IB Mathematics AHL 1.11 The sum of infinite geometric sequences AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question models the concentration of a medicinal drug in a patient named Somsak, who takes the drug repeatedly over time.
For a patient named Somsak, let \( C(t) \), in milligrams per millilitre (mg ml\(^{-1}\)), represent the drug concentration in his blood \( t \) hours after a dose. The drug’s breakdown rate is proportional to its concentration, modeled by the differential equation:
\[ \frac{dC}{dt} = -kC, \quad k \in \mathbb{R}^+. \]
The initial concentration after a dose is \( d \, \text{mg ml}^{-1} \), \( d > 0 \).
(a) Solve the differential equation to show that \( C = de^{-kt} \). [3]
Consider a drug taken by Somsak with \( k = 0.2 \). The first dose is given at \( t = 0 \), with no drug present initially.
(b) Calculate the time, in hours, for the drug concentration to reach 5% of its initial value. [2]
Somsak takes the drug every \( T \) hours in doses that increase the concentration by \( d \, \text{mg ml}^{-1} \). Assume each dose instantly increases the blood concentration.
(c)(i) Show that immediately after Somsak’s third dose, the concentration is \( d (1 + e^{-0.2T} + e^{-0.4T}) \). [3]
(c)(ii) State the concentration immediately after Somsak’s \( n \)th dose. [1]
(d) Show that the concentration after the \( n \)th dose is \( d \cdot \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \). [3]
After Somsak takes the drug for a long time, the concentration must remain within a safe and effective range. Let \( H_n \) be the highest concentration in the interval \((n-1)T \leq t < nT \), and \( L_n \) be the lowest concentration in the same interval.
Define \( H_\infty = \lim_{n \to \infty} H_n \) and \( L_\infty = \lim_{n \to \infty} L_n \).
(e)(i) Find an expression for \( H_\infty \), the long-term highest concentration, in terms of \( d \) and \( T \). [2]
(e)(ii) Find an expression for \( L_\infty \), the long-term lowest concentration, in terms of \( d \) and \( T \). [2]
(f) Demonstrate that:
(i) \( H_\infty – L_\infty = d \); [2]
(ii) \( \ln\left( \frac{H_\infty}{L_\infty} \right) = 0.2T \). [2]
(i) \( H_\infty – L_\infty = d \); [2]
(ii) \( \ln\left( \frac{H_\infty}{L_\infty} \right) = 0.2T \). [2]
For Somsak, the drug is ineffective if the long-term concentration falls below \( 0.06 \, \text{mg ml}^{-1} \) and safe if it never exceeds \( 0.28 \, \text{mg ml}^{-1} \).
(g) Find suitable values for:
(i) \( d \); [2]
(ii) \( T \). [2]
(i) \( d \); [2]
(ii) \( T \). [2]
(h) Using the values of \( d \) and \( T \) from part (g), calculate the proportion of time between Somsak’s first and second doses where the concentration is at least \( 0.06 \, \text{mg ml}^{-1} \). [3]
(i) Suggest a reason why the drug’s label might recommend a different \( T \) than that found in part (g)(ii). [1]
▶️ Answer/Explanation
Markscheme
(a)
Solve \( \frac{dC}{dt} = -kC \):
Separate variables: \( \frac{dC}{C} = -k \, dt \).
Integrate: \( \int \frac{1}{C} \, dC = \int -k \, dt \Rightarrow \ln|C| = -kt + c \).
Exponentiate: \( C = A e^{-kt} \), where \( A = e^c \).
Initial condition: \( C(0) = d \Rightarrow A = d \).
Thus, \( C(t) = d e^{-kt} \).
\( C = d e^{-kt} \) M1A1A1
[3 marks]
Solve \( \frac{dC}{dt} = -kC \):
Separate variables: \( \frac{dC}{C} = -k \, dt \).
Integrate: \( \int \frac{1}{C} \, dC = \int -k \, dt \Rightarrow \ln|C| = -kt + c \).
