HSC Maths Advanced 2024

[3 marks]. Read the gradient (first derivative) and concavity (second derivative) off the curve at each point.

At $x = -3$ the curve rises but is concave down; at $x = 1$ it is a flat point of inflection (both zero); at $x = 5$ it rises and is concave up.

Marker's note. Keep gradient and concavity separate: a positive first derivative means the curve is increasing, a positive second derivative means it is concave up. Read both signs directly from the graph rather than re-sketching the derivative curves.

[3 marks]. Here $a = 50$ and $d = 7$. First find the number of terms from $T_n = a + (n-1)d = 2024$:

Then sum with $S_n = \tfrac{n}{2}(a + l)$:

Marker's note. Identify the series as arithmetic, find the last term's position carefully, then use the reference-sheet sum formula. Many slips came from weak algebra when solving for $n$.

[3 marks]. Read $B = 280$ as the 1985 value of population $K$ from the graph.

The base $1.055 = 1 + 0.055$ is a $5.5\%$ increase; $0.97 = 1 - 0.03$ is a $3\%$ decrease. For $W$ at $x = 50$:

Marker's note. Distinguish the percentage change from the growth factor: $1.055$ is not $105.5\%$ change, and $0.97$ is a $3\%$ decrease, not $97\%$. Identify which curve is $W$ and which is $K$ from the graph before reading $B$.

(a) [1 mark]. Set the curves equal:

Dividing by 2 and factorising, $(x-2)(x+1) = 0$, so $x = 2$ or $x = -1$.

(b) [3 marks]. Between $x = -1$ and $x = 2$ the parabola $y = 5 - x^2$ is the upper curve. The area is

The enclosed area is 9 square units.

Marker's note. Decide which curve is on top before integrating, and show the substitution into the antiderivative with brackets. If a definite integral comes out negative, that signals the curves were the wrong way round.

[3 marks]. Integrate with respect to $t$:

When $t = 0$, $V = 350$, so $C = 350$ and $V = 300t - 3.75t^2 + 350$.

The rate is zero when $300 - 7.5t = 0$, that is $t = 40$ hours. Then

Marker's note. Integrate with respect to $t$ (not $x$), use the initial 350 litres to find $C$, and only then substitute $t = 40$. Keep the working in a clear sequence.

[3 marks]. Address all three features.

• Skewness: Garden A is negatively skewed (the longer tail and whisker point to the lower values), while Garden B is positively skewed.
• Central tendency: the median for Garden A is higher than the median for Garden B, so Garden A flowers are typically taller.
• Spread: the interquartile range (and the range) of Garden A is larger than that of Garden B, so Garden A is more variable.

Marker's note. All three of skewness, centre and spread must be compared explicitly. From a box plot the median is the only measure of centre you can read, so do not claim a mean. Use precise terms such as positively or negatively skewed.

(a) [2 marks]. A smooth curve starting at the origin $(0, 0)$, increasing and concave down, approaching the horizontal asymptote $V = 6.5$ as $t$ increases. Label both axes and the asymptote $V = 6.5$.

(b) [2 marks]. Substitute $t = 1$, $V = 2.6$:

(c) [2 marks]. Differentiate: $\dfrac{dV}{dt} = 6.5k\,e^{-kt}$. At $t = 2$ with $k = 0.5108\ldots$:

Marker's note. In (a) draw a smooth exponential approach to $V = 6.5$ with no straight segments, and label the asymptote. In (b) isolate $e^{-k}$ before taking logs. In (c) differentiate first, keep the full value of $k$, and substitute only at the end.

(a) [1 mark]. $P(\text{miss}) = 1 - 0.15 = 0.85$, so

(b) [2 marks]. $P(\text{at least one goal in } n) = 1 - (0.85)^n$. Require

Taking logarithms (and reversing the inequality, since $\ln 0.85 < 0$):

The player needs at least 10 attempts.

Marker's note. In (a) note $(1 - 0.15)^2$ is not $1 - 0.15^2$. In (b) form $(0.85)^n < 0.2$, use logs to make $n$ the subject, and finish by stating the whole number of attempts.

[5 marks]. Differentiate twice: $y' = 4x^3 - 6x^2$ and $y'' = 12x^2 - 12x$.

Stationary points: $y' = 0$ gives $2x^2(2x - 3) = 0$, so $x = 0$ or $x = \tfrac32$.

• At $x = 0$: $y = 2$ and $y'' = 0$. Testing concavity either side shows $y''$ changes sign, so $(0, 2)$ is a horizontal point of inflection.
• At $x = \tfrac32$: $y = \tfrac{5}{16}$ and $y'' = 9 > 0$, so $\left(\tfrac32, \tfrac{5}{16}\right)$ is a local minimum.

