Past paper practice · Mathematics Advanced

2025 HSC
worked solutions

A clear method for every question. Find a topic, try it on paper, then open the working when you’re ready.

31 questions50 parts100 marks

Use these independently written solutions alongside the official NESA questions. This is an unofficial study resource, not endorsed by NESA.

Section I Multiple choice

Question 11 mark

Total probability

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Probability

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A probability distribution must add to 1. The missing probability is $1-0.4-0.2=0.4$ .

Answer

0.4

Question 21 mark

Exponential growth

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Functions & graphsExponentials & logarithms

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A base greater than 1 gives an increasing exponential curve. It passes through $(0,1)$ , stays positive and bends upwards. Only graph A has these features.

Answer

Question 31 mark

Domain of a square root

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Functions & graphs

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A real square root requires a non-negative radicand: $6-x^2\geq0$ . Thus $x^2\leq6$ . Both endpoints are allowed because the square root of zero exists.

Answer

[-\sqrt6,\sqrt6]

Question 41 mark

Reading a factored cubic

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Functions & graphs

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Rewrite the expression as $5x(x-2)(x-3)$ . Its zeros are 0, 2 and 3, and its leading coefficient is positive. The curve therefore falls to the left and rises to the right, crossing at all three zeros.

Answer

Question 51 mark

Integrating a fractional power

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Integration

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Write the integrand as $(x+5)^{-1/2}$ . Increase the exponent by 1 and divide by the new exponent. There is no extra chain-rule factor because $d(x+5)/dx=1$ .
$\int (x+5)^{-1/2}\,dx=2\sqrt{x+5}+C.$

Answer

2\sqrt{x+5}+C

Question 61 mark

Reflecting a graph twice

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Functions & graphs

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The transformation maps every point $(x,y)$ to $(-x,-y)$ : a half-turn about the origin. The asymptotes become $x=2$ and $y=-1$ , and the vertical intercept becomes $(0,-3)$ .

Answer

Question 71 mark

Relative frequency

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Probability

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The estimated probability left for the other nine faces is $(153-72)/153=81/153$ . Divide this equally among the nine faces:
$P(10)\approx\frac{81}{153\times9}=\frac1{17}.$

Answer

1/17

Question 81 mark

Normal distribution and standard scores

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StatisticsNormal distribution

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Since $\mu=\sigma$ , zero is one standard deviation below the mean: $z=(0-\mu)/\sigma=-1$ . About 68% lies within one standard deviation, so the area above $-1$ is $50\%+34\%=84\%$ .

Answer

D: approximately 84%.

Question 91 mark

Estimating change from a derivative graph

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DifferentiationFunctions & graphs

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Between $x=1$ and $x=1.1$ , the derivative is positive but falls below 2. Hence the increase in $f$ is positive and less than $2(0.1)=0.2$ .
$0<\int_1^{1.1}f\prime(x)\,dx<0.2\quad\Longrightarrow\quad6<f(1.1)<6.2.$
The tangent estimate 6.2 is an upper estimate here; the decreasing slope puts the actual value just below it.

Answer

[6.0,6.2)

Question 101 mark

Stationary points of a composite function

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DifferentiationFunctions & graphsExponentials & logarithms

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The chain rule gives $\frac{d}{dx}f(e^x)=e^x f\prime(e^x)$ . Because $e^x>0$ , a stationary point occurs precisely when $f\prime(e^x)=0$ .
Only the two stationary points of $f$ with positive input can be reached by $e^x$ . Each positive input corresponds to one real value of $x$ .

Answer

C: 2 stationary points.

Section II Written responses

Question 11(a)2 marks

Finding the vertex

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Functions & graphs

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Complete the square: $h=t^2-8t+12=(t-4)^2-4$ . The square is smallest at $t=4$ , giving $h=-4$ .

Answer

(t,h)=(4,-4)

Question 11(b)1 mark

Equal heights on a parabola

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Functions & graphs

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Set $h=12$ : $t^2-8t+12=12$ , so $t(t-8)=0$ . Excluding the supplied solution $t=0$ leaves the other intersection.

Answer

t=8

Question 123 marks

Equation of a tangent

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Differentiation

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Use powers to differentiate: $y=5x^3-2x^{-2}-9$ , so $y\prime=15x^2+4x^{-3}$ . At $x=1$ , the tangent has gradient 19.
The point-gradient equation is $y+6=19(x-1)$ . Expanding gives the tangent.

