VCE Math Methods 2021 Exam 2

Every question from the 2021 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 2021 Examination 1 walkthrough.

How to use this page

• Questions are from the 2021 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 2021 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.
• Study design. This paper was set on the previous Mathematical Methods study design (2016 to 2022); the current one began in 2023. Three things on this paper are no longer in the course: functional relations (Section A Question 5), rectangle approximations to an integral (Question 2a, 2b and 2d, now replaced by the trapezium rule) and the median of a continuous random variable (Question 4f and 4h). Each is flagged where it appears. Everything else 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 (14, 10, 12, 14 and 10 marks). Give exact answers unless told to round, and write down what you entered into CAS.

Section A: Multiple choice

Q1

Period of $y = \tan\!\left(\tfrac{\pi x}{2}\right)$. Answer: B - the period of $\tan(nx)$ is $\tfrac{\pi}{n} = \pi \div \tfrac{\pi}{2} = 2$.

Q2

Which graph is identical to $y = \log_e(x) + \log_e(2x)$, $x > 0$? Answer: C - by the product law, $\log_e(x) + \log_e(2x) = \log_e(2x^2)$.

Q3

A sample of 48 beads has sample proportion 0.125 blue; find the 95% confidence interval. Answer: A - $0.125 \pm 1.96\sqrt{\tfrac{0.125 \times 0.875}{48}} = 0.125 \pm 0.0936$, giving $(0.0314, 0.2186)$.

Q4

Maximum value of $h(x) = (x - 2)e^x$ on $[0, 2]$. Answer: B - $h'(x) = (x - 1)e^x$ gives a minimum of $-e$ at $x = 1$; comparing endpoints, $h(0) = -2$ and $h(2) = 0$, so the maximum is 0.

Q5

How many of $f(x) = f(-x)$, $-f(x) = f(-x)$, $f(x) = -f(x)$ and $\big(f(x)\big)^2 = f(x^2)$ hold for $f(x) = x$? Answer: C - $f(x) = x$ is odd, so $-f(x) = f(-x)$ holds, and $x^2 = x^2$ holds; the other two fail (except at $x = 0$). Study design: functional relations are no longer listed in the current study design, so skip this one.

Q6

Win probability 0.25 per game, 10 independent games; find $\Pr(\text{exactly 4 wins})$. Answer: A - $\binom{10}{4}(0.25)^4(0.75)^6 \approx 0.1460$.

Q7

The tangent to $y = x^3 - ax^2 + 1$ at $x = 1$ passes through the origin. Find $a$. Answer: B - the point is $(1, 2 - a)$ and the gradient is $3 - 2a$; a line through the origin has $y = mx$, so $2 - a = 3 - 2a$, giving $a = 1$.

Q8

Choose the graph of $f'$ for a log-shaped graph with vertical asymptote $x = a$. Answer: E - $f$ is increasing and concave down for $x > a$, so $f'$ is positive and decreasing on $(a, \infty)$, with the same asymptote. (40% correct.)

Q9

Range of $h(x) = f(g(x))$ on $[-5, -1)$, where $g(x) = x + 2$ and $f(x) = x^2 - 4$. Answer: E - $h(x) = (x + 2)^2 - 4$; $x + 2$ runs over $[-3, 1)$, so $(x + 2)^2$ covers $[0, 9]$ and the range is $[-4, 5]$.

Q10

Maximal domain of $h = f + g$, where $f(x) = \sqrt{x + 2}$ and $g(x) = \sqrt{1 - 2x}$. Answer: D - both need non-negative arguments: $x \ge -2$ and $x \le \tfrac12$, so $\left[-2, \tfrac12\right]$.

Q11

Given $\int_0^a f(x)\,dx = k$, find $\int_0^a \big(3f(x) + 2\big)dx$. Answer: A - $3k + \int_0^a 2\,dx = 3k + 2a$.

Q12

Crest proportion $\tfrac35$; smallest $n$ with $\text{sd}(\hat P) < 0.08$. Answer: D - $\sqrt{\tfrac{0.24}{n}} < 0.08$ needs $n > 37.5$, so $n = 38$.

Q13

Average rate of change of $f(n) = 2500(1.004)^n$ over the first 12 months. Answer: B - $\dfrac{f(12) - f(0)}{12} \approx \dfrac{2622.68 - 2500}{12} \approx$ $10.22, closest to $10.20 per month.

