HSC Maths Advanced 2021

Every question from the 2021 HSC Mathematics Advanced exam, with full worked solutions. Section II solutions are tucked behind a Show worked solution toggle, so you can attempt a question first and reveal the model answer when you are ready.

How to use this page

• Questions are from the 2021 HSC Mathematics Advanced exam, copyright NSW Education Standards Authority (NESA). Open the official PDF (button above) for the original stimulus diagrams and graphs.
• Answers are original model responses by ExamExplained (Claude Opus 4.8), written to the official marking guidelines, not copied from NESA's sample answers.
• Each Section II solution shows the mark split and a short Marker's note from the HSC marking feedback.

Structure and timing

100 marks in 180 minutes is about 1.8 minutes per mark.

• Section I (10 marks): 10 multiple-choice. Allow about 15 minutes.
• Section II (90 marks): Questions 11 to 34, short and extended response. Allow about 2 hours and 45 minutes, in proportion to the marks. Show relevant reasoning and calculations, and round only at the final step.

Section I - Multiple choice

Q1

Which of the following is equivalent to $\sin^2 5x$? A. $1 + \cos^2 5x$ B. $1 - \cos^2 5x$ C. $-1 + \cos^2 5x$ D. $-1 - \cos^2 5x$
Answer: B - the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ gives $\sin^2 5x = 1 - \cos^2 5x$.

Q2

A discrete random variable $X$ has $P(X=1) = 0.6$, $P(X=2) = 0.3$, $P(X=3) = 0.1$. What is $E(X)$? A. $0.6$ B. $1.0$ C. $1.5$ D. $2.0$
Answer: C - $E(X) = 1(0.6) + 2(0.3) + 3(0.1) = 1.5$.

Q3

Which represents the domain of $f(x) = \ln(1 - x)$? A. $[1, \infty)$ B. $(1, \infty)$ C. $(-\infty, 1]$ D. $(-\infty, 1)$
Answer: D - a logarithm needs $1 - x > 0$, so $x < 1$.

Q4

A bar graph shows daily downloads of a song over 20 days. Which graph best shows the cumulative number of downloads up to and including each day? (Options A to D are graphs.)
Answer: C - the cumulative total only ever increases, rising more steeply on high-download days and flattening on low-download days.

Q5

Which best represents the graph of $y = 10(0.8)^x$? (Options A to D are graphs.)
Answer: A - exponential decay with a $y$-intercept of 10, decreasing towards the positive $x$-axis as an asymptote.

Q6

A box has 3 peppermint (P) and 5 caramel (C) chocolates; Kim then Sam each eat one without replacement. Probability they choose different centres? A. $\tfrac{15}{64}$ B. $\tfrac{15}{56}$ C. $\tfrac{15}{32}$ D. $\tfrac{15}{28}$
Answer: D - $P(\text{PC}) + P(\text{CP}) = \tfrac38\cdot\tfrac57 + \tfrac58\cdot\tfrac37 = \tfrac{30}{56} = \tfrac{15}{28}$.

Q7

$y = f(x)$ has a local minimum at $x = -2$ and a local maximum at $x = 3$. Which shows the correct relationship between $f''(-2)$, $f(0)$ and $f'(3)$? A. $f(0) < f'(3) < f''(-2)$ B. $f(0) < f''(-2) < f'(3)$ C. $f''(-2) < f'(3) < f(0)$ D. $f''(-2) < f(0) < f'(3)$
Answer: A - at a local minimum the curve is concave up so $f''(-2) > 0$; at a maximum the gradient is zero so $f'(3) = 0$; and $f(0)$ reads as negative from the graph, giving $f(0) < f'(3) < f''(-2)$.

Q8

The graph of $y = f(x)$ is shown. Which could be its equation? A. $y = (1-x)(2+x)^3$ B. $y = (x+1)(x-2)^3$ C. $y = (x+1)(2-x)^3$ D. $y = (x-1)(2+x)^3$
Answer: C - the curve cuts at $x = -1$ and has a flat inflection where it crosses at $x = 2$, and the long-run behaviour matches $(x+1)(2-x)^3$.

Q9

$h(x) = f(g(x))$ with $f$ odd and $g$ even. The tangent at $x = k$ ($k > 0$) is $y = mx + c$. What is the tangent at $x = -k$? A. $y = mx + c$ B. $y = -mx + c$ C. $y = mx - c$ D. $y = -mx - c$
Answer: B - $h$ is even, so its graph is symmetric in the $y$-axis; the tangent at $-k$ is the mirror image, with gradient $-m$ and the same intercept $c$.

Q10

$y = mx$ is a tangent to $y = \cos x$ at $x = a$, as shown. Which statement is true? A. $m < \tfrac{1}{2\pi} < \tfrac1a$ B. $\tfrac{1}{2\pi} < m < \tfrac1a$ C. $\tfrac{1}{2\pi} < \tfrac1a < m$ D. $m < \tfrac1a < \tfrac{1}{2\pi}$
Answer: B - reading the contact point and the small positive gradient from the diagram gives the ordering $\tfrac{1}{2\pi} < m < \tfrac1a$.

