Past paper practice · Mathematics Extension 2

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.

16 questions48 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

Conjugate roots

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Complex numbersPolynomials

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Real coefficients make $3-i$ another root. The roots sum to 10, so the remaining root is $10-(3+i)-(3-i)=4$ .

Answer

a=3-i,\ b=4

Question 21 mark

Vector magnitude

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Vectors

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The length is $\sqrt{(-2)^2+3^2+(-6)^2}=\sqrt{49}$ .

Answer

7

Question 31 mark

Equation of a sphere

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Vectors

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The centre-radius equation is $(x+1)^2+(y-3)^2+z^2=81$ . Expanding gives $x^2+2x+y^2-6y+z^2=71$ .

Answer

Question 41 mark

Intersecting complex regions

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Complex numbers

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The modulus inequality is the closed disc centred at $(2,3)$ with radius 1. The second condition is $y\le x$ , so retain the small cap below the line $y=x$ , including its boundary.

Answer

Question 51 mark

Argument of a quotient

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Complex numbers

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The angle difference is the argument of $3+4i$ , modulo $2\pi$ . Its modulus is 5, so its sine is imaginary part divided by modulus.

Answer

\sin\theta=\frac45

Question 61 mark

Negating quantifiers

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Proof

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Change “for every” to “there exists”, change “there exists” to “for every”, and negate the equality. Keep the stated domains of the variables.

Answer

\exists p>0\text{ such that }\forall q>p,\ q^2\ne p

Question 71 mark

Partial fractions and primitives

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Integration

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The decomposition is $\frac1{7(x+1)}-\frac1{7(x-6)}+\frac1{(x-6)^2}$ . Its primitive contains two logarithms and a term proportional to $(x-6)^{-1}$ .

Answer

C; a primitive is

\frac17\ln|x+1|-\frac17\ln|x-6|-\frac1{x-6}+K

Question 81 mark

Complex triangle area

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Complex numbers

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At the vertex $2z-iz$ , the two side vectors are $iz$ and $-2z$ . They are perpendicular, with lengths $|z|$ and $2|z|$ , so the area is half their product.

Answer

|z|^2

Question 91 mark

Comparing vector lines

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Vectors

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The second direction vector is $-2$ times the first. Also $(3,2,-2)=(2,3,-4)+(1,-1,2)$ , so the second base point lies on the first line. The lines coincide, hence intersect and are parallel.

Answer

Question 101 mark

Counterexample to an inflection claim

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ProofCalculus

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For option C, $f^{\prime\prime}(x)=20x^3+12x^2=4x^2(5x+3)$ . This is zero at 0 but positive on both sides sufficiently close to 0, so concavity does not change there.

Answer

f(x)=x^5+x^4

Section II Written responses

Question 11(a)1 mark

Writing a contrapositive

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Proof

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The contrapositive of “if P then Q” is “if not Q then not P”. For integers, not even means odd.

Answer

If a is odd, then a cubed is odd.

Question 11(b)(i)1 mark

Adding complex numbers

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Complex numbers

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Collect real and imaginary parts: $3(3+2i)+(-1-5i)=9-1+(6-5)i$ .

Answer

8+i

Question 11(b)(ii)1 mark

Conjugate multiplication

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Complex numbers

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Use $\bar z=3-2i$ : $2\bar zw=2(3-2i)(-1-5i)=2(-13-13i)$ .

Answer

-26-26i

Question 11(c)2 marks

Angle between unit vectors

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Vectors

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Both position vectors have length 1. Their dot product gives $\cos\theta=1/\sqrt{10}$ . Hence $\tan^2\theta=(1-\cos^2\theta)/\cos^2\theta$ .

Answer

\tan^2\theta=9

Question 11(d)2 marks

Square roots of a complex number

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Complex numbers

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Put $(x+iy)^2=11+60i$ . Then $x^2-y^2=11$ , $xy=30$ , and $x^2+y^2=|11+60i|=61$ . Thus $x^2=36$ , $y^2=25$ ; the signs agree because $xy>0$ .

Answer

6+5i

and

-6-5i

Question 11(e)2 marks

Product with the conjugate

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Complex numbersProof

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If $z=re^{i\theta}$ , then $\bar z=re^{-i\theta}$ and $|z|=r$ . Multiplying gives $z\bar z=r^2e^{i\theta-i\theta}=r^2$ .

Answer

z\bar z=|z|^2

Question 11(f)3 marks

Completing the square in an integral

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Integration

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Write $6x-x^2=9-(x-3)^2$ . A primitive is $\sin^{-1}((x-3)/3)$ .
At the limits $x=3$ and $x=3/2$ , the inverse sine values are 0 and $-\pi/6$ .

Answer

\frac\pi6

Question 11(g)(i)1 mark

Modulus of a power

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Complex numbers

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$|z|=a\sqrt2$ since $a>0$ . The modulus of the fifth power is the fifth power of the modulus.

