VCE Specialist Mathematics 2024 Exam 2

Every question from the 2024 VCE Specialist Mathematics Examination 2, the technology-active (CAS) paper. Multiple-choice answers come with a one-line reason; Section B solutions sit behind a Show worked solution toggle. For the no-calculator paper, see the 2024 Examination 1 walkthrough.

How to use this page

• Questions are from the 2024 VCE Specialist Mathematics Examination 2, copyright Victorian Curriculum and Assessment Authority (VCAA). Each question is summarised briefly here; open the official examination PDF for the full wording, diagrams and answer options.
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2024 Specialist Mathematics Examination 2 external assessment report (Word document), and every Section B result was recomputed and compared with the report. Both files are listed on the VCAA Specialist Mathematics examinations page.

Structure and timing

Examination 2 is 80 marks in 120 minutes (plus 15 minutes reading time), with a CAS calculator and one bound reference allowed. Take $g = 9.8 \text{ m s}^{-2}$.

• Section A (20 marks): 20 multiple-choice questions. Aim for about 25 minutes.
• Section B (60 marks): 6 extended-response questions (10, 10, 10, 11, 10 and 9 marks).

Section A: Multiple choice

Q1

Contrapositive of "if $m + n \ge 9$ then $m \ge 5$ or $n \ge 5$". Answer: C - negate and swap: if $m < 5$ and $n < 5$, then $m + n < 9$ (the "or" becomes "and").

Q2

$f(x) = \dfrac{x^2 + 3x - 10}{x - 2}$ for $x \ne 2$ and $f(2) = 7$. Which is correct? Answer: A - $\tfrac{(x + 5)(x - 2)}{x - 2} = x + 5$, which equals 7 at $x = 2$, so $f$ is continuous. (37% wrongly chose a point of discontinuity.)

Q3

When does $f(x) = \dfrac{x - h}{(x + 1)(x - 4)}$ have no turning points? Answer: C - $f'(x) = 0$ leads to $x^2 - 2hx + 3h + 4 = 0$, which has no solutions when $(h - 4)(h + 1) < 0$. At $h = -1$ or $h = 4$ the function cancels to a hyperbola, also with no turning points: $-1 \le h \le 4$.

Q4

$\sin(x) = a$ with $x \in \left(\tfrac{3\pi}{2}, 2\pi\right)$. Find $\cos\!\left(\tfrac x2\right)$. Answer: A - $\cos x = \sqrt{1 - a^2} > 0$ in the fourth quadrant, and $\tfrac{x}{2} \in \left(\tfrac{3\pi}{4}, \pi\right)$ where cosine is negative, so $\cos\tfrac x2 = -\sqrt{\tfrac{1 + \cos x}{2}} = -\dfrac{\sqrt{1 + \sqrt{1 - a^2}}}{\sqrt2}$. (27% correct.)

Q5

For $z = 1 + \sqrt3i$, how is $-\bar z$ located? Answer: C - $-\bar z = -1 + \sqrt3i$, the reflection of $z$ in the imaginary axis.

Q6

$z = 3 + ki$. Which $k$ makes $z^2 + 4iz + 3$ purely imaginary? Answer: D - the real part is $12 - 4k - k^2 = 0$, so $k = 2$ or $k = -6$; with $k = 2$ the imaginary part is $24 \ne 0$.

Q7

Solving $\dfrac{dy}{dx} = e^{x - y}\big(\cos(x - y) - \cos(x + y)\big)$. Answer: C - $\cos(x - y) - \cos(x + y) = 2\sin x\sin y$, so separating gives $\int\frac{e^y}{\sin y}\,dy = 2\int e^x\sin x\,dx$.

Q8

Euler's method on $\dfrac{dy}{dx} = xy^2$, $y(0) = 1$, gives $y_3 = 1.126528$. Find $h$. Answer: C - $y_1 = 1$, $y_2 = 1 + h^2$, $y_3 = y_2 + 2h^2y_2^2$; with $h = 0.2$: $1.04 + 0.08(1.0816) = 1.126528$.

Q9

Length of $x = 1 - \cos t$, $y = t - \sin t$, $t \in [0, 2\pi]$. Answer: A - $\dot x^2 + \dot y^2 = \sin^2 t + (1 - \cos t)^2 = 2 - 2\cos t = 4\sin^2\tfrac t2$, and $\sin\tfrac t2 \ge 0$ on this interval, so $L = \int_0^{2\pi}2\sin\tfrac t2\,dt$.

Q10

$x = 5t$, $y = 12t$, $0 \le t \le k$, rotated about the $y$-axis. Surface area? Answer: A - $2\pi\int_0^k 5t\sqrt{5^2 + 12^2}\,dt = 2\pi \times 65 \times \tfrac{k^2}{2} = 65\pi k^2$.