Q14

A value of $k$ for which the average value of $y = \cos\!\left(kx - \tfrac{\pi}{2}\right)$ on $[0, \pi]$ equals that of $y = \sin(x)$. Answer: E - $\cos\!\left(kx - \tfrac{\pi}{2}\right) = \sin(kx)$. The target is $\tfrac{2}{\pi}$; with $k = \tfrac12$, $\tfrac1\pi\int_0^\pi \sin\!\left(\tfrac{x}{2}\right)dx = \tfrac{1}{\pi}\big[-2\cos\!\left(\tfrac{x}{2}\right)\big]_0^\pi = \tfrac{2}{\pi}$.

Q15

Four fair coins; find $\Pr(\text{equal heads and tails} \mid \text{at least one head})$. Answer: D - with $X \sim \text{Bi}\left(4, \tfrac12\right)$, $\dfrac{\Pr(X = 2)}{\Pr(X \ge 1)} = \dfrac{6/16}{15/16} = \dfrac25$.

Q16

$\cos(x) = \tfrac35$ and $\sin^2(y) = \tfrac{25}{169}$ with $x, y \in \left[\tfrac{3\pi}{2}, 2\pi\right]$; find $\sin(x) + \cos(y)$. Answer: A - in the fourth quadrant sine is negative and cosine positive: $\sin(x) = -\tfrac45$ and $\cos(y) = \tfrac{12}{13}$, so the sum is $-\tfrac45 + \tfrac{12}{13} = \tfrac{8}{65}$. (Only 31% correct.)

Q17

$X \sim \text{Bi}(n, 0.1)$; smallest $n$ with $\Pr(X \ge 2) \ge 0.5$. Answer: C - $\Pr(X \ge 2) = 1 - 0.9^n - n(0.1)(0.9)^{n-1}$ is about 0.485 at $n = 16$ and 0.518 at $n = 17$.

Q18

Maximum number of solutions of $f(x - k) = g(x)$, where $f(x) = (2x - 1)(2x + 1)(3x - 1)$ and $g(x) = x\log_e(-x)$ on $(-\infty, 0)$. Answer: D - translating the cubic left (for example $k = -1$) makes it cross the graph of $g$ three times, and exploring other values of $k$ on CAS never gives more than three. (39% correct.)

Q19

Which hybrid function is differentiable for all real $x$? Answer: E - for $f(x) = 4x + 1$ ($x < 0$) and $(2x + 1)^2$ ($x \ge 0$), both pieces equal 1 at $x = 0$ (continuous) and both have gradient 4 there ($4$ and $4(2x + 1)$), so the join is smooth. Option D is continuous but its gradients (2 and 4) do not match. (35% correct.)

Q20

$A$, $B$ independent, $\Pr(A) = p$, $\Pr(B) = p^2$ and $\Pr(A) + \Pr(B) = 1$; find $\Pr(A' \cup B)$. Answer: D - $\Pr(A' \cup B) = 1 - \Pr(A \cap B') = 1 - p(1 - p^2) = 1 - p + p^3$. (39% correct.)

Section B: Extended response

Question 1 (14 marks)

A rectangular sheet of cardboard is $h$ cm wide and $2h$ cm long. Squares of side $x$ cm are cut from each corner and the sides folded up to make an open box. First take $h = 25$.

a

Show that $V_{\text{box}}(x) = 2x(25 - 2x)(25 - x)$. (1 mark)

b

State the domain of $V_{\text{box}}$. (1 mark)

c

Find the derivative of $V_{\text{box}}$ with respect to $x$. (1 mark)

d

Calculate the maximum possible volume of the box and the value of $x$ for which it occurs. (3 marks)

e

Find the percentage of the sheet that is wasted (cut out) when $x = 5$. (2 marks)

Now let $h > 0$, with the length still twice the width.

f. i

State the domain of $V_{\text{box}}$ in terms of $h$. (1 mark)

ii

Find the maximum volume in terms of $h$. (3 marks)

g

For a square sheet of side $h$, show that the maximum volume occurs when $x = \dfrac{h}{6}$. (2 marks)