Section II - Short and extended response

Question 11 (2 marks)

Solve $\displaystyle x + \frac{x - 1}{2} = 9$.

Show worked solution

[2 marks]. Multiply every term by 2 to clear the fraction:

$$2x + (x - 1) = 18 \implies 3x - 1 = 18 \implies 3x = 19.$$

$$x = \frac{19}{3}.$$

Marker's note. Establish a common denominator or multiply each term by 2, then apply inverse operations step by step. The common slip was mishandling the two-term numerator $x - 1$.

Question 12 (5 marks)

A right-angled triangle $XYZ$ is cut from a semicircle with centre $O$. The diameter $XZ$ is 16 cm and $\angle YXZ = 30\degree$. (Stimulus: a semicircle with the right-angled triangle $XYZ$ inscribed, right angle at $Y$ - see the official paper.)
(a) Find the length of $XY$ in centimetres, correct to two decimal places. (2 marks)
(b) Hence find the area of the shaded region in square centimetres, correct to one decimal place. (3 marks)

Show worked solution

(a) [2 marks]. The angle in a semicircle is $90\degree$, so the right angle is at $Y$ and $XZ$ is the hypotenuse. With $\angle YXZ = 30\degree$ and $XY$ adjacent:

$$\cos 30\degree = \frac{XY}{16} \implies XY = 16\cos 30\degree = 13.86 \text{ cm (2 d.p.)}.$$

(b) [3 marks]. The shaded region is the semicircle minus the triangle. The radius is $8$ cm, so the semicircle area is $\tfrac12\pi(8)^2 = 32\pi$. The triangle area, using $A = \tfrac12 ab\sin C$ with $XY = 13.86$, $XZ = 16$ and included angle $30\degree$:

$$\text{Shaded} = \frac12\pi(8)^2 - \frac12 \times 13.86 \times 16 \times \sin 30\degree = 100.53 - 55.44 = 45.1 \text{ cm}^2 \ (1\text{ d.p.}).$$

Marker's note. Identify the hypotenuse and adjacent side correctly for the trig ratio in (a). In (b) take the area as the semicircle minus the triangle, picking the right formula from the reference sheet, and carry full digits through to the final rounding.

Question 13 (3 marks)

Find the exact gradient of the tangent to the curve $y = x\tan x$ at the point where $x = \dfrac{\pi}{3}$.

Show worked solution

[3 marks]. Differentiate using the product rule with $u = x$, $v = \tan x$:

$$\frac{dy}{dx} = x\sec^2 x + \tan x.$$

Substitute $x = \dfrac{\pi}{3}$, where $\cos\dfrac{\pi}{3} = \dfrac12$ so $\sec^2\dfrac{\pi}{3} = 4$, and $\tan\dfrac{\pi}{3} = \sqrt3$:

$$\frac{dy}{dx} = \frac{\pi}{3}\times 4 + \sqrt3 = \frac{4\pi}{3} + \sqrt3.$$

Marker's note. Recognise $y = x\tan x$ as a product, set out $u, u', v, v'$ clearly, then show the substitution of $\tfrac{\pi}{3}$ in radians. Exact answers here contain a surd and $\pi$, so do not switch to a decimal.

Question 14 (2 marks)

The first term of an arithmetic sequence is 5. The sum of the first 43 terms is 2021. What is the common difference?

Show worked solution

[2 marks]. Use $S_n = \dfrac{n}{2}\big(2a + (n-1)d\big)$ with $a = 5$, $n = 43$, $S_{43} = 2021$:

$$2021 = \frac{43}{2}\big(10 + 42d\big) \implies 94 = 10 + 42d \implies 42d = 84.$$

$$d = 2.$$

Marker's note. Choose the sum formula from the reference sheet, substitute carefully, then use clean algebra to make $d$ the subject.

Question 15 (2 marks)

Evaluate $\displaystyle\int_{-2}^{0} \sqrt{2x + 4}\,dx$.

Show worked solution

[2 marks]. Write the integrand as $(2x + 4)^{1/2}$ and integrate, dividing by the new power $\tfrac32$ and the inner coefficient 2:

$$\int_{-2}^{0} (2x+4)^{1/2}\,dx = \left[\frac{1}{2}\cdot\frac{2}{3}(2x+4)^{3/2}\right]_{-2}^{0} = \left[\frac{1}{3}(2x+4)^{3/2}\right]_{-2}^{0}.$$

$$= \frac{1}{3}\big(4^{3/2} - 0\big) = \frac{1}{3}\times 8 = \frac{8}{3}.$$

Marker's note. Express the root as a fractional index, use the reverse chain rule for the linear inside function, then substitute the limits with brackets and simplify the fractional-index value.

Question 16 (3 marks)

For what values of $x$ is $f(x) = x^2 - 2x^3$ increasing?

