VCE Math Methods 2022 Exam 2

Every question from the 2022 VCE Mathematical Methods 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 2022 Examination 1 walkthrough.

How to use this page

• Questions are from the 2022 VCE Mathematical Methods Examination 2, copyright Victorian Curriculum and Assessment Authority (VCAA). Each question is summarised briefly here; open the official examination PDF for the full wording, graphs and answer options.
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2022 Mathematical Methods 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 Mathematical Methods examinations page.
• Redacted part. VCAA has redacted Question 4e.ii from the published paper and report, following the Independent Review into its examination-setting policies, processes and procedures. It is not covered here.
• Study design. This was the last Examination 2 set on the previous Mathematical Methods study design (2016 to 2022); the current one began in 2023. Almost all of the paper is still on the course. The exception is the matrix form of the transformation in Question 2e, flagged where it appears. Anything not flagged is still examinable.

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. That is 1.5 minutes per mark.

• Section A (20 marks): 20 multiple-choice questions. Aim for about 25 minutes.
• Section B (60 marks): 5 extended-response questions (11, 16, 14, 10 and 9 marks). Give exact answers unless told to round, and write down what you entered into CAS.

Section A: Multiple choice

Q1

Period of $f(x) = 3\cos(2x + \pi)$. Answer: B - the period of $\cos(nx + c)$ is $\tfrac{2\pi}{n} = \tfrac{2\pi}{2} = \pi$.

Q2

Horizontal asymptote of $y = \dfrac{1}{(x + 3)^2} + 4$. Answer: A - the fraction tends to 0 as $x \to \pm\infty$, so $y = 4$.

Q3

Gradient of $y = e^{3x}$ where it crosses the vertical axis. Answer: E - $\tfrac{dy}{dx} = 3e^{3x}$, which is 3 at $x = 0$.

Q4

Which function is not continuous on $[0, 5]$? Answer: D - $\tan\!\left(\tfrac{x}{3}\right)$ has an asymptote where $\tfrac{x}{3} = \tfrac{\pi}{2}$, at $x = \tfrac{3\pi}{2} \approx 4.71$, inside the interval.

Q5

Largest $a$ for which $f(x) = x^2 + 3x - 10$ on $(-\infty, a]$ is one-to-one. Answer: C - the parabola's turning point is at $x = -\tfrac32$, so $a = -1.5$.

Q6

Which pair are not inverse functions? Answer: C - the inverse of $f(x) = x^2$, $x < 0$ is $-\sqrt x$, $x > 0$, not $\sqrt x$ (the range of the inverse must be the domain of $f$, which is negative). (47% correct.)

Q7

Choose a possible graph of $f'$ for a positive cubic that passes upward through the origin, has a local maximum just right of the $y$-axis and a local minimum further right. Answer: E - $f$ increases, decreases, then increases, so $f'$ is an upright parabola that is positive at $x = 0$ and has two positive $x$-intercepts.

Q8

Given $\int_0^b f(x)\,dx = 10$ and $\int_0^a f(x)\,dx = -4$ with $0 < a < b$, find $\int_a^b f(x)\,dx$. Answer: E - $10 - (-4) = 14$.

Q9

Shortest distance $d$ from the origin to the point $(x, y)$ on $f(x) = \sqrt{2x + 1}$. Answer: D - $d = \sqrt{x^2 + y^2} = \sqrt{x^2 + 2x + 1} = x + 1$ for $x \ge 0$. (50% correct.)

Q10

55% of 1000 adults surveyed were happy with their physical activity; find the approximate 95% confidence interval (as a percentage). Answer: D - $0.55 \pm 1.96\sqrt{\tfrac{0.55 \times 0.45}{1000}} = 0.55 \pm 0.031$, so $(51.9, 58.1)$.

Q11

Given $\tfrac{d}{dx}\big(x\sin(x)\big) = \sin(x) + x\cos(x)$, find $\tfrac1k\int x\cos(x)\,dx$. Answer: C - integrate both sides: $x\sin(x) = \int\sin(x)\,dx + \int x\cos(x)\,dx$, so $\tfrac1k\int x\cos(x)\,dx = \tfrac1k\left(x\sin(x) - \int\sin(x)\,dx\right) + c$.