Answer

y=19x-25

Question 132 marks

Missing terms in a geometric sequence

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Sequences & series

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Three multiplications by the common ratio take 75 to 2025: $75r^3=2025$ . Thus $r^3=27$ and $r=3$ .
The intermediate terms are $p=75(3)$ and $q=75(3^2)$ .

Answer

p=225,\ q=675

Question 14(a)2 marks

Describing a scatterplot

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StatisticsRegression

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The points form a roughly straight, downward-sloping band. Larger television times tend to be paired with smaller exercise times.

Answer

Approximately linear form; negative direction.

Question 14(b)2 marks

Interpreting regression coefficients

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StatisticsRegression

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The slope −0.7 means that each additional minute of average daily television time is associated with 0.7 fewer predicted minutes of daily exercise.
The intercept 64.3 predicts 64.3 minutes of daily exercise for someone with zero average daily television time. These are predictions from an association, not evidence that television causes the change.

Answer

Slope: −0.7 minutes per minute. Intercept: 64.3 minutes at zero television time.

Question 14(c)1 mark

Making a regression prediction

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StatisticsRegression

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Substitute the television time into the fitted line: $\hat y=64.3-0.7(42)=34.9$ .

Answer

34.9 minutes of exercise per day.

Question 14(d)1 mark

Recognising extrapolation

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StatisticsRegression

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Two hours is 120 minutes, well beyond the observed range (about 0–60 minutes). The observed linear pattern need not continue that far. It would even give $64.3-0.7(120)=-19.7$ minutes, which is impossible.

Answer

The prediction extrapolates far beyond the data and gives an impossible negative duration.

Question 15(a)2 marks

Amplitude and period

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TrigonometryFunctions & graphs

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The amplitude sets $k=2$ . For a sine function the period is $2\pi/a$ , so $2\pi/a=5$ gives $a=2\pi/5$ . Use radians.

Answer

P_1(t)=2\sin(2\pi t/5)

Question 15(b)2 marks

Sketching a sine wave

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TrigonometryFunctions & graphs

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The second wave has amplitude 4 and period $2\pi/(\pi/10)=20$ ms. On the requested interval it traces only the positive half-cycle.
Draw a smooth arch from $(0,0)$ through the maximum $(5,4)$ to $(10,0)$ .

Solution sketch for Question 15(b); key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

A positive half-sine wave with peak

(5,4)

Question 15(c)2 marks

When two functions decrease together

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TrigonometryDifferentiation

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$P_1\prime(t)=(4\pi/5)\cos(2\pi t/5)$ is negative on $(1.25,3.75)$ and $(6.25,8.75)$ .
$P_2\prime(t)=(2\pi/5)\cos(\pi t/10)$ is negative on $(5,10)$ . Intersect the intervals; endpoints with zero gradient are excluded.

Answer

6.25<t<8.75

ms.

Question 16(a)4 marks

Classifying stationary points

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DifferentiationExponentials & logarithms

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Write $f(x)=x^2e^{-x}$ . The product rule gives $f\prime(x)=e^{-x}(2x-x^2)=xe^{-x}(2-x)$ . Since the exponential is positive, the stationary inputs are 0 and 2.
The derivative signs are negative for $x<0$ , positive for $0<x<2$ , and negative for $x>2$ . This makes the first point a minimum and the second a maximum.
$f(0)=0$ and $f(2)=4/e^2\approx0.5413$ .

Answer

Local minimum

(0,0)

; local maximum

(2,4/e^2)

Question 16(b)1 mark

Completing the curve

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Functions & graphsDifferentiation

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Join the left branch smoothly to the minimum at the origin. Rise to $(2,4/e^2)$ , then fall towards the horizontal asymptote $y=0$ from above. The curve touches, rather than crosses, the axis at the origin.

Solution sketch for Question 16(b); key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

A smooth curve through both stationary points, approaching

y=0

x\to\infty

Question 17(a)2 marks

Building a loan recurrence

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Financial mathsSequences & series

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Interest is applied before the repayment, so $A_{n+1}=1.005A_n-5740$ , with $A_0=800000$ .
$A_1=800000(1.005)-5740,$
$A_2=1.005[800000(1.005)-5740]-5740.$

Answer

A_2=800000(1.005)^2-5740(1.005)-5740

Question 17(b)3 marks

Summing the repayments

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Financial mathsSequences & series

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After $n$ months, subtract the accumulated repayments from the accumulated original loan:
$A_n=800000(1.005)^n-5740\sum_{j=0}^{n-1}(1.005)^j.$
The geometric sum is $[(1.005)^n-1]/0.005$ . Since $5740/0.005=1148000$ , collect the coefficients of $(1.005)^n$ .

