Past paper practice · Mathematics Advanced

2026 HSC
worked solutions

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

33 questions52 parts100 marks

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Section I Multiple choice

Question 11 mark

Roots and end behaviour

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Functions & graphs

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The zeros are $-4,1,3$ , so the factors are $(x+4)(x-1)(x-3)$ . The right-hand end rises, requiring a positive leading coefficient.

Answer

(x-1)(x-3)(x+4)

Question 21 mark

Mode of a continuous distribution

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Probability

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The mode is the horizontal coordinate where the density is greatest. The highest point occurs at $x=2\sqrt2$ , rather than at the height $\sqrt2/2$ .

Answer

2\sqrt2

Question 31 mark

Horizontal stretch and reflection

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Functions & graphs

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The domain doubles from $[0,4]$ to $[0,8]$ , giving $f(x/2)$ . The heights change sign, so reflect in the horizontal axis.

Answer

g(x)=-f(x/2)

Question 41 mark

Interpreting correlation

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Statistics

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A positive correlation describes a tendency for both variables to increase together. It does not establish causation or mean that 62% of students follow a particular pattern.

Answer

D: students who study more tend to score higher.

Question 51 mark

Rise above the launch point

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MotionFunctions & graphs

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The starting height is $h(0)=45$ . At the maximum, $h(3)=45+36-18=63$ , so the rise is $63-45$ .

Answer

18\text{ m}

Question 61 mark

The ambiguous sine-rule case

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Trigonometry

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The sine rule gives the whole obtuse angle at G as $180^\circ-68.3^\circ=111.7^\circ$ . The dashed perpendicular already accounts for $55^\circ$ , leaving $111.7^\circ-55^\circ=56.7^\circ$ on the right.

Answer

Question 71 mark

Sum of two independent rolls

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Probability

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The ordered pairs giving a sum of $-3$ are $(-4,1),(1,-4),(-3,0),(0,-3)$ . Multiply within each independent pair, then add.
$P=2\left(\frac3{18}\frac1{18}+\frac4{18}\frac3{18}\right)=\frac{30}{324}=\frac5{54}.$

Answer

5/54

Question 81 mark

Odd functions and signed areas

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IntegrationFunctions & graphs

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Let $A$ be the positive area from 0 to 1 and $B$ the magnitude of the smaller negative area from 1 to a. Then $A>B>0$ . Odd symmetry gives $P=A-B$ , $Q=0$ , $S=-P$ , and $R=-P+A=B$ .
The diagram also shows $A>2B$ , so $R<P$ . Thus $S<Q<R<P$ .

Answer

S,Q,R,P

Question 91 mark

Combining normal tails

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Normal distributionProbability

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By symmetry, $P(Z>1.2)=P(Z<-1.2)$ . Split the latter at $-1.5$ , then add the left tail.
$P(Z<-1.5)+P(Z>1.2)=2P(Z<-1.5)+P(-1.5<Z<-1.2).$

Answer

Question 101 mark

Range of a composite function

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Functions & graphsTrigonometry

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Put $u=\sin x$ , so $-1\le u\le1$ . Complete the square: $u^2-u=(u-1/2)^2-1/4$ . The minimum is $-1/4$ at $u=1/2$ ; the maximum is 2 at $u=-1$ .

Answer

-\tfrac14\le k\le2

Section II Written responses

Question 112 marks

Gradient of a tangent

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Differentiation

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Differentiate to obtain $y\prime=3x^2+3$ , then substitute $x=-1$ .

Answer

m=3(-1)^2+3=6

Question 12(a)2 marks

Interquartile range

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Statistics

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Exclude the middle (seventh) value, 22, when forming the lower and upper halves. Their medians are $Q_1=(15+17)/2=16$ and $Q_3=(27+28)/2=27.5$ .

Answer

\mathrm{IQR}=27.5-16=11.5

Question 12(b)2 marks

Upper outlier fence

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Statistics

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The upper fence is $Q_3+1.5\,\mathrm{IQR}=27.5+1.5(11.5)=44.75$ . The lower fence is $-1.25$ , below the smallest observation. Because m counts cars, it is an integer.

