VCE Math Methods 2024 Exam 2

Every question from the 2024 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 2024 Examination 1 walkthrough.

How to use this page

• Questions are from the 2024 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 2024 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.

• Section A (20 marks): 20 multiple-choice questions. Aim for about 25 minutes.
• Section B (60 marks): 5 extended-response questions (12, 11, 11, 15 and 11 marks). Give exact answers unless told to round, and write down what you entered into CAS.

Section A: Multiple choice

Q1

Asymptotes of $y = \log_e(x + 1) - 3$. Answer: A - a log graph has only a vertical asymptote, here $x = -1$. (20% wrongly added $y = -3$.)

Q2

$g'(x) = x^3 - x$ and $g(0) = 5$; find $g(2)$. Answer: D - $g(x) = \tfrac{x^4}{4} - \tfrac{x^2}{2} + 5$, so $g(2) = 4 - 2 + 5 = 7$.

Q3

A distribution has probabilities $2k, 3k, 5k, 3k, 2k$ for $x = 0$ to $4$; find $\Pr(X < 4 \mid X > 1)$. Answer: C - $\dfrac{\Pr(X = 2) + \Pr(X = 3)}{\Pr(X \ge 2)} = \dfrac{5k + 3k}{5k + 3k + 2k} = \dfrac45$.

Q4

$\int_a^b f = -5$ and $\int_a^c f = 3$ with $a < b < c$; find $\int_b^c 2f(x)\,dx$. Answer: B - $\int_b^c f = 3 - (-5) = 8$, so the answer is $16$.

Q5

Range of $g(f(x))$ where $f(x) = x^2 - 4x$ on $(1, \infty)$ and $g(x) = e^{-x}$. Answer: D - $f$ has range $[-4, \infty)$ on this domain (minimum at $x = 2$), so $e^{-f(x)}$ has range $(0, e^4]$.

Q6

Inverse of $f(x) = \dfrac{2x + 1}{3 - x}$, $x \ne 3$. Answer: B - swapping and solving gives $f^{-1}(x) = \dfrac{3x - 1}{x + 2} = 3 - \dfrac{7}{x + 2}$, with domain $R \setminus \{-2\}$ (the range of $f$).

Q7

Minimum number of rolls of a fair die so that $\Pr(\text{at least one six}) > 0.95$. Answer: C - $1 - \left(\tfrac56\right)^n > 0.95$ needs $n > 16.4$, so 17 rolls.

Q8

Given a table of $f$ and $g$ values at $x = 1, 2, 3$, where must $h = f - g$ have an $x$-intercept? Answer: A - $f(2) = g(2) = 4$, so $h(2) = 0$.

Q9

45% hire a limousine (30% of them get a photo) and 55% are driven by a parent (60% get a photo). Find $\Pr(\text{limousine} \mid \text{photo})$. Answer: C - $\dfrac{0.45 \times 0.3}{0.45 \times 0.3 + 0.55 \times 0.6} = \dfrac{0.135}{0.465} = \dfrac{9}{31}$.

Q10

$f$ and $f'$ are continuous on $[0, 5]$ with $f'(2) < 0$ and $f'(4) > 0$. What must be true? Answer: B - $f$ decreases near $x = 2$ and increases near $x = 4$, so it is not one-to-one and has no inverse. The other statements can fail. (Only 38% correct.)

Q11

7 students in row one, 5 in row two; choose 3 at random. Find $\Pr(\text{exactly 2 from row one})$. Answer: B - $\dfrac{\binom72\binom51}{\binom{12}{3}} = \dfrac{21 \times 5}{220} = \dfrac{21}{44}$.

Q12

Choose the graph of $y = f(2x + 1)$ from a given graph of $y = f(x)$. Answer: A - $f(2x + 1) = f\!\left(2\left(x + \tfrac12\right)\right)$: compress by a factor of $\tfrac12$ from the $y$-axis, then shift left $\tfrac12$. The point $(x, y)$ maps to $\left(\tfrac{x - 1}{2}, y\right)$, so the minimum at $(2, 1)$ moves to $\left(\tfrac12, 1\right)$.