Answer

A_n=1148000-348000(1.005)^n

Question 17(c)2 marks

Solving for a repayment milestone

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Financial mathsExponentials & logarithms

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Rearrange the strict inequality: $1148000-348000(1.005)^n<400000$ gives $(1.005)^n>748000/348000$ .
$n>\frac{\ln(748000/348000)}{\ln(1.005)}\approx153.422.$
The first whole month after this threshold is 154. The balances after months 153 and 154 are about $401,574.11 and $397,841.98 respectively.

Answer

154 months.

Question 182 marks

Range of a composite function

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Functions & graphs

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Composing gives $g(f(x))=3/(x-1)+5$ , with $x\ne1$ . The reciprocal term cannot equal zero, so the output cannot equal 5.
Every other output $y$ is possible: choose $x=1+3/(y-5)$ .

Answer

( -\infty,5)\cup(5,\infty)

, or

\mathbb R\setminus\{5\}

Question 193 marks

Reversing a conditional probability

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ProbabilityConditional probability

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Let $W$ denote a win. The selection probabilities are $1/6,2/6,3/6$ . Weight each win probability by its selection probability:
$P(W)=\tfrac16(0.5)+\tfrac26(0.4)+\tfrac36(0.2)=\frac{19}{60}.$
The joint probability of selecting Amara and winning is $(1/6)(0.5)=5/60$ . Divide it by the probability of winning: $P(A\mid W)=(5/60)/(19/60)$ .

Answer

5/19\approx0.2632

Question 203 marks

Matching a lump sum with an annuity

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Financial maths

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Use the monthly rate $0.06/12=0.005$ and $7(12)=84$ deposits. The lump sum grows to $21000(1.005)^{84}\approx31927.7624$ .
For end-of-month deposits of $M$ , the supplied table gives future value $104.07393M$ . Equating the two future values gives $M=31927.7624/104.07393\approx306.7796$ .

Answer

$306.78 per month.

Question 21(a)2 marks

Mode of a continuous distribution

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ProbabilityContinuous distributions

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The mode is the input where the density is highest. On $[1,e]$ , $f\prime(x)=-1/x^2<0$ , so the density decreases from its left endpoint. Outside that interval it is zero.

Answer

The mode is

x=1

Question 21(b)3 marks

Finding a percentile by integration

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ProbabilityContinuous distributionsIntegration

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The lower quartile leaves probability 0.25 to its left. Accumulate the density from the lower endpoint:
$\int_1^{Q_1}\frac1x\,dx=\ln Q_1=0.25.$
Exponentiating gives $Q_1=e^{1/4}\approx1.284025$ .

Answer

Q_1=1.284

to three decimal places.

Question 222 marks

Proving a trigonometric identity

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Trigonometry

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Use a common denominator on the left. The numerator becomes a perfect square:
$\frac{\sin^4\theta+\cos^4\theta+2\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta}=\frac{(\sin^2\theta+\cos^2\theta)^2}{\sin^2\theta\cos^2\theta}.$
Now use $\sin^2\theta+\cos^2\theta=1$ and the reciprocal definitions. The identity holds wherever both sides are defined, so $\sin\theta\cos\theta\ne0$ .

Answer

1/(\sin^2\theta\cos^2\theta)=\sec^2\theta\,\operatorname{cosec}^2\theta

Question 23(a)4 marks

Converting a tail count to a weight

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StatisticsNormal distribution

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There are $12600/21=600$ males. The upper-tail proportion is $15/600=0.025$ .
Using the 95% empirical rule, the upper 2.5% begins about two standard deviations above the mean. Thus $x\approx76.2+2(6.8)=89.8$ .
If using an inverse-normal calculator instead, $z_{0.975}\approx1.960$ gives $x\approx89.53$ kg. The empirical-rule estimate is the one used in the official marking guidelines.

Answer

Approximately

89.8

kg using the empirical rule.

Question 23(b)1 mark

Comparing normal upper tails

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StatisticsNormal distribution

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There are 12000 females, so 300 would again be 2.5%. But their mean is lower, making $x-\mu$ larger, and their standard deviation is smaller. Both changes increase the standard score of this same cutoff.
The female upper-tail proportion is therefore smaller than the male upper-tail proportion. The expected count above the cutoff is less than 300.

Answer

No; fewer than 300 females are expected above the cutoff.

