Exponential functions and exponential models - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Exponential functions and exponential models for Edexcel A-Level Maths (Paper 2). 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.

18 questions
3–10 marks each
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18 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 with equation \(y = \mathrm{e}^{ax + b}\), where \(a\) and \(b\) are constants, passes through the points \((1, \mathrm{e}^5)\) and \((3, \mathrm{e})\).

Find \(a\) and \(b\).

Model answer

At \((1, \mathrm{e}^5)\): \(\mathrm{e}^{a + b} = \mathrm{e}^5\), so \(a + b = 5\)
At \((3, \mathrm{e})\): \(\mathrm{e}^{3a + b} = \mathrm{e}^1\), so \(3a + b = 1\)
Subtracting: \(2a = -4\), so \(a = -2\)
\(b = 5 - a = 7\)

Explanation

Taking natural logarithms of \(y = \mathrm{e}^{ax + b}\) gives \(\ln y = ax + b\), so each point gives a linear equation in \(a\) and \(b\), with \(\ln \mathrm{e}^5 = 5\) and \(\ln \mathrm{e} = 1\). Solving them simultaneously gives \(a = -2\) and \(b = 7\). A common error is to write \(\ln \mathrm{e} = 0\), giving \(3a + b = 0\).

Mark scheme

1 mark
M1 Uses the points to form two linear equations, e.g. by taking natural logarithms: \(a + b = 5\) and \(3a + b = 1\); or correct final answers of \(a = -2\) and \(b = 7\) with no working
Correct answers of \(a = -2\) and \(b = 7\) with no working score 3 marks. Writing \(\mathrm{e}^{a + b} = \mathrm{e}^5\) and comparing powers is equivalent.
1 mark
A1 \(a = -2\)
1 mark
A1 \(b = 7\)
Depends on the M mark.

Question 2

5 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The curve \(C_1\) has equation \(y = 2\mathrm{e}^{2x}\) and the curve \(C_2\) has equation \(y = 5\mathrm{e}^{x} + 12\).

Show that \(C_1\) and \(C_2\) meet at exactly one point, and find the exact coordinates of this point.

Model answer

At an intersection: \(2\mathrm{e}^{2x} = 5\mathrm{e}^{x} + 12\)
\(2(\mathrm{e}^{x})^2 - 5\mathrm{e}^{x} - 12 = 0\)
\((2\mathrm{e}^{x} + 3)(\mathrm{e}^{x} - 4) = 0\)
\(\mathrm{e}^{x} = 4\) or \(\mathrm{e}^{x} = -\dfrac{3}{2}\)
\(\mathrm{e}^{x} > 0\) for all \(x\), so \(\mathrm{e}^{x} = -\dfrac{3}{2}\) has no solution. Hence there is exactly one value of \(x\), and the curves meet at exactly one point.
\(\mathrm{e}^{x} = 4\), so \(x = \ln 4\) and \(y = 2\mathrm{e}^{2\ln 4} = 2 \times 4^2 = 32\)
The point is \((\ln 4, 32)\).

Explanation

Because \(\mathrm{e}^{2x} = (\mathrm{e}^{x})^2\), the equation is a quadratic in \(\mathrm{e}^{x}\), which factorises. Only one root is possible because an exponential is always positive, and this is what shows there is exactly one point. Common errors are writing \(\mathrm{e}^{2x}\) as \(2\mathrm{e}^{x}\) and keeping \(\ln\left(-\frac{3}{2}\right)\) as a second answer.

