VCE Math Methods 2023 Exam 2

Every question from the 2023 VCE Mathematical Methods Examination 2, the technology-active (CAS) paper and the first under the current study design. 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 2023 Examination 1 walkthrough.

How to use this page

• Questions are from the 2023 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 2023 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.

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. In 2023 the multiple-choice questions had five options (A to E).

• Section A (20 marks): 20 multiple-choice questions. Aim for about 25 minutes.
• Section B (60 marks): 5 extended-response questions (11, 11, 12, 15 and 11 marks).

Section A: Multiple choice

Q1

Amplitude and period of $f(x) = -\tfrac12\sin(3x + 2\pi)$. Answer: E - amplitude is a positive distance, $\tfrac12$; period is $\tfrac{2\pi}{3}$.

Q2

Axis of symmetry of $y = ax^2 + 2bx + c$. Answer: A - $x = -\dfrac{2b}{2a} = -\dfrac{b}{a}$.

Q3

$p$ and $q$ have domains $[-2, 3)$ and $(-1, 5]$. Domain of $p + q$? Answer: E - the intersection, $(-1, 3)$.

Q4

For which $k$ does $kx + 5y = k + 5$, $4x + (k + 1)y = 0$ have infinitely many solutions? Answer: B - the determinant $k(k+1) - 20 = (k + 5)(k - 4)$ is zero at $k = -5$ or $4$. At $k = -5$ both equations reduce to $y = x$ (infinitely many); at $k = 4$ they are parallel with no solution. So $k \in \{-5\}$.

Q5

Which function has a horizontal tangent at $(0, 0)$? Answer: D - $y = x^{4/3}$ has $y' = \tfrac43x^{1/3}$, which is 0 at $x = 0$. The others have undefined or infinite gradients at the origin.

Q6

$\int_3^{10} f = C$ and $\int_7^{10} f = D$. Find $\int_7^3 f(x)\,dx$. Answer: D - $\int_3^7 f = C - D$, and reversing the terminals gives $D - C$.

Q7

$f(x) = \log_e x$ and $g(x) = \sqrt{1 - x}$, $x < 1$. Domain of the derivative of $f \circ g$? Answer: C - $\log_e\sqrt{1 - x}$ has derivative $-\dfrac{1}{2(1 - x)}$, defined for all $x < 1$: $(-\infty, 1)$.

Q8

$n$ green and $m$ red balls, 8 draws with replacement. $\Pr(\text{at least one green})$? Answer: C - $1 - \Pr(\text{all red}) = 1 - \left(\dfrac{m}{n + m}\right)^8$.

Q9

$f$ is $\tan\!\left(\tfrac{x}{2}\right)$ on $[4, 2\pi)$ and $\sin(ax)$ on $[2\pi, 8]$. Find $a$ so $f$ is continuous and smooth at $2\pi$. Answer: C - continuity needs $\sin(2\pi a) = \tan(\pi) = 0$; smoothness needs $a\cos(2\pi a) = \tfrac12\sec^2(\pi) = \tfrac12$. Only $a = -\tfrac12$ gives $-\tfrac12\cos(-\pi) = \tfrac12$.

Q10

A pdf is $\tfrac{x - 1}{20}$ on $[1, 6)$ and $\tfrac{9 - x}{12}$ on $[6, 9]$. Find $k$ with $\Pr(X < k) = 0.35$. Answer: B - $\int_1^k\tfrac{x - 1}{20}\,dx = \tfrac{(k - 1)^2}{40} = 0.35$ gives $(k - 1)^2 = 14$, so $k = 1 + \sqrt{14} \approx 4.74$ (in $[1, 6)$).

Q11

$f(-2) = -7$, $g(-2) = 8$, $f'(-2) = 3$, $g'(-2) = 2$. Gradient of $y = f(x)g(x)$ at $x = -2$? Answer: E - product rule: $3 \times 8 + (-7) \times 2 = 10$. (Only 22% correct; 51% chose 6.)

Q12

A pmf has probabilities $k^2, 3k, k, -k^2 - 4k + 1$ at $x = -1, 0, 1, 2$. Maximum possible mean? Answer: E - $\text{E}(X) = -3k^2 - 7k + 2$, and every probability must lie in $[0, 1]$, which forces $0 \le k \le \sqrt5 - 2$. The mean decreases on this interval, so its maximum is 2, at $k = 0$.

