How many SCSA Mathematics Methods questions cover Further differentiation and applications?

AusGrader has 290 SCSA Mathematics Methods questions on Further differentiation and applications, all with instant AI grading and detailed marking feedback.

WACE Mathematics Methods — Further differentiation and applications Questions & Answers

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 $h \to 0$ .

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

Q11

2020

QCAA

Paper 1

3 marks

Q11

Determine the derivative of each of the following with respect to $x$ .

Q11a

1 mark

$y = \frac{1}{\sin(x)}$

Reveal Answer

$f'(x) = \frac{-\cos(x)}{(\sin(x))^2}$

Marking Criteria

Descriptor	Marks
Correctly determines the derivative	1

Q11b

2 marks

$y = x^2 \times e^{-x}$

Express your answer in factorised form.

Reveal Answer

$f'(x) = -x^2 e^{-x} + 2x e^{-x}$

$= x e^{-x}(-x + 2)$

Marking Criteria

Descriptor	Marks
Correctly determines the derivative in expanded form	1
Determines factorised form of derivative	1

Q9

2024

QCAA

Paper 1

1 mark

Q9

1 mark

At a certain location, the temperature (°C) can be modelled by the function $T = 5\sin\left(\frac{\pi}{12}x\right) + 23$ , where $x$ is the number of hours after sunrise.

Determine the rate of change of temperature (°C/hour) when $x = 4$

A

$\frac{5\pi}{48}$

B

$\frac{5\pi}{24}$

C

$\frac{5\pi\sqrt{3}}{24}$

D

$\frac{5\pi\sqrt{3}}{6}$

Reveal Answer

A

$\frac{5\pi}{48}$

This incorrect value is half of the correct answer, which may result from a calculation error during the multiplication of fractions or evaluating the trigonometric ratio.

B

$\frac{5\pi}{24}$

Correct Answer

The rate of change is the derivative $T'(x) = 5 \cdot \frac{\pi}{12}\cos\left(\frac{\pi}{12}x\right)$ . Evaluating at $x=4$ gives $\frac{5\pi}{12}\cos\left(\frac{\pi}{3}\right) = \frac{5\pi}{12}\left(\frac{1}{2}\right) = \frac{5\pi}{24}$ .

C

$\frac{5\pi\sqrt{3}}{24}$

This answer results from evaluating $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ instead of $\cos\left(\frac{\pi}{3}\right)$ in the derivative, or incorrectly assuming the derivative of sine is sine.

D

$\frac{5\pi\sqrt{3}}{6}$

This option is incorrect and likely results from misapplying the chain rule or arithmetic errors when combining the constants.

Q1

2025

VCAA

Paper 1

3 marks

Q1a

1 mark

Let $y = x^2 \cos(x)$ .

Find $\frac{dy}{dx}$ .

Reveal Answer

\begin{align*} \frac{dy}{dx} &= 2x\cos(x) + x^2 \cdot (-\sin(x))\\ &= 2x\cos(x) - x^2\sin(x) \\ &= x(2\cos(x) - x\sin(x)) \end{align*}

Marking Criteria

Descriptor	Marks
Correctly finds the derivative using the product rule, e.g., $2x\cos(x) - x^2\sin(x)$ or equivalent.	1

Q1b

2 marks

Let $f(x) = 6\sqrt{x+1} + 5$ .

Find the gradient of the tangent to $y = f(x)$ at $x = 8$ .

Reveal Answer

$f(x) = 6\sqrt{x+1} + 5 = 6(x+1)^{\frac{1}{2}} + 5$

$f'(x) = 6 \cdot \frac{1}{2} \cdot (x+1)^{-\frac{1}{2}} = 3(x+1)^{-\frac{1}{2}} = \frac{3}{\sqrt{x+1}}$

$f'(8) = \frac{3}{\sqrt{8+1}} = 1$

Marking Criteria

Descriptor	Marks
Correctly finds the derivative $f'(x) = 3(x+1)^{-\frac{1}{2}}$ or equivalent.	1
Correctly evaluates the derivative at $x=8$ to find the gradient is $1$ .	1

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.

Q19

2020

QCAA

Paper 1

6 marks

Q19

6 marks

A horizontal point of inflection is a point of inflection that is also a stationary point.

Determine the value/s of $k$ for which the graph of $f(x) = \frac{\ln(x)}{k} - \frac{kx}{x+1}$ has only one horizontal point of inflection.