Question 244 marks

A tangent shared by two curves

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DifferentiationExponentials & logarithms

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The line has gradient 1. For the logarithmic curve, $y\prime=e/x$ , so tangency requires $e/x=1$ , hence $x=e$ . Its height is $e\ln e=e$ .
At that same point the quadratic has derivative $2ae=1$ , giving $a=1/(2e)$ . Substituting the point into the quadratic gives $e=ae^2+c=e/2+c$ .

Answer

a=1/(2e),\quad c=e/2

Question 25(a)2 marks

Finding an antiderivative by differentiation

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DifferentiationIntegration

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Apply the product rule to $x\cos x$ , remembering the minus sign in front:
$\frac{d}{dx}(\sin x-x\cos x)=\cos x-(\cos x-x\sin x)=x\sin x.$

Answer

The derivative is

x\sin x

, as required.

Question 25(b)2 marks

Evaluating a signed integral

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IntegrationTrigonometry

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Part (a) supplies the antiderivative. At the upper limit, $\sin(2025\pi)=0$ and $\cos(2025\pi)=-1$ , because 2025 is odd.
$\int_0^{2025\pi}x\sin x\,dx=[\sin x-x\cos x]_0^{2025\pi}=2025\pi.$

Answer

2025\pi

Question 25(c)2 marks

Adding areas rather than signed integrals

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IntegrationSequences & series

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Area counts every lobe positively. There are 2025 lobes, so sum the supplied absolute areas:
$A=\pi\sum_{n=1}^{2025}(2n-1)=\pi\frac{2025(1+4049)}2=2025^2\pi.$
This is much larger than the signed integral because the regions below the axis no longer cancel regions above it.

Answer

4100625\pi

square units.

Question 26(a)2 marks

Expressing area in one variable

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OptimisationFunctions & graphs

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The perimeter constraint is $2\pi r+3x=100$ , so $r=(100-3x)/(2\pi)$ . The triangle area is $\tfrac12x^2\sin60^\circ=\sqrt3 x^2/4$ .
Add the circle area $\pi r^2=(100-3x)^2/(4\pi)$ and factor out $1/4$ .

Answer

A(x)=\tfrac14\left(\sqrt3 x^2+\frac{(100-3x)^2}{\pi}\right)

Question 26(b)3 marks

Maximising a quadratic on an interval

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OptimisationFunctions & graphs

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The allowable interval is $0\leq x\leq100/3$ , including the limiting all-circle and all-triangle cases. The coefficient of $x^2$ is $(\sqrt3+9/\pi)/4>0$ , so the quadratic opens upwards. Its maximum on this interval must occur at an endpoint.
$A(0)=\frac{2500}{\pi}\approx795.77,\qquad A(100/3)=\frac{2500\sqrt3}{9}\approx481.13.$
The larger endpoint value is at $x=0$ , leaving all 100 cm for the circle.

Answer

Maximum area

2500/\pi\text{ cm}^2

, when all the wire forms the circle.

Question 27(a)2 marks

Two trapezoids

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IntegrationNumerical methods

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Two equal strips over $[0,2]$ have width 1. The three heights are $1,1/2,1/4$ . Count the middle height twice:
$T=\frac12\left(1+2\cdot\frac12+\frac14\right)=\frac98.$

Answer

9/8=1.125

square units.

Question 27(b)2 marks

Exact exponential area

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IntegrationExponentials & logarithms

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Write $(1/2)^x=e^{-x\ln2}$ . Its antiderivative is $-2^{-x}/\ln2$ .
$A=\left[-\frac{2^{-x}}{\ln2}\right]_0^2=\frac{1-1/4}{\ln2}=\frac3{4\ln2}.$

Answer

3/(4\ln2)

square units.

Question 27(c)2 marks

Turning an area bound into an inequality

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IntegrationExponentials & logarithmsNumerical methods

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The curve is strictly concave up, since $y\prime\prime=(\ln2)^2 2^{-x}>0$ . Each trapezoid lies above it, so $3/(4\ln2)<9/8$ .
As $\ln2>0$ , rearrange to obtain $\ln2>2/3$ , then $\ln(2\sqrt2)=\tfrac32\ln2>1=\ln e$ . The logarithm is increasing, so the ordering also holds for its arguments.

Answer

e<2\sqrt2

Question 28(a)2 marks

Area of a circular segment

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Trigonometry

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The smaller segment is the sector minus its isosceles triangle. In radians its area is $\tfrac12(10)^2(\theta-\sin\theta)$ .
It occupies a quarter of the circle: $50(\theta-\sin\theta)=25\pi$ . Divide by 50 and rearrange.

