How many VCAA Mathematical Methods questions cover Calculus?

AusGrader has 457 VCAA Mathematical Methods questions on Calculus, all with instant AI grading and detailed marking feedback.

VCE Mathematical Methods — Calculus 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 $h \to 0$ .

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

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

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 $[0, 2]$ .

Q3

2024

QCAA

Paper 2

1 mark

Q3

1 mark

The derivative of the function $f(x)$ is given by $f'(x) = \sin(2x)$ . It is known that $f\left(\frac{\pi}{2}\right) = 4$ .
Determine $f(x)$ .

A

$-\cos(2x) + 3$

B

$\cos(2x) + 5$

C

$-\frac{1}{2}\cos(2x) + 3.5$

D

$\frac{1}{2}\cos(2x) + 4.5$

Reveal Answer

A

$-\cos(2x) + 3$

This option fails to apply the reverse chain rule (u-substitution). The integral of $\sin(kx)$ is $-\frac{1}{k}\cos(kx)$ , so you must divide by the coefficient $2$ .

B

$\cos(2x) + 5$

This option uses the wrong sign for the antiderivative and misses the chain rule factor. The integral of $\sin(x)$ is $-\cos(x)$ , not $\cos(x)$ , and the result must be divided by $2$ .

C

$-\frac{1}{2}\cos(2x) + 3.5$

Correct Answer

Integrating $f'(x) = \sin(2x)$ yields $f(x) = -\frac{1}{2}\cos(2x) + C$ . Using the condition $f(\frac{\pi}{2}) = 4$ , we solve $4 = -\frac{1}{2}\cos(\pi) + C$ to find $C = 3.5$ .

D

$\frac{1}{2}\cos(2x) + 4.5$

This option has the wrong sign for the cosine term. Since the derivative of $\cos(x)$ is $-\sin(x)$ , the antiderivative of $\sin(x)$ must be negative.

Q11

2024

QCAA

Paper 2

4 marks

Q11

4 marks

State the trapezoidal rule and use it with six strips to determine an approximate value of the definite integral for the curve of $f(x) = 4(x-3)^2$ from $x = 0$ to $x = 3$ . Show all substitutions made into the rule.

Reveal Answer

$w=\dfrac{b-a}{n}$

$=\dfrac{3-0}{6}$

$=\dfrac{1}{2}$

$A\approx \dfrac{1}{2}w\left[f(x_0)+2f(x_1)+2f(x_2)+\ldots+2f(x_n)\right]$

$=\dfrac{1}{2}\times \dfrac{1}{2}\left[36+2\times 25+2\times 16+2\times 9+2\times 4+2\times 1+0\right]$

$=36.5\ \text{units}^2$

Marking Criteria

Descriptor	Marks
Correctly determines the rectangle width	1
Correctly states the trapezoidal rule	1
Substitutes appropriate values into the trapezoidal rule	1
Determines the approximate value of the definite integral	1

Q2

2024

VCAA

Paper 2

1 mark

Q2

1 mark

A function $g : R \rightarrow R$ has the derivative $g'(x) = x^3 - x$ .

Given that $g(0) = 5$ , the value of $g(2)$ is

A

2

B

3

C

5

D

7

Reveal Answer

A

2

Incorrect. Evaluating the antiderivative $g(x) = \frac{x^4}{4} - \frac{x^2}{2} + 5$ at $x=2$ yields 7, not 2.

B

3

Incorrect. This might result from ignoring the constant of integration $C=5$ and calculating $g(2) = 4 - 2 = 2$ , then making an arithmetic error.

C

5

Incorrect. This is the value of the initial condition $g(0)$ , not the requested value $g(2)$ .

D

7

Correct Answer

Correct. Integrating $g'(x)$ gives $g(x) = \frac{x^4}{4} - \frac{x^2}{2} + C$ . Substituting $g(0) = 5$ gives $C = 5$ , and evaluating $g(2)$ yields $\frac{16}{4} - \frac{4}{2} + 5 = 7$ .

Q14

2025

VCAA

Paper 2

1 mark

Q14

1 mark

Let $f$ be the probability density function for a continuous random variable $X$ , where

f(x) = \begin{cases} k\sin(x) & 0 \le x < \frac{\pi}{4} \\ k\cos(x) & \frac{\pi}{4} \le x \le \frac{\pi}{2} \\ 0 & \text{otherwise} \end{cases}

and $k$ is a positive real number.

The value of $k$ is

A

$\frac{1}{\sqrt{2}}$

B

$\frac{1}{2 - \sqrt{2}}$

C

$\sqrt{2} + 2$

D

$2 - \sqrt{2}$

Reveal Answer

A

$\frac{1}{\sqrt{2}}$

Incorrect. This value does not make the total area under the probability density function equal to 1, likely resulting from an error in evaluating the trigonometric integrals.