Show worked solution

[3 marks]. A function is increasing when $f'(x) > 0$. Differentiate:

$$f'(x) = 2x - 6x^2 = 2x(1 - 3x).$$

This concave-down parabola is positive between its roots $x = 0$ and $x = \tfrac13$:

$$2x(1 - 3x) > 0 \implies 0 < x < \frac{1}{3}.$$

Marker's note. State the condition $f'(x) > 0$, differentiate the cubic, factorise, then solve the quadratic inequality with a negative leading term using a sketch or sign diagram. Give the answer in interval or inequality form.

Question 17 (4 marks)

For 17 inland towns, height above sea level $x$ (m) and average maximum daily temperature $y$ (degrees C) were recorded. The regression line is $y = 29.2 - 0.011x$ with $r = -0.494$. (Stimulus: a scatterplot with a fitted regression line - see the official paper.)
(a) (i) Predict the average maximum daily temperature for a town 540 m above sea level, correct to one decimal place. (1 mark)
(a) (ii) Interpret the value of the gradient $-0.011$ in this context. (2 marks)
(b) A second regression of temperature against latitude $x$ (degrees south) for the same towns gives $y = 45.6 - 0.683x$ with $r = -0.897$. For another inland town 540 m above sea level and at latitude 28 degrees south, which measurement, height or latitude, would be better to use to predict its average maximum daily temperature? Give a reason. (1 mark)

Show worked solution

(a)(i) [1 mark]. Substitute $x = 540$:

$$y = 29.2 - 0.011 \times 540 = 23.3 \text{ degrees C}.$$

(a)(ii) [2 marks]. For each one-metre increase in height above sea level, the model predicts the average maximum daily temperature falls by about $0.011$ degrees C.

(b) [1 mark]. Latitude is the better predictor here, because its correlation with temperature ($r = -0.897$) is stronger than that of height above sea level ($r = -0.494$).

Marker's note. In (a)(i) substitute and round, not leaving a bald number. In (a)(ii) describe both the rise and the run using the variable names. In (b) compare the strength of correlation, recalling that $r = -0.897$ is stronger than $r = -0.494$, and keep the one-mark reason simple.

Question 18 (3 marks)

In triangle $ABC$, $AC = 25$ cm, $BC = 16$ cm, $\angle BAC = 28\degree$ and $\angle ABC$ is obtuse. (Stimulus: triangle $ABC$, not to scale - see the official paper.) Find the size of the obtuse angle $ABC$ correct to the nearest degree.

Show worked solution

[3 marks]. Apply the sine rule with each side opposite its angle:

$$\frac{\sin\angle ABC}{25} = \frac{\sin 28\degree}{16} \implies \sin\angle ABC = \frac{25\sin 28\degree}{16} = 0.7336\ldots$$

The acute solution is $47.18\degree$, but $\angle ABC$ is obtuse, so take the supplement:

$$\angle ABC = 180\degree - 47.18\degree = 133\degree \text{ (nearest degree)}.$$

Marker's note. Substitute into the sine rule correctly, find the acute angle, then apply the ambiguous case via $\sin\theta = \sin(180\degree - \theta)$ to obtain the obtuse angle.

Question 19 (3 marks)

Without using calculus, sketch the graph of $y = 2 + \dfrac{1}{x + 4}$, showing the asymptotes and the $x$ and $y$ intercepts.

Show worked solution

[3 marks]. This is the basic hyperbola $y = \tfrac1x$ shifted left 4 and up 2.

• Asymptotes: vertical $x = -4$ and horizontal $y = 2$.
• $y$-intercept ($x = 0$): $y = 2 + \tfrac14 = \tfrac94$.
• $x$-intercept ($y = 0$): $0 = 2 + \tfrac{1}{x+4}$ gives $\tfrac{1}{x+4} = -2$, so $x + 4 = -\tfrac12$ and $x = -\tfrac92$.

The two branches sit in the regions set by the asymptotes, the upper branch through $\left(0, \tfrac94\right)$ and the lower branch through $\left(-\tfrac92, 0\right)$.

Marker's note. Find both intercepts and both asymptotes, then draw the branches approaching but not crossing the asymptotes. Use a ruler for the axes and asymptotes on a sufficiently large diagram and watch the signs.

Question 20 (2 marks)

For what values of $x$, in the interval $0 \le x \le \dfrac{\pi}{4}$, does the line $y = 1$ intersect the graph of $y = 2\sin 4x$?

Show worked solution

[2 marks]. Setting $2\sin 4x = 1$ gives $\sin 4x = \tfrac12$. Over $0 \le x \le \tfrac{\pi}{4}$ the angle $4x$ runs through $0 \le 4x \le \pi$, where $\sin 4x = \tfrac12$ at

$$4x = \frac{\pi}{6}, \ \frac{5\pi}{6} \implies x = \frac{\pi}{24}, \ \frac{5\pi}{24}.$$

Marker's note. Form $2\sin 4x = 1$, transform the domain to $0 \le 4x \le \pi$, and find both related angles in radians. The question asks for the values of $x$, not the number of solutions.