Q12

A bag has 3 red and $x$ black pens; two are drawn without replacement. Find $\Pr(\text{one of each colour})$. Answer: A - $2 \times \dfrac{3}{3 + x} \times \dfrac{x}{2 + x} = \dfrac{6x}{(2 + x)(3 + x)}$ (red then black, or black then red).

Q13

Maximal domain of $f(x) = \log_e\!\left(\tfrac{x + a}{x - a}\right)$, $a > 0$. Answer: C - $\tfrac{x + a}{x - a} > 0$ when the numerator and denominator have the same sign: $x < -a$ or $x > a$, that is $R \setminus [-a, a]$. (39% correct; 40% chose $R \setminus (-a, a)$, which wrongly includes $x = \pm a$.)

Q14

$X$ has pdf $f(x) = \tfrac29xe^{-\frac19x^2}$ for $x \ge 0$. Find $\text{E}(X)$ to three decimal places. Answer: B - $\int_0^\infty x \cdot \tfrac29xe^{-\frac{x^2}{9}}\,dx \approx 2.659$ (CAS).

Q15

Maximal domain of $f(x) = \sqrt{x^2 - 2x - 3}$. Answer: E - need $(x - 3)(x + 1) \ge 0$: $(-\infty, -1] \cup [3, \infty)$.

Q16

$f(x) = \tfrac13x^3 + mx^2 + nx + p$ has turning points at $x = -3$ and $x = 1$ and passes through $(3, 4)$. Find $m, n, p$. Answer: B - $f'(x) = x^2 + 2mx + n = (x + 3)(x - 1) = x^2 + 2x - 3$, so $m = 1$, $n = -3$; then $f(3) = 9 + 9 - 9 + p = 4$ gives $p = -5$.

Q17

$g$ is continuous on $[a, b]$, its average rate of change over $[a, b]$ is positive, and its instantaneous rate of change at $x = \tfrac{a + b}{2}$ is negative. What must $g$ be? Answer: A - $g(b) > g(a)$ overall but $g$ is decreasing at the midpoint, so it rises and falls and some $y$-values occur twice: $g$ is many-to-one. (39% correct.)

Q18

$X \sim \text{Bi}(20, 0.88)$ and $\Pr(X \ge 16 \mid X \ge a) = 0.9175$. Find $a$. Answer: B - since $a < 16$ in every option, this is $\dfrac{\Pr(X \ge 16)}{\Pr(X \ge a)}$; testing the options on CAS, $a = 12$ gives 0.9175.

Q19

An open box is made from an $a$ by $b$ sheet by cutting squares of side $x$ from the corners. Find $x$ for maximum volume. Answer: D - $V = x(a - 2x)(b - 2x)$ and $V'(x) = 12x^2 - 4(a + b)x + ab = 0$ gives $x = \dfrac{a + b \pm \sqrt{a^2 - ab + b^2}}{6}$; the maximum is at the smaller root. (34% correct.)

Q20

$\theta \sim \text{N}(42, 8^2)$ degrees and $d = 50\sin(2\theta)$. Find $\Pr(d > 40)$. Answer: A - $\sin(2\theta) > 0.8$ gives $26.57° < \theta < 63.43°$, and $\Pr(26.57 < \theta < 63.43) \approx 0.969$. (30% correct.)