Reveal Answer

$f'(x) = \frac{1}{kx} - (\frac{k(x + 1) - kx}{(x + 1)^2})$
Stationary points $f'(x) = 0$

$0 = \frac{1}{kx} - (\frac{k(x + 1) - kx}{(x + 1)^2})$

$0 = \frac{1}{kx} - \frac{k}{(x + 1)^2}$

$0 = x^2 + (2 - k^2)x + 1$ (i)

The quadratic has real roots when discriminant $\ge 0$

$(2 - k^2)^2 - 4 \ge 0$
$2 - k^2 \ge \pm 2$
There is only ONE phi $\therefore$
$2 - k^2 = \pm 2$
$k = 0$ (not valid)
and $k^2 = 4$ so
$k = 2, -2$
Sub into (i) to determine the x-ordinate of the stationary point.
$\rightarrow x = 1$
For $k = 2$
$\therefore f''(x) = \frac{-1}{2x^2} + \frac{4}{(x + 1)^3}$
$f''(1) = \frac{-1}{2} + \frac{4}{8}$
$f''(1) = 0$
For $k = -2$
$f''(x) = \frac{1}{2x^2} - \frac{4}{(x + 1)^3}$
$f''(1) = \frac{1}{2} - \frac{4}{8}$
$f''(1) = 0$
For each $k$ value, $x = 1$ is the x-ordinate of both a stationary point ( $f'(x) = 0$ ) and a point of inflection ( $f''(x) = 0$ )
There is a point of horizontal inflection at $x = 1$ when $k = \pm 2$

Marking Criteria

Descriptor	Marks
correctly determines the first derivative	1
correctly determines the quadratic equation to identify the stationary point/s	1
determines valid and non-valid solutions of k	1
determines x-ordinate of stationary point	1
determines values of second derivative for both values of k	1
shows logical organisation communicating key steps	1

Q8

2024

SCSA

Paper 2

11 marks

Q8

John is given a prescription for blood pressure tablets by his doctor. Each tablet contains five milligrams of the active medication. The mass of active medication, $A$ , remaining in John's body $t$ hours after taking a single tablet is given by

$A(t) = 5e^{-0.0173t}$

where $A$ is in milligrams.

Q8a

1 mark

Calculate the mass of medication remaining in John's body 10 hours after taking a single tablet.

Reveal Answer

Setting $t = 10$ gives

$A(10) = 5e^{-0.0173(10)} \approx 4.21 \text{ mg}$

Marking Criteria

Descriptor	Marks
obtains correct mass of medication including units	1

Q8b

2 marks

After how many hours will the mass of medication remaining in John's body have halved?

Reveal Answer

Setting $A = 2.5$ gives

$2.5 = 5e^{-0.0173(t)}$

$\Rightarrow 0.5 = e^{-0.0173(t)}$

$\Rightarrow t = -\frac{1}{0.0173}\ln(0.5) = 40.066 \approx 40 \text{ hours}$

Marking Criteria

Descriptor	Marks
correctly substitutes $A = 2.5$ into the equation	1
correctly solves for $t$	1

Q8c

3 marks

Determine at what rate the mass of medication remaining in John's body is decreasing 24 hours after taking a single tablet.

Reveal Answer

The derivative of $A$ is

$A'(t) = -0.0865 e^{-0.0173t}$

When $t = 24$

$A'(24) = -0.0865e^{-0.0173(24)} = -0.057 \text{ mg/hr}$

The mass is decreasing at a rate of $0.057 \text{ mg/hr}$ .

Marking Criteria

Descriptor	Marks
correctly differentiates $A$	1
substitutes $t = 24$ to obtain correct rate of increase	1
converts to a rate of decrease	1

Q8d

2 marks

If a tablet is taken every $T$ hours, the mass of medication, $B$ , remaining in John's body immediately after taking a tablet will be given by

$B(T) = \frac{5}{1 - e^{-0.0173T}}$

where $B$ is in milligrams.

How frequently should John take a tablet so that the mass of medication remaining in his body immediately after taking each tablet is 8.85 mg?

Reveal Answer

Setting $B = 8.85$ gives

$8.85 = \frac{5}{1 - e^{-0.0173T}}$

$\Rightarrow 1 - e^{-0.0173T} = \frac{5}{8.85}$

$\Rightarrow e^{-0.0173T} = \frac{3.85}{8.85}$

$\Rightarrow T = -\frac{1}{0.0173}\ln\left(\frac{3.85}{8.85}\right) = 48.112 \approx 48 \text{ hours}$

John should take one tablet every 48 hours.

