Exponential functions and exponential models - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 19 exam-style questions on Exponential functions and exponential models for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

The curve \(C\) has equation \(y = A\mathrm{e}^{kx}\), where \(A\) and \(k\) are constants.

The curve passes through the point (0, 5), and the gradient of \(C\) at this point is \(-15\).

Find \(A\) and \(k\).

Model answer

At (0, 5): \(5 = A\mathrm{e}^0 = A\), so \(A = 5\)
\(\dfrac{dy}{dx} = Ak\mathrm{e}^{kx} = 5k\mathrm{e}^{kx}\)
At \(x = 0\): \(5k = -15\), so \(k = -3\)

Explanation

Since \(\mathrm{e}^0 = 1\), the curve meets the \(y\)-axis at \(y = A\), so \(A = 5\). The gradient of \(\mathrm{e}^{kx}\) is \(k\mathrm{e}^{kx}\), so the gradient at \(x = 0\) is \(Ak = 5k\), giving \(k = -3\). A common error is to differentiate \(\mathrm{e}^{kx}\) as if it were a power of \(x\).

Mark scheme

1 mark
B1 \(A = 5\); or correct final answers of \(A = 5\) and \(k = -3\) with no working
Correct answers of \(A = 5\) and \(k = -3\) with no working score 3 marks.
1 mark
M1 Differentiates to \(\dfrac{dy}{dx} = Ak\mathrm{e}^{kx}\) and sets \(Ak = -15\) at \(x = 0\) (with their \(A\))
The derivative must be of the form \(Ak\mathrm{e}^{kx}\). \(\dfrac{dy}{dx} = kx\mathrm{e}^{kx - 1}\) or \(A\mathrm{e}^{kx}\) scores M0.
1 mark
A1 \(k = -3\)
cao

Question 2

5 marks

The table shows typical values of the air pressure, \(p\) hectopascals (hPa), at a height of \(h\) km above sea level.

By calculating the ratios of the pressures at consecutive heights in the table, explain why a model of the form \(p = 1013\mathrm{e}^{-kh}\), where \(k\) is a positive constant, may be suitable.

Using the pressure when \(h = 4\), find the value of \(k\). Give your answer correct to 3 significant figures.

The summit of Ben Nevis is 1345 m above sea level. Use the model to estimate the air pressure at the summit of Ben Nevis, giving your answer to 3 significant figures.

Table with columns "Height above sea level, h (km)", "Air pressure, p (hPa)". Height above sea level, h (km): 0; Air pressure, p (hPa): 1013 | Height above sea level, h (km): 1; Air pressure, p (hPa): 899 | Height above sea level, h (km): 2; Air pressure, p (hPa): 795 | Height above sea level, h (km): 3; Air pressure, p (hPa): 701 | Height above sea level, h (km): 4; Air pressure, p (hPa): 617.

Model answer

Ratios: \(\dfrac{899}{1013} = 0.887\), \(\dfrac{795}{899} = 0.884\), \(\dfrac{701}{795} = 0.882\), \(\dfrac{617}{701} = 0.880\)
The ratios are approximately equal, so the pressure is multiplied by roughly the same factor for each extra kilometre of height. This is exponential decay, so a model of the form \(p = 1013\mathrm{e}^{-kh}\) may be suitable.
When \(h = 4\): \(1013\mathrm{e}^{-4k} = 617\)
\(\mathrm{e}^{-4k} = \dfrac{617}{1013}\)
\(-4k = \ln\dfrac{617}{1013} \Rightarrow k = \dfrac{1}{4}\ln\dfrac{1013}{617} = 0.124\) (3 s.f.)
At the summit, \(h = 1.345\): \(p = 1013\mathrm{e}^{-0.124 \times 1.345} = 857\) hPa (3 s.f.)

Explanation

In an exponential model the pressure is multiplied by the same factor, \(\mathrm{e}^{-k}\), for each kilometre of height, so roughly equal ratios of consecutive pressures support it. Using \(h = 4\) gives \(\mathrm{e}^{-4k} = \dfrac{617}{1013}\) and \(k = 0.124\), and substituting \(h = 1.345\) (in km) gives about 857 hPa. Common errors are working out differences instead of ratios, and substituting 1345 for \(h\).

