693 \). A1
[2 marks]
(b)
Time: 36 hours = \( 1.5 \) days. \( p \approx 0.693 \) (from (a)). M1
Function: \( R(t) = 100e^{-0.693t} \). Substitute: \( R(1.5) = 100e^{-0.693 \times 1.5} \).
Exponent: \( -0.693 \times 1.5 = -1.0395 \), so \( e^{-1.0395} \approx 0.353553 \).
Result: \( 100 \times 0.353553 \approx 35.3553 \). Rounded: 35.4%. A1
[2 marks]
Time: 36 hours = \( 1.5 \) days. \( p \approx 0.693 \) (from (a)). M1
Function: \( R(t) = 100e^{-0.693t} \). Substitute: \( R(1.5) = 100e^{-0.693 \times 1.5} \).
Exponent: \( -0.693 \times 1.5 = -1.0395 \), so \( e^{-1.0395} \approx 0.353553 \).
Result: \( 100 \times 0.353553 \approx 35.3553 \). Rounded: 35.4%. A1
[2 marks]
(c)
Function: \( R(t) = 100e^{-pt} \), \( p \approx 0.693 \), \( t \geq 0 \).
Since \( e^{-pt} > 0 \) for all \( t \geq 0 \), \( R(t) > 0 \). A1
The exponential function never reaches zero, so some information is always retained. A1
[1 mark]
Function: \( R(t) = 100e^{-pt} \), \( p \approx 0.693 \), \( t \geq 0 \).
Since \( e^{-pt} > 0 \) for all \( t \geq 0 \), \( R(t) > 0 \). A1
The exponential function never reaches zero, so some information is always retained. A1
[1 mark]
(d)
The domain \( t \geq 0 \) implies infinite time, but large \( t \) values are not realistic as people do not live forever (e.g., beyond a human lifespan). A1
[1 mark]
The domain \( t \geq 0 \) implies infinite time, but large \( t \) values are not realistic as people do not live forever (e.g., beyond a human lifespan). A1
[1 mark]
Total Marks: 6
Question
Stars are classified by their brightness. The brightest stars in the sky have a magnitude of \(1\). The magnitude, \(m\), of another star can be modelled as a function of its brightness, \(b\), relative to a star of magnitude \(1\), as shown by the equation
\(m=1-2.5\log_{10}(b)\)
The star called Acubens has a brightness of \(0.0525\).
(a) Find the magnitude of Acubens. [2]
Ceres has a magnitude of \(7\) and is the least bright star visible without magnification.
(b) Find the brightness of Ceres. [2]
(c) Find how many times brighter Acubens is compared to Ceres. [2]
▶️Answer/Explanation
Markscheme
(a)
\[ m=1-2.5\log_{10}(0.0525) =1-2.5(-1.27996\ldots)=\boxed{4.20}\ (4.19960\ldots). \] M1 A1
[2 marks]
(b)
\(7=1-2.5\log_{10}(b)\Rightarrow \log_{10}(b)=-2.4\Rightarrow b=10^{-2.4}=\boxed{0.00398107\ldots}.\) M1 A1
[2 marks]
(c)
Brightness ratio \(=\dfrac{0.0525}{0.00398107\ldots}=\boxed{13.2\text{ (approx.)}}\). M1 A1
[2 marks]
Total Marks: 6