Q13

$f(x) = \tfrac{x}{2} + \tfrac{2}{x}$, $x > 0$, is dilated by 3 from the $y$-axis then translated 1 down. Find the local minimum of the image. Answer: A - $f$ has its minimum at $(2, 2)$; the dilation gives $(6, 2)$ and the translation gives $(6, 1)$.

Q14

A pdf is $\tfrac{x}{6} + k$ on $[-3, 0)$ and $-\tfrac{x}{2} + k$ on $[0, 1]$. Find $\Pr(X < 0.5)$. Answer: B - total area $4k - 1 = 1$ gives $k = \tfrac12$; then $\Pr(X \ge 0.5) = \int_{0.5}^{1}\left(\tfrac12 - \tfrac{x}{2}\right)dx = \tfrac{1}{16}$, so $\Pr(X < 0.5) = \tfrac{15}{16}$.

Q15

Points of inflection of $y = 2 - \tan\!\left(\pi\left(x - \tfrac14\right)\right)$. Answer: A - they sit where the tangent term is 0: $x - \tfrac14 = k$, so $\left(k + \tfrac14, 2\right)$, $k \in Z$.

Q16

$f(4) = 25$ and $f'(4) = 15$; find the gradient of $y = \sqrt{f(x)}$ at $x = 4$. Answer: D - chain rule: $\dfrac{f'(4)}{2\sqrt{f(4)}} = \dfrac{15}{10} = \dfrac32$. (Only 36% correct.)

Q17

An algorithm tests $c = 18, 17, \ldots, 1$ and prints $c$ if $f(c) = 0$, then $-c$ if $f(-c) = 0$, for $f(x) = x^3 - 2x^2 - 9x + 18$. What is printed, in order? Answer: D - $f(x) = (x - 2)(x - 3)(x + 3)$. At $c = 3$ it prints 3 then $-3$; at $c = 2$ it prints 2. So 3, $-3$, 2. (27% correct.)

Q18

A trapezium has parallel sides $x$ and $3x$ and slant sides 10. Find $x$ for maximum area. Answer: B - height $\sqrt{100 - x^2}$, area $A = 2x\sqrt{100 - x^2}$; $A'(x) = 0$ gives $100 - 2x^2 = 0$, so $x = 5\sqrt2$.

Q19

$X$ is normal with $\Pr(X < 10) = \Pr(X > 18) = 0.2$. Find $\Pr(X < 12)$. Answer: C - by symmetry $\mu = 14$; $\tfrac{10 - 14}{\sigma} = -0.8416$ gives $\sigma \approx 4.753$, so $\Pr(X < 12) \approx 0.337$.

Q20

$f(x) = f(x + 2)$ for all $x$ and the average value on $[0, 2]$ is $k$. Find $\int_2^6 f(x)\,dx$. Answer: C - $\int_0^2 f = 2k$, and $[2, 6]$ covers two full periods, so the integral is $4k$.

Section B: Extended response

Question 1 (12 marks)

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

a

State the values of $x$ for which $f(x) = 0$. (1 mark)

b

Find the values of $a$ for which the graph has i. exactly three $x$-intercepts (2 marks)

ii

exactly four $x$-intercepts. (1 mark)

c

Let $g(x) = (x + 1)^2(x - 2)^2$ (the case $a = 1$). i. Find $g'(x)$. (1 mark)

ii

Find the coordinates of the local maximum of $g$. (1 mark)

iii

Find the values of $x$ for which $g'(x) > 0$. (1 mark)

iv

Find where the tangents to $y = g(x)$ at $x = \tfrac{1 - \sqrt3}{2}$ and $x = \tfrac{1 + \sqrt3}{2}$ intersect. (2 marks)

d

Let $h(x) = (x + 1)(x - 1)(x + 2)(x - 2)$ (the case $a = -1$). i. Using translations only, map $h$ so that its local maximum matches that of $g$. (1 mark)

ii

Using a dilation and translations, map $h$ so that both its local minimums match those of $g$. (2 marks)

Show worked solution

a. [1 mark]. $x = -1,\ -a,\ 2,\ 2a$.

b. i. [2 marks]. Exactly three intercepts means exactly one pair of these four values coincide.