Marking Criteria

Descriptor	Marks
correctly substitutes $B = 8.85$ into the equation	1
correctly solves for $T$	1

Q8e

3 marks

Use the increments formula to approximate the change in $B$ if the time between taking tablets increased by 30 minutes from the time determined in part (d).

Reveal Answer

The derivative of $B$ with respect to $T$ is given by

$\frac{dB}{dT} = -\frac{0.0865e^{-0.0173T}}{\left(1 - e^{-0.0173T}\right)^2}$

and when $T = 48$

$\frac{dB}{dT}(48) \approx -0.118$

An increase of 30 minutes corresponds to $\delta T = 0.5$ . Using the increments formula gives

$\delta B \approx -0.118 \times 0.5$

$= -0.059 \text{ mg}$

Marking Criteria

Descriptor	Marks
correctly differentiates $B$ with respect to $T$	1
states that $\delta T = 0.5$	1
correctly uses the increments formula to estimate the change in $B$	1

Q11

2024

QCAA

Paper 1

6 marks

Q11a

2 marks

Determine the second derivative of $y = x^3 - 3x^2$ .

Reveal Answer

$y = x^3 - 3x^2$
$\frac{dy}{dx} = 3x^2 - 6x$
$\frac{d^2x}{dy^2} = 6x - 6$

Marking Criteria

Descriptor	Marks
Correctly determines the first derivative	1
Determines the second derivative	1

Q11b

1 mark

Use your result from Question 11a) to calculate the value of the second derivative when $x = -1$ .

Reveal Answer

$\frac{d^2y}{dx^2} = 6x - 6$
When $x = -1$
$\frac{d^2y}{dx^2} = 6 \times -1 - 6$
$= -12$

Marking Criteria

Descriptor	Marks
Determines the value of the second derivative	1

Q11c

3 marks

Determine the $x$ - and $y$ -coordinates of the point on the graph of $y = x^3 - 3x^2$ for which the rate of change of the first derivative is zero.

Reveal Answer

$\frac{d^2y}{dx^2} = 0$
$6x - 6 = 0$
$6x = 6$
$x = 1$
Substitute into the graph equation:
$y = (1)^3 - 3(1)^2$
$y = -2$
Coordinates are $(1, -2)$ .

Marking Criteria

Descriptor	Marks
Equates the second derivative to zero	1
Determines the x-coordinate of the point	1
Determines the y-coordinate of the point	1

Q14

2021

SCSA

Paper 2

5 marks

Q14

5 marks

Question 14

The displacement in metres, $x(t)$ , of a power boat $t$ seconds after it was launched is given by:

$x(t)=\dfrac{5t(t^2-15t+48)}{6},\quad t\ge 0$

How far has the power boat travelled before its acceleration is zero?

Reveal Answer

$x(t) = \frac{5t(t^2 - 15t + 48)}{6}, \quad t \ge 0$

$v(t) = \frac{dx}{dt} = \frac{5t^2 - 50t + 80}{2}$

$a(t) = \frac{d^2x}{dt^2} = 5t - 25$

$5t - 25 = 0 \implies \therefore t = 5$

$\text{Distance travelled} = \int_0^5 \left| \frac{5t^2 - 50t + 80}{2} \right| dt = \frac{245}{3} \approx 81.7 \text{ metres}$

Marking Criteria

Descriptor	Marks
determines an expression for velocity	1
determines an expression for acceleration	1
equates acceleration to zero and determines $t$	1
shows integration expression for distance travelled	1
determines how far the power boat has travelled	1

Q15

2022

QCAA

Paper 1

4 marks

Q15

4 marks

The derivative of a function is given by $f'(x) = e^x(x-4)$ .
Determine the interval on which the graph of $f(x)$ is both decreasing and concave up.

Reveal Answer

The function is decreasing when $f'(x) < 0$ and concave up when $f''(x) > 0$
$f'(x) = (x-4)e^x < 0$ when $x < 4$

$f''(x) = (x-4)e^x + e^x = e^x(x-3) > 0$ when $x > 3$

Therefore, the function is decreasing and concave up when $3 < x < 4$

Marking Criteria

Descriptor	Marks
correctly describes conditions when the function is decreasing and concave up	1
correctly determines the interval where f(x) is decreasing	1
correctly determines the interval where f(x) is concave up	1
determines interval when function is decreasing and concave up	1

Q17

2023

QCAA

Paper 1

6 marks

Q17

6 marks

A chemical is added to the water in a swimming pool at 10:00 am to prevent algae. The amount of chemical absorbed into the water over time $t$ (hours) is represented by

$A = 10t^2 - 4t^3, \quad 0 \leq t \leq 1\frac{2}{3}$

Determine the time of day when the rate of absorption of the chemical is at its maximum. Use calculus techniques to verify that your time corresponds to a maximum rate.