Answer

\theta=\sin\theta+\pi/2

Question 28(b)2 marks

Estimating an angle graphically

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TrigonometryNumerical methods

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Add the line $y=\theta$ to the supplied graph. Its intersection with $y=\sin\theta+\pi/2$ is near $\theta=2.3$ radians (numerically about 2.310).
Arc length is $s=r\theta$ , so $s\approx10(2.3)=23$ metres. Use the curved boundary, not the straight fence length.

Solution sketch for Question 28(b); key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

Approximately 23 m.

Question 29(a)1 mark

Using an angle of elevation

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Trigonometry

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In the vertical right triangle through $Y$ , $\tan60^\circ=h/OY$ . Since $\tan60^\circ=\sqrt3$ , rearrange for the horizontal distance.

Answer

OY=h/\sqrt3

Question 29(b)3 marks

Combining two triangles

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Trigonometry

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The 45° elevation at $F$ gives $OF=h$ . On the ground, $\angle OYF=45^\circ$ : west and south-west differ by 45°. Apply the cosine rule:
$h^2=\left(\frac h{\sqrt3}\right)^2+4^2-2\left(\frac h{\sqrt3}\right)4\cos45^\circ.$
Simplifying gives $h^2+2\sqrt6h-24=0$ . Its positive root is $h=\sqrt{30}-\sqrt6\approx3.027736$ . Discard the negative root because height is positive.

Answer

h=3.03

km to two decimal places.

Question 29(c)3 marks

Finding a bearing from components

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Trigonometry

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Use east and north as positive directions and place $Y$ at $(0,0)$ . Then $F=(-2\sqrt2,-2\sqrt2)$ and $O=(-h/\sqrt3,0)$ .
From $F$ to $O$ , the eastward component is $2\sqrt2-h/\sqrt3$ and the northward component is $2\sqrt2$ . Both are positive, so the bearing is in the north-east quadrant.
$\beta=\tan^{-1}\!\left(\frac{2\sqrt2-h/\sqrt3}{2\sqrt2}\right)\approx20.905^\circ.$
Use the unrounded height from part (b), and express the final bearing with three digits.

Answer

Bearing

021^\circ

Question 303 marks

Translating a parabola

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Functions & graphs

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A shift right by $k$ replaces $x$ with $x-k$ ; a shift up by $k$ adds $k$ . Thus the new curve is $y=(x-k-1)(x-k-5)+k$ .
Substitute $(6,11)$ : $11=(5-k)(1-k)+k=k^2-5k+5$ . This gives $k^2-5k-6=(k-6)(k+1)=0$ . Only the positive solution satisfies the condition.

Answer

k=6

Question 313 marks

Counting solutions as a parameter changes

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TrigonometryFunctions & graphs

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Set $u=px$ . Because $p>0$ , the input interval becomes $0\leq u\leq2\pi p$ . The first three non-negative solutions of $\cos u=1/2$ are $\pi/3,5\pi/3,7\pi/3$ .
Exactly two solutions means including the second but stopping before the third: $5\pi/3\leq2\pi p<7\pi/3$ . Divide by $2\pi$ .
The lower endpoint is included because it admits the second solution; the upper endpoint is excluded because it would admit a third.

Answer

\frac56\leq p<\frac76

2025 HSCworked solutions

Choose your topics

Section I Multiple choice

Total probability

Exponential growth

Domain of a square root

Reading a factored cubic

Integrating a fractional power

Reflecting a graph twice

Relative frequency

Normal distribution and standard scores

Estimating change from a derivative graph

Stationary points of a composite function

Section II Written responses

Finding the vertex

Equal heights on a parabola

Equation of a tangent

Missing terms in a geometric sequence

Describing a scatterplot

Interpreting regression coefficients

Making a regression prediction

Recognising extrapolation

Amplitude and period

Sketching a sine wave

When two functions decrease together

Classifying stationary points

Completing the curve

Building a loan recurrence

Summing the repayments

Solving for a repayment milestone

Range of a composite function

Reversing a conditional probability

Matching a lump sum with an annuity

Mode of a continuous distribution

Finding a percentile by integration

Proving a trigonometric identity

Converting a tail count to a weight

Comparing normal upper tails

A tangent shared by two curves

Finding an antiderivative by differentiation

Evaluating a signed integral

Adding areas rather than signed integrals

Expressing area in one variable

Maximising a quadratic on an interval

Two trapezoids

Exact exponential area

Turning an area bound into an inequality

Area of a circular segment

Estimating an angle graphically

Using an angle of elevation

Combining two triangles

Finding a bearing from components

Translating a parabola

Counting solutions as a parameter changes

No matching question parts

2025 HSC
worked solutions