QCAA Mathematical Methods — Differentiation of exponential and logarithmic functions Questions & Answers

Q: How many QCAA Mathematical Methods questions cover Differentiation of exponential and logarithmic functions?

AusGrader has 186 QCAA Mathematical Methods questions on Differentiation of exponential and logarithmic functions, all with instant AI grading and detailed marking feedback.

Q5

2020

QCAA

Paper 1

1 mark

Q5

1 mark

The equation of the tangent to the curve $f(t) = te^t$ at $t = 1$ is

A

$y = et$

B

$y = 2et - e$

C

$y = et - e^2 + 1$

D

$y = 2et - 2e^2 + 1$

Reveal Answer

A

$y = et$

This option implies a slope of $e$ , but the derivative at $t=1$ is $f'(1) = e^1 + 1\cdot e^1 = 2e$ .

B

$y = 2et - e$

Correct Answer

The point of tangency is $(1, e)$ and the slope is $f'(1) = 2e$ . Using the point-slope form $y - e = 2e(t - 1)$ simplifies to $y = 2et - e$ .

C

$y = et - e^2 + 1$

This option uses the incorrect slope $e$ instead of $2e$ , likely failing to apply the product rule correctly when finding the derivative.

D

$y = 2et - 2e^2 + 1$

While the slope $2e$ is correct, the y-intercept is wrong. Expanding $y - e = 2e(t - 1)$ results in a constant term of $-e$ , not $-2e^2 + 1$ .

Q8

2023

QCAA

Paper 2

1 mark

Q8

1 mark

The number of koalas in a conservation park is modelled by $N = 15 \ln(7t + 1)$ , $t \geq 1$ , where $t$ represents the time (years) since the park opened. There were 20 koalas in the park when it opened.

Determine the approximate rate of change in the number of koalas when $t = 3$ .

A

46

B

26

C

25

D

5

Reveal Answer

A

46

This is the value of the function $N(3) = 15 \ln(22) \approx 46$ . This represents the number of koalas (or the population increase) at year 3, rather than the rate at which the population is changing.

B

26

This value appears to be the result of calculating $N(3) - 20 \approx 26$ . This subtracts the initial population from the model's value at $t=3$ , which does not represent the instantaneous rate of change.

C

25

This is an incorrect value. It does not correspond to the derivative at $t=3$ or the function value, likely resulting from a calculation error.

D

5

Correct Answer

The rate of change is found by taking the derivative $\frac{dN}{dt}$ . Using the chain rule, $\frac{dN}{dt} = 15 \cdot \frac{1}{7t+1} \cdot 7 = \frac{105}{7t+1}$ . Evaluating at $t=3$ gives $\frac{105}{22} \approx 4.77$ , which rounds to 5.

Q6

2023

QCAA

Paper 1

1 mark

Q6

1 mark

Substitutions for $h$ are used to estimate the limit of $\frac{a^h - 1}{h}$ as $h \to 0$ . Which sequence is the most appropriate?

A

$-4, -2, -1, -0.5, -0.25, -0.125 \dots$

B

$-0.05, -0.1, -0.2, -0.4, -0.8 \dots$

C

$2, 1, 0, -1, -2, -3 \dots$

D

$1, 2, 3, 4, 5, 6 \dots$

Reveal Answer

A

$-4, -2, -1, -0.5, -0.25, -0.125 \dots$

Correct Answer

This sequence is appropriate because the magnitude of the values decreases ( $|-4|, |-2|, |-1|, \dots$ ), meaning $h$ is getting progressively closer to $0$ .

B

$-0.05, -0.1, -0.2, -0.4, -0.8 \dots$

This sequence is incorrect because the values are moving away from $0$ ( $-0.05, -0.1, -0.2, \dots$ ), which does not help estimate the limit as $h \to 0$ .

C

$2, 1, 0, -1, -2, -3 \dots$

This sequence includes $0$ , where the expression $\frac{a^h - 1}{h}$ is undefined (division by zero), and subsequent terms move away from $0$ .

D

$1, 2, 3, 4, 5, 6 \dots$

This sequence consists of increasing integers moving away from $0$ , which would be used to investigate the limit as $h \to \infty$ , not as $h \to 0$ .