Q11

From a velocity-time graph (40 m/s east at $t = 0$, down to $-20$ at $t = 100$, back to 0 at $t = 150$), how far east of $O$ is the particle at $t = 150$? Answer: B - signed area: $\tfrac12\left(\tfrac{200}{3}\right)(40) - \tfrac12\left(\tfrac{100}{3}\right)(20) - \tfrac12(50)(20) = \tfrac{4000}{3} - \tfrac{1000}{3} - 500 = 500$ m.

Q12

$x = e^{(k - 1)t}$, $k > 1$. Acceleration when $x = k + 1$? Answer: C - $\ddot x = (k - 1)^2x = (k - 1)^2(k + 1) = (k^2 - 1)(k - 1)$.

Q13

The angle between $2\mathbf i - \mathbf j + 2\mathbf k$ and $2\mathbf i + m\mathbf j + 6\mathbf k$ is $\cos^{-1}\tfrac{13}{21}$, $m > 0$. Find $m$. Answer: B - $\dfrac{16 - m}{3\sqrt{40 + m^2}} = \dfrac{13}{21}$ is satisfied by $m = 3$: $\tfrac{13}{3 \times 7}$.

Q14

$|\mathbf r| = 9$, $\mathbf s = 2\mathbf i - 2\mathbf j + \mathbf k$, and the vector resolute of $\mathbf r$ on $\mathbf s$ is $-4\mathbf i + 4\mathbf j - 2\mathbf k = -2\mathbf s$. Scalar resolute of $\mathbf s$ on $\mathbf r$? Answer: B - $\tfrac{\mathbf r\cdot\mathbf s}{|\mathbf s|^2} = -2$ gives $\mathbf r\cdot\mathbf s = -18$, so the scalar resolute is $\tfrac{-18}{9} = -2$. (36% correct.)

Q15

$\mathbf r(t) = \sin t\,\mathbf i + \cos 2t\,\mathbf j$. Describe the motion. Answer: A - $y = 1 - 2\sin^2 t = 1 - 2x^2$; starting at $(0, 1)$, it reaches $(1, -1)$ at $t = \tfrac{\pi}{2}$, reverses, reaches $(-1, -1)$ at $t = \tfrac{3\pi}{2}$, and returns to $(0, 1)$ after $2\pi$ seconds. (44% chose C, which has the wrong period.)

Q16

How many times in $\left(0, \tfrac{\pi}{2}\right)$ is the velocity of particle 1 perpendicular to the position of particle 2? Answer: A - the dot product simplifies to $\cos^2t - \sin^2t + \cos 2t = 2\cos 2t$, which is zero only at $t = \tfrac{\pi}{4}$.

Q17

Shortest distance between the parallel lines through $(1, 3, 1)$ and $(-2, 1, 3)$ with direction $(1, 1, 1)$. Answer: B - $\dfrac{|(-3, -2, 2)\times(1, 1, 1)|}{\sqrt3} = \dfrac{|(-4, 5, -1)|}{\sqrt3} = \dfrac{\sqrt{42}}{\sqrt3} = \sqrt{14}$.

Q18

Where does $\mathbf r = (1 - 2t, 1 + t, -2 + 3t)$ meet $3x - 2y + 4z = 5$? Answer: D - substituting gives $4t - 7 = 5$, so $t = 3$ and the point is $(-5, 4, 7)$.

Q19

A type II error occurs when... Answer: A - the null hypothesis is not rejected even though the alternative is true.

Q20

Avocado mass $M \sim N(200, 7.5^2)$ and flesh $F = 0.7M$. Probability four avocados give more than 570 g of flesh? Answer: B - the total has mean $0.7 \times 800 = 560$ and variance $4 \times 0.49 \times 56.25 = 110.25$ (sd 10.5), so $\Pr\!\left(Z > \tfrac{10}{10.5}\right) \approx 0.1705$.

Section B: Extended response

Question 1 (10 marks)

Let $f(x) = \dfrac{x^4 - x^2 + 1}{1 - x^2}$.

a

Sketch $y = f(x)$, labelling vertical asymptotes and stationary points. (3 marks)

b

The region bounded by $y = f(x)$ and the lines $y = 1$ and $y = 6$ is rotated about the $y$-axis. i. Write a definite integral in $y$ only for the volume. (2 marks)

ii

Find the volume to one decimal place. (1 mark)

c

For $g(x) = \dfrac{x^4 + b}{1 - x^2}$, find $b$ so that the graph has no asymptotes. (1 mark)

d

Given $g'(x) = \dfrac{-2x\big((x^2 - 1)^2 - (b + 1)\big)}{(1 - x^2)^2}$, find the values of $b$ for which $g$ has exactly i. one stationary point (1 mark)

ii

three stationary points (1 mark)

iii

five stationary points. (1 mark)