Points of inflection: $y'' = 0$ gives $12x(x - 1) = 0$, so $x = 0$ or $x = 1$. The concavity changes at both, giving inflections at $(0, 2)$ (already found) and $(1, 1)$.

Marker's note. Solve $y' = 0$ with the null-factor law, justify the nature of each stationary point (note the one at $x = 0$ is a horizontal inflection, not a minimum or maximum), justify the change in concavity for the inflections, and draw a clean calculus-based sketch.

(a) [1 mark]. In right-angled triangle $TCA$, $\tan 35\degree = \dfrac{40}{AC}$, so

(b) [3 marks]. Similarly $BC = \dfrac{40}{\tan 30\degree} = 69.28$ m. In triangle $ACB$, $AB = 100$, so by the cosine rule:

giving $\angle ACB = 104\degree$ (nearest degree). Since $A$ is due east of $C$ (bearing $090\degree$), the bearing of $B$ from $C$ is

Marker's note. In (a) a simple trig ratio is more efficient than the sine rule, and remember to divide because the unknown is in the denominator. In (b) compute $BC$, apply the cosine rule with all substitutions shown, then add to the $090\degree$ east bearing to finish the question.

[3 marks]. Three valid observations about the snakes (not merely the data):

• Females tend to be longer than males at the same age.
• Both sexes keep growing well past 4 years of age (maturity), though the rate of length increase slows for older animals.
• For young anacondas (under about 4 years), females appear to grow in length faster than males.

Marker's note. Give three distinct observations about the anacondas themselves, using clear language such as "increasing at a decreasing rate". The plot shows many different snakes, not one snake over time, so do not extrapolate beyond about 10 years.

(a) [3 marks]. Differentiate:

The denominator is always positive, so $f''(x) > 0$ when $2 - 2x^2 > 0$, that is $x^2 < 1$, i.e. $-1 < x < 1$. Hence $f$ is concave up there.

(b) [2 marks]. By symmetry the shaded area is twice the area from $0$ to $1$. With $h = 0.25$ and the trapezoidal rule:

Doubling for the full region:

(c) [1 mark]. It is an overestimate. From part (a) the curve is concave up on $-1 < x < 1$, so each trapezoidal chord lies above the curve, and the trapezia enclose more area than the region.

Marker's note. Use the second derivative (via the quotient rule or the reference sheet) and solve the inequality with a graph or sign argument. Remember to double the half-interval result in (b), and link the concave-up shape to the chords lying above the curve in (c).

(a) [2 marks]. For $x = 70$: $z = \dfrac{70 - 58}{15} = 0.8$. From the table $P(Z < 0.8) = 0.7881$, so

(b) [1 mark]. The score 46 gives $z = \dfrac{46 - 58}{15} = -0.8$. By the symmetry of the normal curve about the mean, the area from 46 to 58 equals the area from 58 to 70, so the total from 46 to 70 is $2 \times 28.81\%$.

(c) [2 marks]. The top 10% start at the 90th percentile. From the table $P(Z < 1.3) = 0.9032$, closest to $0.90$, so $z \approx 1.3$:

So a candidate needs an approximate minimum score of about 77 (78 is also acceptable).

Marker's note. A $z$-score is not a probability; subtract 0.5 to get the area above the mean. In (b) argue from symmetry and the negative $z$-score. In (c) pick the table $z$ nearest the 90th percentile, then substitute back.

(a) [3 marks]. Monthly rate $= 0.5\% = 0.005$. The first deposit earns interest for 24 months, the last for 1 month, so the total is a geometric series:

This is geometric with first term $80(1.005)$, ratio $1.005$, 24 terms:

(b) [1 mark]. The factor multiplies the $80 deposit to give the future value of the annuity:

Marker's note. Build the series term by term, showing enough terms to reveal the common ratio, and note deposits are at the start of each month. In (b) connect the table factor to the geometric-series result by dividing the total by the deposit.

(a) [2 marks]. The total area under a probability density function is 1. The graph is a right-angled triangle with base $h$ and height 1, so

(b) [2 marks]. With $h = 2$, $f(x) = 1 - \tfrac{x}{2}$. The cumulative distribution function is

with $F(x) = 0$ for $x < 0$ and $F(x) = 1$ for $x > 2$. The graph rises from $(0, 0)$ as a concave-down curve to $(2, 1)$, then stays flat at $y = 1$.

(c) [2 marks]. The median $m$ satisfies $F(m) = 0.5$:

Since $0 \le m \le 2$, take $m = 2 - \sqrt2 \approx 0.586$.

Marker's note. In (a) use area $= 1$ (the triangle), taking care with the algebra. In (b) integrate from the lowest domain value and sketch all three pieces, including the flat $y = 1$ for $x > 2$. In (c) set $F(x) = 0.5$ (not $x = 0.5$), solve the quadratic, and reject the root outside $0 \le x \le 2$.