Q13

Three Newton iterations on $x^3 + 3x - 3$ from $x_0 = 1$. Answer: C - $x_1 = 0.83333$, $x_2 = 0.81785$, $x_3 = 0.81773$.

Q14

How many tangents to $y = x(3x - 1)(x + 3)(x + 1)$ pass through its positive $x$-intercept? Answer: D - the tangent at $\left(\tfrac13, 0\right)$ itself, plus two more from points near $x = -2.22$ and $x = -0.45$: three in total.

Q15

$X \sim N(100, 20^2)$ and $Y \sim N(80, 10^2)$; which diagram shows both pdfs? Answer: A - $Y$ (solid) is centred further left and, with half the standard deviation, is narrower and twice as tall.

Q16

$f(x) = e^{x - 1}$ and $f(x)g(x) = e^{(x - 1)^2}$. Find $g$. Answer: B - $g(x) = e^{(x-1)^2 - (x - 1)} = e^{(x - 1)(x - 2)}$.

Q17

A cylinder is formed from an $x$ by $y$ sheet. Volume in terms of $x$ and $y$? Answer: B - the circumference is $y$, so $r = \tfrac{y}{2\pi}$; the two end circles use $4r$ of the length, so $h = x - \tfrac{2y}{\pi}$. Then $V = \pi r^2h = \dfrac{\pi xy^2 - 2y^3}{4\pi^2}$. (28% correct.)

Q18

Number of local minima of $f(x) = \sin(ax)$ on $[-a\pi, a\pi]$, $a$ a positive integer. Answer: E - $ax$ runs over an interval of length $2a^2\pi$, which is $a^2$ full periods, each with one minimum: $a^2$.

Q19

When does $x^2 + (4k + 3)x + 4k^2 - \tfrac94 = 0$ have one positive and one negative solution? Answer: D - the product of the roots must be negative: $4k^2 - \tfrac94 < 0$, so $-\tfrac34 < k < \tfrac34$.

Q20

$f(x) = \log_e\!\left(x + \tfrac{1}{\sqrt2}\right)$, $g(x) = \sin(x)$ for $x < 5$. Largest interval on which both $f \circ g$ and $g \circ f$ exist? Answer: A - $f \circ g$ needs $\sin x > -\tfrac{1}{\sqrt2}$ and $g \circ f$ needs $x > -\tfrac{1}{\sqrt2}$; starting from $-\tfrac{1}{\sqrt2}$, $\sin x$ first drops to $-\tfrac{1}{\sqrt2}$ at $\tfrac{5\pi}{4}$. So $\left(-\tfrac{1}{\sqrt2}, \tfrac{5\pi}{4}\right)$.

From the report. The hardest multiple-choice items were Q11 (22% correct), Q17 (28%), Q12, Q14 and Q18 (29% each), and Q20 (30%).

Section B: Extended response

Question 1 (11 marks)

Let $f(x) = x(x - 2)(x + 1)$, $x \in R$.

a

State the coordinates of all axial intercepts. (1 mark)

b

Find the coordinates of the stationary points. (2 marks)

c

Let $g(x) = x - 2$. i. Find the values of $x$ for which $f(x) = g(x)$. (1 mark)

ii

Write an expression using definite integrals for the area of the regions bounded by $f$ and $g$. (2 marks)

iii

Hence find the total area, to two decimal places. (1 mark)

d

Let $h(x) = (x - a)(x - b)^2$ with $h(x) = f(x) + k$. Find the possible values of $a$ and $b$. (4 marks)

Show worked solution

a. [1 mark]. $(-1, 0)$, $(0, 0)$ and $(2, 0)$.

b. [2 marks]. $f(x) = x^3 - x^2 - 2x$, so $f'(x) = 3x^2 - 2x - 2 = 0$ gives $x = \tfrac{1 \pm \sqrt7}{3}$. Substituting:

$$\left(\frac{1 - \sqrt7}{3},\ \frac{14\sqrt7 - 20}{27}\right) \quad\text{and}\quad \left(\frac{1 + \sqrt7}{3},\ \frac{-14\sqrt7 - 20}{27}\right).$$

c. i. [1 mark]. $x^3 - x^2 - 3x + 2 = 0$ factorises as $(x - 2)(x^2 + x - 1) = 0$, so $x = 2$ or $x = \dfrac{-1 \pm \sqrt5}{2}$.