• $-a = 2$ gives $a = -2$: roots $-1, 2, 2, -4$. Three intercepts.
• $2a = -1$ gives $a = -\tfrac12$: roots $-1, \tfrac12, 2, -1$. Three intercepts.
• $-a = 2a$ gives $a = 0$: roots $-1, 0, 2, 0$. Three intercepts.
• $-a = -1$ (or $2a = 2$) gives $a = 1$: roots $-1, -1, 2, 2$. Only two intercepts.

So $a \in \left\{-2, -\tfrac12, 0\right\}$.

ii. [1 mark]. All four roots distinct: $a \in R \setminus \left\{-2, -\tfrac12, 0, 1\right\}$.

c. i. [1 mark].

$$g'(x) = 2(x + 1)(x - 2)^2 + 2(x + 1)^2(x - 2) = 2(x + 1)(x - 2)(2x - 1).$$

ii. [1 mark]

The local maximum is at $x = \tfrac12$: $g\!\left(\tfrac12\right) = \left(\tfrac32\right)^2\left(-\tfrac32\right)^2 = \tfrac{81}{16}$. So $\left(\tfrac12, \tfrac{81}{16}\right)$.

iii. [1 mark]

$g'$ is a positive cubic with roots $-1, \tfrac12, 2$, so $g'(x) > 0$ for $x \in \left(-1, \tfrac12\right) \cup (2, \infty)$.

iv. [2 marks]

$g(x) = \big((x+1)(x-2)\big)^2$ is symmetric about $x = \tfrac12$, and the two given $x$-values are symmetric about $\tfrac12$, so the tangents meet on $x = \tfrac12$. Finding both tangent equations with CAS and solving them simultaneously gives the point $\left(\tfrac12, \tfrac{27}{4}\right)$.

d. i. [1 mark]

$h(x) = x^4 - 5x^2 + 4$ has its local maximum at $(0, 4)$; $g$ has its at $\left(\tfrac12, \tfrac{81}{16}\right)$. Translate $\tfrac12$ unit to the right and $\tfrac{17}{16}$ units up.

ii. [2 marks]

$h$ has local minimums at $\left(\pm\tfrac{\sqrt{10}}{2}, -\tfrac94\right)$, a distance $\sqrt{10}$ apart; $g$ has them at $(-1, 0)$ and $(2, 0)$, a distance 3 apart. One valid sequence:

1. Dilate by a factor of $\tfrac{3}{\sqrt{10}} = \tfrac{3\sqrt{10}}{10}$ from the $y$-axis (minimums move to $\left(\pm\tfrac32, -\tfrac94\right)$).
2. Translate $\tfrac12$ unit to the right (minimums at $(-1, -\tfrac94)$ and $(2, -\tfrac94)$).
3. Translate $\tfrac94$ units up.

From the report. In b.i some values were often missing; in b.ii many did not exclude $a = 1$ (31% correct). In c.ii the exact $\tfrac{81}{16}$ was required, not 5.06. In c.iv a common wrong answer was the local maximum $\left(\tfrac12, \tfrac{81}{16}\right)$. In d.i many translated left and down, or used $\tfrac{81}{16}$ as the vertical translation. Part d.ii was poorly done (5% full marks): the horizontal dilation and translation must be in the right order, while the vertical translation can come at any point.