Show worked solution

a. [1 mark]. The base is $(25 - 2x)$ by $(50 - 2x)$ and the height is $x$:

$$V_{\text{box}}(x) = x(25 - 2x)(50 - 2x) = x(25 - 2x) \times 2(25 - x) = 2x(25 - 2x)(25 - x).$$

b. [1 mark]. Need $x > 0$ and $25 - 2x > 0$ (then $25 - x > 0$ too): the domain is $(0, 12.5)$.

c. [1 mark]. Expanding, $V_{\text{box}}(x) = 4x^3 - 150x^2 + 1250x$, so

$$V_{\text{box}}'(x) = 12x^2 - 300x + 1250.$$

d. [3 marks]. Solve $12x^2 - 300x + 1250 = 0$:

$$x = \frac{300 \pm \sqrt{90\,000 - 60\,000}}{24} = \frac{25}{2} \pm \frac{25\sqrt3}{6}.$$

Only $x = \tfrac{25}{2} - \tfrac{25\sqrt3}{6} = \tfrac{25(3 - \sqrt3)}{6} \approx 5.28$ is in the domain. Substituting (CAS):

$$V_{\text{max}} = \frac{15\,625\sqrt3}{9} \text{ cm}^3 \approx 3007 \text{ cm}^3.$$

e. [2 marks]. Four squares of area $5^2$ are cut from a $25 \times 50$ sheet:

$$\frac{4 \times 25}{25 \times 50} \times 100\% = \frac{100}{1250} \times 100\% = 8\%.$$

f. i. [1 mark]. $V_{\text{box}}(x) = x(h - 2x)(2h - 2x)$ needs $0 < x < \tfrac{h}{2}$: the domain is $\left(0, \tfrac{h}{2}\right)$.

ii. [3 marks]. Solve $V_{\text{box}}'(x) = 0$ with CAS: $x = \tfrac{h(3 \pm \sqrt3)}{6}$. Only $x = \tfrac{h(3 - \sqrt3)}{6} \approx 0.21h$ lies in $\left(0, \tfrac{h}{2}\right)$ (the other is about $0.79h$). Substituting:

$$V_{\text{max}} = \frac{\sqrt3h^3}{9}.$$

(Check: $h = 25$ gives $\tfrac{15\,625\sqrt3}{9}$, matching part d.)

g. [2 marks]. For a square sheet, $V(x) = x(h - 2x)^2$ on $\left(0, \tfrac{h}{2}\right)$. By the product and chain rules:

$$V'(x) = (h - 2x)^2 - 4x(h - 2x) = (h - 2x)(h - 6x).$$

$V'(x) = 0$ gives $x = \tfrac{h}{2}$ or $x = \tfrac{h}{6}$. Since $x = \tfrac{h}{2}$ is not in the domain (the volume would be 0), and $V'$ changes from positive to negative at $x = \tfrac{h}{6}$, the maximum volume occurs when $x = \dfrac{h}{6}$.

From the report. In part a some could not identify the dimensions or left out brackets. In part b $(0, 25)$ was often seen, as was the wrong bracket $(0, 12.5]$. In part d exact values were needed; some found $x$ but not the volume, some substituted the wrong root $\tfrac{25\sqrt3}{6} + \tfrac{25}{2}$, and transcription errors were common. In part e some divided by the volume instead of the area, and 0.08% was often seen. In f.i some gave the rule instead of the domain, or $(0, 25h)$. In f.ii some used the wrong $x$-value and got a negative volume. In part g, a "show that", the domain had to be used to reject $x = \tfrac{h}{2}$, and some did not show enough working.

Question 2 (10 marks)

Four rectangles of equal width, with heights taken at the right endpoints, approximate the area under $y = x^2$ from $x = 0$ to $x = 1$.

a

State the width of each rectangle. (1 mark)

b

Find the total area of the four rectangles. (1 mark)

c

Find the area between $y = x^2$, the $x$-axis and $x = 1$. (2 marks)

d

Using a given graph of $f$ on $[-3, 3]$, approximate $\int_{-2}^{2} f(x)\,dx$ with four right-endpoint rectangles of equal width. (1 mark)

e

Find the area of the region enclosed by $y = x^2$ and $y = \sqrt x$. (1 mark)

f

$y = x^2$ is transformed to $y = ax^2$, $a \in (0, 2]$. Find the values of $a$ such that the area of the region(s) bounded by $y = ax^2$, $y = \sqrt x$, $x = 0$ and $x = a$ is $\tfrac13$, correct to two decimal places. (4 marks)

Study design. Parts a, b and d use rectangle (left or right endpoint) approximations, which were in the previous study design. The current course approximates definite integrals with the trapezium rule instead, so treat a, b and d as optional. Parts c, e and f are fully on the current course.

