HSC Maths Advanced 2025

(a) [2 marks]. The axis of symmetry is at $t = -\dfrac{b}{2a} = \dfrac{8}{2} = 4$. Then

The turning point is $(4, -4)$.

(b) [1 mark]. The two values of $t$ giving the same height are symmetric about $t = 4$. Since one is $t = 0$, the other is $t = 8$.

Marker's note. Use the axis of symmetry to find $t$, then substitute back for $h$, keeping the vertex distinct from the intercepts. In part (b) the symmetry of the parabola gives the second value directly.

[3 marks]. Write the middle term with a negative index, $y = 5x^3 - 2x^{-2} - 9$, and differentiate:

At $x = 1$ the gradient is $y' = 15 + 4 = 19$. Using $y - y_1 = m(x - x_1)$ at $(1, -6)$:

Marker's note. Differentiate the negative-index term carefully, evaluate $f'(1)$ as the gradient (not by solving $f'(x) = a$), then substitute the point and gradient into the straight-line form.

[2 marks]. With first term $a = 75$ and the fourth term $ar^3 = 2025$:

Then $p = ar = 75 \times 3 = 225$ and $q = ar^2 = 75 \times 9 = 675$.

Marker's note. Recognise the sequence as geometric and use the $n$th-term formula from the reference sheet to find $r$ before reading off $p$ and $q$.

(a) [2 marks]

Form: linear. Direction: negative (as television time increases, exercise time tends to decrease).

(b) [2 marks]

Slope: for each extra minute per day watching television, the predicted exercise falls by $0.7$ minutes per day. The $y$-intercept: a person who watches no television is predicted to exercise $64.3$ minutes per day.

(c) [1 mark]

Substitute $x = 42$:

(d) [1 mark]. Two hours is $120$ minutes, which gives $y = 64.3 - 0.7 \times 120 = -19.7$, a negative time that is impossible. The value $x = 120$ also lies outside the range of the data, so the line is not reliable there.

Marker's note. Use the precise statistical terms (linear, negative) and interpret the slope and intercept in context with their values, not in general. In part (d) note both that the prediction is negative and that $x = 120$ is outside the dataset.

(a) [2 marks]. The amplitude is the coefficient $k$, so $k = 2$. The period is $\dfrac{2\pi}{a} = 5$, so $a = \dfrac{2\pi}{5}$. Hence

(b) [2 marks]. $P_2(t) = 4\sin\!\left(\dfrac{\pi}{10}t\right)$ has amplitude 4 and period $\dfrac{2\pi}{\pi/10} = 20$, so only the first half of a sine curve appears on $0 \le t \le 10$: it rises from $(0, 0)$ to a maximum of 4 at $t = 5$, then falls back to $0$ at $t = 10$.

(c) [2 marks]. $P_1(t)$ has period 5, so on $0 < t < 10$ it is decreasing between its maximum and minimum, that is for $1.25 < t < 3.75$ and $6.25 < t < 8.75$. $P_2(t)$ is decreasing only for $5 < t < 10$. Both are decreasing where these overlap:

Marker's note. The amplitude is the coefficient of the sine term, and the period is $\tfrac{2\pi}{a}$, so only half of $P_2$ fits the domain. Find where both graphs fall by inspection, exclude the stationary points, and write the answer with correct interval notation.

(a) [4 marks]. By the quotient rule with $u = x^2$, $v = e^x$:

Since $e^x > 0$, $f'(x) = 0$ when $x = 0$ or $x = 2$.

So $(0, 0)$ is a minimum and, since $f(2) = \dfrac{4}{e^2} \approx 0.54$, $\left(2, \dfrac{4}{e^2}\right)$ is a maximum.

(b) [1 mark]. Complete the curve with a minimum at $(0, 0)$ and a maximum at $(2, 0.54)$, decreasing back towards the $x$-axis for large $x$ (the $x$-axis is a horizontal asymptote as $x \to \infty$).

Marker's note. Apply the quotient rule and factorise the $e^x$ in the numerator before dividing through. Find both $x$ and $f(x)$ for each stationary point, and use a sign table or the second derivative to classify them.

(a) [2 marks]. Starting from $A_0 = 800\,000$, each month the balance grows by the factor $1.005$ then loses the $5740 repayment:

(b) [3 marks]. Continuing the pattern,

The bracket is a geometric series with $n$ terms, first term 1, ratio $1.005$:

So

(c) [2 marks]. Require $A_n < 400\,000$:

Taking logarithms,

so the balance is first below $400 000 after 154 months.

Marker's note. Build $A_2$ from $A_1$ before expanding. In part (b) sum the geometric series with the reference-sheet formula and collect like terms. In part (c) make $(1.005)^n$ the subject, use logs, and round up to a whole number of months.

[2 marks]. The composite is

The fraction $\dfrac{3}{x - 1}$ takes every real value except 0, so adding 5 gives every real value except 5. The range is all real $y$ with $y \ne 5$.

Marker's note. Form the composite, recognise it as the hyperbola $\tfrac{3}{x-1}$ shifted up 5 units (horizontal asymptote $y = 5$), and state the range with correct notation excluding $y = 5$.