ii. [2 marks]. $f$ is above $g$ between the first two intersections and below between the last two:

$$A = \int_{\frac{-1 - \sqrt5}{2}}^{\frac{-1 + \sqrt5}{2}}\big(f(x) - g(x)\big)dx + \int_{\frac{-1 + \sqrt5}{2}}^{2}\big(g(x) - f(x)\big)dx.$$

iii. [1 mark]. $A = \tfrac{25\sqrt5}{8} - \tfrac{25}{24} \approx 5.95$ square units.

d. [4 marks]. $h$ is a vertical translation of $f$ that touches the $x$-axis at $x = b$, so $b$ must be the $x$-value of a turning point. Equate coefficients of $(x - a)(x - b)^2 = x^3 - (a + 2b)x^2 + (2ab + b^2)x - ab^2$ with $x^3 - x^2 - 2x + k$:

$$a + 2b = 1, \qquad 2ab + b^2 = -2.$$

Substituting $a = 1 - 2b$: $3b^2 - 2b - 2 = 0$, so $b = \tfrac{1 \pm \sqrt7}{3}$ and $a = 1 - 2b$:

$$a = \frac{1 - 2\sqrt7}{3},\ b = \frac{1 + \sqrt7}{3} \qquad\text{or}\qquad a = \frac{1 + 2\sqrt7}{3},\ b = \frac{1 - \sqrt7}{3}.$$

From the report. In part b exact coordinates were needed, not just $x$-values. In c.ii many gave only one integral or subtracted the wrong way round; in c.iii 5.94 (a rounding error) was a common wrong answer. Part d was poorly done (13% full marks): equating coefficients worked best, and many gave only one pair of values.

Question 2 (11 marks)

An observation wheel of radius 60 m turns anticlockwise once every 30 minutes; its lowest point $A$ is 15 m above the ground, and $B$ is level with the centre $P$. A pod starting at $A$ has height $h(t) = -60\cos(bt) + c$ after $t$ minutes.

a

Show that $b = \tfrac{\pi}{15}$ and $c = 75$. (2 marks)

b

Find the average height of a pod as it travels from $A$ to $B$, to two decimal places. (2 marks)

c

Find the average rate of change of height from $A$ to $B$, in metres per minute. (1 mark)

d

The wheel stops after 15 minutes for 5 minutes, then turns at double speed for 7.5 minutes; the height is $w(t) = h(t)$ for $0 \le t < 15$, $k$ for $15 \le t < 20$, and $h(mt + n)$ for $20 \le t \le 27.5$. i. State $k$ and $m$. (1 mark)

ii

Find all possible values of $n$. (2 marks)

iii

Sketch $w$, showing the coordinates of the endpoints. (3 marks)

Show worked solution

a. [2 marks]. The period is 30 minutes, so $\tfrac{2\pi}{b} = 30$ and $b = \tfrac{\pi}{15}$. At $t = 0$ the pod is at $A$: $h(0) = -60 + c = 15$, so $c = 75$.

b. [2 marks]. $A$ to $B$ is a quarter turn, $t = 0$ to $7.5$:

$$\frac{1}{7.5}\int_0^{7.5}\left(75 - 60\cos\!\left(\frac{\pi t}{15}\right)\right)dt = 75 - \frac{120}{\pi} \approx 36.80 \text{ m}.$$

c. [1 mark]

$\dfrac{h(7.5) - h(0)}{7.5} = \dfrac{75 - 15}{7.5} = 8$ m per minute.

d. i. [1 mark]

At $t = 15$ the pod is at the top: $k = h(15) = 135$. Double speed means $m = 2$.

ii. [2 marks]

Continuity at $t = 20$ requires $h(40 + n) = 135$, so $\cos\!\left(\tfrac{\pi(40 + n)}{15}\right) = -1$:

$$\frac{\pi(40 + n)}{15} = \pi + 2p\pi \implies n = 30p - 25 = 5 + 30q, \quad q \in Z.$$

iii. [3 marks]. Three pieces: a half-cosine rise from $(0, 15)$ to $(15, 135)$; a horizontal segment at 135 from $t = 15$ to $t = 20$; then a half-cosine fall, twice as fast, from $(20, 135)$ to $(27.5, 15)$. Endpoints $(0, 15)$ and $(27.5, 15)$.