Show worked solution

a. [1 mark]. $\dfrac{1 - 0}{4} = 0.25$.

b. [1 mark]. Right endpoints $0.25, 0.5, 0.75, 1$:

$$0.25\left(0.25^2 + 0.5^2 + 0.75^2 + 1^2\right) = \frac14\left(\frac{1 + 4 + 9 + 16}{16}\right) = \frac{30}{64} = \frac{15}{32}.$$

c. [2 marks].

$$\int_0^1 x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac13.$$

d. [1 mark]. Width 1, right endpoints $x = -1, 0, 1, 2$. Reading the graph, $f(-1) = 6$, $f(0) = 2$, $f(1) = -4$ and $f(2) = -6$:

$$1 \times (6 + 2 - 4 - 6) = -2.$$

The rectangles below the axis count as negative because this approximates a definite integral, not an area.

e. [1 mark]. The curves meet at $x = 0$ and $x = 1$, with $\sqrt x$ on top:

$$\int_0^1\left(\sqrt x - x^2\right)dx = \frac23 - \frac13 = \frac13.$$

f. [4 marks]. The curves meet where $ax^2 = \sqrt x$, that is $x^{\frac32} = \tfrac1a$, so $x = a^{-\frac23}$. Compare this with the right boundary $x = a$:

Case $a \le 1$. Then $a^{-\frac23} \ge 1 \ge a$, so $\sqrt x \ge ax^2$ on all of $[0, a]$ and there is one region:

$$\int_0^a\left(\sqrt x - ax^2\right)dx = \frac23a^{\frac32} - \frac{a^4}{3} = \frac13.$$

CAS gives $a \approx 0.77$ and $a = 1$.

Case $a > 1$. The curves cross at $x = a^{-\frac23}$, which is less than 1 and so less than $a$, giving two regions:

$$\int_0^{a^{-2/3}}\left(\sqrt x - ax^2\right)dx + \int_{a^{-2/3}}^{a}\left(ax^2 - \sqrt x\right)dx = \frac13.$$

CAS gives $a \approx 1.13$.

So $a \approx 0.77$, $a = 1.00$ or $a \approx 1.13$.

From the report. Part a was done very well (96%). In part b an exact answer was needed; 0.47 lost the mark. In part c the definite integral had to be shown for full marks, and some rounded to 0.3. Part d was poorly answered (16%): a common wrong answer was $6 + 2 + 4 + 6 = 18$, treating every rectangle as positive. In part f many found 0.77 or 1.13 but not both; others found $x = a^{-2/3}$ but did not set up the integrals, and $\int_0^a\left(ax^2 - \sqrt x\right)dx = \tfrac13$, giving $a = 1.46$, was often seen.

Question 3 (12 marks)

Let $q(x) = \log_e\left(x^2 - 1\right) - \log_e(1 - x)$.

a

State the maximal domain and the range of $q$. (2 marks)

b. i

Find the equation of the tangent to the graph of $q$ at $x = -2$. (1 mark)

ii

Find the equation of the line perpendicular to the graph of $q$ at $x = -2$ that passes through $(-2, 0)$. (1 mark)

Let $p(x) = e^{-2x} - 2e^{-x} + 1$.

c

Explain why $p$ is not a one-to-one function. (1 mark)

d

Find the gradient of the tangent to the graph of $p$ at $x = a$. (1 mark)

e

The line $y = x + 2$ and the tangent to $p$ at $x = a$ meet at an acute angle $\theta$. Find the value(s) of $a$ for which $\theta = 60°$, correct to two decimal places. (3 marks)

f

Find the $x$-coordinate where $y = x + 2$ meets the graph of $p$, and hence the area bounded by $y = x + 2$, the graph of $p$ and the $x$-axis, both correct to three decimal places. (3 marks)