[3 marks]. The ratio $1 : 2 : 3$ gives selection probabilities $P(A) = \tfrac16$, $P(B) = \tfrac13$, $P(C) = \tfrac12$. The total probability of a win is

By conditional probability,

Marker's note. Convert the ratio to probabilities, use a tree diagram, and apply the conditional-probability formula triggered by the word given.

[3 marks]. The monthly rate is $r = \dfrac{6\%}{12} = 0.5\%$ over $n = 7 \times 12 = 84$ periods. Lin's future value is

Yemi's deposits accumulate to (deposit) $\times$ (table factor). The $0.5\%$, 84-period factor is $104.07393$, so

Marker's note. Convert the rate and number of periods to the monthly compounding period, use the future value formula for Lin's lump sum, then divide by the correct table factor to find Yemi's monthly deposit.

(a) [2 marks]. On $1 \le x \le e$, $f(x) = \dfrac1x$ is a decreasing function, so its greatest value occurs at the left endpoint $x = 1$, where $f(1) = 1$ (compared with $f(e) = \tfrac1e \approx 0.37$). The mode is therefore $x = 1$.

(b) [3 marks]. The 25th percentile $Q_1$ satisfies an area of $0.25$ from the lower limit:

Hence $Q_1 = e^{0.25} = 1.284$ (3 d.p.).

Marker's note. In (a) justify the mode from the decreasing shape and the endpoint values, using precise language. In (b) set the definite integral equal to $0.25$, integrate to $\ln x$, then solve in exponential form.

[2 marks]. Start with the left-hand side. The printed $+2$ in the numerator is the cross term $2\sin^2\theta\cos^2\theta$, which completes the square:

Hence the left-hand side is

Marker's note. Work from one side only, take a common denominator, and recognise that $\sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta = (\sin^2\theta + \cos^2\theta)^2 = 1$.

(a) [4 marks]. The number of male sheep is $\dfrac{12\,600}{21} = 600$. The 15 heaviest males are

of the males. From the empirical rule, $2.5\%$ of a normal distribution lies above $z = 2$ (two standard deviations above the mean). So

(b) [1 mark]. There are $12\,600 - 600 = 12\,000$ females, and $300$ of them is $2.5\%$, the same proportion as the males. But the females have a smaller mean and smaller standard deviation, so $x = 89.8$ kg sits at a $z$-score larger than 2 for the females. The proportion above $x$ is therefore less than $2.5\%$, so fewer than 300 females would be expected to weigh more than $x$ kg.

Marker's note. In (a) find the number of males from the ratio, convert 15 to a percentage, match $2.5\%$ to $z = 2$, and substitute into the $z$-score formula. In (b) compare $z$-scores: a smaller mean and standard deviation push $x$ further into the tail.

[4 marks]. At the point of contact each curve has gradient 1 (the gradient of $y = x$).

For $y = e\ln x$, $\dfrac{dy}{dx} = \dfrac{e}{x}$. Setting this equal to 1 gives $x = e$, and then $y = e\ln e = e$, so the point of contact is $(e, e)$.

For $y = ax^2 + c$, $\dfrac{dy}{dx} = 2ax$. At $x = e$ the gradient is 1:

The curve passes through $(e, e)$:

So $a = \dfrac{1}{2e}$ and $c = \dfrac{e}{2}$.

Marker's note. Use that the tangent has the same gradient as each curve at the contact point: differentiate $e\ln x$ to locate $(e, e)$, then use the gradient and the point on the parabola to solve for $a$ and $c$. Remember $e$ is a number.

(a) [2 marks]. By the product rule on $x\cos x$:

(b) [2 marks]. Part (a) gives the antiderivative directly:

Since $\sin(2025\pi) = 0$ and $\cos(2025\pi) = -1$ (odd multiple of $\pi$):

(c) [2 marks]. The total area is the sum of the absolute values $|A_n|$ for $n = 1$ to $2025$:

This is an arithmetic series with $a = 1$, $l = 4049$, $n = 2025$ terms:

Marker's note. In (a) apply the product rule and simplify. In (b) the word hence means use part (a); check the calculator is in radians. In (c) the total area sums the absolute values $|A_n|$, an arithmetic series, so identify $a$, $d$, $n$ and $l$ before substituting.

(a) [2 marks]. The circumference of the circle plus the triangle's perimeter is 100:

The area of the circle is $\pi r^2 = \pi\left(\dfrac{100 - 3x}{2\pi}\right)^2 = \dfrac{(100 - 3x)^2}{4\pi}$. The equilateral triangle has area $\dfrac12 x^2\sin 60\degree = \dfrac{\sqrt3}{4}x^2$. Adding,

(b) [3 marks]. $A(x)$ is a quadratic in $x$ with a positive coefficient of $x^2$ (both $\dfrac{\sqrt3}{4}$ and $\dfrac{9}{4\pi}$ are positive), so it is concave up and any stationary point is a minimum. The maximum therefore occurs at an endpoint of the domain $0 \le x \le \dfrac{100}{3}$ (the triangle cannot use more than the whole wire).