Mark scheme

1 mark
M1 Calculates at least two ratios of consecutive pressures, e.g. \(\dfrac{899}{1013} = 0.887\ldots\), \(\dfrac{795}{899} = 0.884\ldots\); or correct final answers of \(k = 0.124\) and 857 hPa with no working
Answers of \(k = 0.124\) and 857 hPa with no working score 3 marks (the two marks for \(k\) and the B1 for the estimate). Accept the reciprocal ratios (1.13, 1.13, 1.13, 1.14).
1 mark
A1 All four ratios (0.887, 0.884, 0.882, 0.880, or their reciprocals) with the conclusion that they are approximately equal, so the pressure is multiplied by roughly the same factor for each extra kilometre of height, as in an exponential model
Accept ratios to 2 s.f. or better. The conclusion must refer to the ratios being (approximately) constant. Differences of consecutive pressures (114, 104, 94, 84) are not ratios: M0A0.
1 mark
M1 Sets \(1013\mathrm{e}^{-4k} = 617\) and takes natural logarithms: \(-4k = \ln\dfrac{617}{1013}\)
Not dependent on the first M mark.
1 mark
A1 \(k =\) awrt 0.124, i.e. in [0.1235, 0.1245)
The value is 0.12395... Do not accept 0.12 as the final answer.
1 mark
B1 awrt 857 hPa, i.e. in [856.5, 857.5)
B1ft on their \(k\): \(1013\mathrm{e}^{-1.345k}\). The value is 857.4 with \(k = 0.124\) or with the unrounded \(k\). Using \(h = 1345\) (metres instead of kilometres) gives a value close to 0: B0. Units are not required.

Question 3

7 marks

Ella invests £3000 in a savings account. The amount, £\(A\), in the account \(t\) years after she invests the money is modelled by
\[A = 3000\mathrm{e}^{0.035t}\]
Find, according to the model, the time taken for the amount in the account to double. Give your answer in years, correct to 3 significant figures.

Show that, according to the model, the amount in the account increases by approximately 3.56% each year.

A different account pays interest of 3.6% of the amount in the account once at the end of each year. Determine which account would give Ella more money after 10 years, and by how much.

Model answer

Doubling: \(3000\mathrm{e}^{0.035t} = 6000 \Rightarrow \mathrm{e}^{0.035t} = 2\)
\(t = \dfrac{\ln 2}{0.035} = 19.8\) years (3 s.f.)
\(\dfrac{A(t + 1)}{A(t)} = \dfrac{3000\mathrm{e}^{0.035(t + 1)}}{3000\mathrm{e}^{0.035t}} = \mathrm{e}^{0.035} = 1.0356\ldots\)
So each year the amount is multiplied by 1.0356..., an increase of approximately 3.56%.
After 10 years: first account \(3000\mathrm{e}^{0.35} = 4257.20\), i.e. £4257.20; second account \(3000 \times 1.036^{10} = 4272.86\), i.e. £4272.86
\(4272.86 - 4257.20 = 15.66\), so the second account gives more, by £15.66.

Explanation

Doubling means \(\mathrm{e}^{0.035t} = 2\), solved with natural logarithms. In the model the amount is multiplied by \(\mathrm{e}^{0.035} = 1.0356\) every year, which is an annual increase of 3.56%, slightly less than 3.6%, so the second account gives more after 10 years. Common errors are using 1.36 or 0.036 as the yearly multiplier and not saying which account gives more.

Mark scheme

1 mark
M1 Sets \(3000\mathrm{e}^{0.035t} = 6000\), so \(\mathrm{e}^{0.035t} = 2\), and uses logarithms: \(t = \dfrac{\ln 2}{0.035}\). An answer with no working scores 0 marks
An answer with no working scores 0 marks for the show-that. SC: an answer of 19.8 years with no working scores M1 A1, and correct amounts and a correct conclusion for the 10-year comparison with no working score M1 A1 A1. For the doubling time, accept \(\mathrm{e}^{0.035t} = 2\) solved with logarithms to any base.
1 mark
A1 awrt 19.8 years, i.e. in [19.75, 19.85)
The value is 19.804...
1 mark
M1 Finds the yearly growth factor: \(\dfrac{A(t + 1)}{A(t)} = \mathrm{e}^{0.035}\), or \(\dfrac{A(1)}{A(0)} = \dfrac{3000\mathrm{e}^{0.035}}{3000}\)
Accept a numerical comparison of two consecutive years, e.g. \(\dfrac{3106.86}{3000}\).
1 mark
A1* \(\mathrm{e}^{0.035} = 1.0356\ldots\), so the amount increases by approximately 3.56% each year (cso)
Accept \(\mathrm{e}^{0.035}\) as 1.0356 or more accurately (at least 4 d.p.); a link to 3.56% is required.
1 mark
M1 Finds both amounts after 10 years: \(3000\mathrm{e}^{0.35}\) and \(3000 \times 1.036^{10}\)
Accept the amounts to the nearest pound as evidence. Using \(1.36^{10}\) or \(3000 \times 0.036 \times 10\) for the second account scores M0.
1 mark
A1 awrt £4257.20 and awrt £4272.86, i.e. in [4257.195, 4257.205) and [4272.855, 4272.865)
Both needed. Accept values to the nearest pound (£4257 and £4273).
1 mark
A1 The second account (3.6% added yearly) gives more, by £15.66
The account must be identified, and the difference is a sum of money, so units (£ or pounds) are required. Accept awrt 15.66, i.e. in [15.655, 15.665), or £15.7, or £16 from amounts rounded to the nearest pound.