Section B: Extended response

Question 1 (11 marks)

Let $f(x) = \dfrac{x^2}{12}$. The tangent to $f$ at a point $M$ has gradient $-2$.

a

State the equation of the axis of symmetry of the graph of $f$. (1 mark)

b

State the derivative of $f$ with respect to $x$. (1 mark)

c

Find the equation of the tangent to $f$ at $M$. (2 marks)

d. i

Find the equation of the line perpendicular to the tangent passing through $M$. (1 mark)

ii

This perpendicular line also cuts $f$ at $N$. Find the area enclosed by this line and the curve $y = f(x)$. (2 marks)

e

Another parabola is $g(x) = \dfrac{x^2}{4a^2}$, $a > 0$. The tangent to $g$ at $x = -b$ ($b > 0$) and the line perpendicular to it there are drawn; the shaded region lies between that perpendicular line and $g$. Find $b$, in terms of $a$, such that the shaded area is a minimum. (4 marks)

Show worked solution

a. [1 mark]

$x = 0$ (the parabola's turning point is at the origin).

b. [1 mark]

$f'(x) = \dfrac{x}{6}$.

c. [2 marks]

Solve $f'(x) = -2$: $\tfrac{x}{6} = -2$, so $x = -12$ and $f(-12) = \tfrac{144}{12} = 12$. So $M = (-12, 12)$ and the tangent is

$$y - 12 = -2(x + 12) \implies y = -2x - 12.$$

d. i. [1 mark]. Perpendicular gradient $\tfrac12$, through $(-12, 12)$:

$$y - 12 = \frac12(x + 12) \implies y = \frac{x}{2} + 18.$$

ii. [2 marks]. Solve $\tfrac{x^2}{12} = \tfrac{x}{2} + 18$: $x^2 - 6x - 216 = 0$, so $(x + 12)(x - 18) = 0$ and $N$ has $x = 18$. The line is above the parabola between the two points:

$$\int_{-12}^{18}\left(\frac{x}{2} + 18 - \frac{x^2}{12}\right)dx = 375 \text{ square units}.$$

e. [4 marks]. $g'(x) = \tfrac{x}{2a^2}$, so the tangent at $x = -b$ has gradient $-\tfrac{b}{2a^2}$ and the perpendicular line has gradient $\tfrac{2a^2}{b}$. Through $\left(-b, \tfrac{b^2}{4a^2}\right)$:

$$y_n = \frac{2a^2}{b}(x + b) + \frac{b^2}{4a^2}.$$

Solving $y_n = g(x)$ with CAS gives $x = -b$ and $x = \dfrac{8a^4 + b^2}{b}$. The shaded area is

$$A(b) = \int_{-b}^{\frac{8a^4 + b^2}{b}}\big(y_n - g(x)\big)dx = \frac{\left(4a^4 + b^2\right)^3}{3a^2b^3}.$$

(Expanded, the numerator is $64a^{12} + 48a^8b^2 + 12a^4b^4 + b^6$.) Solve $A'(b) = 0$ for $b > 0$:

$$\frac{d}{db}\ln A = \frac{6b}{4a^4 + b^2} - \frac{3}{b} = 0 \implies 2b^2 = 4a^4 + b^2 \implies b^2 = 4a^4 \implies b = 2a^2.$$

$A \to \infty$ as $b \to 0^+$ and as $b \to \infty$, so this is a minimum.

From the report. Part a was not answered well for a 1-mark part: an equation was required, and the common errors were $y = 0$, 0, $(0, 0)$, "$y$-axis" and $-\tfrac{b}{2a} = 0$. In parts c and d.i an equation was required; some gave only the expression $-2x - 12$, some found $f'(-2)$ instead of solving $f'(x) = -2$, and $y = \tfrac{x}{2} + 12$ was the most common wrong perpendicular line. Those who used technology made fewer algebraic errors. In d.ii some subtracted the wrong way, used the tangent line, or had the wrong terminals. In part e (55% scored 0) many could not find the perpendicular line, some integrated $y_n$ without subtracting $g(x)$, and a common incorrect answer was $b = -2a^2$.

