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IB Mathematics SL 1.1 Operations with numbers AI SL Paper 2 - Exam Style Questions - New Syllabus

IB DP Maths AI SL Paper 2 - All Topics

Question

Dr. Miller is conducting an experiment on the growth of a particular species of bacteria.
The population of the bacteria, \(P\), can be modelled by the function \[ P(t)=1200\times k^{\,t},\qquad t\ge 0, \] where \(t\) is the number of hours since the experiment began, and \(k\) is a positive constant.
(a) (i) Write down the value of \(P(0)\).
(ii) Interpret what this value means in this context. [2]
3 hours after the experiment began, the population of the bacteria is \(18\ 750\).
(b) Find the value of \(k\). [2]
(c) Find the population of the bacteria 1 hour and 30 minutes after the experiment began. [2]
Dr. Miller conducts a second experiment with a different species of bacteria. The population of this bacteria, \(S\), can be modelled by the function \[ S(t)=5000\times 1.65^{\,t},\qquad t\ge 0, \] where \(t\) is the number of hours since both experiments began.
(d) Find the value of \(t\) when the two populations of bacteria are equal. [2]
It takes 2 hours and \(m\) minutes for the number of bacteria in the second experiment to reach \(19\ 000\).
(e) Find the value of \(m\), giving your answer as an integer value. [4]
The bacteria in the second experiment are growing inside a container. Dr. Miller models the volume of each bacterium in the second experiment to be \(1\times 10^{-18}\text{ m}^3\), and the available volume inside the container is \(2.1\times 10^{-5}\text{ m}^3\).
(f) Determine how long it would take for the bacteria to fill the container. [3]
▶️ Answer/Explanation
Markscheme (with detailed working)

(a)

(i) \(P(0)=\boxed{1200}\). A1
(ii) This is the initial population of the bacteria at the start of the experiment. A1

(b)

\(P(3)=1200\,k^{3}=18\,750\Rightarrow k^{3}=15.625\Rightarrow \boxed{k=2.5}.\) A1 A1

(c)

\(t=1.5\) h: \(P(1.5)=1200\,(2.5)^{1.5}=4743.41\ldots\approx \boxed{4740}\) (to 3 s.f.).
(Accept exact/unsimplified calculator value; do not penalize if not an integer.) A1 A1

(d)

Solve \(1200\,(2.5)^{t}=5000\,(1.65)^{t}\). \(\ \Rightarrow\ (2.5/1.65)^{t}=5000/1200\). \(\ \Rightarrow\ t=\dfrac{\ln(25/16.5)}{\ln(2.5/1.65)}=\boxed{3.43\ \text{hours}}\ (3.43456\ldots).\) M1 A1

(e)

\(5000\cdot 1.65^{\,t}=19\,000\Rightarrow t=\dfrac{\ln(19\,000/5000)}{\ln 1.65}=2.66586\ldots\ \text{h}.\)
With \(t=2+\dfrac{m}{60}\): \(\dfrac{m}{60}=0.66586\ldots\Rightarrow m=39.95\ldots\approx \boxed{40\ \text{minutes}}\) (or \(2\text{ h }40\text{ min}\)). M1 A1 M1 A1

(f)

Container fits \(\dfrac{2.1\times 10^{-5}}{1\times 10^{-18}}=2.1\times 10^{13}\) bacteria. Set \(5000\cdot 1.65^{\,t}=2.1\times 10^{13}\) and solve for \(t\): \[ t=\frac{\ln(2.1\times 10^{13}/5000)}{\ln 1.65}=\boxed{44.2\ \text{hours}}\ (44.2480\ldots). \] M1 M1 A1
Total Marks: 15
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