Mark scheme

1 mark
M1 Sets \(2\mathrm{e}^{2x} = 5\mathrm{e}^{x} + 12\) and forms a 3-term quadratic in \(\mathrm{e}^{x}\): \(2(\mathrm{e}^{x})^2 - 5\mathrm{e}^{x} - 12 = 0\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the question demands working, so \((\ln 4, 32)\) with no working scores 0 marks. Allow \(u = \mathrm{e}^{x}\): \(2u^2 - 5u - 12 = 0\).
1 mark
M1 (dM1) Solves their quadratic by factorising, the formula or completing the square: \((2\mathrm{e}^{x} + 3)(\mathrm{e}^{x} - 4) = 0\)
Depends on the previous M mark. Roots stated from a calculator with no factorising, formula or completing the square seen score M0.
1 mark
A1 \(\mathrm{e}^{x} = 4\) or \(\mathrm{e}^{x} = -\dfrac{3}{2}\)
Both values are needed for this mark.
1 mark
B1 States that \(\mathrm{e}^{x} = -\dfrac{3}{2}\) has no solution because \(\mathrm{e}^{x} > 0\) for all \(x\), so there is exactly one point of intersection
Requires the reason and the conclusion. "\(\ln\left(-\tfrac{3}{2}\right)\) is undefined" or "you cannot take the log of a negative number" is an acceptable reason. Do not accept "\(\mathrm{e}^x\) cannot be negative" with no conclusion about the number of points.
1 mark
A1 \((\ln 4, 32)\)
Depends on both M marks. Both coordinates are needed and must be exact: accept \((2\ln 2, 32)\); \((1.39, 32)\) scores A0.

Question 3

6 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Using \(a^x = \mathrm{e}^{x\ln a}\), show that the gradient of the curve with equation \(y = a^x\), where \(a\) is a positive constant, at the point \((0, 1)\) is \(\ln a\).

The curves \(C_1\) and \(C_2\) have equations \(y = 2^x\) and \(y = 3^x\) respectively. There is exactly one value of \(x\) at which \(C_1\) and \(C_2\) have the same gradient. Find this value of \(x\), giving your answer in an exact form.

State, with a reason, which of \(C_1\) and \(C_2\) is steeper for all values of \(x\) greater than this value.

Model answer

\(y = a^x = \mathrm{e}^{x\ln a}\), where \(\ln a\) is a constant, so \(\dfrac{dy}{dx} = \ln a \times \mathrm{e}^{x\ln a}\)
At \((0, 1)\): \(\dfrac{dy}{dx} = \ln a \times \mathrm{e}^{0} = \ln a\), as required
In general \(\dfrac{dy}{dx} = a^x\ln a\), so the gradients of \(C_1\) and \(C_2\) are equal when \(2^x\ln 2 = 3^x\ln 3\)
\(\dfrac{3^x}{2^x} = \dfrac{\ln 2}{\ln 3}\), so \(\left(\dfrac{3}{2}\right)^x = \dfrac{\ln 2}{\ln 3}\)
\(x\ln\dfrac{3}{2} = \ln\left(\dfrac{\ln 2}{\ln 3}\right)\), so \(x = \dfrac{\ln\left(\frac{\ln 2}{\ln 3}\right)}{\ln\frac{3}{2}}\) \((= -1.14\ldots)\)
The ratio of the gradients of \(C_2\) and \(C_1\) is \(\left(\dfrac{3}{2}\right)^x\dfrac{\ln 3}{\ln 2}\), which increases as \(x\) increases and equals 1 at this value, so for all greater values of \(x\) it is greater than 1: \(C_2\) is steeper.

Explanation

Writing \(a^x\) as \(\mathrm{e}^{x\ln a}\) and using the chain rule gives the gradient \(a^x\ln a\), which is \(\ln a\) at \((0, 1)\). Setting \(2^x\ln 2 = 3^x\ln 3\) and collecting the powers gives \(\left(\frac{3}{2}\right)^x = \frac{\ln 2}{\ln 3}\), which is solved exactly with logarithms. A common error is to differentiate \(a^x\) as \(xa^{x - 1}\), which treats it as a power of \(x\).