Question 21 (2 marks)

The graph of $y = f(x)$ is shown, with a turning point at $(4, -8)$ and $x$-intercepts at $x = 2$ and $x = 6$. (Stimulus: a cubic-style curve, not to scale - see the official paper.) Sketch the graph of $y = 4f(2x)$ showing the $x$-intercepts and the coordinates of the turning points.

Show worked solution

[2 marks]. The transformation $y = 4f(2x)$ is a horizontal dilation by factor $\tfrac12$ (replace $x$ by $2x$) and a vertical dilation by factor 4.

• $x$-intercepts halve: $x = 2 \to x = 1$ and $x = 6 \to x = 3$.
• The turning point $(4, -8)$ moves to $\left(\tfrac42, 4\times(-8)\right) = (2, -32)$.
• The maximum turning point at the origin stays at the origin.

So the new curve has $x$-intercepts at $x = 1$ and $x = 3$, a minimum at $(2, -32)$, and the same maximum at the origin.

Marker's note. Separate the horizontal dilation (factor $\tfrac12$, so $(6,y)\to(3,y)$) from the vertical dilation (factor 4, so $(x,-8)\to(x,-32)$), and label the transformed intercepts and turning points.

Question 22 (4 marks)

A random variable is normally distributed with mean 0 and standard deviation 1. A table gives $P(0 < Z < z)$ for $z = 0.1$ to $0.6$, namely $0.0398$, $0.0793$, $0.1179$, $0.1554$, $0.1915$, $0.2257$. (Stimulus: a standard normal curve with the area from 0 to $z$ shaded - see the official paper.)
(a) Using the table, find the probability that a value lies between $0.1$ and $0.5$. (1 mark)
(b) Birth weights are normally distributed with mean 3300 grams and standard deviation 570 grams. By first calculating a $z$-score, find how many babies, out of 1000 born, are expected to have a birth weight greater than 3528 grams. (3 marks)

Show worked solution

(a) [1 mark]. The area between $0.1$ and $0.5$ is the difference of the two tabulated areas:

$$P(0.1 < Z < 0.5) = 0.1915 - 0.0398 = 0.1517.$$

(b) [3 marks]. The $z$-score for 3528 grams is

$$z = \frac{3528 - 3300}{570} = 0.4.$$

The area beyond $z = 0.4$ is $0.5 - 0.1554 = 0.3446$, so the expected number out of 1000 is

$$1000 \times 0.3446 = 344.6 \approx 345 \text{ babies}.$$

Marker's note. In (a) recognise that the area between two $z$-scores needs a subtraction. In (b) compute the $z$-score, sketch the curve, take the right-tail area by subtracting from $0.5$, then multiply by 1000. Use the table rather than the empirical rule.

Question 23 (4 marks)

A population $P$, initially 5000, varies as $P = 5000\,b^{-t/10}$, where $b$ is a positive constant and $t$ is time in years, $t \ge 0$. The population is 1250 after 20 years. Find the value of $t$, correct to one decimal place, for which the instantaneous rate of decrease is 30 people per year.

Show worked solution

[4 marks]. First find $b$. At $t = 20$, $P = 1250$:

$$1250 = 5000\,b^{-2} \implies b^{-2} = \frac14 \implies b = 2 \ (b > 0).$$

So $P = 5000(2)^{-t/10}$. Differentiate, using $\dfrac{d}{dt}a^{kt} = a^{kt}\ln a \cdot k$:

$$\frac{dP}{dt} = 5000\times\left(-\frac{1}{10}\right)\ln 2\times 2^{-t/10} = -500\ln 2 \times 2^{-t/10}.$$

A decrease of 30 people per year means $\dfrac{dP}{dt} = -30$:

$$-30 = -500\ln 2 \times 2^{-t/10} \implies 2^{t/10} = \frac{500\ln 2}{30} = 11.55\ldots$$

$$\frac{t}{10}\ln 2 = \ln 11.55\ldots \implies t = 35.3 \text{ years (1 d.p.)}.$$

Marker's note. Use index rules to get $b = 2$, then the rule for differentiating $ab^x$ with $b \ne e$. Set the derivative to $-30$ (negative, since the population is decreasing) and apply the logarithm laws to solve for $t$.

Question 24 (3 marks)

The curve $y = \dfrac{3}{x - 1}$ meets the line $y = \dfrac{3}{2}x$ at $(2, 3)$. The region bounded by the curve, the line, the $x$-axis and the line $x = 4$ is shaded. (Stimulus: the curve, the line through the origin and the boundary $x = 4$ - see the official paper.) Find the exact area of the shaded region.

Show worked solution

[3 marks]. Split the region into a triangle (under the line, from $x = 0$ to $x = 2$) plus the area under the curve from $x = 2$ to $x = 4$.