[4 marks]. Monthly rate $r = \dfrac{2.4\%}{12} = 0.002$. The deposit must equal the present value (at the start) of both streams of withdrawals.

Treat the withdrawals as a base $1200 per month over the whole 25 years (300 months), plus an extra $800 per month over the first 15 years (180 months), since $2000 = $1200 + $800. Using the table factors for $r = 0.002$:

Adding gives the minimum deposit:

Marker's note. Convert the rate to monthly $r = 0.002$ and read the correct factors for 180 and 300 periods. Split the withdrawals into a base $1200 stream over 25 years plus an extra $800 stream over the first 15 years; the table is there to do the present-value arithmetic.

(a) [2 marks]. By the product rule with $u = x^2$, $v = \tan x$:

(b) [3 marks]. Expand the integrand and use $\tan^2 x = \sec^2 x - 1$:

The bracket is exactly the derivative from part (a), so

Marker's note. Apply the product rule correctly in (a), watching the notation $\sec^2 x$. In (b) expand the square, substitute the identity $\tan^2 x = \sec^2 x - 1$, then recognise part (a)'s result inside the integrand and write the answer sequentially.

(a) [2 marks]. The centre line is the mean of the extremes and the amplitude is half their difference:

(b) [1 mark]. The period of $\cos\!\left(\tfrac{\pi t}{24}\right)$ is $\dfrac{2\pi}{\pi/24} = 48$, so one revolution takes 48 seconds.

(c) [4 marks]. Set $A(t) = B(t)$:

Equal cosines mean the angles are equal or negatives (mod $2\pi$). The "equal" case is impossible, so use $\dfrac{\pi t}{24} = -\dfrac{\pi}{24}(t - 6)$ (and the $2\pi$-shifted version). The first gives

Allowing for the other quadrant ($2\pi - \tfrac{\pi t}{24} = \tfrac{\pi}{24}(t - 6)$) gives $t = 27$. The heights are

So the carriages are level at heights of about 4.37 m and 37.63 m.

Marker's note. Keep amplitude ($k = 18$) and vertical shift ($c = 21$) distinct. In (c) use that equal cosines allow angles differing by sign and by $2\pi$, so look for solutions in more than one quadrant; a sketch of the two curves helps confirm both times.

[4 marks]. The tangent and normal meet at the point of contact:

So the point of contact is $(-2, -1)$.

For $y = ax^2 + bx + c$, $y' = 2ax + b$. The minimum at $x = -4$ gives $y'(-4) = 0$:

The tangent gradient at $x = -2$ is 2, so $y'(-2) = 2$:

Finally, the curve passes through $(-2, -1)$:

So $a = \tfrac12$, $b = 4$, $c = 5$.

Marker's note. Use that the tangent and normal cross at the point of contact to get $(-2, -1)$. Form two equations from $y'(-4) = 0$ and $y'(-2) = 2$, solve simultaneously for $a$ and $b$, then use the point to find $c$, and check the result.

[3 marks]. A limiting sum exists when $-1 < x < 1$ (and $x \ne 0$), and

The given function rearranges to the same expression:

So $S$ is the graph $y = \dfrac{x}{1 - x}$ restricted to $-1 < x < 1$. At $x = -1$, $S = -\tfrac12$; as $x \to 1^-$, $S \to +\infty$; and $S$ is increasing on this interval. Hence

Marker's note. Find the limiting sum and the restriction $-1 < x < 1$ from the reference sheet, show the given graph equals $S$, then read the range off the sketch with the asymptote labelled. The endpoint $x = -1$ gives the boundary value $-\tfrac12$.

(a) [3 marks]. The perimeter is the two arcs plus the two straight radial sides $x$:

The area is the difference of the two sectors:

From $(\star)$, $(2 + x)\theta = \dfrac{2A}{x}$. Substituting into $P$:

(b) [3 marks]. Minimise: $P'(x) = 2 - \dfrac{2A}{x^2} = 0$ gives $x^2 = A$, so $x = \sqrt{A}$ (taking $x > 0$). Since $P''(x) = \dfrac{4A}{x^3} > 0$, this is a minimum.

Substitute $x = \sqrt A$ into $(\star)$ with $A = \tfrac12\theta\sqrt A\,(2 + \sqrt A)$:

Since $\sqrt A > 0$, the denominator $2 + \sqrt A > \sqrt A$, so $\dfrac{2\sqrt A}{2 + \sqrt A} < \dfrac{2\sqrt A}{\sqrt A} = 2$. Hence $\theta < 2$.

Marker's note. Use the radian arc-length and sector-area formulae to build $P$ and $A$, make $\theta$ the subject of the area equation, and substitute neatly to show the result. In (b) confirm $x = \sqrt A$ is a minimum with the second derivative, then bound $\theta$ algebraically. Write every step clearly on a "show" question.