From the report. In part a some wrote the period as $b$ instead of $\tfrac{2\pi}{b}$. In part b many used the wrong terminals (such as $\tfrac{1}{60}\int_0^{60}$) or found the average rate of change instead. In part c $-8$ was a common wrong answer. In d.i $m = \tfrac12$ was often seen. Part d.ii needed a general solution (12% full marks). In d.iii endpoint coordinates were often missing and some graphs were straight lines rather than curves.

Question 3 (12 marks)

Let $g(x) = 2^x + 5$.

a

State $\displaystyle\lim_{x \to -\infty} g(x)$. (1 mark)

b

Find $k$ such that $g'(x) = k \times 2^x$. (1 mark)

c. i

Find the equation of the tangent to $g$ at $(a, g(a))$. (1 mark)

ii

Hence find the equation of the tangent to $g$ that passes through the origin, to three decimal places. (2 marks)

Let $h(x) = 2^x - x^2$.

d

Find the point of inflection of $h$, to two decimal places. (1 mark)

e

Find the largest interval on which $h$ is strictly decreasing, to two decimal places. (1 mark)

f

Apply Newton's method to $h$ with $x_0 = 0$; give $x_1, x_2, x_3$ to three decimal places. (2 marks)

g

Explain why a solution of $\log_e(2) \times 2^x - 2x = 0$ should not be used as $x_0$. (1 mark)

h

Find the positive $n$ for which $f(x) = n^x - x^n$ has a local minimum on the $x$-axis. (2 marks)

Show worked solution

a. [1 mark]

As $x \to -\infty$, $2^x \to 0$, so the limit is 5.

b. [1 mark]

$g'(x) = \log_e(2)\,2^x$, so $k = \log_e(2)$.

c. i. [1 mark]

$y = \log_e(2)\,2^a(x - a) + 2^a + 5$.

ii. [2 marks]

The tangent passes through $(0, 0)$: $0 = -a\log_e(2)\,2^a + 2^a + 5$. Solving with CAS gives $a \approx 2.61785$, so the gradient is $\log_e(2)\,2^a \approx 4.255$ and the tangent is $y = 4.255x$.

d. [1 mark]

$h''(x) = (\log_e 2)^2\,2^x - 2 = 0$ gives $x \approx 2.058$, and $h(2.058) \approx -0.071$. The point of inflection is $(2.06, -0.07)$.

e. [1 mark]

$h'(x) = \log_e(2)\,2^x - 2x \le 0$ between its zeros, $x \approx 0.485$ and $x \approx 3.212$. The largest interval is $[0.49, 3.21]$ (square brackets: the endpoints are included).

f. [2 marks]

$x_{n+1} = x_n - \dfrac{h(x_n)}{h'(x_n)}$ with $x_0 = 0$:

$x_1$	$x_2$	$x_3$
$-1.443$	$-0.897$	$-0.773$

g. [1 mark]. $\log_e(2)\,2^x - 2x$ is $h'(x)$. At such an $x_0$ the tangent to $h$ is horizontal, so $h'(x_0) = 0$ and the Newton step $\tfrac{h(x_0)}{h'(x_0)}$ is undefined (the tangent never meets the $x$-axis).

h. [2 marks]. Need $f(x) = 0$ and $f'(x) = 0$ at the same point: $n^x = x^n$ and $n^x\log_e n = nx^{n-1}$. Dividing gives $x\log_e n = n$, and taking logs of the first gives $x\log_e n = n\log_e x$. So $\log_e x = 1$, $x = e$, and then $e\log_e n = n$, whose only solution is $n = e$. (Check: $f(x) = e^x - x^e$ has $f(e) = 0$, $f'(e) = 0$ and $f''(e) = e^{e-1} > 0$.)

From the report. In part a, 6 was a common wrong answer. In c.i an equation was required and there were many transcription errors. In c.ii many found the tangent at $x = 0$ instead, or gave a non-zero intercept from CAS rounding. In part d many gave the stationary points $(0.49, 1.16)$ and $(3.21, -1.05)$ instead. In part e round brackets were marked wrong, and some gave the intervals where $h$ is increasing. In part f answers were needed to three decimal places. Part h was very poorly done (3% full marks): many set $f'(x) = 0$ but did not combine it with $f(x) = 0$, or gave a decimal such as 2.7 instead of the exact $n = e$.