B

$\frac{1}{2 - \sqrt{2}}$

Correct Answer

Correct. For $f(x)$ to be a valid probability density function, its integral over all $x$ must equal 1. Evaluating $\int_0^{\pi/4} k\sin(x)dx + \int_{\pi/4}^{\pi/2} k\cos(x)dx = 1$ yields $k(2 - \sqrt{2}) = 1$ , which gives $k = \frac{1}{2 - \sqrt{2}}$ .

C

$\sqrt{2} + 2$

Incorrect. This might result from incorrectly rationalizing the denominator or making an arithmetic error when solving $k(2 - \sqrt{2}) = 1$ .

D

$2 - \sqrt{2}$

Incorrect. This is the value of the integral when $k=1$ . Since the total area must be 1, $k$ must be the reciprocal of this value.

Q17

2024

QCAA

Paper 1

3 marks

Q17

3 marks

A community group that uses social media created a new post on the internet on a day when they had 1000 members. The rate of change in their number of members (members/day) is given by $f'(t) = 3e^{0.5t}$ , where $t$ represents days after the new post.
Determine the time it will take for the community group to achieve seven times the initial number of members. Express your answer in the form $a \ln(b)$ .

Reveal Answer

7 times the members is 7000.
Let m be the time when 7000 members is reached.
The required change in members is 6000.
$6000 = \int_0^m 3e^{0.5t} dt$
$= \left[ 3 \times \frac{1}{0.5}e^{0.5t} \right]_0^m$
$= \left[ 6e^{0.5t} \right]_0^m$
$= \left[ 6e^{0.5m} - 6e^0 \right]$
$= \left[ 6e^{0.5m} - 6 \right]$
$6000 = 6e^{0.5m} - 6$
$6006 = 6e^{0.5m}$
$\frac{6006}{6} = e^{0.5m}$
$0.5m = \ln 1001$
$m = 2\ln 1001 \text{ days}$

Marking Criteria

Descriptor	Marks
Correctly uses the initial conditions to determine the increase	1
Correctly determines the integral	1
Determines the number of days required	1

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 $3e^{3(0)} = 3$ .

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.

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

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

Q1

2025

QCAA

Paper 2

1 mark

Q1

1 mark

Determine $\int 10.4x^3 \, dx$

A

$2.6x^4 + c$

B

$6.4x^4 + c$

C

$14.4x^4 + c$

D

$41.6x^4 + c$

Reveal Answer

A

$2.6x^4 + c$

Correct Answer

Correct. Using the power rule for integration, we increase the exponent by 1 and divide the coefficient by the new exponent: $\frac{10.4}{3+1}x^{3+1} + c = 2.6x^4 + c$ .

B

$6.4x^4 + c$

Incorrect. This option incorrectly subtracts the new exponent (4) from the coefficient 10.4 instead of dividing by it.

C

$14.4x^4 + c$

Incorrect. This option incorrectly adds the new exponent (4) to the coefficient 10.4 instead of dividing by it.

D

$41.6x^4 + c$

Incorrect. This option incorrectly multiplies the coefficient 10.4 by the new exponent (4) instead of dividing by it, confusing the integration rule with the differentiation rule.

Q14

2022

QCAA

Paper 1

6 marks

Q14

The rate that water fills an empty vessel is given by $\frac{dV}{dt} = 0.25e^{0.25t}$ (in litres per hour), $0 \le t \le 8\ln(6)$ , where $t$ is time (in hours).

Q14a

2 marks

Determine the function that represents the volume of water in the vessel (in litres).

Reveal Answer

$V = \int 0.25e^{0.25t} dt$
$= e^{0.25t} + c$
when $t = 0, V = 0$
$\therefore 0 = e^{0.25 \times 0} + c$
$\therefore c = -1$
$\therefore V = e^{0.25t} - 1$

Marking Criteria

Descriptor	Marks
correctly determines the integral of the function V(t)	1
determines the value of c	1

Q14b

2 marks

The vessel is full when $t = 8\ln(6)$ . Determine the volume of water, to the nearest litre, the vessel can hold when full.

Reveal Answer

$V(8\ln(6)) = e^{0.25 \times 8 \ln(6)} - 1$
$= 36 - 1$

$= 35$ litres

Marking Criteria

Descriptor	Marks
determines the simplified exponential term	1
determines number of litres	1

Q14c

2 marks

Use information from the table and the trapezoidal rule to determine the approximate volume of water in the vessel after three hours.

$t$	$\frac{dV}{dt}$
0	0.25
1	0.32
2	0.41
3	0.53

Reveal Answer

Using trapezoidal rule
Volume after 3 hours $= \frac{1}{2}(0.25+0.53+2(0.32+0.41))$

Volume after 3 hours $= 1.12$ litres

Marking Criteria

Descriptor	Marks
establishes expression for approximate number of litres of water in vessel after 3 hours	1
determines approximate number of litres	1

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