Show worked solution

a. [2 marks]. Need $x^2 - 1 > 0$ (so $x < -1$ or $x > 1$) and $1 - x > 0$ (so $x < 1$). Together: $x < -1$. On this domain

$$q(x) = \log_e\left(\frac{(x - 1)(x + 1)}{1 - x}\right) = \log_e\big(-(x + 1)\big),$$

and $-(x + 1)$ takes every value in $(0, \infty)$. Domain $(-\infty, -1)$, range $R$.

b. i. [1 mark]

$q(-2) = \log_e(1) = 0$ and $q'(x) = \dfrac{1}{x + 1}$, so $q'(-2) = -1$. The tangent is $y = -(x + 2)$, that is $y = -x - 2$.

ii. [1 mark]

Perpendicular gradient $1$, through $(-2, 0)$: $y = x + 2$.

c. [1 mark]

$p(x) = \left(e^{-x} - 1\right)^2$ has a minimum of 0 at $x = 0$, decreasing before it and increasing after it. So it fails the horizontal line test: for example, $p(x) = \tfrac14$ at both $x = \log_e 2$ and $x = -\log_e\!\left(\tfrac32\right)$. Two $x$-values give the same $y$-value, so $p$ is many-to-one.

d. [1 mark]

$$p'(a) = -2e^{-2a} + 2e^{-a} = 2\left(e^{a} - 1\right)e^{-2a}.$$

e. [3 marks]. The line $y = x + 2$ makes an angle of $45°$ with the positive $x$-axis. A tangent at $60°$ to it must be inclined at $45° + 60° = 105°$ or $45° - 60° = -15°$, so its gradient is

$$\tan(105°) = -(2 + \sqrt3) \approx -3.732 \quad\text{or}\quad \tan(-15°) = -(2 - \sqrt3) \approx -0.268.$$

Both are negative, so $a < 0$ (where $p' < 0$). Solve $p'(a) =$ each value with CAS:

$$a \approx -0.67 \quad\text{or}\quad a \approx -0.11.$$

f. [3 marks]. Solve $e^{-2x} - 2e^{-x} + 1 = x + 2$ with CAS: $x \approx -0.750$. From the diagram the region runs from the line's $x$-intercept $x = -2$ to the minimum of $p$ at $x = 0$, with the line on the left and $p$ on the right:

$$\int_{-2}^{-0.750}(x + 2)\,dx + \int_{-0.750}^{0}p(x)\,dx \approx 1.038 \text{ square units}.$$

From the report. In part a some gave only the domain, or wrote it as $(-\infty, 1)$, $(-\infty, 1]$ or $(-1, -\infty)$. In part b equations, not expressions, were needed, and both were quick with technology. In part c some said there are two $x$-values for every $y$-value (not true) or confused this with the vertical line test. In part d some gave the answer in terms of $x$, or the equation of the tangent. Part e was not answered well (83% scored 0); some found one of $-0.67$ and $-0.11$ but not both. In part f the $x$-coordinate was needed to three decimal places ($-0.75$ lost the mark), and some could not set up the two integrals.

Question 4 (14 marks)

Ball speeds $W$ (m/s) from a table tennis machine are normally distributed with mean 10 and standard deviation 0.8.

a. Find $\Pr(W \ge 11)$, correct to three decimal places. (1 mark)

b. Find the speed $k$ that 80% of ball speeds are below, correct to one decimal place. (1 mark)

On a new height setting, 8% of balls miss the table. $\hat P$ is the proportion that miss in random samples of 25 balls.

c. Find the mean and standard deviation of $\hat P$. (2 marks)

d. Use the binomial distribution to find $\Pr\left(\hat P > 0.1\right)$, correct to three decimal places. (2 marks)

The spin $X$ (revolutions per second) has pdf $f(x) = \dfrac{x}{500}$ for $0 \le x < 20$, $f(x) = \dfrac{50 - x}{750}$ for $20 \le x \le 50$, and 0 elsewhere.

e

Find the maximum possible spin. (1 mark)

f

Find the median spin, correct to one decimal place. (2 marks)

g

Find the standard deviation of the spin, correct to one decimal place. (3 marks)

h

The new pdf $g(x) = af\!\left(\tfrac{x}{b}\right)$ has median spin 30. Find $a$ and $b$, correct to two decimal places. (2 marks)

Study design. Parts f and h rely on the median of a continuous random variable. That was in the previous study design, but the current one lists only the mean, variance and standard deviation, so you can skip f and h. The rest of the question is on the current course.