Question 2 (11 marks)

A room's temperature (°C) $t$ hours after a heater is switched on is modelled by $f(t) = 12 + 30t$ for $0 \le t \le \tfrac13$ and $f(t) = 22$ for $t > \tfrac13$.

a

Express $f'(t)$ as a hybrid function. (2 marks)

b

Find the average rate of change of temperature between $t = 0$ and $t = \tfrac12$, in °C per hour. (1 mark)

c

A second model is $g(t) = 22 - 10e^{-6t}$, $t \ge 0$. i. Find $g'(t)$. (1 mark)

ii

Find $t$ when $g'(t) = 10$, to three decimal places. (1 mark)

d

Find $t \in (0, 1)$ where the two models give equal temperatures, to two decimal places. (1 mark)

e

Find $t \in (0, 1)$ where the difference between the models is greatest, to two decimal places. (1 mark)

f

The heater's power (kW) is $p(t) = 1.5$ for $0 \le t \le 0.4$ and $p(t) = 0.3 + Ae^{-10t}$ for $t > 0.4$; energy used (kWh) is the area under $y = p(t)$. i. Given $p$ is continuous, show that $A = 1.2e^4$. (1 mark)

ii

Find how long it takes to use 0.5 kWh. (1 mark)

iii

Find how long it takes to use 1 kWh, to two decimal places. (2 marks)

Show worked solution

a. [2 marks]. Differentiate each piece. The derivative does not exist at the corner $t = \tfrac13$, so leave it out of the domain:

$$f'(t) = \begin{cases} 30, & 0 < t < \tfrac13 \\ 0, & t > \tfrac13 \end{cases}$$

b. [1 mark]

$\dfrac{f\!\left(\tfrac12\right) - f(0)}{\tfrac12 - 0} = \dfrac{22 - 12}{\tfrac12} = 20$ °C per hour.

c. i. [1 mark]

$g'(t) = 60e^{-6t}$.

ii. [1 mark]

$60e^{-6t} = 10$ gives $t = \tfrac{\log_e 6}{6} \approx 0.299$.

d. [1 mark]

For $t > \tfrac13$, $f(t) = 22 > g(t)$, so solve $12 + 30t = 22 - 10e^{-6t}$ with CAS: $t \approx 0.27$.

e. [1 mark]

The difference is $|f(t) - g(t)|$. Two candidates:

• $g - f$ is largest where $\tfrac{d}{dt}(g - f) = 60e^{-6t} - 30 = 0$, at $t = \tfrac{\log_e 2}{6} \approx 0.1155$, where $g - f \approx 1.53$.
• $f - g$ is largest at the corner $t = \tfrac13$, where $f - g = 10e^{-2} \approx 1.35$.

The greatest difference is at $t \approx 0.12$ hours.

f. i. [1 mark]

Continuity at $t = 0.4$: $0.3 + Ae^{-10 \times 0.4} = 1.5$, so $Ae^{-4} = 1.2$ and $A = 1.2e^4$.

ii. [1 mark]

In the first 0.4 hours energy is $1.5t$ kWh. $1.5t = 0.5$ gives $t = \tfrac13$ hour (which is less than 0.4).

iii. [2 marks]

By $t = 0.4$ the heater has used $1.5 \times 0.4 = 0.6$ kWh, so a further 0.4 kWh is needed:

$$\int_{0.4}^{a}\left(0.3 + 1.2e^4e^{-10t}\right)dt = 0.4 \implies a \approx 1.33 \text{ hours}.$$

From the report. In part a many included $t = \tfrac13$ in the domain. In part b a common wrong answer was 30 (from substituting $t = \tfrac12$ into $12 + 30t$), and some found an average value instead. Part e was poorly done (29% correct): the common wrong answer $\tfrac13$ is where $f - g$ is greatest, but the greatest difference is where $g - f$ peaks. In f.ii and f.iii some solved $p(t) = 0.5$ or $p(t) = 1$ instead of using area.

