How many VCAA Mathematical Methods questions cover Algebra, number and structure?

AusGrader has 115 VCAA Mathematical Methods questions on Algebra, number and structure, all with instant AI grading and detailed marking feedback.

VCE Mathematical Methods — Algebra, number and structure Questions & Answers

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

2025

QCAA

Paper 1

6 marks

Q15

6 marks

Determine all possible solutions for the equation $(\log_2(x))^2 = \log_2(x^2) + 8$ .

Evaluate the reasonableness of one solution.

Reveal Answer

\begin{align*} (\log_2(x))^2 &= \log_2(x^2) + 8\\ (\log_2(x))^2 - \log_2(x^2) - 8 &= 0\\ (\log_2(x))^2 - 2\log_2(x) - 8 &= 0 \end{align*}

Let $y = \log_2 x$

\begin{align*} y^2 - 2y - 8 &= 0\\ (y - 4)(y + 2) &= 0 \end{align*}

Solving this gives $x = 16 \text{ or } x = \frac{1}{4}$ .

Substitute $x = 16$ into the original equation to evaluate its reasonableness.

$\text{LHS} = (\log_2(16))^2 = 16$

$\text{RHS} = \log_2(16^2) + 8 = 16$

Therefore, $x = 16$ is a reasonable solution.

Marking Criteria

Descriptor	Marks
correctly applies an appropriate logarithmic law to convert the $x^2$ term to a term in $x$	1
realises the need to solve a quadratic equation	1
determines a solution	1
determines a second solution	1
substitutes the chosen solution into the equation	1
evaluates the reasonableness of one possible solution	1

Q1

2023

QCAA

Paper 1

1 mark

Q1

1 mark

$e^{\ln(x)}$ is equivalent to

A

0

B

1

C

$x$

D

$\frac{1}{x}$

Reveal Answer

A

0

This is incorrect. The expression evaluates to $x$ , not the constant 0. Furthermore, $\ln(x)$ is undefined at $x=0$ .

B

1

This is incorrect. The expression would only equal 1 if $x=1$ (since $\ln(1)=0$ and $e^0=1$ ), but it is not equal to 1 for all values of $x$ .

C

$x$

Correct Answer

This is correct. The natural logarithm function $\ln(x)$ and the exponential function $e^x$ are inverse functions, so $e^{\ln(x)} = x$ for all $x > 0$ .

D

$\frac{1}{x}$

This is incorrect. This result corresponds to $e^{-\ln(x)} = e^{\ln(1/x)} = \frac{1}{x}$ , not the original expression.

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 $y = 6$ .

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

Q4

2025

VCAA

Paper 2

1 mark

Q4

1 mark

Consider the system of equations below containing the parameter $k$ , where $k \in R$ .

\begin{align*} kx + 3y &= k^2\\ 2x + (2k + 1)y &= 6 - 2k \end{align*}

Find the value(s) of $k$ for which this system has no real solutions.

A

$k = -2$ only

B

$k = \frac{3}{2}$ only

C

$k = -2$ or $\frac{3}{2}$

D

$k \in R \setminus \left\{-2, \frac{3}{2}\right\}$

Reveal Answer

A

$k = -2$ only

Correct Answer

Setting the ratios of the $x$ and $y$ coefficients equal gives $k = -2$ and $k = \frac{3}{2}$ . For $k = -2$ , the ratio of the constants is different from the coefficient ratio, resulting in parallel lines and no solution.

B

$k = \frac{3}{2}$ only

When $k = \frac{3}{2}$ , the ratio of the constants equals the ratio of the coefficients, meaning the lines are coincident and the system has infinitely many solutions, not zero.

C

$k = -2$ or $\frac{3}{2}$

While both values make the lines parallel, $k = \frac{3}{2}$ makes the lines coincident (infinitely many solutions). Only $k = -2$ makes them distinct and parallel (no solutions).

D

$k \in R \setminus \left\{-2, \frac{3}{2}\right\}$

For all values of $k$ other than $-2$ and $\frac{3}{2}$ , the lines have different slopes and intersect at exactly one point, yielding a unique solution rather than no solution.

Q16

2023

QCAA

Paper 1

5 marks

Q16

5 marks

Solve for $x$ in the equation $4 + 7e^{-2x} = 3e^{2x}$ .

Reveal Answer

$4 + 7e^{-2x} = 3e^{2x}$
$4e^{2x} + 7 = 3e^{4x}$
$3e^{4x} - 4e^{2x} - 7 = 0$
$(3e^{2x} - 7)(e^{2x} + 1) = 0$
$3e^{2x} - 7 = 0$ $e^{2x} + 1 = 0$
$3e^{2x} = 7$ $e^{2x} = -1$
$e^{2x} = \frac{7}{3}$ not possible
$\ln \left(\frac{7}{3}\right) = 2x$
$x = \frac{1}{2}\ln \left(\frac{7}{3}\right)$

Marking Criteria

Descriptor	Marks
Correctly removes the negative index in the equation	1
Rearranges equation to equate to zero	1
Factorises the equation	1
Rejects the non-feasible solution	1
Determines a feasible solution for x	1

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

Q10

2025

VCAA

Paper 2

1 mark

Q10

1 mark

Consider $f : R \rightarrow R, \; f(x) = 2x^2 + x - 1$ and $g : R \rightarrow R, \; g(x) = \sin(x)$ .

