IB Mathematics AHL 2.9 modelling of Exponential and other functions AI HL Paper 2- Exam Style Questions- New Syllabus

Question

A scientist is conducting an experiment on the growth of a certain species of bacteria. The population of the bacteria, \( P \), can be modelled by the function:

\( P(t) = 1200 \times k^t \), \( t \geq 0 \)

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.

3 hours after the experiment began, the population of the bacteria is 18 750.
(b) Find the value of \( k \).

(c) Find the population of the bacteria 1 hour and 30 minutes after the experiment began.

The scientist conducts a second experiment with a different species of bacteria.
The population of this bacteria, \( S \), can be modelled by the function \( S(t) = 5000 \times 1.65^t \), \( t \geq 0 \), 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.

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.

▶️ Answer/Explanation

Markscheme

(a)(i)
\( P(0) = 1200 \times k^0 \)
\( = 1200 \times 1 \)
Result:
1200

(a)(ii)
The initial population of the bacteria
Result:
the initial population of the bacteria

(b)
    \( 1200 \times k^3 = 18750 \) (A1)
    \( k^3 = \frac{18750}{1200} \)
    \( = 15.625 \)
    \( k = (15.625)^{1/3} \)
    \( = 2.5 \)
Result:
\( k = 2.5 \)

(c)
    \( t = 1.5 \) hours
    \( P(1.5) = 1200 \times 2.5^{1.5} \) (A1)
    \( 2.5^{1.5} = 2.5 \times \sqrt{2.5} \)
    \( \sqrt{2.5} \approx 1.5811 \)
    \( 2.5 \times 1.5811 \approx 3.9528 \)
    \( 1200 \times 3.9528 \approx 4743.36 \)
    Note: Do not penalize if final answer is not given as an integer. Award (A1)A0 for an answer of 3950 (3949.14…) from use of 1.3 in the exponent, but only if working is shown.
Result:
4740

(d)
    Equating \( P(t) \) and \( S(t) \) or each to a common variable (M1)
    \( 1200 \times 2.5^t = 5000 \times 1.65^t \)
    \( \frac{1200 \times 2.5^t}{5000} = 1.65^t \)
    \( \frac{2.5^t}{4.1667} = 1.65^t \)
    \( \left(\frac{2.5}{1.65}\right)^t = \frac{25}{6} \)
    \( \left(\frac{25}{16.5}\right)^t = \frac{25}{6} \)
    \( t \ln\left(\frac{25}{16.5}\right) = \ln\left(\frac{25}{6}\right) \)
    \( t \approx \frac{\ln(25/6)}{\ln(25/16.5)} \)
    \( \ln(25/6) \approx 1.4469 \), \( \ln(25/16.5) \approx 0.4209 \)
    \( t \approx \frac{1.4469}{0.4209} \approx 3.4346 \)
Result:
\( t = 3.43 \) hours

(e)
    Method 1:
    \( 5000 \times 1.65^t = 19000 \) (M1)
    \( 1.65^t = \frac{19000}{5000} \)
    \( = 3.8 \)
    \( t \ln(1.65) = \ln(3.8) \)
    \( \ln(1.65) \approx 0.5008 \), \( \ln(3.8) \approx 1.3350 \)
    \( t \approx \frac{1.3350}{0.5008} \approx 2.6659 \)
    \( t – 2 = 0.6659 \) hours
    \( m = 0.6659 \times 60 \approx 39.954 \) minutes
    Rounded to nearest integer: \( m = 40 \) minutes
    Method 2:
    Equating expression for \( S(t) \) to 19000 (M1)
    \( 5000 \times 1.65^{2 + m/60} = 19000 \)
    \( 1.65^{2 + m/60} = 3.8 \)
    \( 2 + \frac{m}{60} = 2.6659 \) (A1)
    \( \frac{m}{60} = 0.6659 \)
    \( m \approx 39.954 \) minutes
    Rounded to nearest integer: \( m = 40 \) minutes
    Note: Award (M1)(A1)(M1)A0 for an answer of 39.9521… or 39 with or without working.
Result:
\( m = 40 \) minutes

IB Mathematics AHL 2.9 modelling of Exponential and other functions AI HL Paper 2- Exam Style Questions- New Syllabus

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