Q9

2021

QCAA

Paper 2

1 mark

Q9

1 mark

The graphs of the functions $f(x) = 2e^x + 5$ and $g(x) = \frac{3}{e^x}$ intersect at point A. Determine the coordinates of point A.

A

(1.609, 15)

B

(1.099, 1)

C

(0.4065, 2)

D

(-0.693, 6)

Reveal Answer

A

(1.609, 15)

Incorrect. While this point lies on the graph of $f(x)$ (since $x \approx \ln(5)$ gives $f(x) = 15$ ), $g(1.609) \approx 0.6$ . The intersection point must satisfy both equations.

B

(1.099, 1)

Incorrect. This point lies on the graph of $g(x)$ (since $x \approx \ln(3)$ gives $g(x) = 1$ ), but $f(1.099) \approx 11$ . The functions are not equal at this x-value.

C

(0.4065, 2)

Incorrect. This point lies on the graph of $g(x)$ (since $x \approx \ln(1.5)$ gives $g(x) = 2$ ), but $f(0.4065) \approx 8$ . The y-values differ significantly.

D

(-0.693, 6)

Correct Answer

Correct. Set $2e^x + 5 = \frac{3}{e^x}$ and multiply by $e^x$ to get the quadratic $2(e^x)^2 + 5e^x - 3 = 0$ . Factoring gives $(2e^x - 1)(e^x + 3) = 0$ . Since $e^x > 0$ , $e^x = \frac{1}{2}$ , which yields $x = \ln(0.5) \approx -0.693$ and $y = 6$ .

Q3

2022

VCAA

Paper 2

1 mark

Q3

1 mark

The gradient of the graph of $y = e^{3x}$ at the point where the graph crosses the vertical axis is equal to

A

$0$

B

$\frac{1}{e}$

C

$1$

D

$e$

E

$3$

Reveal Answer

A

$0$

Incorrect. This might result from confusing the x-coordinate of the y-intercept ( $x=0$ ) with the gradient itself.

B

$\frac{1}{e}$

Incorrect. This value does not match the derivative evaluated at the y-intercept.

C

$1$

Incorrect. This is the y-coordinate of the y-intercept ( $y=e^0=1$ ), not the gradient. It could also result from forgetting the chain rule and incorrectly assuming the derivative is $e^{3x}$ .

D

$e$

Incorrect. This value does not correspond to the derivative evaluated at $x=0$ .

E

$3$

Correct Answer

Correct. The gradient is found using the derivative $\frac{dy}{dx} = 3e^{3x}$ . The graph crosses the vertical axis at $x=0$ , so evaluating the derivative gives $3e^{3(0)} = 3$ .

Q15

2020

SCSA

Paper 2

9 marks

Q15

A chef needs to use an oven to boil 100 mL of water in five minutes for a new experimental recipe. The temperature of the water must reach 100 °C in order to boil. The temperature, $T$ , of 100 mL of water $t$ minutes after being placed in an oven set to $T_0$ °C can be modelled by the equation below.

$T(t) = T_0 - 175e^{-0.07t}$

In a preliminary experiment, the chef placed a 100 mL bowl of water into an oven that had been heated to $T_0 = 200$ °C.

Q15a

1 mark

What is the temperature of the water at the moment it is placed into the oven?

Reveal Answer

$T(0) = 200 - 175e^{-0.07(0)} = 25 \ ^\circ\text{C}$

Marking Criteria

Descriptor	Marks
states correct temperature	1

Q15b

1 mark

What is the temperature of the water five minutes after being placed in the oven?

Reveal Answer

$T(5) = 200 - 175e^{-0.07(5)} = 76.68 \ ^\circ\text{C}$

Marking Criteria

Descriptor	Marks
states correct temperature	1

Q15c

2 marks

What change could be made to the temperature at which the oven is set in order to achieve the five-minute boiling requirement?

Reveal Answer

$100 = T_0 - 175e^{-0.07(5)}$
$T_0 = 100 + 175e^{-0.07(5)} \approx 223 \ ^\circ\text{C}$

Marking Criteria

Descriptor	Marks
states correct equation to be solved	1
solves for $T_0$ , giving changed temperature	1

Q15d

2 marks

Assume that $T_0$ is still 200 °C.

Determine the rate of increase in temperature of the water five minutes after being placed in the oven. Give your answer rounded to two decimal places.