Question 3 (11 marks)

Monthly online sales $y$ ($ million) are plotted for months $t = 1, 2, \ldots$ from January 2021.

a. i

A cubic $p(t) = at^3 + bt^2 + ct + d$ on $(0, 12]$ has a local minimum at $(2, 2500)$ and a local maximum at $(11, 4400)$. Find $a, b, c, d$ to two decimal places. (3 marks)

ii

$q(t) = p(t - h) + k$ on $(12, 24]$ has a local maximum at $(23, 4750)$. Find $h$ and $k$. (2 marks)

b

A second model is $f(t) = 3000 + 30t + 700\cos\!\left(\tfrac{\pi t}{6}\right) + 400\cos\!\left(\tfrac{\pi t}{3}\right)$ on $(0, 36]$. i. Complete the graph for $t \in (24, 36]$ and label the endpoint. (2 marks)

ii

Every 12 months the model's sales increase by $n$ million dollars. Find $n$. (1 mark)

iii

Find $f'(t)$. (1 mark)

iv

Find the maximum instantaneous rate of change (nearest million dollars per month) and the values of $t$ in $(0, 36]$ where it occurs, to one decimal place. (2 marks)

Show worked solution

a. i. [3 marks]. Four conditions give four equations:

$$p(2) = 2500,\quad p(11) = 4400,\quad p'(2) = 0,\quad p'(11) = 0.$$

That is $8a + 4b + 2c + d = 2500$, $1331a + 121b + 11c + d = 4400$, $12a + 4b + c = 0$ and $363a + 22b + c = 0$. Solving with CAS:

$$a \approx -5.21,\quad b \approx 101.65,\quad c \approx -344.03,\quad d \approx 2823.18.$$

ii. [2 marks]. A translation moves the local maximum $(11, 4400)$ to $(23, 4750)$: $h = 23 - 11 = 12$ and $k = 4750 - 4400 = 350$.

b. i. [2 marks]. The two cosine terms have periods 12 and 6, so the shape over $(24, 36]$ repeats the shape over $(12, 24]$, shifted up by 360. The endpoint is

$$f(36) = 3000 + 1080 + 700\cos(6\pi) + 400\cos(12\pi) = 5180, \quad\text{so } (36, 5180).$$

ii. [1 mark]. $f(t + 12) - f(t) = 30 \times 12 = 360$ (the cosine terms repeat), so $n = 360$.

iii. [1 mark].

$$f'(t) = 30 - \frac{700\pi}{6}\sin\!\left(\frac{\pi t}{6}\right) - \frac{400\pi}{3}\sin\!\left(\frac{\pi t}{3}\right).$$

iv. [2 marks]. Maximise $f'(t)$ on $(0, 36]$ with CAS: the maximum rate is about 725 million dollars per month, occurring at $t \approx 10.2$, $22.2$ and $34.2$ (12 months apart).

From the report. In a.i marks were lost for no working, only two correct equations, setting $p'(2) = 2500$, rounding 101.646 to 101.64, or forgetting $d$. Part a.ii was poorly answered: only the maximum needed translating, and some translated the minimum instead ($h = 21$, $k = 2250$). In b.ii a common wrong answer was 30, found by calculating $f'(12)$. In b.iv many gave extra or only one $t$-value, left out the maximum rate, or maximised $f$ instead of $f'$.

Question 4 (15 marks)

Luggage mass $X$ (kg) has pdf $f(x) = \dfrac{1}{67\,500}x^2(30 - x)$ for $0 \le x \le 30$. Luggage over 23 kg is "heavy".

a

Write a definite integral for the probability that a piece is heavy. (1 mark)

b. i

Find the mean of $X$. (1 mark)

ii

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

iii

Given a piece is heavier than the mean, find the probability it is heavy, to three decimal places. (2 marks)

c

10% of travellers check no bags, 40% check one and 50% check two; use 0.234 as the probability a piece is heavy. $W$ is the number of heavy pieces a traveller checks in. i. Show that $\Pr(W = 2) = 0.027$. (1 mark)

ii

Complete the distribution of $W$ to three decimal places. (2 marks)

d

$\hat P$ is the proportion of heavy pieces in random samples of 35. i. Find $\Pr(\hat P > 0.2)$ to three decimal places. (2 marks)

ii

Without a normal approximation, find the probability that $\hat P$ is within one standard deviation of its mean. (2 marks)

e. i

A sample of 50 has 10 heavy pieces. Find an approximate 90% confidence interval for $p$. (1 mark)

ii

A second sample of 50 gives a wider 90% interval. State the minimum and maximum possible number of heavy pieces. (1 mark)