Mark scheme

1 mark
M1 Differentiates \(\mathrm{e}^{x\ln a}\) with the chain rule, treating \(\ln a\) as a constant: \(\dfrac{dy}{dx} = \ln a \times \mathrm{e}^{x\ln a}\) \((= a^x\ln a)\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the question demands working, so a value of \(x\) with no working scores 0 marks. Writing \(\dfrac{dy}{dx} = xa^{x - 1}\) scores M0.
1 mark
A1* At \(x = 0\): \(\dfrac{dy}{dx} = \ln a \times \mathrm{e}^{0} = \ln a\), cso
A conclusion is needed.
1 mark
M1 Sets the gradients equal: \(2^x\ln 2 = 3^x\ln 3\)
Independent of the first two marks: may use the result \(\dfrac{d}{dx}\left(a^x\right) = a^x\ln a\). Setting the gradients equal only at \(x = 0\) (\(\ln 2 = \ln 3\)) scores M0.
1 mark
M1 (dM1) Collects the powers and takes logarithms: \(\left(\dfrac{3}{2}\right)^x = \dfrac{\ln 2}{\ln 3}\), so \(x\ln\dfrac{3}{2} = \ln\left(\dfrac{\ln 2}{\ln 3}\right)\)
Depends on the previous M mark. Alternative: takes logarithms first, \(x\ln 2 + \ln(\ln 2) = x\ln 3 + \ln(\ln 3)\).
1 mark
A1 \(x = \dfrac{\ln\left(\frac{\ln 2}{\ln 3}\right)}{\ln\frac{3}{2}}\)
Accept any exact equivalent, e.g. \(\dfrac{\ln(\ln 2) - \ln(\ln 3)}{\ln 3 - \ln 2}\) or \(\log_{1.5}\left(\dfrac{\ln 2}{\ln 3}\right)\). A decimal only (\(-1.14\)) scores A0; isw after a correct exact answer.
1 mark
B1 \(C_2\) (\(y = 3^x\)) is steeper, because \(\left(\dfrac{3}{2}\right)^x\) increases with \(x\), so for greater values of \(x\), \(\left(\dfrac{3}{2}\right)^x > \dfrac{\ln 2}{\ln 3}\) and \(3^x\ln 3 > 2^x\ln 2\)
Requires a reason linked to the gradients, e.g. the ratio of the gradients, \(\left(\tfrac{3}{2}\right)^x\dfrac{\ln 3}{\ln 2}\), increases with \(x\) and equals 1 at this value. "\(3 > 2\)" alone scores B0.

Question 4

8 marks

Radon is a radioactive gas. In a cellar, the source of radon is sealed off at time \(t = 0\), when the concentration of radon in the air is 950 becquerels per cubic metre (\(\text{Bq m}^{-3}\)). A model for the concentration, \(R\) \(\text{Bq m}^{-3}\), \(t\) days later is
\[R = 950\mathrm{e}^{-kt}, \quad t \geqslant 0\]
where \(k\) is a positive constant. The half-life of radon is 3.8 days.

Show that \(k = \dfrac{5\ln 2}{19}\).

The UK target level for radon in homes is 100 \(\text{Bq m}^{-3}\). Find, according to the model, the time for the concentration to fall to this level, in days to 3 significant figures.

In a second, identical cellar, a fan that also removes radon is switched on at \(t = 0\). A refined model for this cellar is \(R = 950\mathrm{e}^{-kt}\mathrm{e}^{-ct}\), where \(c\) is a positive constant with \(\mathrm{e}^{-1.2c} = \dfrac{1}{2}\) (the fan alone would halve the concentration every 1.2 days).

Show that, in the refined model, the concentration halves every 0.912 days.

Find, according to the refined model, the time for the concentration in the second cellar to fall to 100 \(\text{Bq m}^{-3}\), in hours to 2 significant figures.