Reveal Answer

$T'(t) = 12.25e^{-0.07t}$
$T'(5) = 12.25e^{-0.07(5)} = 8.63 \ ^\circ\text{C/min}$

Marking Criteria

Descriptor	Marks
states correct derivative of $T$ with respect to $t$	1
calculates correct rate	1

Q15e

3 marks

Explain what happens to the rate of change in the temperature of the water as time increases and how this relates to the temperature of the water.

Reveal Answer

As time increases, the rate of change in the temperature of the water $\rightarrow 0$ .
The temperature of the water $\rightarrow$ the constant value of $T_0$ .

Marking Criteria

Descriptor	Marks
states that the rate of change in the temperature $\rightarrow 0$	1
states the water temperature approaches a constant	1
states the water temperature approaches $T_0$	1

Q5

2025

VCAA

Paper 1

4 marks

Q5a

2 marks

Solve $e^{2x} - 8e^x + 7 = 0$ for $x$ .

Reveal Answer

\begin{align*} e^{2x} - 8e^x + 7 &= 0\\ (e^x - 1)(e^x - 7) &= 0\\ e^x = 1 \quad &\text{or} \quad e^x = 7\\ x = 0 \quad &\text{or} \quad x = \log_e(7) \end{align*}

Marking Criteria

Descriptor	Marks
Factorises the equation correctly	1
Solves for $x$ correctly	1

Q5b

2 marks

Let $g(x) = e^{2x} - 8e^x + 7$ , where $x \in R$ .

The function $g(x)$ has exactly one stationary point, a local minimum.

Find the largest value of $a$ such that when $g$ is restricted to the domain $(-\infty, a]$ it has an inverse function.

Reveal Answer

Find turning point $x$ -value

\begin{align*} g'(x) = 2e^{2x} - 8e^x &= 0\\ 2e^x(e^x - 4) &= 0 \end{align*}

As $e^x > 0, \quad e^x = 4 \text{ only}, \therefore x = \log_e(4)$

So $a = \log_e(4) = 2\log_e(2)$ .

Marking Criteria

Descriptor	Marks
Finds the derivative and sets to 0	1
Solves for $x$ correctly	1

Q2

2025

QCAA

Paper 1

1 mark

Q2

1 mark

State the domain of the function $y = \ln(x)$

A

$0 < x < \infty$

B

$0 \le x < \infty$

C

$-\infty < x < 0$

D

$-\infty < x < \infty$

Reveal Answer

A

$0 < x < \infty$

Correct Answer

The natural logarithm function $y = \ln(x)$ is only defined for strictly positive real numbers, meaning $x$ must be greater than 0.

B

$0 \le x < \infty$

The natural logarithm is undefined at $x = 0$ because there is no real power to which $e$ can be raised to yield 0.

C

$-\infty < x < 0$

The natural logarithm is not defined for negative numbers in the real number system.

D

$-\infty < x < \infty$

This represents all real numbers, which is the domain of an exponential function like $y = e^x$ , but the domain of $y = \ln(x)$ is restricted to strictly positive values.

Q2

2024

QCAA

Paper 1

1 mark

Q2

1 mark

Determine $\frac{dy}{dx}$ for the function $y = e^{\sin(x)}$

A

$\cos(x) e^{\sin(x)}$

B

$\sin(x) e^{\cos(x)}$

C

$e^{\sin(x)}$

D

$e^{\cos(x)}$

Reveal Answer

A

$\cos(x) e^{\sin(x)}$

Correct Answer

This is correct. By applying the chain rule $\frac{d}{dx} e^{u} = e^u \frac{du}{dx}$ with $u = \sin(x)$ , the derivative is $e^{\sin(x)} \cdot \cos(x)$ .

B

$\sin(x) e^{\cos(x)}$

This is incorrect. It confuses the derivative of the exponent with the exponent itself. The term $e^{\sin(x)}$ should remain, multiplied by the derivative of the sine function.

C

$e^{\sin(x)}$

This is incorrect because it ignores the chain rule. While the derivative of $e^x$ is $e^x$ , the derivative of a composite function $e^{u(x)}$ requires multiplying by $u'(x)$ .

D

$e^{\cos(x)}$

This is incorrect. You cannot simply differentiate the exponent in place; the chain rule requires preserving the original exponential term $e^{\sin(x)}$ and multiplying by the derivative of the exponent.