Show worked solution

a. [1 mark]

$\Pr(X > 23) = \displaystyle\int_{23}^{30}\frac{1}{67\,500}x^2(30 - x)\,dx$ ($\approx 0.234$).

b. i. [1 mark]

$\text{E}(X) = \displaystyle\int_0^{30}x\,f(x)\,dx = 18$ kg.

ii. [2 marks]

$\text{var}(X) = \displaystyle\int_0^{30}x^2 f(x)\,dx - 18^2 = 360 - 324 = 36$, so $\text{sd}(X) = 6$ kg.

iii. [2 marks]

$$\Pr(X > 23 \mid X > 18) = \frac{\Pr(X > 23)}{\Pr(X > 18)} = \frac{\int_{23}^{30}f(x)\,dx}{\int_{18}^{30}f(x)\,dx} \approx \frac{0.23393}{0.5248} \approx 0.446.$$

c. i. [1 mark]. Two heavy pieces needs two bags, both heavy: $\Pr(W = 2) = 0.5 \times 0.234^2 = 0.027378 \approx 0.027$.

ii. [2 marks].

• $\Pr(W = 0) = 0.1 + 0.4(0.766) + 0.5(0.766)^2 = 0.699778 \approx 0.700$
• $\Pr(W = 1) = 0.4(0.234) + 0.5 \times 2(0.234)(0.766) = 0.272844 \approx 0.273$

$w$	0	1	2
$\Pr(W = w)$	0.700	0.273	0.027

d. i. [2 marks]. Let $Y \sim \text{Bi}(35, 0.234)$ be the number of heavy pieces. $\hat P > 0.2$ means $Y > 7$:

$$\Pr(\hat P > 0.2) = \Pr(Y \ge 8) \approx 0.595.$$

ii. [2 marks]. $\text{sd}(\hat P) = \sqrt{\tfrac{0.234 \times 0.766}{35}} \approx 0.0716$, so "within one standard deviation" is $0.1624 < \hat P < 0.3056$, that is $5.69 < Y < 10.69$:

$$\Pr(6 \le Y \le 10) \approx 0.684.$$

e. i. [1 mark]. $\hat p = 0.2$ and $z \approx 1.6449$:

$$0.2 \pm 1.6449\sqrt{\frac{0.2 \times 0.8}{50}} \implies (0.107,\ 0.293).$$

ii. [1 mark]. The width is proportional to $\sqrt{\hat p(1 - \hat p)}$, which is larger than $\sqrt{0.2 \times 0.8}$ exactly when $0.2 < \hat p < 0.8$. With 50 pieces that is 11 to 39 heavy pieces: minimum 11, maximum 39.

From the report. In b.ii some gave the variance or forgot to square the mean. In b.iii some divided the probabilities the wrong way round (an answer above 1), used 0.5 in the denominator instead of $\Pr(X > 18)$, or tried a normal distribution. Part c.ii was poorly done (11% full marks). In d.i a common wrong answer was $\Pr(Y \ge 7) \approx 0.743$; in d.ii students often found the standard deviation but then used the wrong integer range, or used a normal approximation. Part e.ii had only 8% correct: many found 11 but gave 50, or 40, as the maximum.

Question 5 (11 marks)

Let $f(x) = \sin(x)$ and $g(x) = \sin(2x)$; the graphs of $g \circ f$ and $f \circ g$ are shown for $x \in [0, 2\pi]$.