Question 2 (16 marks)

Fox and rabbit populations on an island vary periodically with the same period, $t$ in weeks. The graph shows a fox minimum at $(20, 700)$ and a fox maximum at $(100, 2500)$. Rabbits: $r(t) = 1700\sin\!\left(\dfrac{\pi t}{80}\right) + 2500$.

a. i

State the initial population of rabbits. (1 mark)

ii

State the minimum and maximum population of rabbits. (1 mark)

iii

State the number of weeks between maximum populations of rabbits. (1 mark)

Foxes: $f(t) = a\sin\big(b(t - 60)\big) + 1600$.

b

Show that $a = 900$ and $b = \dfrac{\pi}{80}$. (2 marks)

c

Find the maximum combined population of foxes and rabbits, to the nearest whole number. (1 mark)

d

What is the number of weeks between the times when the combined population is a maximum? (1 mark)

The foxes are better modelled by the image of $y = \sin(t)$ under $Q\left(\begin{bmatrix} t \\ y \end{bmatrix}\right) = \begin{bmatrix} \frac{90}{\pi} & 0 \\ 0 & 900 \end{bmatrix}\begin{bmatrix} t \\ y \end{bmatrix} + \begin{bmatrix} 60 \\ 1600 \end{bmatrix}$.

e. Using this fox model, find the average combined population during the first 300 weeks, to the nearest whole number. (4 marks)

Over a longer time the rabbits are modelled by $s(t) = 1700e^{-0.003t}\sin\!\left(\dfrac{\pi t}{80}\right) + 2500$, $t \ge 0$.

f

Find the average rate of change between the first two times when the rabbit population is at a maximum, correct to one decimal place. (2 marks)

g

Find the time $t > 40$, in weeks, when the rate of change of the rabbit population is at its greatest positive value, to the nearest whole number. (2 marks)

h

Over time, the rabbit population approaches a particular value. State this value. (1 mark)

Study design. Part e gives the transformation $Q$ as a matrix, which belongs to the previous study design; the current course does not require matrix notation for transformations. The transformation itself (dilations from both axes and a translation) is still examinable, so read $Q$ as the words in the solution below and part e is still good practice.

Show worked solution

a. i. [1 mark]

$r(0) = 1700\sin(0) + 2500 = 2500$ rabbits.

ii. [1 mark]

$2500 \pm 1700$: minimum 800 and maximum 4200 rabbits.

iii. [1 mark]

The period is $\dfrac{2\pi}{\pi/80} = 160$ weeks.

b. [2 marks]

The fox amplitude is half the distance from minimum to maximum, and the period is twice the time from minimum to maximum:

$$a = \frac{2500 - 700}{2} = 900, \qquad \text{period} = 2(100 - 20) = 160 = \frac{2\pi}{b} \implies b = \frac{2\pi}{160} = \frac{\pi}{80}.$$

Check: the centre $\tfrac{2500 + 700}{2} = 1600$ matches, and $f(20) = 900\sin\!\left(-\tfrac{\pi}{2}\right) + 1600 = 700$, a minimum as required.

c. [1 mark]

Maximise $r(t) + f(t)$ with CAS: the maximum combined population is about 5339.46, so 5339. (It is not $4200 + 2500$, because the two maximums happen at different times.)

d. [1 mark]

Both populations have period 160 weeks, so the combined population does too: 160 weeks.

e. [4 marks]

$Q$ maps $(t, y)$ to $\left(\tfrac{90}{\pi}t + 60,\ 900y + 1600\right)$: a dilation by $\tfrac{90}{\pi}$ from the $y$-axis, a dilation by 900 from the $t$-axis, then a translation of 60 right and 1600 up. Writing $t' = \tfrac{90}{\pi}t + 60$ and $y' = 900y + 1600$, so $t = \tfrac{\pi(t' - 60)}{90}$ and $y = \tfrac{y' - 1600}{900}$, the image of $y = \sin(t)$ is

$$y' = 900\sin\!\left(\frac{\pi(t' - 60)}{90}\right) + 1600.$$

The average value of the combined population over $[0, 300]$ is

$$\frac{1}{300}\int_0^{300}\left(900\sin\!\left(\frac{\pi(t - 60)}{90}\right) + 1600 + r(t)\right)dt \approx 4142.$$

f. [2 marks]. Solve $s'(t) = 0$ with CAS: the first two maximums are at $t \approx 38.058$ and $t \approx 198.058$ (160 weeks apart). Then

$$\frac{s(198.058\ldots) - s(38.058\ldots)}{198.058\ldots - 38.058\ldots} \approx \frac{3435.7 - 4012.2}{160} \approx -3.6 \text{ rabbits per week}.$$