Exponentiate: \( C = A e^{-kt} \), where \( A = e^c \).
Initial condition: \( C(0) = d \Rightarrow A = d \).
Thus, \( C(t) = d e^{-kt} \).
\( C = d e^{-kt} \) M1A1A1
[3 marks]
(b)
For \( k = 0.2 \), find time when \( C(t) = 0.05d \):
\( 0.05d = d e^{-0.2t} \Rightarrow 0.05 = e^{-0.2t} \).
\( \ln(0.05) = -0.2t \Rightarrow -2.995732 \approx -0.2t \).
\( t = \frac{2.995732}{0.2} \approx 14.97866 \approx 15.0 \, \text{h} \).
Time \( \approx 15.0 \, \text{h} \) M1A1
[2 marks]
For \( k = 0.2 \), find time when \( C(t) = 0.05d \):
\( 0.05d = d e^{-0.2t} \Rightarrow 0.05 = e^{-0.2t} \).
\( \ln(0.05) = -0.2t \Rightarrow -2.995732 \approx -0.2t \).
\( t = \frac{2.995732}{0.2} \approx 14.97866 \approx 15.0 \, \text{h} \).
Time \( \approx 15.0 \, \text{h} \) M1A1
[2 marks]
(c)(i)
After the third dose at \( t = 2T \):
1st dose (at \( t = 0 \)) decays for \( 2T \): \( d e^{-0.2 \cdot 2T} = d e^{-0.4T} \).
2nd dose (at \( t = T \)) decays for \( T \): \( d e^{-0.2T} \).
3rd dose (at \( t = 2T \)) adds \( d \).
Total: \( C = d + d e^{-0.2T} + d e^{-0.4T} = d (1 + e^{-0.2T} + e^{-0.4T}) \).
\( C = d (1 + e^{-0.2T} + e^{-0.4T}) \) M1A1A1
[3 marks]
After the third dose at \( t = 2T \):
1st dose (at \( t = 0 \)) decays for \( 2T \): \( d e^{-0.2 \cdot 2T} = d e^{-0.4T} \).
2nd dose (at \( t = T \)) decays for \( T \): \( d e^{-0.2T} \).
3rd dose (at \( t = 2T \)) adds \( d \).
Total: \( C = d + d e^{-0.2T} + d e^{-0.4T} = d (1 + e^{-0.2T} + e^{-0.4T}) \).
\( C = d (1 + e^{-0.2T} + e^{-0.4T}) \) M1A1A1
[3 marks]
(c)(ii)
After the \( n \)th dose at \( t = (n-1)T \):
\( C = d (1 + e^{-0.2T} + e^{-0.4T} + \dots + e^{-0.2(n-1)T}) \).
\( C = d \sum_{i=0}^{n-1} e^{-0.2iT} \) A1
[1 mark]
After the \( n \)th dose at \( t = (n-1)T \):
\( C = d (1 + e^{-0.2T} + e^{-0.4T} + \dots + e^{-0.2(n-1)T}) \).
\( C = d \sum_{i=0}^{n-1} e^{-0.2iT} \) A1
[1 mark]
(d)
Sum the geometric series: \( C = d (1 + e^{-0.2T} + e^{-0.4T} + \dots + e^{-0.2(n-1)T}) \).
First term: 1, ratio: \( r = e^{-0.2T} \), \( n \) terms:
\( \sum_{i=0}^{n-1} (e^{-0.2T})^i = \frac{1 – (e^{-0.2T})^n}{1 – e^{-0.2T}} = \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \).
Thus, \( C = d \cdot \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \).
\( C = d \cdot \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \) M1A1A1
[3 marks]
Sum the geometric series: \( C = d (1 + e^{-0.2T} + e^{-0.4T} + \dots + e^{-0.2(n-1)T}) \).