Answer

The greatest possible value is

m=44

Question 13(a)2 marks

Finding an arithmetic sequence

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Sequences & series

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With first term a and common difference d, $a+2d=30$ and $a+5d=42$ . Subtracting gives $3d=12$ , so $d=4$ and $a=30-8$ .

Answer

22 seats.

Question 13(b)2 marks

Summing the seats

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Sequences & series

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Use $S_n=\frac n2[2a+(n-1)d]$ with $a=22,d=4,n=9$ .

Answer

S_9=\tfrac92(44+32)=342

seats.

Question 142 marks

Range and period

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TrigonometryFunctions & graphs

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Since $-1\le\cos(2x)\le1$ , multiplying by 3 and adding 1 gives $-2\le h(x)\le4$ . The angular multiplier 2 halves the usual cosine period.

Answer

Range

[-2,4]

; period

\pi

Question 152 marks

Differentiating an exponential composite

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DifferentiationExponentials & logarithms

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For $a^{u(x)}$ , the derivative is $a^{u(x)}\ln(a)u\prime(x)$ . Here $u=x^2+3$ , so $u\prime=2x$ .

Answer

\displaystyle\frac{dy}{dx}=2x\ln4\;4^{x^2+3}

Question 16(a)1 mark

Reading a scatterplot

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Statistics

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Read the vertical coordinate of the lowest plotted point. It lies just above 6.0, at 6.1.

Answer

6.1.

Question 16(b)2 marks

Interpreting a regression slope

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Statistics

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The slope compares the predicted change in the public’s score with a one-unit increase in the critics’ score.

Answer

For each extra point in the critics’ score, the predicted public score increases by 0.54 points on average.

Question 16(c)1 mark

Adding a point to a regression

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Statistics

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The new point is below the fitted line and to the left of the data centre. It pulls the left end of the fitted line downward, making the positive slope steeper.

Answer

The gradient increases.

Question 17(a)2 marks

Fitting exponential decay

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Exponentials & logarithms

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At $t=0$ , $40=k a^0$ , so $k=40$ . At $t=1$ , $32=40a$ .

Answer

k=40,\quad a=0.8

Question 17(b)1 mark

Interpreting the decay factor

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Exponentials & logarithms

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A factor of 0.8 means that each year the maximum capacity is 80% of its value one year earlier.

Answer

The maximum storage capacity falls by 20% per year.

Question 18(a)2 marks

Intersections of a line and parabola

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Functions & graphs

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Equate the two formulas: $7-3x=-x^2+x+12$ . Rearranging gives $x^2-4x-5=(x-5)(x+1)=0$ .

Answer

x=-1

x=5

Question 18(b)2 marks

Area between two curves

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Integration

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The parabola is above the line between the intersections, so integrate their difference.
$A=\int_{-1}^{5}(-x^2+4x+5)\,dx=\left[-\frac{x^3}{3}+2x^2+5x\right]_{-1}^{5}=36.$

Answer

36 square units.

Question 193 marks

Sketching a derivative

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DifferentiationFunctions & graphs

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The original curve decreases up to $x=2$ , so the derivative starts below the axis. It is zero at the local minimum and positive afterwards.
The slope is greatest at the inflection point $x=5$ , giving a local maximum of $f\prime$ . The slope then decreases towards 0 from above as f approaches its horizontal asymptote. The vertical scale is not specified.

Solution sketch for Question 19; key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

A curve crossing at

x=2

, peaking above the axis at

x=5

, then approaching

y=0

from above.

Question 20(a)1 mark

Multiplying along a probability tree

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Probability

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Follow the walk branch and then the incorrect-uniform branch: $\frac3{10}\times\frac1{10}$ .

Answer

3/100=0.03

Question 20(b)2 marks

At least one success

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Probability

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Using the independent-selection model intended by the tree, the probability a student does not meet both conditions is $1-0.03=0.97$ . Subtract the probability neither student does from 1.