Question 4

8 marks

A biologist grows yeast in a sugar solution. Two models for the number of yeast cells, \(N\) thousand, \(t\) hours after the start of the experiment are

Model A: \(N = 40\mathrm{e}^{0.5t}\)
Model B: \(N = \dfrac{2000}{1 + 49\mathrm{e}^{-0.5t}}\)

Show that the two models predict the same initial number of yeast cells.

Find the number of yeast cells predicted by each model when \(t = 10\).

State the number of yeast cells that model B predicts in the long term. Given that the sugar solution can support at most 2 million yeast cells, explain which model is more suitable.

Using model B, find the exact value of \(t\) at which \(N = 1000\).

Model answer

When \(t = 0\): model A \(N = 40\mathrm{e}^0 = 40\); model B \(N = \dfrac{2000}{1 + 49} = 40\). Both predict 40 thousand yeast cells initially.
When \(t = 10\): model A \(N = 40\mathrm{e}^5 = 5936.5\ldots\), about 5940 thousand (5.94 million)
model B \(N = \dfrac{2000}{1 + 49\mathrm{e}^{-5}} = 1503.5\ldots\), about 1500 thousand (1.50 million)
As \(t \to \infty\), \(\mathrm{e}^{-0.5t} \to 0\), so model B gives \(N \to 2000\): 2 million yeast cells.
Model B is more suitable: it levels off at 2 million, the most the solution can support, whereas model A increases without limit and already predicts about 5.94 million at \(t = 10\).
Model B: \(\dfrac{2000}{1 + 49\mathrm{e}^{-0.5t}} = 1000 \Rightarrow 1 + 49\mathrm{e}^{-0.5t} = 2 \Rightarrow \mathrm{e}^{-0.5t} = \dfrac{1}{49}\)
\(-0.5t = \ln\dfrac{1}{49} \Rightarrow t = 2\ln 49 = 4\ln 7\)

Explanation

Both models start at 40 thousand because \(\mathrm{e}^0 = 1\), but model A grows without limit while in model B the term \(49\mathrm{e}^{-0.5t}\) tends to 0, so \(N\) levels off at 2000 thousand, which matches the 2 million the solution can support. To solve model B equal to 1000, rearrange to \(\mathrm{e}^{-0.5t} = \dfrac{1}{49}\) before taking logarithms. Common errors are forgetting that \(N\) is in thousands and taking logarithms of each term of the denominator separately.

Mark scheme

1 mark
B1 Shows that both models give \(N = 40\) when \(t = 0\): \(40\mathrm{e}^0 = 40\) and \(\dfrac{2000}{1 + 49} = 40\), so 40 000 cells. The show-that needs working (an answer with no working scores 0 marks for it); correct final answers with no working score up to 5 marks, as stated in the comments
An answer with no working scores 0 marks for the show-that, where both substitutions must be seen. Correct answers with no working score 5 marks: 5940 thousand and 1500 thousand (M1 A1), 2 million (B1), model B with a valid reason (B1, as the reason is the working) and \(t = 2\ln 49\) (1 mark, the final A1).
1 mark
M1 Substitutes \(t = 10\) into both models: \(40\mathrm{e}^{5}\) and \(\dfrac{2000}{1 + 49\mathrm{e}^{-5}}\)
1 mark
A1 Model A: awrt 5940 thousand (about 5.94 million), i.e. in [5935, 5945); model B: awrt 1500 thousand (about 1.50 million), i.e. in [1495, 1505)
Both needed. The values are 5936.5... and 1503.5... (thousand); accept 5 940 000 and 1 500 000 (or 1 504 000) cells. Answers of 5940 and 1500 with no indication of thousands are condoned.
1 mark
B1 2 million (2000 thousand) in the long term
From \(\mathrm{e}^{-0.5t} \to 0\) as \(t \to \infty\). Accept 2000 if it is clear that \(N\) is in thousands.
1 mark
B1 Model B is more suitable, with a reason, e.g. it levels off at 2 million, the most the solution can support, whereas model A grows without limit (it predicts about 5.94 million cells at \(t = 10\))
Requires model B and a reason that refers to the limit of 2 million. Do not accept "model B because the numbers are smaller".
1 mark
M1 Sets \(\dfrac{2000}{1 + 49\mathrm{e}^{-0.5t}} = 1000\) and rearranges to \(\mathrm{e}^{-0.5t} = \dfrac{1}{49}\)
1 mark
M1 (dM1) Uses logarithms: \(-0.5t = \ln\dfrac{1}{49}\)
1 mark
A1 \(t = 2\ln 49\)
Accept \(4\ln 7\) or any exact equivalent. The decimal 7.78 on its own scores A0 as the exact value is required.

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