Question 4 (15 marks)

Tennis ball diameters $D$ are normal with mean 6.7 cm and standard deviation 0.1 cm. A ball fits through the container opening if its diameter is less than 6.95 cm.

a

Find $\Pr(D > 6.8)$ to four decimal places. (1 mark)

b

Find the minimum diameter of a ball larger than 90% of all balls, to two decimal places. (1 mark)

c

Find the probability a ball fits through the opening, to four decimal places. (1 mark)

d

Of 4 random balls, find the probability that at least 3 fit, to four decimal places. (2 marks)

e

Grade A means a diameter between 6.54 and 6.86 cm. Given a ball fits, find the probability it is grade A, to four decimal places. (2 marks)

f

Keeping the mean, find the standard deviation needed so that more than 99% of balls are grade A, to two decimal places. (2 marks)

g

A sample of 32 balls gives a confidence interval $(0.7382, 0.9493)$ for the proportion of grade A balls. Find the level of confidence, to the nearest per cent. (2 marks)

h

Grade A serving speed $V$ has pdf $f(v) = \tfrac{1}{6\pi}\sin\!\left(\sqrt{\tfrac{v - 30}{3}}\right)$ for $30 \le v \le 3\pi^2 + 30$. Find $\Pr(V > 50)$ to four decimal places. (1 mark)

i

Find the exact mean of $V$. (1 mark)

j

Grade B speed $W$ has pdf $g(w) = af\!\left(\tfrac{w}{b}\right)$. If $\text{E}(W) = 2\pi^2 + 8$, find $a$ and $b$. (2 marks)

Show worked solution

a. [1 mark]

$\Pr(D > 6.8) = \Pr(Z > 1) \approx 0.1587$.

b. [1 mark]

The 90th percentile: $6.7 + 1.2816 \times 0.1 \approx 6.83$ cm.

c. [1 mark]

$\Pr(D < 6.95) = \Pr(Z < 2.5) \approx 0.9938$.

d. [2 marks]

Let $X \sim \text{Bi}(4, 0.99379)$ be the number that fit. $\Pr(X \ge 3) \approx 0.9998$.

e. [2 marks]

Every grade A ball fits (6.86 < 6.95), so

$$\Pr(A \mid \text{fits}) = \frac{\Pr(6.54 < D < 6.86)}{\Pr(D < 6.95)} \approx \frac{0.89040}{0.99379} \approx 0.8960.$$

f. [2 marks]. The grade A interval is $6.7 \pm 0.16$. Need $\Pr(-\tfrac{0.16}{\sigma} < Z < \tfrac{0.16}{\sigma}) > 0.99$, so $\tfrac{0.16}{\sigma} > 2.5758$ and $\sigma < 0.0621$. To two decimal places, $\sigma = 0.06$ cm.

g. [2 marks]. The centre is $\hat p = \tfrac{0.7382 + 0.9493}{2} = 0.84375 = \tfrac{27}{32}$ and the margin is $0.10555$:

$$z = \frac{0.10555}{\sqrt{\frac{0.84375 \times 0.15625}{32}}} \approx 1.644, \qquad \Pr(-1.644 < Z < 1.644) \approx 0.90.$$

The confidence level is 90%.

h. [1 mark]

$\Pr(V > 50) = \displaystyle\int_{50}^{3\pi^2 + 30}f(v)\,dv \approx 0.1345$.

i. [1 mark]

$\text{E}(V) = \displaystyle\int_{30}^{3\pi^2 + 30}v\,f(v)\,dv = 3\pi^2 + 12$.

j. [2 marks]

The graph of $g$ is $f$ dilated by $b$ from the vertical axis and by $a$ from the horizontal axis, so its area is $ab$, which must be 1. The dilation by $b$ scales the mean by $b$:

$$b\left(3\pi^2 + 12\right) = 2\pi^2 + 8 \implies b = \frac23, \qquad a = \frac1b = \frac32.$$

From the report. In part a, 0.1586 (a rounding error) was sometimes seen. In part b a common error was solving $\Pr(D > a) = 0.9$, giving 6.57. In part d students were expected to state $n$ and $p$. In part f, $\sigma = 0.06$ or any smaller value was accepted. In part g many calculated $\hat p$ wrongly, and some found the right $z$ but then answered 95%. Part j was very poorly done (6% full marks): many recognised $ab = 1$ but could not find the values, and those using simultaneous equations often forgot to scale the terminals by $b$.