Answer

1-0.97^2=0.0591

Question 21(a)2 marks

Quotient rule with a logarithm

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DifferentiationExponentials & logarithms

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Apply the quotient rule, keeping the entire denominator squared.
$g\prime(x)=\frac{(\ln x)(1)-x(1/x)}{(\ln x)^2}=\frac{\ln x-1}{(\ln x)^2}.$

Answer

\displaystyle g\prime(x)=\frac{\ln x-1}{(\ln x)^2}

Question 21(b)3 marks

Checking endpoints for a global maximum

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Differentiation

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The stationary point is $x=e$ . The derivative changes from negative to positive there, so it is a minimum.
Compare the endpoint values: $g(2)=2/\ln2\approx2.885$ , while $g(5)=5/\ln5\approx3.107$ .

Answer

The global maximum is

5/\ln5\approx3.107

, attained at

x=5

Question 22(a)1 mark

Rewriting a rational function

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Functions & graphs

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Split the numerator as $x+2=(x+1)+1$ , then divide each term by $x+1$ .

Answer

\displaystyle\frac{x+2}{x+1}=1+\frac1{x+1}

, for

x\ne-1

Question 22(b)2 marks

Translating a reciprocal graph

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Functions & graphs

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Replacing x by $x+1$ shifts $y=1/x$ left by one unit. Adding 1 shifts it up by one unit.

Answer

Translate one unit left and one unit up.

Question 234 marks

Bearings, area and the cosine rule

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Trigonometry

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Write $OA=OB=r$ . The area formula gives $80=\tfrac12 r^2\sin45^\circ$ , so $r=\sqrt{160/\sin45^\circ}\approx15.0424\text{ km}$ .
The included angle $BOC=253^\circ-101^\circ=152^\circ$ . Apply the cosine rule: $BC=\sqrt{r^2+25^2-2r(25)\cos152^\circ}\approx38.93\text{ km}$ .

Answer

BC\approx39\text{ km}

, to the nearest kilometre.

Question 24(a)2 marks

Finding a continuous decay constant

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Exponentials & logarithms

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If 75% has decayed, 25% remains: $250=1000e^{10.54k}$ . Taking logarithms gives $k=\ln(0.25)/10.54$ .

Answer

k\approx-0.1315\text{ year}^{-1}

Question 24(b)2 marks

Time from a decay rate

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DifferentiationExponentials & logarithms

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Differentiate to get $A\prime(t)=1000k e^{kt}$ . Set this to $-9.48$ and use the unrounded value $k=\ln(0.25)/10.54$ .
$t=\frac{\ln\!\left(\frac{-9.48}{1000k}\right)}{k}\approx19.996.$

Answer

20 years, to the nearest year.

Question 252 marks

Normal symmetry about the mean

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Normal distributionProbability

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The upper boundary is $5.89-4.82=1.07$ kg above the mean, exactly matching the distance $4.82-3.75=1.07$ kg below it. Symmetry makes the upper tail probability 0.116.
The area from the mean to the upper boundary is therefore $0.5-0.116$ .

Answer

0.384

Question 26(a)1 mark

Direction changes from velocity

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Motion

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Solve $1-e^{\cos(\pi t)}=0$ , giving $\cos(\pi t)=0$ . Only $t=0.5$ lies in the given interval; velocity changes from negative to positive there.

Answer

t=0.5\text{ s}

Question 26(b)2 marks

Acceleration using the chain rule

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MotionDifferentiation

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Differentiating velocity gives $a(t)=\pi e^{\cos(\pi t)}\sin(\pi t)$ . At $t=0.5$ , the exponential factor and sine factor are both 1.