• When $x = 0$ (all wire to the circle): $r = \dfrac{50}{\pi}$ and $A = \pi\left(\dfrac{50}{\pi}\right)^2 = \dfrac{2500}{\pi} \approx 796 \text{ cm}^2$.
• When $x = \dfrac{100}{3}$ (all wire to the triangle): $A = \dfrac{\sqrt3}{4}\left(\dfrac{100}{3}\right)^2 \approx 481 \text{ cm}^2$.

Since $796 > 481$, the maximum area occurs at $x = 0$, that is when all the wire is used for the circle.

Marker's note. In (a) make $r$ the subject of the wire equation and show each step neatly on a show question. In (b) note the positive leading coefficient gives a minimum turning point, so test the two endpoints of the domain; calculus is not needed.

(a) [2 marks]. With strip width $h = 1$ and ordinates at $x = 0, 1, 2$: $y(0) = 1$, $y(1) = \tfrac12$, $y(2) = \tfrac14$. Two applications of the trapezoidal rule give

(b) [2 marks]. Integrate $\left(\tfrac12\right)^x = 2^{-x}$:

(c) [2 marks]. The curve $y = \left(\tfrac12\right)^x$ is concave up, so the trapezoidal estimate overestimates the exact area:

Rearranging (both sides positive), $24 < 36\ln 2$, so $2 < 3\ln 2 = \ln 8$. Hence $e^2 < 8$, giving $e < \sqrt8 = 2\sqrt2$.

Marker's note. In (a) apply the trapezoidal rule with the three ordinates from the reference sheet. In (b) integrate the exponential and convert $\ln\tfrac12 = -\ln 2$. In (c) the concave-up curve makes the trapezoidal value larger, so write the inequality the correct way and manipulate to $e < 2\sqrt2$.

(a) [2 marks]. The smaller segment's area is the sector minus the triangle, and this equals $\tfrac14$ of the circle:

(b) [2 marks]. The solution is the intersection of $y = \theta$ with $y = \sin\theta + \dfrac{\pi}{2}$, which the graph gives as $\theta \approx 2.3$ radians. The arc length of the smaller segment is

Marker's note. In (a) build the segment area as sector minus triangle, set it equal to $\tfrac14$ of the circle, and simplify using radian formulae from the reference sheet. In (b) read $\theta = 2.3$ from the intersection with $y = \theta$, not the $y$-value, then use $l = r\theta$.

(a) [1 mark]. In right-angled triangle $TOY$, $\tan 60\degree = \dfrac{h}{OY}$, so

(b) [3 marks]. In right-angled triangle $TOF$, the angle of elevation is $45\degree$, so the triangle is isosceles and $OF = h$. Since $Y$ is due east of $O$ and $F$ is south-west of $Y$, the angle $\angle OYF = 45\degree$. Applying the cosine rule in triangle $FOY$ with $OY = \dfrac{h}{\sqrt3}$, $YF = 4$:

Solving this quadratic for $h > 0$ gives $h = 3.03$ km (2 d.p.).

(c) [3 marks]. In triangle $FOY$, use the sine rule to find $\angle OFY$:

giving $\angle OFY \approx 24\degree$. Since $F$ is south-west of $Y$, the direction from $F$ to $Y$ is north-east ($045\degree$), and $O$ lies $24\degree$ west of that, so the bearing of $O$ from $F$ is $045\degree - 24\degree = 021\degree$.

Marker's note. In (a) use the right-angled tangent ratio, not the sine rule. In (b) spot the isosceles triangle $TOF$ so $OF = h$, then apply the cosine rule and solve the quadratic, rejecting the negative root. In (c) find $\angle OFY$ by the sine rule and combine it with the north-east direction to $Y$ to read the bearing.

[3 marks]. Translating right by $k$ replaces $x$ with $x - k$; translating up by $k$ adds $k$:

Substitute $(6, 11)$:

So $k = 6$ or $k = -1$. Since $k$ is positive, $k = 6$.

Marker's note. Translate horizontally by replacing $x$ with $x - k$ and vertically by adding $k$, substitute the point, solve the quadratic, and reject the negative root.

[3 marks]. The graph $y = \cos px$ is $y = \cos x$ dilated horizontally by a factor of $\dfrac1p$. The solutions of $\cos\phi = \tfrac12$ are $\phi = \dfrac{\pi}{3}, \dfrac{5\pi}{3}, \dfrac{7\pi}{3}, \ldots$, that is $px = \dfrac{\pi}{3} + 2\pi n$ and $px = \dfrac{5\pi}{3} + 2\pi n$.

For exactly two solutions in $0 \le x \le 2\pi$, the value $px$ ranges over $0$ to $2\pi p$. The second crossing $x = \dfrac{5\pi}{3p}$ must fall within the domain and the third crossing $x = \dfrac{7\pi}{3p}$ must fall outside it:

Marker's note. Treat $\cos px$ as a horizontal dilation of $\cos x$, list the first crossings of $y = \tfrac12$, and find the range of $p$ for which exactly the second crossing (but not the third) falls in the domain.