The triangle has base 2 and height 3 (the point $(2, 3)$), so its area is $\tfrac12\times 2 \times 3 = 3$. The area under the curve:

$$\int_2^4 \frac{3}{x - 1}\,dx = 3\big[\ln(x - 1)\big]_2^4 = 3(\ln 3 - \ln 1) = 3\ln 3.$$

Adding gives the exact shaded area:

$$3 + 3\ln 3 \text{ square units}.$$

Marker's note. Interpret the shaded area as a triangle plus the area under the hyperbola. The primitive of $\dfrac{3}{x-1}$ is logarithmic; show the substitution of the limits, then add the two parts.

Question 25 (3 marks)

A table of future value interest factors for an annuity of $1 is given (for 2, 4, 6, 8, 10 periods at rates from $0.25\%$ to $1.25\%$). Simone deposits $1000 at the end of each year for 8 years at $0.75\%$ per annum, compounded annually. After the 8th deposit she makes no more deposits, and the balance earns $1.25\%$ per annum, compounded annually, for a further two years. Find the amount in the account at the end of ten years.

Show worked solution

[3 marks]. The $0.75\%$, 8-period factor from the table is $8.2132$, so the balance just after the 8th deposit is

$$1000 \times 8.2132 = \$8213.20.$$

This sum then earns $1.25\%$ per annum for two more years (no further deposits), so apply compound interest:

$$8213.20 \times (1.0125)^2 = \$8419.81.$$

Marker's note. Read the correct $8.2132$ factor from the table and multiply by the $1000 deposit; do not rebuild the series by hand when a future-value table is provided. Then apply two years of compound interest at the second rate.

Question 26 (5 marks)

A particle is shot vertically upwards from 100 metres above ground level. Its height $y$ metres above the ground after $t$ seconds is $y(t) = -5t^2 + 70t + 100$.
(a) Find the maximum height above ground level reached by the particle. (2 marks)
(b) Find the velocity, in metres per second, immediately before it hits the ground, leaving your answer in the form $a\sqrt{b}$ where $a$ and $b$ are integers. (3 marks)

Show worked solution

(a) [2 marks]. Maximum height occurs when $y'(t) = 0$:

$$y'(t) = -10t + 70 = 0 \implies t = 7.$$

Then

$$y(7) = -5(7)^2 + 70(7) + 100 = -245 + 490 + 100 = 345 \text{ m}.$$

(b) [3 marks]. The particle hits the ground when $y(t) = 0$:

$$-5t^2 + 70t + 100 = 0 \implies t^2 - 14t - 20 = 0 \implies t = \frac{14 \pm \sqrt{196 + 80}}{2} = 7 \pm \sqrt{69}.$$

Since $t > 0$, take $t = 7 + \sqrt{69}$. The velocity is $y'(t) = -10t + 70$:

$$y'(7 + \sqrt{69}) = -10(7 + \sqrt{69}) + 70 = -10\sqrt{69} \text{ m/s}.$$

So the velocity just before impact is $-10\sqrt{69}$ m/s, that is $a = -10$, $b = 69$.

Marker's note. In (a) solve $y'(t) = 0$ for the time, then substitute back; finding $t$ is only a step. In (b) solve $y(t) = 0$ with the quadratic formula keeping $t > 0$ in surd form, then substitute into the derivative and simplify, retaining the surd.

Question 27 (8 marks)

Kenzo's solar charger power is $P(t) = 400\sin\dfrac{\pi t}{12}$, $0 \le t \le 12$, where $t$ is hours after sunrise.
(a) Sketch the graph of $P$ for $0 \le t \le 12$. (2 marks)
(b) Energy is $E = \displaystyle\int_a^b P(t)\,dt$. Show that $E = \dfrac{4800}{\pi}\left(\cos\dfrac{a\pi}{12} - \cos\dfrac{b\pi}{12}\right)$. (2 marks)
(c) A phone call needs at least 300 units of energy. Kenzo woke 3 hours after sunrise with no energy and began charging immediately. Find the least time he must wait before he can make a call, to the nearest minute. (3 marks)
(d) The next day he woke 6 hours after sunrise with no energy and began charging. Would it take more, less or the same time to reach 300 units, compared with part (c)? Explain by referring to the graph in part (a). (1 mark)

Show worked solution

(a) [2 marks]. The amplitude is 400 and the period is $\dfrac{2\pi}{\pi/12} = 24$ hours, so over $0 \le t \le 12$ the graph is the first half-period of a sine wave: it starts at the origin, rises smoothly to a maximum of 400 at $t = 12$, concave down throughout. Label both axes.