Keep full accuracy until the end; rounding the times early changes the answer.

g. [2 marks]. The rate of change $s'(t)$ is greatest where $s''(t) = 0$ (and $s'$ is positive). The first two solutions of $s''(t) = 0$ with $t > 40$ are $t \approx 76.1$ (where $s'$ is most negative) and $t \approx 156.1$ (where $s'$ is most positive, about 41.8 rabbits per week). Because the oscillation dies away, later peaks of $s'$ are smaller. So $t \approx 156$ weeks.

h. [1 mark]. $e^{-0.003t} \to 0$, so $s(t) \to 2500$ rabbits.

From the report. Part a.i was done very well. In a.ii some gave coordinates such as $(120, 800)$ and $(40, 4200)$ instead of the values, or used 700 and 4000; in a.iii 80 was a common wrong answer. Part b is a "show that", so working was needed. Part c was not answered well (38%): 5340 (a rounding error) was common, as was adding the two separate maximums. In part d an exact answer was needed; 160.1 was common. In part e many did the transformation but then forgot to add $r(t)$, subtracted it, gave 1600, or found an average rate of change instead of an average value. In part f some rounded too early, used $\tfrac{s(200) - s(40)}{200 - 40}$, or used $r(t)$. In part g many solved $\tfrac{ds}{dt} = 0$; common wrong answers were 41.8 (the rate itself) and 76 weeks (the greatest negative rate). In part h, 0 was a common wrong answer.

Question 3 (14 marks)

Mika flips a fair coin five times; $X$ is the number of heads.

a. i

Find $\Pr(X = 5)$. (1 mark)

ii

Find $\Pr(X \ge 2)$. (1 mark)

iii

Find $\Pr(X \ge 2 \mid X < 5)$, correct to three decimal places. (2 marks)

iv

Find the expected value and the standard deviation of $X$. (2 marks)

The height $H$ (m) reached by a flip has pdf $f(h) = ah^2 + bh + c$ for $1.5 \le h \le 3$, and 0 elsewhere.

b. i

State the value of $\displaystyle\int_{1.5}^{3}f(h)\,dh$. (1 mark)

ii

Given $\Pr(H \le 2) = 0.35$ and $\Pr(H \ge 2.5) = 0.25$, find $a$, $b$ and $c$. (3 marks)

iii

The ceiling is 3 m above the floor. The minimum distance $D$ between the coin and the ceiling has pdf $g(d) = f(rd + s)$. Find $r$ and $s$. (1 mark)

Bella's coin lands heads with probability $p$. She flips it 25 times; $\hat P$ is the sample proportion of heads.

c. i

Is $\hat P$ discrete or continuous? Justify your answer. (1 mark)

ii

If $\hat p = 0.4$, find an approximate 95% confidence interval for $p$, correct to three decimal places. (1 mark)

iii

With $\hat p = 0.4$, how many flips would be needed to halve the width of the interval in part c.ii? (1 mark)

Show worked solution

a. i. [1 mark]

$X \sim \text{Bi}\left(5, \tfrac12\right)$, so $\Pr(X = 5) = \left(\tfrac12\right)^5 = \tfrac{1}{32} = 0.03125$.

ii. [1 mark]

$\Pr(X \ge 2) = 1 - \Pr(X = 0) - \Pr(X = 1) = 1 - \tfrac{1}{32} - \tfrac{5}{32} = \tfrac{26}{32} = \tfrac{13}{16} = 0.8125$.

iii. [2 marks]

$$\Pr(X \ge 2 \mid X < 5) = \frac{\Pr(2 \le X \le 4)}{\Pr(X \le 4)} = \frac{\frac{13}{16} - \frac{1}{32}}{1 - \frac{1}{32}} = \frac{25/32}{31/32} = \frac{25}{31} \approx 0.806.$$

iv. [2 marks]

$\text{E}(X) = np = \tfrac52$ and $\text{sd}(X) = \sqrt{np(1 - p)} = \sqrt{\tfrac54} = \tfrac{\sqrt5}{2}$.