Q15

2020

QCAA

Paper 1

4 marks

Q15

Solve the following equations.

Q15a

1 mark

$4e^x = 100$

Reveal Answer

$e^x = 25$
$x = ln(25)$

Marking Criteria

Descriptor	Marks
correctly determines $x = ln(25)$	1

Q15b

3 marks

$2 \log_4 x - \log_4 (x - 1) = 1$

Reveal Answer

Using log laws
$log_4(\frac{x^2}{x-1}) = 1$
Change from log to index form
$\frac{x^2}{x-1} = 4$
$x^2 - 4x + 4 = 0$
$x = 2$

Marking Criteria

Descriptor	Marks
correctly establishes equation using log laws	1
correctly establishes the quadratic equation	1
determines x	1

Q5

2025

QCAA

Paper 2

1 mark

Q5

1 mark

Which statement best describes a feature of the graph of the exponential function $y = e^x$ , $x \in R$ ?

A

$\lim_{x \to \infty} (e^x) = e$

B

When $x = 0$ , $y = e$

C

The graph has an asymptote with the equation $x = 0$

D

The gradient of the graph has the same value as the function at all points on the graph.

Reveal Answer

A

$\lim_{x \to \infty} (e^x) = e$

Incorrect. As $x$ approaches infinity, the value of $e^x$ grows without bound, meaning the limit is infinity, not $e$ .

B

When $x = 0$ , $y = e$

Incorrect. Any non-zero base raised to the power of 0 equals 1, so when $x = 0$ , $y = e^0 = 1$ , rather than $e$ .

C

The graph has an asymptote with the equation $x = 0$

Incorrect. The function $y = e^x$ has a horizontal asymptote at $y = 0$ as $x$ approaches negative infinity, not a vertical asymptote at $x = 0$ .

D

The gradient of the graph has the same value as the function at all points on the graph.

Correct Answer

Correct. A unique property of the natural exponential function $y = e^x$ is that its derivative is also $e^x$ , meaning the gradient at any point equals the function's value.

Q18

2021

QCAA

Paper 2

3 marks

Q18

3 marks

The number of animals in a population (in thousands) is modelled by the function $P$ such that
$P(t) = \frac{100}{1+4e^{-t}}, \text{ where } t \text{ is in years.}$
Determine the number of animals in the population when the population is growing the fastest.

Reveal Answer

The population is increasing most rapidly at the maximum value of $P'(t)$
$t = 1.386$
$\therefore P(1.386) \approx 49.993$
There are approximately 50 000.

Marking Criteria

Descriptor	Marks
Correctly identifies the conditions for the most rapid increase	1
Determines when population growing the fastest	1
Determines population at this time	1

Q13

2024

QCAA

Paper 2

8 marks

Q13

The number of termites in a particular nest can be modelled by $N(t) = \frac{A}{2 + e^{-t}}$ , where $A$ is a constant and $t$ represents time (months) since the nest first became a visible mound above ground level.
It is estimated that when the mound first became visible, the population was $3 \times 10^5$ termites.

Q13a

1 mark

Determine the value of $A$ .

Reveal Answer

$N(0)=\dfrac{A}{2+e^{-0}}$

$3\times 10^5=\dfrac{A}{2+1}$

$A=3\times 3\times 10^5$

$A=9\times 10^5$

Marking Criteria

Descriptor	Marks
Correctly determines the value of $A$	1

Q13b

2 marks

Determine the number of termites in the nest half a year after the mound became visible.

Reveal Answer

$N(t)=\dfrac{9\times 10^5}{2+e^{-t}}$

$N(6)=\dfrac{9\times 10^5}{2+e^{-6}}$

$=449\ 442.9711$

$=449\ 443\ \text{termites}$

Marking Criteria

Descriptor	Marks
Correctly determines $t=6$	1
Determines the number of termites	1

Q13c

2 marks

Determine the time in months after the mound became visible for the initial population to increase by 130 000 termites. Express the time as a decimal.

Reveal Answer

$300\ 000+130\ 000=430\ 000$

$N(t)=\dfrac{9\times 10^5}{2+e^{-t}}$

$430\ 000=\dfrac{9\times 10^5}{2+e^{-t}}$

Using a GDC:

$t=2.3749\ \text{months}$

Marking Criteria

Descriptor	Marks
Correctly determines the required number of termites	1
Determines the time required	1

Q13d

2 marks

Develop a formula for the rate of change in the number of termites at any time after the mound became visible. Express your formula as a fraction.