(b) [2 marks]. Integrate $P(t) = 400\sin\dfrac{\pi t}{12}$, where the primitive of $\sin\dfrac{\pi t}{12}$ is $-\dfrac{12}{\pi}\cos\dfrac{\pi t}{12}$:

$$E = \int_a^b 400\sin\frac{\pi t}{12}\,dt = 400\left[-\frac{12}{\pi}\cos\frac{\pi t}{12}\right]_a^b = -\frac{4800}{\pi}\left(\cos\frac{b\pi}{12} - \cos\frac{a\pi}{12}\right).$$

$$= \frac{4800}{\pi}\left(\cos\frac{a\pi}{12} - \cos\frac{b\pi}{12}\right).$$

(c) [3 marks]. Charging starts at $a = 3$ and he needs $E = 300$:

$$300 = \frac{4800}{\pi}\left(\cos\frac{3\pi}{12} - \cos\frac{b\pi}{12}\right) \implies \frac{300\pi}{4800} = \cos\frac{\pi}{4} - \cos\frac{b\pi}{12}.$$

Since $\dfrac{300\pi}{4800} = \dfrac{\pi}{16}$ and $\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt2}$:

$$\cos\frac{b\pi}{12} = \frac{1}{\sqrt2} - \frac{\pi}{16} = 0.5108\ldots \implies b = \frac{12}{\pi}\cos^{-1}(0.5108\ldots) = 3.95\ldots$$

So $b \approx 3$ hours $57$ minutes, and the waiting time after $a = 3$ is about $57$ minutes (to the nearest minute).

(d) [1 mark]. It would take less time. Starting at $t = 6$ means charging around the peak of $P(t)$, where the power is greatest, so 300 units of energy accumulate more quickly than starting at $t = 3$.

Marker's note. In (b) factor out the 400, find the correct primitive and show the limit substitution. In (c) substitute $a = 3$ and $E = 300$, then solve the trig equation in radians. In (d) refer to the graph: the power peaks at $t = 6$, so charging is faster there.

Question 28 (6 marks)

The region bounded by $f(x) = 8 - 2^x$ and the coordinate axes is shown. (Stimulus: the decreasing curve $y = 8 - 2^x$ with the region under it shaded - see the official paper.)
(a) Show that the exact area of the shaded region is $24 - \dfrac{7}{\ln 2}$. (3 marks)
(b) A new function $g(x)$ is found by taking $y = -f(-x)$ and translating it 5 units to the right. Sketch $y = g(x)$ showing the $x$-intercept and the asymptote. (2 marks)
(c) Hence find the exact value of $\displaystyle\int_2^5 g(x)\,dx$. (1 mark)

Show worked solution

(a) [3 marks]. The curve cuts the $x$-axis when $8 - 2^x = 0$, that is $2^x = 8$, so $x = 3$. The area is

$$\int_0^3 (8 - 2^x)\,dx = \left[8x - \frac{2^x}{\ln 2}\right]_0^3 = \left(24 - \frac{8}{\ln 2}\right) - \left(0 - \frac{1}{\ln 2}\right).$$

$$= 24 - \frac{8}{\ln 2} + \frac{1}{\ln 2} = 24 - \frac{7}{\ln 2}.$$

(b) [2 marks]. Reflecting $f$ in both axes ($y = -f(-x)$) gives $y = -(8 - 2^{-x}) = 2^{-x} - 8$, then shifting 5 right gives

$$g(x) = 2^{-(x - 5)} - 8 = 2^{5 - x} - 8.$$

As $x \to \infty$, $g(x) \to -8$ (horizontal asymptote $y = -8$); it cuts the $x$-axis where $2^{5 - x} = 8$, that is $5 - x = 3$, so $x = 2$.

(c) [1 mark]. Over $2 \le x \le 5$ the graph of $g$ is the original shaded region reflected, lying below the $x$-axis, so the integral is the negative of the area in part (a):

$$\int_2^5 g(x)\,dx = -\left(24 - \frac{7}{\ln 2}\right) = \frac{7}{\ln 2} - 24.$$

Marker's note. In (a) find the $x$-intercept, use the reference-sheet integral of $a^x$, and substitute limits with brackets. In (b) apply both reflections then the translation, labelling the intercept and asymptote. In (c) read hence as use part (a): the answer is just its negative, since the region has the same area below the axis.

Question 29 (5 marks)

On the day Megan was born, her grandfather deposited $5000 into an account earning $3\%$ per annum compounded annually. On each birthday after this he deposited $1000, making his final deposit on Megan's 17th birthday (18 deposits in total). Let $A_n$ be the amount on Megan's $n$th birthday after the deposit is made.
(a) Show that $A_3 = \$8554.54$. (2 marks)
(b) On her 17th birthday, just after the final deposit, Megan has $30 025.83 (you are not required to show this). She leaves the money to earn $3\%$ per annum compounded annually. On her 18th birthday, and on each birthday after, she withdraws $2000. How many withdrawals of $2000 will Megan be able to make? (3 marks)

Show worked solution

(a) [2 marks]. Build the balance year by year. The opening balance is $A_0 = 5000$, and each year the balance grows by $3\%$ then a $1000 deposit is added:

$$A_1 = 5000(1.03) + 1000 = 6150,$$

$$A_2 = 6150(1.03) + 1000 = 7334.50,$$

$$A_3 = 7334.50(1.03) + 1000 = \$8554.54.$$

(b) [3 marks]. Let $B_n$ be the balance after $n$ withdrawals, starting from $B_0 = 30\,025.83$. Each year the balance grows by $3\%$ and $2000 is withdrawn, so

$$B_n = 30\,025.83(1.03)^n - 2000\big[1 + 1.03 + \cdots + (1.03)^{n-1}\big] = 30\,025.83(1.03)^n - 2000\,\frac{(1.03)^n - 1}{0.03}.$$