b. i. [1 mark]

The total area under a pdf is 1, and $[1.5, 3]$ is the whole of the non-zero part, so the integral is 1.

ii. [3 marks]

Three conditions give three linear equations in $a$, $b$ and $c$:

$$\int_{1.5}^{3}f(h)\,dh = 1, \qquad \int_{1.5}^{2}f(h)\,dh = 0.35, \qquad \int_{2.5}^{3}f(h)\,dh = 0.25.$$

Solving with CAS:

$$a = -\frac45, \qquad b = \frac{17}{5}, \qquad c = -\frac{167}{60}.$$

(Check: $f(h) \ge 0$ on $[1.5, 3]$; for example $f(1.5) \approx 0.52$ and $f(3) \approx 0.22$.)

iii. [1 mark]

$D = 3 - H$, so $H = 3 - D$ and $g(d) = f(3 - d) = f(-d + 3)$. Hence $r = -1$ and $s = 3$.

c. i. [1 mark]

Discrete: $\hat P$ can take only the countable set of values $0, \tfrac{1}{25}, \tfrac{2}{25}, \ldots, 1$.

ii. [1 mark]

$0.4 \pm 1.96\sqrt{\tfrac{0.4 \times 0.6}{25}}$ (or the CAS confidence interval command) gives $(0.208, 0.592)$.

iii. [1 mark]

The width is proportional to $\tfrac{1}{\sqrt n}$, so halving it needs $\sqrt n$ to double: $n = 4 \times 25 = 100$ flips.

From the report. Exact answers were needed in a.i and a.ii (0.0313 and 0.813 lost the mark, as did 0.3125 from a transcription slip). In a.iii many had the right denominator but used $\Pr(2 \le X \le 5)$ on top. In a.iv some gave the variance, or $\tfrac{\sqrt5}{4}$ or 1.118; use $np$ and $\sqrt{np(1 - p)}$ rather than a table. Part b.i was not answered well (38%): many gave the expression $\tfrac{63a}{8} + \tfrac{27b}{8} + \tfrac{3c}{2}$ instead of 1. In b.ii exact values were required ($c = -2.783$ lost marks). Part b.iii had only 6% correct, and $r = 1$, $s = 3$ was a common error. In c.ii some did not write an interval, and working by formula wasted time when CAS gives it directly. In c.iii (28% correct) common wrong answers were 0, 10, 11, 50 and 101.

Question 4 (10 marks)

Let $f: \left(-\tfrac12, \tfrac12\right) \to R$, $f(x) = \log_e\!\left(x + \tfrac12\right) - \log_e\!\left(\tfrac12 - x\right)$.

a

State the range of $f(x)$. (1 mark)

b. i

Find $f'(0)$. (2 marks)

ii

State the maximal domain over which $f$ is strictly increasing. (1 mark)

c

Show that $f(x) + f(-x) = 0$. (1 mark)

d

Find the domain and the rule of $f^{-1}$. (3 marks)

e

Let $h(x) = \dfrac1k\left(\log_e\!\left(x + \tfrac12\right) - \log_e\!\left(\tfrac12 - x\right)\right)$ on $\left(-\tfrac12, \tfrac12\right)$, $k > 0$, with inverse $h^{-1}(x) = \dfrac{e^{kx} - 1}{2\left(e^{kx} + 1\right)}$. The area of the regions bounded by $h$ and $h^{-1}$ is $A(k)$ (you do not need to find it).

i

Determine the range of values of $k$ such that $A(k) > 0$. (1 mark)

ii

Redacted by VCAA (see above).