First term: 1, ratio: \( r = e^{-0.2T} \), \( n \) terms:
\( \sum_{i=0}^{n-1} (e^{-0.2T})^i = \frac{1 – (e^{-0.2T})^n}{1 – e^{-0.2T}} = \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \).
Thus, \( C = d \cdot \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \).
\( C = d \cdot \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \) M1A1A1
[3 marks]
(e)(i)
\( H_n = d \cdot \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \). As \( n \to \infty \), \( e^{-0.2nT} \to 0 \):
\( H_\infty = \lim_{n \to \infty} H_n = \frac{d}{1 – e^{-0.2T}} \).
\( H_\infty = \frac{d}{1 – e^{-0.2T}} \) M1A1
[2 marks]
\( H_n = d \cdot \frac{1 – e^{-0.2nT}}{1 – e^{-0.2T}} \). As \( n \to \infty \), \( e^{-0.2nT} \to 0 \):
\( H_\infty = \lim_{n \to \infty} H_n = \frac{d}{1 – e^{-0.2T}} \).
\( H_\infty = \frac{d}{1 – e^{-0.2T}} \) M1A1
[2 marks]
(e)(ii)
\( L_n = H_n e^{-0.2T} \). As \( n \to \infty \):
\( L_\infty = H_\infty e^{-0.2T} = \frac{d}{1 – e^{-0.2T}} \cdot e^{-0.2T} = \frac{d}{e^{0.2T} – 1} \).
\( L_\infty = \frac{d}{e^{0.2T} – 1} \) M1A1
[2 marks]
\( L_n = H_n e^{-0.2T} \). As \( n \to \infty \):
\( L_\infty = H_\infty e^{-0.2T} = \frac{d}{1 – e^{-0.2T}} \cdot e^{-0.2T} = \frac{d}{e^{0.2T} – 1} \).
\( L_\infty = \frac{d}{e^{0.2T} – 1} \) M1A1
[2 marks]
(f)(i)
\( H_\infty – L_\infty = \frac{d}{1 – e^{-0.2T}} – \frac{d}{e^{0.2T} – 1} \).
Let \( q = e^{-0.2T} \):
\( H_\infty – L_\infty = \frac{d}{1 – q} – \frac{d}{\frac{1}{q} – 1} = \frac{d}{1 – q} – \frac{d q}{1 – q} = \frac{d (1 – q)}{1 – q} = d \).
\( H_\infty – L_\infty = d \) M1A1
[2 marks]
\( H_\infty – L_\infty = \frac{d}{1 – e^{-0.2T}} – \frac{d}{e^{0.2T} – 1} \).
Let \( q = e^{-0.2T} \):
\( H_\infty – L_\infty = \frac{d}{1 – q} – \frac{d}{\frac{1}{q} – 1} = \frac{d}{1 – q} – \frac{d q}{1 – q} = \frac{d (1 – q)}{1 – q} = d \).
\( H_\infty – L_\infty = d \) M1A1
[2 marks]
(f)(ii)
\( \frac{H_\infty}{L_\infty} = \frac{\frac{d}{1 – e^{-0.2T}}}{\frac{d}{e^{0.2T} – 1}} = \frac{e^{0.2T} – 1}{1 – e^{-0.2T}} = \frac{e^{0.2T} – 1}{\frac{1 – e^{0.2T}}{e^{0.2T}}} = e^{0.2T} \).
\( \ln\left( \frac{H_\infty}{L_\infty} \right) = \ln(e^{0.2T}) = 0.2T \).
\( \ln\left( \frac{H_\infty}{L_\infty} \right) = 0.2T \) M1A1
[2 marks]
\( \frac{H_\infty}{L_\infty} = \frac{\frac{d}{1 – e^{-0.2T}}}{\frac{d}{e^{0.2T} – 1}} = \frac{e^{0.2T} – 1}{1 – e^{-0.2T}} = \frac{e^{0.2T} – 1}{\frac{1 – e^{0.2T}}{e^{0.2T}}} = e^{0.2T} \).