Megan can keep withdrawing while $B_n > 0$. Setting $B_n = 0$:

$$0.03\times 30\,025.83\,(1.03)^n = 2000\big[(1.03)^n - 1\big] \implies 900.77\,(1.03)^n = 2000(1.03)^n - 2000.$$

$$1099.23\,(1.03)^n = 2000 \implies (1.03)^n = 1.8194\ldots \implies n = \frac{\ln 1.8194\ldots}{\ln 1.03} \approx 20.25.$$

Since the balance is still positive at $n = 20$ but not at $n = 21$, Megan can make 20 withdrawals of $2000.

Marker's note. In (a) a show question needs the sequential working for $A_1$, $A_2$ and $A_3$, noting the first $1000 deposit earns no interest in its first year. In (b) build the series for $B_n$, sum it with the geometric formula, set $B_n = 0$, and solve the resulting exponential equation carefully.

Question 30 (2 marks)

The number of hours light bulbs work before failing is modelled by $X$ with cumulative distribution function $F(x) = 1 - e^{-0.01x}$ for $x \ge 0$ (and 0 for $x < 0$). Jane promises a bulb will work longer than exactly $99\%$ of all bulbs. Find how long, according to her promise, a bulb should work, in hours, rounded to two decimal places.

Show worked solution

[2 marks]. Jane's promised time $x$ is the value below which $99\%$ of bulbs fail, so $F(x) = 0.99$:

$$1 - e^{-0.01x} = 0.99 \implies e^{-0.01x} = 0.01.$$

Take logarithms:

$$-0.01x = \ln(0.01) \implies x = \frac{\ln(0.01)}{-0.01} = 460.52 \text{ hours (2 d.p.)}.$$

Marker's note. Recognise that $F(x)$ is the cumulative function, so set $F(x) = 0.99$ (not $f(x)$), then solve the exponential equation with logarithms.

Question 31 (4 marks)

By considering the equation of the tangent to $y = x^2 - 1$ at the point $(a, a^2 - 1)$, find the equations of the two tangents to $y = x^2 - 1$ which pass through $(3, -8)$.

Show worked solution

[4 marks]. For $y = x^2 - 1$, $y' = 2x$, so the gradient at $x = a$ is $2a$. The tangent at $(a, a^2 - 1)$ is

$$y - (a^2 - 1) = 2a(x - a) \implies y = 2ax - a^2 - 1.$$

This tangent passes through $(3, -8)$:

$$-8 = 2a(3) - a^2 - 1 \implies a^2 - 6a - 7 = 0 \implies (a - 7)(a + 1) = 0.$$

So $a = 7$ or $a = -1$. Substituting each back:

$$a = 7: \ y = 14x - 49 - 1 = 14x - 50, \qquad a = -1: \ y = -2x - 1 - 1 = -2x - 2.$$

The two tangents are $y = 14x - 50$ and $y = -2x - 2$.

Marker's note. Find the gradient $2a$, write the tangent at $x = a$, substitute $(3, -8)$ to form a quadratic in $a$, solve it, then substitute both values back to give the two tangent equations. A sketch helps confirm two tangents exist.

Question 32 (4 marks)

In a city, adult female and adult male heights are each normally distributed. For two females: 175 cm is taller than $97.5\%$ of females, and 160.6 cm is taller than $16\%$ of females. The female mean and standard deviation are $\mu$ and $\sigma$; the male mean is $1.05\mu$ and standard deviation $1.1\sigma$. A selected male is taller than $84\%$ of adult males. By first labelling the normal curve with the two female heights, calculate the height of the selected male, in centimetres, correct to two decimal places. (Stimulus: a blank normal distribution curve to be labelled - see the official paper.)

Show worked solution

[4 marks]. By the empirical rule, $97.5\%$ corresponds to 2 standard deviations above the female mean and $16\%$ corresponds to 1 standard deviation below it. So

$$\mu + 2\sigma = 175, \qquad \mu - \sigma = 160.6.$$

Subtracting, $3\sigma = 14.4$, so $\sigma = 4.8$ and $\mu = 160.6 + 4.8 = 165.4$ cm.

The male mean and standard deviation are

$$\mu_M = 1.05 \times 165.4 = 173.67, \qquad \sigma_M = 1.1 \times 4.8 = 5.28.$$

Being taller than $84\%$ of males means 1 standard deviation above the male mean, so

$$\text{height} = 173.67 + 5.28 = 178.95 \text{ cm (2 d.p.)}.$$

Marker's note. Convert the female percentages to $z$-scores using the empirical rule, solve for $\mu$ and $\sigma$, scale to the male mean and standard deviation, then add one standard deviation for the $84\%$ male.