Show worked solution

a. [1 mark]. As $x \to -\tfrac12^+$, $f(x) \to -\infty$, and as $x \to \tfrac12^-$, $f(x) \to \infty$; $f$ is continuous, so the range is $R$.

b. i. [2 marks].

$$f'(x) = \frac{1}{x + \frac12} + \frac{1}{\frac12 - x}, \qquad f'(0) = 2 + 2 = 4.$$

ii. [1 mark]. Both terms of $f'(x)$ are positive on the whole domain, so $f$ is strictly increasing on $\left(-\tfrac12, \tfrac12\right)$.

c. [1 mark]. Replace $x$ with $-x$:

$$f(-x) = \log_e\!\left(-x + \tfrac12\right) - \log_e\!\left(\tfrac12 + x\right) = -\left[\log_e\!\left(x + \tfrac12\right) - \log_e\!\left(\tfrac12 - x\right)\right] = -f(x),$$

so $f(x) + f(-x) = 0$.

d. [3 marks]. Swap $x$ and $y$: $x = \log_e\!\left(\dfrac{y + \frac12}{\frac12 - y}\right)$, so

$$e^x = \frac{2y + 1}{1 - 2y} \implies e^x - 2ye^x = 2y + 1 \implies y = \frac{e^x - 1}{2\left(e^x + 1\right)}.$$

The domain of $f^{-1}$ is the range of $f$:

$$f^{-1}: R \to R, \quad f^{-1}(x) = \frac{e^x - 1}{2\left(e^x + 1\right)} = \frac12 - \frac{1}{e^x + 1}.$$

e. i. [1 mark]. Both graphs pass through the origin and are reflections of each other in $y = x$. They enclose regions only if they cross again, which happens when $h$ is flatter than $y = x$ at the origin. Since $h'(0) = \tfrac{f'(0)}{k} = \tfrac4k$, we need $\tfrac4k < 1$, that is $k > 4$. (For $k \le 4$ the only intersection is the origin and $A(k) = 0$.)

From the report. In part a some gave the domain instead of the range; $(-26.2, 26.2)$ was a common error. In b.i some did not substitute $x = 0$, and $f'(0) = 0$ was a common wrong answer. In b.ii common wrong answers included $\left(-\tfrac12, 0\right) \cup \left(0, \tfrac12\right)$, $\left[-\tfrac12, \tfrac12\right]$ and $\left(0, \tfrac12\right)$. In part c some substituted $-x$ incorrectly or tried a particular value of $x$. In part d many swapped $x$ and $y$ correctly, but some gave $\tfrac12\tan\!\left(\tfrac{x}{2}\right)$ instead of CAS's $\tfrac12\tanh\!\left(\tfrac{x}{2}\right)$ (tanh is not in the study design but is a correct form), left out the domain, or found $\tfrac{1}{f(x)}$. Part e.i had 6% correct; $k > 0$ and $4 < k < 33$ were common wrong answers.

Question 5 (9 marks)

$g(x) = f\big(\sin(2x)\big)$, where $f$ is differentiable. A table gives $f\!\left(\tfrac12\right) = -2$, $f\!\left(\tfrac{\sqrt2}{2}\right) = 5$, $f\!\left(\tfrac{\sqrt3}{2}\right) = 3$ and $f'\!\left(\tfrac12\right) = 7$, $f'\!\left(\tfrac{\sqrt2}{2}\right) = 0$, $f'\!\left(\tfrac{\sqrt3}{2}\right) = \tfrac19$.

a. Find $g\!\left(\tfrac{\pi}{6}\right)$. (1 mark)

$g'(x) = 2\cos(2x) \cdot f'\big(\sin(2x)\big)$.

b

Show that $g'\!\left(\tfrac{\pi}{6}\right) = \tfrac19$. (1 mark)

c

Find the equation of the tangent to $g$ at $x = \tfrac{\pi}{6}$. (2 marks)

d

Find the average value of $g'(x)$ between $x = \tfrac{\pi}{8}$ and $x = \tfrac{\pi}{6}$. (2 marks)

e

Find four solutions of $g'(x) = 0$ for $x \in [0, \pi]$. (3 marks)

Show worked solution

a. [1 mark]. $g\!\left(\tfrac{\pi}{6}\right) = f\!\left(\sin\tfrac{\pi}{3}\right) = f\!\left(\tfrac{\sqrt3}{2}\right) = 3$.

b. [1 mark].