Answer

a(0.5)=\pi\text{ m/s}^2

Question 26(c)2 marks

Trapezoidal estimate of distance

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MotionIntegration

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Distance integrates speed, so use the absolute values of the velocities. The sign change is at a table point, allowing the trapezoidal rule to be applied across both pieces.
$D\approx\frac{0.25}{2}[1.72+2(1.03+0+0.51)+0.63]=0.67875.$

Answer

Approximately

0.679\text{ m}

Question 274 marks

Changing annuity contributions

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Financial mathematicsSequences & series

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The half-yearly rate is $i=0.01$ , and there are 10 end-of-period payments. Let $s_n=((1.01)^n-1)/0.01$ . Paying $750 throughout gives $750s_{10}$ . Doubling the last m payments adds a further $750s_m$ .
Thus $750(10.4622+s_m)\ge14000$ , or $s_m\ge8.20447$ . The table gives $s_8=8.2857$ , so eight doubled payments suffice.
Seven doubled payments give $s_7\approx7.2135$ , which is too small. Therefore at least eight of the ten payments must be $1500.

Answer

A maximum of 2 contributions of $750.

Question 284 marks

Three distinct cubic roots

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DifferentiationFunctions & graphs

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Differentiate: $y\prime=12x^2-6x-6=6(2x+1)(x-1)$ . The stationary points occur at $x=-1/2$ and $x=1$ .
The derivative signs $+,-,+$ show a local maximum at $-1/2$ with height $d+7/4$ , and a local minimum at 1 with height $d-5$ .
For three distinct crossings, the maximum must be strictly positive and the minimum strictly negative: $d+7/4>0$ and $d-5<0$ . Equality would give a repeated root and only two intercepts.

Answer

-\tfrac74<d<5

Question 29(a)2 marks

Sketching the cotangent graph

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TrigonometryFunctions & graphs

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Use $\cot x=\cos x/\sin x$ . It has vertical asymptotes at $x=0,\pi,2\pi$ and zeros at $x=\pi/2,3\pi/2$ .
Each branch decreases from $+\infty$ to $-\infty$ between successive asymptotes. Neither endpoint is in the domain.

Solution sketch for Question 29(a); key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

Two decreasing branches, with period

\pi

Question 29(b)2 marks

Domain of a logarithmic composite

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TrigonometryExponentials & logarithms

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The logarithm needs $\cot x>0$ . Sine and cosine have the same sign in the first and third quadrants. Exclude both zeros of cotangent and its asymptotes.

Answer

\displaystyle x\in(0,\pi/2)\cup(\pi,3\pi/2)

Question 29(c)2 marks

Derivative of log cotangent

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DifferentiationTrigonometry

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Use the chain rule and $(\cot x)\prime=-\csc^2x$ .
$\frac{d}{dx}\ln(\cot x)=\frac{-\csc^2x}{\cot x}=-\frac{1}{\sin x\cos x}=-\csc x\sec x.$

Answer

-\csc x\sec x

, wherever

\ln(\cot x)

is defined.

Question 30(a)3 marks

Pyramid volume in terms of an angle

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Trigonometry

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In the right triangle AOM, the height is $MO=\ell\sin\theta$ and the half-diagonal of the square is $AO=\ell\cos\theta$ .
The square’s full diagonal is $2\ell\cos\theta$ , so its area is half the diagonal squared: $2\ell^2\cos^2\theta$ . Multiply base area by height and divide by 3.

Answer

\displaystyle V=\frac{2\ell^3}{3}\sin\theta\cos^2\theta

Question 30(b)4 marks

Optimising the pyramid volume

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DifferentiationTrigonometry

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For fixed $\ell>0$ , differentiate the angle-dependent factor: $V\prime=\frac{2\ell^3}{3}\cos\theta(1-3\sin^2\theta)$ .
For a non-degenerate pyramid, $0<\theta<\pi/2$ , so the interior stationary point satisfies $\sin\theta=1/\sqrt3$ . The derivative is positive before this angle and negative after it, giving the maximum.

Answer

\theta=\arcsin(1/\sqrt3)\approx0.62\text{ radians}

Question 31(a)1 mark

Integrating a cosine

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IntegrationTrigonometry

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The derivative of $\sin(2x)$ is $2\cos(2x)$ , so divide by 2.