a. i

Find the local maximum of $y = (g \circ f)(x)$ with $x \in \left[0, \tfrac{\pi}{2}\right]$, to one decimal place. (1 mark)

ii

State the range of $g \circ f$ on $[0, 2\pi]$. (1 mark)

b. i

Find the derivative of $f \circ g$. (1 mark)

ii

Show that $\cos\big(\sin(2x)\big) = 0$ has no real solutions. (2 marks)

iii

Find the $x$-values of the stationary points of $f \circ g$ on $[0, 2\pi]$. (1 mark)

iv

Find the range of $f \circ g$ on $[0, 2\pi]$. (1 mark)

c. i

Write a single definite integral for the area between $y = (f \circ g)(x)$ and $y = (g \circ f)(x)$ on $[0, 2\pi]$. (1 mark)

ii

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

d

Let $f_1:(0, 2\pi) \to R$, $f_1(x) = \sin(x)$. Find all $x \in (0, 2\pi)$ for which $f_1 \circ g$ is defined. (2 marks)

Show worked solution

a. i. [1 mark]

$(g \circ f)(x) = \sin(2\sin x)$. Its derivative $2\cos(x)\cos(2\sin x)$ is zero on $\left[0, \tfrac{\pi}{2}\right)$ when $2\sin x = \tfrac{\pi}{2}$, so $x = \sin^{-1}\!\left(\tfrac{\pi}{4}\right) \approx 0.9$ and $y = 1$. The local maximum is $(0.9, 1)$.

ii. [1 mark]

$2\sin x$ takes every value in $[-2, 2]$, which includes $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, so $\sin(2\sin x)$ reaches both $-1$ and $1$. Range $[-1, 1]$.

b. i. [1 mark]

$(f \circ g)(x) = \sin(\sin 2x)$, so $(f \circ g)'(x) = 2\cos(2x)\cos\big(\sin(2x)\big)$.

ii. [2 marks]

$\cos(\theta) = 0$ requires $\theta = \tfrac{\pi}{2} + k\pi$, so $|\theta| \ge \tfrac{\pi}{2} > 1$. But $\theta = \sin(2x)$ always lies in $[-1, 1]$. So $\cos\big(\sin(2x)\big) = 0$ has no real solutions.

iii. [1 mark]

From parts i and ii, stationary points need $\cos(2x) = 0$: $x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$.

iv. [1 mark]

$\sin(2x) \in [-1, 1]$ and $\sin$ is increasing there, so the range is $\big[-\sin(1), \sin(1)\big]$.

c. i. [1 mark]

The graphs cross only at $x = 0, \pi, 2\pi$, with $g \circ f$ above on $(0, \pi)$. One single integral is

$$\int_0^{2\pi}\Big|\sin(2\sin x) - \sin(\sin 2x)\Big|\,dx \quad\text{or}\quad 2\int_0^{\pi}\Big(\sin(2\sin x) - \sin(\sin 2x)\Big)dx.$$

ii. [1 mark]. Area $\approx 4.97$ square units.

d. [2 marks]. $f_1 \circ g$ is defined when the output of $g$ lies in the domain of $f_1$, $(0, 2\pi)$. Since $\sin(2x) \le 1 < 2\pi$, we just need $\sin(2x) > 0$, that is $2x \in (0, \pi) \cup (2\pi, 3\pi)$:

$$x \in \left(0, \frac{\pi}{2}\right) \cup \left(\pi, \frac{3\pi}{2}\right).$$

From the report. In a.ii many wrote $[1, -1]$ or used round brackets, and $[0, 1]$ was a common wrong answer. Part b.ii is a "show that", and many could not find $\cos^{-1}(0)$ correctly, giving 0 or $\pi$. In b.iii some gave fewer than four values or decimals. In b.iv an exact answer with square brackets was needed. In c.i a single integral was required and many wrote two; in c.ii 2.48 (half the area) was a common wrong answer. In part d (9% full marks) many knew the range of $g$ must sit inside the domain of $f_1$ but could not turn that into intervals.

General advice from the 2024 report

• Exact values were required but decimals given in 1c.ii, 1d.i, 2f.ii, 5b.iii and 5b.iv.
• Interval notation: write the smaller endpoint first and choose round or square brackets deliberately.
• Show your set-up on multi-mark parts; answers alone lost marks in 2f.iii, 3a and 4b.ii.
• Read what is asked: an average rate of change is not an average value, and a maximum rate of change is a value of $f'$, not $f$.