Reveal Answer

The rate of absorption is given by:
$\frac{dA}{dt} = 20t - 12t^2$

$\frac{d^2A}{dt^2} = 20 - 24t$

$\therefore 20 - 24t = 0$
$t = \frac{20}{24} = \frac{5}{6}$ hours

Verify this corresponds to a maximum rate.
Using the second derivative test, we investigate the sign of the derivative of $\frac{d^2A}{dt^2}$ , i.e. $\frac{d^3A}{dt^3}$
$\frac{d^3A}{dt^3} = -24$
This is negative, therefore the rate of absorption is a maximum.
The time the chemical is increasing most rapidly since delivery is $\frac{5}{6}$ hours.
$= \frac{5}{6} \times 60$
$= 50$ minutes
The required time is 10:50 am.

Marking Criteria

Descriptor	Marks
Correctly determines the first derivative	1
Determines the second derivative	1
Equates the second derivative to zero	1
Determines time when second derivative is zero	1
Performs a calculus test to confirm the time corresponds to a maximum for dA/dt	1
Determines the time for maximum rate of absorption in minutes	1

Q13

2024

VCAA

Paper 2

1 mark

Q13

1 mark

The function $f : (0, \infty) \rightarrow R, f(x) = \frac{x}{2} + \frac{2}{x}$ is mapped to the function $g$ with the following sequence of transformations:

dilation by a factor of 3 from the $y$ -axis
translation by 1 unit in the negative direction of the $y$ -axis.

The function $g$ has a local minimum at the point with the coordinates

A

$(6, 1)$

B

$\left(\frac{2}{3}, 1\right)$

C

$(2, 5)$

D

$\left(2, -\frac{1}{3}\right)$

Reveal Answer

A

$(6, 1)$

Correct Answer

The transformed function is $g(x) = f\left(\frac{x}{3}\right) - 1 = \frac{x}{6} + \frac{6}{x} - 1$ . Setting the derivative $g'(x) = \frac{1}{6} - \frac{6}{x^2}$ to zero gives $x = 6$ , and evaluating $g(6)$ yields a minimum value of $1$ .

B

$\left(\frac{2}{3}, 1\right)$

This is the local minimum if the function was incorrectly dilated by a factor of $1/3$ from the $y$ -axis, which would use $f(3x)$ instead of $f\left(\frac{x}{3}\right)$ .

C

$(2, 5)$

This is the local minimum if the function was dilated by a factor of 3 from the $x$ -axis (evaluating $3f(x) - 1$ ) rather than from the $y$ -axis.

D

$\left(2, -\frac{1}{3}\right)$

This is the local minimum if the function was dilated by a factor of $1/3$ from the $x$ -axis (evaluating $\frac{1}{3}f(x) - 1$ ) rather than from the $y$ -axis.

Q4

2021

VCAA

Paper 2

1 mark

Q4

1 mark

The maximum value of the function $h : [0, 2] \rightarrow R, h(x) = (x - 2)e^x$ is

A

$-e$

B

0

C

1

D

2

E

$e$

Reveal Answer

A

$-e$

This is the minimum value of the function on the interval, which occurs at the critical point $x = 1$ since $h(1) = -e$ .

B

0

Correct Answer

To find the maximum, we evaluate $h(x)$ at the critical point $x=1$ and endpoints $x=0, 2$ . Comparing $h(0) = -2$ , $h(1) = -e$ , and $h(2) = 0$ , the maximum value is $0$ .

C

1

This is the $x$ -value of the critical point (found by setting $h'(x) = (x-1)e^x = 0$ ), not the maximum value of the function itself.

D

2

This is the $x$ -value at which the maximum occurs, but the question asks for the maximum value of the function, which is $h(2) = 0$ .

E

$e$

This value does not correspond to the function evaluated at any critical point or endpoint within the given interval $[0, 2]$ .

SCSA Mathematics Methods Further differentiation and applications

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