Show worked solution

a. [1 mark]

$\Pr(W \ge 11) \approx 0.106$ (normal cdf, or $\Pr(Z \ge 1.25)$).

b. [1 mark]

Inverse normal: $\Pr(W < k) = 0.8$ gives $k \approx 10.67$, so $k \approx 10.7$ m/s.

c. [2 marks]

$$\text{E}\left(\hat P\right) = p = 0.08 = \frac{2}{25}, \qquad \text{sd}\left(\hat P\right) = \sqrt{\frac{0.08 \times 0.92}{25}} = \sqrt{\frac{46}{15\,625}} = \frac{\sqrt{46}}{125} \approx 0.0543.$$

d. [2 marks]. Let $X \sim \text{Bi}(25, 0.08)$ count the misses, so $\hat P = \tfrac{X}{25}$. Then $\hat P > 0.1$ means $X > 2.5$, that is $X \ge 3$:

$$\Pr(X \ge 3) \approx 0.323.$$

e. [1 mark]. The pdf is non-zero up to $x = 50$, so the maximum spin is 50 revolutions per second.

f. [2 marks]. First, $\Pr(X < 20) = \int_0^{20}\tfrac{x}{500}\,dx = \tfrac{400}{1000} = 0.4 < 0.5$, so the median $m$ is in the second piece:

$$\int_m^{50}\frac{50 - x}{750}\,dx = \frac{(50 - m)^2}{1500} = \frac12 \implies (50 - m)^2 = 750 \implies m = 50 - 5\sqrt{30} \approx 22.6.$$

g. [3 marks]. Define the hybrid pdf in CAS, then

$$\text{E}(X) = \int_0^{50}x f(x)\,dx = \frac{70}{3}, \qquad \text{sd}(X) = \sqrt{\int_0^{50}x^2 f(x)\,dx - \left(\frac{70}{3}\right)^2} \approx 10.3.$$

h. [2 marks]. $g(x) = af\!\left(\tfrac{x}{b}\right)$ dilates the pdf by a factor $b$ from the $y$-axis and $a$ from the $x$-axis. Every $x$-value, including the median, is multiplied by $b$:

$$b(50 - 5\sqrt{30}) = 30 \implies b \approx 1.33.$$

A dilation by $b$ horizontally multiplies the area by $b$, so to keep the total area 1, $a = \tfrac1b \approx 0.75$. (Equivalently, solve $\int_0^{30}af\!\left(\tfrac{x}{b}\right)dx = \tfrac12$ and $\int_0^{50b}af\!\left(\tfrac{x}{b}\right)dx = 1$ together.)

$$a \approx 0.75, \quad b \approx 1.33.$$

From the report. In part a a common error was 0.228; in part b some rounded down to 10.6. In part c exact answers were required, and $\text{E}\left(\hat P\right) = 2$ and $\text{sd}\left(\hat P\right) = \tfrac{\sqrt{46}}{25}$ (the values for $X$, not $\hat P$) were often seen. In part d some wrote $\Pr(X > 3)$ instead of $\Pr(X \ge 3)$, or gave $n$ and $p$ but no answer. Only 21% got part e; a common wrong answer was 0.04. In part f some found the mean, not the median; students who wrote the integral with $f(x)$ and defined the hybrid function on their CAS did better. In part g some gave the variance. Part h was poorly done (86% scored 0): most could not set up the equations, and the terminals were often wrong.

Question 5 (10 marks)

Let $f(x) = \sin\!\left(\dfrac{x}{2}\right) + \cos(2x)$, $x \in R$ (a graph is given).

a

State the period of $f$. (1 mark)

b

State the minimum value of $f$, correct to three decimal places. (1 mark)

c

Find the smallest positive value of $h$ for which $f(h - x) = f(x)$. (1 mark)

Consider $g_a(x) = \sin\!\left(\dfrac{x}{a}\right) + \cos(ax)$, where $a$ is a positive integer.

d

State the value of $a$ such that $g_a(x) = f(x)$ for all $x$. (1 mark)

e. i

Find an antiderivative of $g_a$ in terms of $a$. (1 mark)

ii

Use a definite integral to show that the area bounded by $g_a$ and the $x$-axis over $[0, 2a\pi]$ is equal above and below the $x$-axis for all values of $a$. (3 marks)

f

Explain why the maximum of $g_a$ cannot be greater than 2 and the minimum cannot be less than $-2$, for all values of $a$. (1 mark)

g

Find the greatest possible minimum value of $g_a$. (1 mark)