Question 33 (8 marks)

The time in hours to complete a puzzle (maximum six hours) is modelled by $X$ with probability density function $f(x) = \dfrac{Ax}{x^2 + 4}$ for $0 \le x \le 6$ (where $A > 0$), and 0 otherwise. The graph has a local maximum. (Stimulus: the density curve rising to a local maximum then falling over $0 \le x \le 6$ - see the official paper.)
(a) Show that $A = \dfrac{2}{\ln 10}$. (2 marks)
(b) Show that the mode of $X$ is two hours. (2 marks)
(c) Show that $P(X < 2) = \log_{10} 2$. (2 marks)
(d) IQ scores are normally distributed with mean 100 and standard deviation 15. It is known that $80\%$ of people with an IQ greater than 130 complete the puzzle in less than two hours. A person chosen at random completes the puzzle in less than two hours. What is the probability that this person has an IQ greater than 130? Give your answer correct to three decimal places. (2 marks)

Show worked solution

(a) [2 marks]. A density must integrate to 1 over its support:

$$\int_0^6 \frac{Ax}{x^2 + 4}\,dx = \frac{A}{2}\big[\ln(x^2 + 4)\big]_0^6 = \frac{A}{2}\big(\ln 40 - \ln 4\big) = \frac{A}{2}\ln 10.$$

Setting this equal to 1 gives $\dfrac{A}{2}\ln 10 = 1$, so $A = \dfrac{2}{\ln 10}$.

(b) [2 marks]. The mode is the maximum of $f$. By the quotient rule,

$$f'(x) = \frac{A(x^2 + 4) - Ax(2x)}{(x^2 + 4)^2} = \frac{A(4 - x^2)}{(x^2 + 4)^2}.$$

Setting $f'(x) = 0$ needs $4 - x^2 = 0$, so $x = 2$ (taking $x > 0$). This gives the maximum, so the mode is two hours.

(c) [2 marks]. Integrate the density from 0 to 2:

$$P(X < 2) = \frac{A}{2}\big[\ln(x^2 + 4)\big]_0^2 = \frac{A}{2}(\ln 8 - \ln 4) = \frac{A}{2}\ln 2.$$

Substitute $A = \dfrac{2}{\ln 10}$:

$$P(X < 2) = \frac{1}{\ln 10}\times\ln 2 = \frac{\ln 2}{\ln 10} = \log_{10} 2.$$

(d) [2 marks]. First, $P(\text{IQ} > 130) = P\!\left(Z > \dfrac{130 - 100}{15}\right) = P(Z > 2) = 2.5\%$. By conditional probability,

$$P(\text{IQ} > 130 \mid X < 2) = \frac{P(\text{IQ} > 130 \cap X < 2)}{P(X < 2)} = \frac{2.5\% \times 80\%}{\log_{10} 2}.$$

$$= \frac{0.025 \times 0.8}{0.30103\ldots} = 0.066 \text{ (3 d.p.)}.$$

Marker's note. In (a) state that the area is 1 and integrate to a logarithm, using $\ln 40 - \ln 4 = \ln 10$. In (b) use the quotient rule keeping $A$, set the numerator to zero. In (c) link back to part (a) and use the change-of-base law. In (d) build the conditional probability with the part (c) value as the denominator.

Question 34 (3 marks)

A discrete random variable has the probability distribution shown, where $n$ is a finite positive integer: $X$ takes values $r, r^2, r^3, \ldots, r^n$ with $P(X = r^k) = r^{n - k + 1}$. (Stimulus: the probability distribution table - see the official paper.) Show that $E(X) = n(2r - 1)$.

Show worked solution

[3 marks]. First, the probabilities must sum to 1. They form a geometric series with first term $r$, ratio $r$ and $n$ terms:

$$r + r^2 + \cdots + r^n = \frac{r(r^n - 1)}{r - 1} = 1 \implies r^{n+1} - r = r - 1 \implies r^{n+1} = 2r - 1. \quad (\star)$$

Now compute $E(X) = \sum x\,P(X = x)$. The value $r^k$ has probability $r^{n - k + 1}$, so each product is

$$r^k \times r^{n - k + 1} = r^{n + 1}.$$

There are $n$ such terms, each equal to $r^{n+1}$, so

$$E(X) = n\,r^{n+1} = n(2r - 1) \quad\text{by } (\star).$$

Marker's note. Use that the probabilities sum to 1 (a geometric series) to obtain $r^{n+1} = 2r - 1$, then notice each term of $E(X)$ collapses to $r^{n+1}$, giving $n$ equal terms. Show every step on this show question.

General marker feedback

Stronger responses across the paper: showed full reasoning and substitution, especially on show questions; read each stem for key words such as show, hence and increasing; used the reference sheet for formulae, derivatives and standard integrals; worked confidently in radians for the trigonometry; kept exact answers in surd, logarithmic or $\pi$ form when asked; rounded only at the final step and carried full digits through; distinguished clearly between $f(x)$, $f'(x)$ and $f''(x)$, and between gradient and correlation; read future-value tables rather than rebuilding financial series by hand; and drew large, clearly labelled graphs showing intercepts, asymptotes and turning points.