Reveal Answer

$N(t)=\dfrac{9\times 10^5}{2+e^{-t}}$

$N(t)=(9\times 10^5)(2+e^{-t})^{-1}$

$\dfrac{dN}{dt}=-(9\times 10^5)(2+e^{-t})^{-2}\times -e^{-t}$

$\dfrac{dN}{dt}=\dfrac{(9\times 10^5)}{e^t(2+e^{-t})^2}$

Marking Criteria

Descriptor	Marks
Shows application of the chain rule	1
Determines a formula for the number of termites expressed as a fraction	1

Q13e

1 mark

Determine the rate of change in the number of termites five months after the mound became visible.

Reveal Answer

Using the formula developed in d) or a GDC:

$N'(5)=1505.8744$

$=1506\ \text{termites/month}$

Marking Criteria

Descriptor	Marks
Determines the rate of change	1

Q14

2024

QCAA

Paper 2

6 marks

Q14

A football coach offered a 12-day intensive training clinic. During the clinic, the height that each player could kick a football was monitored.
One player's kick heights could be modelled by $H(t) = \log_{10}(10t + 10) + 5$ , $0 \le t \le 12$ , where $H(t)$ is vertical height (m) and $t$ is the time (days) spent in training.

Q14a

1 mark

Determine the initial height that the player could kick the ball.

Reveal Answer

$H(0)=6\ \text{m}$

Marking Criteria

Descriptor	Marks
Correctly determines the initial height	1

Q14b

1 mark

Determine the training time needed for the player to be able to kick the ball to a height of 7 m.

Reveal Answer

Using a GDC:

$t=9\ \text{days}$

Marking Criteria

Descriptor	Marks
Correctly determines the time required	1

Q14c

2 marks

Determine the overall improvement in kick height achieved by completing the clinic.

Reveal Answer

Initially, the kick height was 6 metres.

At the end of the course, $t=12$ , the kick height increased
to 7.1139 metres.

The kick height has increased by 1.1139 metres during the
course.

Marking Criteria

Descriptor	Marks
Correctly determines the kick height at the end of the course	1
Determines the overall kick height improvement	1

Q14d

1 mark

Determine the rate of change in kick height when $t = 1.5$ days.

Reveal Answer

Using a GDC:

$H'(1.5)=0.17372\ \text{m/day}$

Marking Criteria

Descriptor	Marks
Correctly determines the derivative value when $t=1.5$	1

Q14e

1 mark

Determine the training time (as a decimal) when the rate of change in kick height is 0.09 m/day.

Reveal Answer

Using a GDC:

Graph derivative function and $y=0.09$

Find point of intersection.

$H'(t)=0.09$

$t=3.82549\ \text{days}$

Marking Criteria

Descriptor	Marks
Correctly determines the time as a decimal	1

Q5

2024

VCAA

Paper 2

1 mark

Q5

1 mark

Consider the functions $f : (1, \infty) \rightarrow R, f(x) = x^2 - 4x$ and $g : R \rightarrow R, g(x) = e^{-x}$ .

The range of the composite function $g(f(x))$ is

A

$(0, e^3)$

B

$(0, e^3]$

C

$(0, e^4)$

D

$(0, e^4]$

Reveal Answer

A

$(0, e^3)$

This incorrectly assumes the minimum of $f(x)$ occurs at the boundary $x=1$ (giving $f(1)=-3$ ), missing the true minimum at the vertex $x=2$ .

B

$(0, e^3]$

This incorrectly uses $f(1)=-3$ as the minimum value of the inner function, failing to account for the vertex of the parabola at $x=2$ .

C

$(0, e^4)$

The maximum value $e^4$ is achieved at $x=2$ , which is included in the domain $(1, \infty)$ , so the interval must be closed at $e^4$ .

D

$(0, e^4]$

Correct Answer

The range of $f(x) = (x-2)^2 - 4$ on $(1, \infty)$ is $[-4, \infty)$ . Applying the strictly decreasing function $g(x) = e^{-x}$ to this range yields $(0, e^4]$ .

QCAA Mathematical Methods Differentiation of exponential and logarithmic functions

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