Answer

\displaystyle\int\cos(2x)\,dx=\tfrac12\sin(2x)+C

Question 31(b)3 marks

Recovering a positive function

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IntegrationExponentials & logarithms

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Since f is positive, $\int f\prime(x)/f(x)\,dx=\ln f(x)$ . Hence $\ln f(x)=\tfrac12\sin(2x)+C$ .
Use $f(0)=1$ to obtain $C=0$ , then exponentiate.

Answer

f(x)=e^{\frac12\sin(2x)}

Question 32(a)1 mark

Definite integral with a variable bound

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Integration

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Apply the fundamental theorem using $F\prime(t)=\sqrt{\tan t}$ . Evaluate F at the upper bound and subtract its value at the lower bound.

Answer

g(x)=F(1)-F(x^3)

Question 32(b)2 marks

Differentiating a variable lower bound

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DifferentiationIntegration

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The term $F(1)$ is constant. Differentiate $-F(x^3)$ by the chain rule, then substitute $F\prime(x^3)=\sqrt{\tan(x^3)}$ .

Answer

g\prime(x)=-3x^2\sqrt{\tan(x^3)}

Question 33(a)2 marks

Loan balance after three repayments

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Financial mathematicsSequences & series

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Write $r=1.005$ for monthly interest and $q=1.1$ for payment growth. The first three balances are $A_1=83000r-760$ , $A_2=83000r^2-760r-760q$ , and $A_3=83000r^3-760r^2-760qr-760q^2$ .

Answer

A_3=83000(1.005)^3-760[(1.005)^2+(1.005)(1.1)+(1.1)^2]

Question 33(b)3 marks

Summing increasing loan repayments

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Financial mathematicsSequences & series

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After n repayments, $A_n=83000r^n-760\sum_{j=0}^{n-1}r^{n-1-j}q^j$ , where $r=1.005$ and $q=1.1$ .
The geometric sum equals $(q^n-r^n)/(q-r)$ . Since $760/(1.1-1.005)=8000$ , simplify to $83000r^n-8000(q^n-r^n)$ .

Answer

A_n=91000(1.005)^n-8000(1.1)^n

Question 33(c)3 marks

The final loan repayment

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Financial mathematicsSequences & series

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The scheduled formula first reaches a non-positive balance when $n\ge\ln(91000/8000)/\ln(1.1/1.005)\approx26.919$ . Thus 26 full repayments are made and the final payment is in month 27.
The amount owing after repayment 26 is $A_{26}=91000(1.005)^{26}-8000(1.1)^{26}\approx8254.41$ . Add one month’s interest rather than paying the full scheduled 27th instalment: $1.005A_{26}\approx8295.68$ .

Answer

The final payment is $8295.68.

2026 HSCworked solutions

Choose your topics

Section I Multiple choice

Roots and end behaviour

Mode of a continuous distribution

Horizontal stretch and reflection

Interpreting correlation

Rise above the launch point

The ambiguous sine-rule case

Sum of two independent rolls

Odd functions and signed areas

Combining normal tails

Range of a composite function

Section II Written responses

Gradient of a tangent

Interquartile range

Upper outlier fence

Finding an arithmetic sequence

Summing the seats

Range and period

Differentiating an exponential composite

Reading a scatterplot

Interpreting a regression slope

Adding a point to a regression

Fitting exponential decay

Interpreting the decay factor

Intersections of a line and parabola

Area between two curves

Sketching a derivative

Multiplying along a probability tree

At least one success

Quotient rule with a logarithm

Checking endpoints for a global maximum

Rewriting a rational function

Translating a reciprocal graph

Bearings, area and the cosine rule

Finding a continuous decay constant

Time from a decay rate

Normal symmetry about the mean

Direction changes from velocity

Acceleration using the chain rule

Trapezoidal estimate of distance

Changing annuity contributions

Three distinct cubic roots

Sketching the cotangent graph

Domain of a logarithmic composite

Derivative of log cotangent

Pyramid volume in terms of an angle

Optimising the pyramid volume

Integrating a cosine

Recovering a positive function

Definite integral with a variable bound

Differentiating a variable lower bound

Loan balance after three repayments

Summing increasing loan repayments

The final loan repayment

No matching question parts

2026 HSC
worked solutions