Question 5 (11 marks)

Let $f(x) = e^x + e^{-x}$ and $g(x) = \tfrac12f(2 - x)$.

a. Complete a sequence of transformations mapping $f$ to $g$, starting with a dilation of factor $\tfrac12$ from the $x$-axis. (2 marks)

$g_1$ and $g_2$ have the rule of $g$ on distinct domains, with $g_1$ strictly increasing and $g_2$ strictly decreasing.

b

Give the domain and range of the inverse of $g_1$. (2 marks)

c. i

The graphs of $y = x$, $y = g(x)$ and the inverses of $g_1$ and $g_2$ meet at $P$ and $Q$. Find $P$ and $Q$ to two decimal places. (1 mark)

ii

Find the area of the region bounded by $g$ and the inverses of $g_1$ and $g_2$, to two decimal places. (2 marks)

Let $h(x) = \tfrac1kf(k - x)$, $k > 0$.

d

The turning point of $h$ always lies on $y = 2x^n$, $n$ an integer. Find $n$. (1 mark)

e

With $h_1$ the restriction of $h$ to $[k, \infty)$, find the smallest $k$ (to two decimal places) such that $h$ intersects the inverse of $h_1$. (1 mark)

f

When $k = 5$ the graphs of $h$ and the inverse of $h_1$ meet twice. Find the area between them, to two decimal places. (2 marks)

Show worked solution

a. [2 marks]

After the dilation: reflect in the $y$-axis, then translate 2 units in the positive $x$-direction. (Check: $f(x) \to f(-x) \to f(-(x - 2)) = f(2 - x)$.) Because $f$ is even, a translation of 2 units right on its own also works.

b. [2 marks]

$g(x) = \tfrac12\left(e^{2 - x} + e^{x - 2}\right)$ has its minimum $g(2) = 1$, so $g_1$ can be taken on $[2, \infty)$ with range $[1, \infty)$. The inverse of $g_1$ swaps these: domain $[1, \infty)$, range $[2, \infty)$. (The report also accepted open brackets or a smaller domain, since a maximal domain was not asked for.)

c. i. [1 mark]

$P$ and $Q$ lie on $y = x$: solve $g(x) = x$ with CAS. $P(1.27, 1.27)$ and $Q(4.09, 4.09)$.

ii. [2 marks]

The region is symmetric about $y = x$, so it is twice the area between $y = x$ and $y = g(x)$:

$$A = 2\int_{1.2747\ldots}^{4.0852\ldots}\big(x - g(x)\big)dx \approx 5.56 \text{ square units}.$$

d. [1 mark]

$f$ has its minimum at $(0, 2)$, so $h$ has its turning point at $x = k$ with $h(k) = \tfrac{2}{k}$. The point $\left(k, \tfrac2k\right)$ lies on $y = 2x^{-1}$, so $n = -1$.

e. [1 mark]

$h_1$ is increasing, so it first meets its inverse by touching the line $y = x$. Using CAS (for example, a slider on $k$, or solving $h(x) = x$ and $h'(x) = 1$ together) gives $k \approx 1.27$.

f. [2 marks]

With $k = 5$, the graphs of $h$ and $h_1^{-1}$ meet at $x \approx 1.4509$ and $x \approx 8.7816$, and $h_1^{-1}$ is above $h$ between them:

$$A = \int_{1.4509\ldots}^{8.7816\ldots}\Big(h_1^{-1}(x) - h(x)\Big)dx \approx 43.91 \text{ square units}.$$

From the report. In part a the wording and order mattered; "reflect in the $y$-axis then translate 2 units left" was a common wrong answer. In part b the domain and range were often swapped. In c.ii some forgot to double the integral and gave 2.78. Parts d to f were very poorly done (4% correct on part e): in d, $n = 1$ was a common wrong answer, and in f the common wrong terminal was $x \approx 2.468$, where $h$ meets $y = x$ rather than $h_1^{-1}$.

General advice from the 2023 report

• Exact versus rounded. Many parts needed exact values, others three decimal places; set your CAS float correctly and read the output carefully.
• Average value is not average rate of change (Questions 2b and 2c).
• Brackets matter: the largest interval on which a function is strictly decreasing includes its endpoints.
• Write definite integrals down when asked, and state $n$ and $p$ for binomial calculations.