\( \ln\left( \frac{H_\infty}{L_\infty} \right) = \ln(e^{0.2T}) = 0.2T \).
\( \ln\left( \frac{H_\infty}{L_\infty} \right) = 0.2T \) M1A1
[2 marks]
(g)(i)
Given \( L_\infty \geq 0.06 \), \( H_\infty \leq 0.28 \). From (f)(i):
\( H_\infty – L_\infty = d \Rightarrow d = 0.28 – 0.06 = 0.22 \, \text{mg ml}^{-1} \).
\( d = 0.22 \, \text{mg ml}^{-1} \) M1A1
[2 marks]
Given \( L_\infty \geq 0.06 \), \( H_\infty \leq 0.28 \). From (f)(i):
\( H_\infty – L_\infty = d \Rightarrow d = 0.28 – 0.06 = 0.22 \, \text{mg ml}^{-1} \).
\( d = 0.22 \, \text{mg ml}^{-1} \) M1A1
[2 marks]
(g)(ii)
From (f)(ii): \( \ln\left( \frac{0.28}{0.06} \right) = 0.2T \).
\( \frac{0.28}{0.06} = \frac{14}{3} \approx 4.666\overline{6} \).
\( \ln(4.666\overline{6}) \approx 1.54045 \Rightarrow 0.2T \approx 1.54045 \Rightarrow T \approx \frac{1.54045}{0.2} \approx 7.702 \, \text{h} \).
\( T \approx 7.70 \, \text{h} \) M1A1
[2 marks]
From (f)(ii): \( \ln\left( \frac{0.28}{0.06} \right) = 0.2T \).
\( \frac{0.28}{0.06} = \frac{14}{3} \approx 4.666\overline{6} \).
\( \ln(4.666\overline{6}) \approx 1.54045 \Rightarrow 0.2T \approx 1.54045 \Rightarrow T \approx \frac{1.54045}{0.2} \approx 7.702 \, \text{h} \).
\( T \approx 7.70 \, \text{h} \) M1A1
[2 marks]
(h)
Using \( d = 0.22 \), \( T \approx 7.702 \), first interval \( [0, T] \): \( C(t) = 0.22 e^{-0.2t} \).
Solve \( C(t) \geq 0.06 \):
\( 0.22 e^{-0.2t} \geq 0.06 \Rightarrow e^{-0.2t} \geq \frac{0.06}{0.22} \approx 0.272727 \).
\( -0.2t \geq \ln(0.272727) \approx -1.29928 \Rightarrow t \leq \frac{1.29928}{0.2} \approx 6.4964 \, \text{h} \).
Proportion: \( \frac{6.4964}{7.702} \approx 0.843 \).
Proportion \( \approx 0.843 \ (84.3\%) \) M1A1A1
[3 marks]
Using \( d = 0.22 \), \( T \approx 7.702 \), first interval \( [0, T] \): \( C(t) = 0.22 e^{-0.2t} \).
Solve \( C(t) \geq 0.06 \):
\( 0.22 e^{-0.2t} \geq 0.06 \Rightarrow e^{-0.2t} \geq \frac{0.06}{0.22} \approx 0.272727 \).
\( -0.2t \geq \ln(0.272727) \approx -1.29928 \Rightarrow t \leq \frac{1.29928}{0.2} \approx 6.4964 \, \text{h} \).
Proportion: \( \frac{6.4964}{7.702} \approx 0.843 \).
Proportion \( \approx 0.843 \ (84.3\%) \) M1A1A1
[3 marks]
(i)
The label may recommend \( T \approx 8 \, \text{h} \) (three times daily) for easier patient adherence.
Practical scheduling for adherence A1
[1 mark]
The label may recommend \( T \approx 8 \, \text{h} \) (three times daily) for easier patient adherence.
Practical scheduling for adherence A1
[1 mark]
Total Marks: 28