Question 2 (10 marks)

a

Write $|z - z_1| = |z - z_2|$, with $z_1 = 1 + 2i$ and $z_2 = 4$, in the form $y = mx + c$. (2 marks)

b

The segment from $z_1$ to $z_2$ is a diameter of a circle. Write the circle as $|z - z_c| = r$. (2 marks)

c

Sketch $|z - (1 + 2i)| = 2$, labelling its imaginary-axis intercepts. (2 marks)

d

A ray starts at $z = 2 - i$ and passes through $z = -2 + 3i$, cutting this circle. i. Sketch the ray. (1 mark)

ii

Write its equation as $\text{Arg}(z - z_0) = \theta$. (1 mark)

e

Find the area of the minor segment cut off by the ray. (2 marks)

Question 3 (10 marks)

Pollutant enters a pond at $\dfrac{dV}{dt} = \dfrac{8t}{240 + 5t^4}$ m³/day, spreading as a thin disc of radius $r(t)$ m and constant depth 1 mm.

a

Find the maximum rate of entry and when it occurs. (1 mark)

b

At what rate is the radius increasing when $t = 4$, given $r = 6.54$ m then? Answer to two decimal places. (3 marks)

c. i

Use $u = \sqrt5t^2$ to write $\int\frac{8t}{240 + 5t^4}\,dt$ in terms of $u$. (1 mark)

ii

Hence find $V$ in the form $\dfrac{1}{a\sqrt b}\arctan\!\left(\dfrac{t^c}{d\sqrt b}\right)$. (1 mark)

d

What surface area does the pollutant approach? Answer in m² to two decimal places. (2 marks)

e

Clean-up starts after five days, removing 0.05 m³/day while pollutant keeps entering. After how many days from the start of clean-up is the pond free of pollutant (one decimal place)? (2 marks)

Question 4 (11 marks)

A yacht moves along $\mathbf r_Y(t) = 3\sec(t)\,\mathbf i + 2\tan(t)\,\mathbf j$, $\tfrac{2\pi}{3} \le t \le \tfrac{4\pi}{3}$ (metres, minutes).

a

Show that the path is $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$. (1 mark)

b

Sketch the path, labelling endpoints and the direction of motion. (2 marks)

c. i

Write the square of the speed in terms of $\sec(t)$. (1 mark)

ii

Find when the minimum speed occurs. (1 mark)

iii

State the minimum speed. (1 mark)

iv

State the yacht's position then. (1 mark)

d. i

Write a definite integral for the distance travelled. (1 mark)

ii

Find it to one decimal place. (1 mark)

e

A drone is at $\mathbf r_D(t) = (2 - 3t)\mathbf i + (4t - 1)\mathbf j + (6 - t)\mathbf k$. Find the shortest distance from the drone to the yacht during the yacht's journey, to one decimal place. (2 marks)

Question 5 (10 marks)

$A(1, -2, 3)$ and $B(2, -5, -1)$.

a

Find a vector equation of the line through $A$ and $B$. (1 mark)

b

Find the shortest distance from $L_1: \mathbf r_1(t) = 2\mathbf i + \mathbf j - 3\mathbf k + t(-\mathbf i + 2\mathbf j + \mathbf k)$ to $A$, in the form $\dfrac{a\sqrt b}{c}$. (3 marks)

c

Find the Cartesian equation of the plane through $A$, $B$ and $C(0, 2, -5)$. (3 marks)

d

The plane $2x - 3y + 4z = 12$ meets the axes at three points. i. Find them. (1 mark)

ii

Find the area of the triangle they form, as $m\sqrt n$. (2 marks)

Question 6 (9 marks)

Bottles are filled with volume normally distributed, mean $\mu$ mL and $\sigma = 4.2$ mL; the target is $\mu = 1000$. A sample of nine bottles has mean 997.5 mL, and the machine is paused if the mean is significantly less than 1000 mL at the 5% level.

a

State $H_0$ and $H_1$. (1 mark)

b. i

Find the $p$ value to three decimal places. (1 mark)

ii

Should the machine be paused? Give a reason. (1 mark)

c

If the true mean is 997 mL, find the probability of a type II error for this test, to two decimal places. (2 marks)

d

The machine is paused if $\bar X < a$ or $\bar X > b$, where each tail has probability 0.01. Find $a$ and $b$ to one decimal place. (1 mark)

A new machine gives a sample of 50 bottles with mean 1005 mL; assume $\sigma = 4$ mL.

e

Find a 95% confidence interval for the mean, to one decimal place. (1 mark)

f

Of 40 such 95% intervals, how many would be expected to contain the true mean? (1 mark)

g

Find the minimum sample size so the sample mean is within 1 mL of the true mean with 95% confidence. (1 mark)