$$g'\!\left(\frac{\pi}{6}\right) = 2\cos\!\left(\frac{\pi}{3}\right) \cdot f'\!\left(\sin\frac{\pi}{3}\right) = 2 \times \frac12 \times f'\!\left(\frac{\sqrt3}{2}\right) = 1 \times \frac19 = \frac19.$$

c. [2 marks]. Through $\left(\tfrac{\pi}{6}, 3\right)$ with gradient $\tfrac19$:

$$y - 3 = \frac19\left(x - \frac{\pi}{6}\right) \implies y = \frac{x}{9} - \frac{\pi}{54} + 3 = \frac{x}{9} + \frac{162 - \pi}{54}.$$

d. [2 marks]. The average value of $g'$ is the integral of $g'$ divided by the interval length, and $\int g'(x)\,dx = g(x)$:

$$\frac{1}{\frac{\pi}{6} - \frac{\pi}{8}}\int_{\pi/8}^{\pi/6}g'(x)\,dx = \frac{24}{\pi}\left[g\!\left(\frac{\pi}{6}\right) - g\!\left(\frac{\pi}{8}\right)\right] = \frac{24}{\pi}\left(3 - f\!\left(\frac{\sqrt2}{2}\right)\right) = \frac{24}{\pi}(3 - 5) = -\frac{48}{\pi}.$$

(Here $g\!\left(\tfrac{\pi}{8}\right) = f\!\left(\sin\tfrac{\pi}{4}\right) = f\!\left(\tfrac{\sqrt2}{2}\right) = 5$.)

e. [3 marks]. $g'(x) = 2\cos(2x) \cdot f'\big(\sin(2x)\big) = 0$ when either factor is 0.

• $\cos(2x) = 0$ with $2x \in [0, 2\pi]$: $2x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}$, so $x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}$.
• $f'\big(\sin(2x)\big) = 0$: the table has $f'\!\left(\tfrac{\sqrt2}{2}\right) = 0$, so $\sin(2x) = \tfrac{\sqrt2}{2}$, giving $2x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}$ and $x = \tfrac{\pi}{8}, \tfrac{3\pi}{8}$.

Four solutions: $x = \dfrac{\pi}{8}, \dfrac{\pi}{4}, \dfrac{3\pi}{8}, \dfrac{3\pi}{4}$.

From the report. In part a a common wrong answer was $\tfrac{\sqrt3}{2}$ (stopping at $\sin\tfrac{\pi}{3}$). In part b some did not show enough working, or ignored the factor $2\cos\!\left(\tfrac{\pi}{3}\right)$. In part c an equation was needed; some used the point $\left(\tfrac{\pi}{6}, \tfrac{\sqrt3}{2}\right)$ or $\left(\tfrac{\pi}{6}, \tfrac19\right)$. Part d was poorly done (70% scored 0): some substituted into $g'$ instead of $g$, and $\tfrac{24}{\pi}(3 - 5)$ miscalculated as $\tfrac{24}{\pi} - 2$ was common. In part e many found $\tfrac{\pi}{4}$ and $\tfrac{3\pi}{4}$ but did not also solve $f'\big(\sin(2x)\big) = 0$, and some gave values outside $[0, \pi]$.

General advice from the 2022 report

• Answer every part of the question: the domain as well as the rule (4d), and the standard deviation as well as the mean (3a.iv).
• Show working on multi-mark parts, usually the equation you solved as well as the answer; answers alone lost marks in 1d.ii, 2g, 3b.ii, 4b.i and 5e.
• Exact or approximate? Read each part's instruction; many lost marks by rounding when an exact answer was needed, or by rounding too early (5339.46 rounds to 5339, not 5340).
• Equations, not expressions, when an equation is asked for (1a, 1c, 1d.i, 5c).
• Average value is not average rate of change. 2e and 5d asked for average values; 2f asked for an average rate of change.
• Use technology efficiently: tangent and perpendicular lines, confidence intervals and sliders can all be done directly on CAS.
• Answer in context: the minimum and maximum rabbit populations (not coordinates) in 2a.ii, and a time (not a population) in 2g.