Model answer

Half-life 3.8 days: \(950\mathrm{e}^{-3.8k} = 475 \Rightarrow \mathrm{e}^{-3.8k} = \dfrac{1}{2} \Rightarrow -3.8k = -\ln 2\)
\(k = \dfrac{\ln 2}{3.8} = \dfrac{10\ln 2}{38} = \dfrac{5\ln 2}{19}\)
\(950\mathrm{e}^{-kt} = 100 \Rightarrow \mathrm{e}^{-kt} = \dfrac{2}{19} \Rightarrow t = \dfrac{\ln 9.5}{k} = \dfrac{19\ln 9.5}{5\ln 2} = 12.34\dots\)
So the time is 12.3 days (3 s.f.).
Refined model: \(R = 950\mathrm{e}^{-(k + c)t}\), with \(c = \dfrac{\ln 2}{1.2}\)
Halving time \(T\): \(\mathrm{e}^{-(k + c)T} = \dfrac{1}{2} \Rightarrow T = \dfrac{\ln 2}{k + c} = \dfrac{\ln 2}{\frac{\ln 2}{3.8} + \frac{\ln 2}{1.2}} = \dfrac{1}{\frac{5}{19} + \frac{5}{6}} = \dfrac{1}{\frac{125}{114}} = \dfrac{114}{125} = 0.912\) days
Time to fall to 100: \(t = \dfrac{\ln 9.5}{k + c} = 0.912 \times \dfrac{\ln 9.5}{\ln 2} = 2.962\dots\) days
\(2.962\dots \times 24 = 71.09\dots\), so about 71 hours (2 s.f.)

Explanation

The half-life gives \(\mathrm{e}^{-3.8k} = \dfrac{1}{2}\), so \(k = \dfrac{\ln 2}{3.8}\), and the time to reach 100 comes from taking natural logarithms of \(\mathrm{e}^{-kt} = \dfrac{100}{950}\). In the refined model the exponents add, so the decay constants add: \(\dfrac{1}{T} = \dfrac{1}{3.8} + \dfrac{1}{1.2}\), giving a halving time of 0.912 days, and the time to reach 100 falls to about 71 hours. Common errors are adding the half-lives instead of the decay constants and forgetting to convert days into hours.

Mark scheme

1 mark
M1 Uses the half-life: \(\mathrm{e}^{-3.8k} = \dfrac{1}{2}\) (or \(950\mathrm{e}^{-3.8k} = 475\)) and takes logarithms, \(-3.8k = \ln\dfrac{1}{2}\). An answer with no working scores 0 marks for the show-that parts
Correct answers of 12.3 days and 71 hours with no working score 4 marks: the two show-that results need working. For the M1, accept exact or decimal working.
1 mark
A1* \(k = \dfrac{\ln 2}{3.8} = \dfrac{10\ln 2}{38} = \dfrac{5\ln 2}{19}\)
cso. Working only with the decimal 0.182 is A0*: the exact form must be reached.
1 mark
M1 Solves \(950\mathrm{e}^{-kt} = 100\): \(t = \dfrac{\ln 9.5}{k}\) (\(= \dfrac{19\ln 9.5}{5\ln 2}\))
Using their \(k\), exact or decimal.
1 mark
A1 12.3 days (awrt 12.3, i.e. \([12.25, 12.35)\))
\(12.342\dots\)
1 mark
M1 Combines the exponents, \(R = 950\mathrm{e}^{-(k + c)t}\) with \(c = \dfrac{\ln 2}{1.2}\), and sets \(\mathrm{e}^{-(k + c)T} = \dfrac{1}{2}\): \(T = \dfrac{\ln 2}{k + c} = \dfrac{1}{\frac{1}{3.8} + \frac{1}{1.2}}\)
Adding the half-lives (\(3.8 + 1.2\)) or averaging them is M0. Accept exact working, or decimals such as \(k + c = 0.760\) (awrt 0.760, i.e. \([0.7595, 0.7605)\)).
1 mark
A1* \(T = \dfrac{1}{\frac{5}{19} + \frac{5}{6}} = \dfrac{114}{125} = 0.912\) days
cso. Showing \(0.912\) from rounded decimals (e.g. \(\dfrac{\ln 2}{0.760}\)) is acceptable if the working is complete and correct to enough accuracy, e.g. \(k + c = 0.7600\dots\)
1 mark
M1 Finds the time to fall to 100: \(t = \dfrac{\ln 9.5}{k + c}\) (\(= 0.912 \times \dfrac{\ln 9.5}{\ln 2}\)) and converts days to hours (\(\times 24\))
Independent of the previous A mark. Accept their \(k + c\).
1 mark
A1 71 hours
From \(2.962\dots\) days \(= 71.09\dots\) hours. Accept awrt 71, i.e. \([70.5, 71.5)\), including a more accurate value such as 71.1 hours. An answer in days only (2.96) scores A0.

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