The inequality $(f \circ g)(x) > 0$ is satisfied when

A

$\sin(x) \le -1$

B

$-1 < \sin(x) < 0$

C

$\frac{1}{2} < \sin(x) \le 1$

D

$0 < \sin(x) < \frac{1}{2}$

Reveal Answer

A

$\sin(x) \le -1$

Substituting $y = \sin(x)$ , the inequality becomes $(2y-1)(y+1) > 0$ . If $\sin(x) = -1$ , the expression equals 0, and $\sin(x) < -1$ is impossible for real numbers.

B

$-1 < \sin(x) < 0$

For $-1 < \sin(x) < 0$ , the factored expression $(2\sin(x)-1)(\sin(x)+1)$ is negative, meaning $(f \circ g)(x) < 0$ .

C

$\frac{1}{2} < \sin(x) \le 1$

Correct Answer

Substituting $y = \sin(x)$ gives the quadratic inequality $2y^2 + y - 1 > 0$ , which factors to $(2y-1)(y+1) > 0$ . This is satisfied when $y < -1$ or $y > \frac{1}{2}$ , and since $\sin(x) \ge -1$ , the only valid solution is $\frac{1}{2} < \sin(x) \le 1$ .

D

$0 < \sin(x) < \frac{1}{2}$

For $0 < \sin(x) < \frac{1}{2}$ , the expression $(2\sin(x)-1)(\sin(x)+1)$ evaluates to a negative number, failing the inequality $(f \circ g)(x) > 0$ .

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

Q20

2021

QCAA

Paper 2

3 marks

Q20

3 marks

The random variable $B$ is normally distributed with a mean of 0 and a standard deviation of 1.

Determine the probability that the quadratic equation $x^2 + 3x + 2B = 0$ has real roots.

Reveal Answer

The quadratic has real roots when
$b^2 - 4ac \ge 0$
$\therefore 9 - 8B \ge 0$
$\therefore 9 \ge 8B$
$\therefore \frac{9}{8} \ge B$
Using the standard normal distribution (the given distribution)
$P(B \le \frac{9}{8})$
$= 0.8697$

Marking Criteria

Descriptor	Marks
Correctly identifies the need to use the discriminant	1
Correctly determines the range of values for $B$	1
Determines probability	1

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

Q4

2025

QCAA

Paper 2

1 mark

Q4

1 mark

An object moves in a straight line such that its velocity (m s $^{-1}$ ) is given by $v(t) = t^2 - e^{-t}$ for $0 \le t \le 5$ , where $t$ represents time (s).

Determine the time, $t$ , when the acceleration of the object is 2 m s $^{-2}$ .

A

0.768

B

1.157

C

3.865

D

4.135

Reveal Answer

A

0.768

Correct Answer

Acceleration is the derivative of velocity, so $a(t) = v'(t) = 2t + e^{-t}$ . Setting $2t + e^{-t} = 2$ and solving for $t$ using a calculator yields $t \approx 0.768$ s.

B

1.157

This is the solution to $2t - e^{-t} = 2$ . This error occurs if you incorrectly differentiate the velocity function by leaving the sign of the exponential term unchanged, rather than applying the chain rule to get $+e^{-t}$ .

C

3.865

This value does not satisfy the acceleration equation $2t + e^{-t} = 2$ . To find the correct time, you must first find the derivative of the velocity function and set it equal to 2.

D

4.135

This value is incorrect and does not satisfy $a(t) = 2$ . Remember that acceleration is the first derivative of velocity with respect to time, giving $a(t) = 2t + e^{-t}$ .

Q17

2022

QCAA

Paper 1

4 marks

Q17

4 marks

Determine the value of $b$ given $\int_a^b 3x^2 dx = 117$ and $\int_a^{b-1} 3x^2 dx = 56$ for $b > 1$ .

Reveal Answer

Using first integral
$F(b) - F(a) = 117$
$b^3 - a^3 = 117$ ... Equation I

Using second integral
$(b-1)^3 - a^3 = 56$ ... Equation II
Equation I – Equation II
$b^3 - (b-1)^3 = 61$
$b^3 - (b^3 - 3b^2 + 3b - 1) = 61$
$3b^2 - 3b - 60 = 0$
$b^2 - b - 20 = 0$
$(b-5)(b+4) = 0$
$b = -4, 5$
Given $b > 1$
$\therefore b = 5$

Marking Criteria

Descriptor	Marks
correctly establishes a formula for one of the integrals	1
determines equation in b	1
determines values of b	1
evaluates the reasonableness of solutions	1

Q4

2025

QCAA

Paper 1

1 mark

Q4

1 mark

Solve for $x$ in the equation $\ln(x) + \ln(2) = \ln(8)$ .

A

3

B

4

C

6

D

10

Reveal Answer

A

3

This is incorrect. A student might choose this by confusing the equation with an exponential one like $2^x = 8$ , rather than applying the properties of logarithms.

B

4

Correct Answer

This is correct. Using the product rule for logarithms, $\ln(x) + \ln(2) = \ln(2x)$ . Setting $2x = 8$ yields $x = 4$ .

C

6

This is incorrect. It results from the common misconception that $\ln(x) + \ln(y) = \ln(x+y)$ , which would incorrectly lead to $x+2=8$ and $x=6$ .

D

10

This is incorrect. A student might choose this by incorrectly adding the arguments ( $x = 8 + 2 = 10$ ) instead of using the product rule for logarithms.

VCAA Mathematical Methods Algebra, number and structure

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