Show worked solution

a. [1 mark]

$\sin\!\left(\tfrac{x}{2}\right)$ has period $4\pi$ and $\cos(2x)$ has period $\pi$. The period of the sum is the lowest common multiple, $4\pi$.

b. [1 mark]

Find the minimum over one full period, $[0, 4\pi]$, with CAS: the minimum value is $-1.722$. (Evaluating at a point that merely looks lowest, such as $x = -\tfrac{\pi}{2}$, gives $-1.707$, which is not the minimum.)

c. [1 mark]

$f(h - x) = f(x)$ means the graph is symmetric about the line $x = \tfrac{h}{2}$. The graph is symmetric about $x = \pi$, and indeed

$$f(2\pi - x) = \sin\!\left(\pi - \frac{x}{2}\right) + \cos(4\pi - 2x) = \sin\!\left(\frac{x}{2}\right) + \cos(2x) = f(x).$$

There is no line of symmetry at $x = 0$ (since $f(-x) \ne f(x)$), so the smallest positive value is $h = 2\pi$.

d. [1 mark]. $a = 2$.

e. i. [1 mark].

$$\int g_a(x)\,dx = -a\cos\!\left(\frac{x}{a}\right) + \frac{1}{a}\sin(ax) \; (+c).$$

ii. [3 marks]. Evaluate the definite integral over $[0, 2a\pi]$:

$$\int_0^{2a\pi}g_a(x)\,dx = \left[-a\cos\!\left(\frac{x}{a}\right) + \frac{\sin(ax)}{a}\right]_0^{2a\pi} = \left(-a\cos(2\pi) + \frac{\sin\left(2a^2\pi\right)}{a}\right) - (-a + 0) = \frac{\sin\left(2a^2\pi\right)}{a}.$$

Since $a$ is a positive integer, $2a^2\pi$ is a multiple of $2\pi$, so $\sin\left(2a^2\pi\right) = 0$ and the integral is 0. The definite integral is (area above the axis) minus (area below the axis), so the two areas are equal for every $a$.

f. [1 mark]. For every $a$, $\sin\!\left(\tfrac{x}{a}\right)$ and $\cos(ax)$ each lie between $-1$ and $1$. So their sum can be at most $1 + 1 = 2$ and at least $-1 - 1 = -2$.

g. [1 mark]. For $a = 1$, $g_1(x) = \sin(x) + \cos(x) = \sqrt2\sin\!\left(x + \tfrac{\pi}{4}\right)$, whose minimum is $-\sqrt2$. For larger $a$ the minimum is lower (about $-1.722$ for $a = 2$, and close to $-2$ for $a \ge 3$). The greatest possible minimum is $-\sqrt2$.

From the report. In part a, $2\pi$ was a common wrong answer. In part b some gave the coordinates of the turning point instead of the value, rounded to $-1.72$ or dropped the sign; many evaluated $f\!\left(-\tfrac{\pi}{2}\right) \approx -1.707$, which is not the minimum. Part c needed an exact answer (21% correct); 6.28 was common. In e.i some differentiated instead, or got the signs wrong. In e.ii some could not interpret $\tfrac{\sin(2a^2\pi)}{a}$. In part f (13% correct) many explained only the maximum. Only 2% answered part g correctly; $-2$ was the common wrong answer.

General advice from the 2021 report

• Use brackets correctly, in rules and in intervals; practise domains and ranges and choose round or square brackets deliberately.
• Exact answers were required in 1d, 2c, 2e, 4c, 5a, 5c and 5g. Where rounding is asked for, give exactly the number of decimal places requested ($-0.750$, not $-0.75$; $-1.722$, not $-1.72$).
• Give everything asked for: the maximum volume as well as the $x$-value in 1d, the range as well as the domain in 3a, and equations (not expressions) in 3b.
• Use CAS efficiently: define a hybrid function once and reuse it, and find tangent and perpendicular lines with technology. Copy results carefully; transcription errors cost marks.
• Explanations (3c, 5f) must refer to the specific function in the question.