Answer

|z^5|=4\sqrt2\,a^5

Question 11(g)(ii)2 marks

Principal argument of a power

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Complex numbers

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$\arg z=\pi/4$ , so an argument of $z^5$ is $5\pi/4$ . Subtract $2\pi$ to enter the principal range $(-\pi,\pi]$ .

Answer

\operatorname{Arg}(z^5)=-\frac{3\pi}4

Question 12(a)(i)3 marks

Exponential form of a trigonometric identity

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Complex numbersProof

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Expand the fourth power. Since $(2i)^4=16$ , the left side is $\frac1{16}(e^{4i\theta}-4e^{2i\theta}+6-4e^{-2i\theta}+e^{-4i\theta})$ .
Pair conjugate exponentials using the supplied identity: $e^{4i\theta}+e^{-4i\theta}=2\cos4\theta$ and $e^{2i\theta}+e^{-2i\theta}=2\cos2\theta$ .

Answer

\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^4=\frac18(\cos4\theta-4\cos2\theta+3)

Question 12(a)(ii)2 marks

Integrating a fourth power of sine

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IntegrationTrigonometry

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Euler’s sine identity identifies the fourth power in part (i) as $\sin^4\theta$ . Integrate the three terms separately, dividing by each inner frequency.

Answer

\int\sin^4\theta\,d\theta=\frac{3\theta}8-\frac{\sin2\theta}4+\frac{\sin4\theta}{32}+C

Question 12(b)2 marks

Constructing a real polynomial

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PolynomialsComplex numbers

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Include the conjugate zeroes $1-6i$ and $3+4i$ . The conjugate pairs give $(x-1)^2+36$ and $(x-3)^2+16$ .

Answer

One example is

P(x)=(x-2)(x^2-2x+37)(x^2-6x+25)

Question 12(c)2 marks

Applying the arithmetic-geometric mean inequality

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ProofInequalities

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Apply the given inequality to each factor: $a+7b\ge2\sqrt{7ab}$ , $b+2c\ge2\sqrt{2bc}$ , and $c+14a\ge2\sqrt{14ac}$ .
All factors are positive, so multiplication preserves the inequalities. Their product is at least $8\sqrt{196a^2b^2c^2}=112abc$ .

Answer

(a+7b)(b+2c)(c+14a)\ge112abc

Question 12(d)3 marks

Recovering displacement from acceleration

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MechanicsSimple harmonic motion

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The graph has period 4 and acceleration amplitude $\pi^2$ , starting at its maximum. Thus $\ddot x=\pi^2\cos(\pi t/2)$ .
Integrating gives $\dot x=2\pi\sin(\pi t/2)+C_1$ . Initial rest gives $C_1=0$ ; integrating again gives $x=-4\cos(\pi t/2)+C_2$ . Use $x(0)=-2$ to obtain $C_2=2$ .

Answer

x(t)=2-4\cos\left(\frac{\pi t}{2}\right)

Question 12(e)3 marks

Exponential substitution

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Integration

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Put $u=e^x$ , so $du=e^x\,dx$ . The integral becomes $\int(1+u)/(1+u^2)\,du$ .
Split it into $\int du/(1+u^2)+\int u\,du/(1+u^2)$ , then substitute back.

Answer

\tan^{-1}(e^x)+\frac12\ln(1+e^{2x})+C

Question 13(a)2 marks

Difference of squares pattern

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Proof

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Every positive multiple of 4 is $4n$ for a positive integer $n$ . The displayed pattern suggests consecutive terms differing by 2: $(n+1)^2-(n-1)^2=4n$ .
Both bases are non-negative integers, so this supplies a difference of integer squares for every such multiple.

Answer

4n=(n+1)^2-(n-1)^2

, for every integer

n\ge1

Question 13(b)2 marks

Distance in one oscillation

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Simple harmonic motionMechanics

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Complete the square: $v^2=16-2(x-2)^2$ . The turning points occur at $v=0$ , giving $x=2\pm2\sqrt2$ , so the amplitude is $2\sqrt2$ .
One full period traverses four amplitudes, regardless of the starting point.

Answer

8\sqrt2\text{ m}

Question 13(c)3 marks

Tangent half-angle substitution

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IntegrationTrigonometry

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Let $u=\tan(\theta/2)$ . Then $\cos\theta=(1-u^2)/(1+u^2)$ , $\sin\theta=2u/(1+u^2)$ , and $d\theta=2\,du/(1+u^2)$ .
The integral simplifies to $\int du/(u^2-2u+4)=\int du/((u-1)^2+3)$ .

Answer

\frac1{\sqrt3}\tan^{-1}\left(\frac{\tan(\theta/2)-1}{\sqrt3}\right)+C

, on each interval where the substitution is continuous.

Question 13(d)3 marks

Vertical motion with quadratic resistance

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MechanicsDifferential equations

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During descent, measure distance $s$ downwards from the highest point and speed $v\ge0$ . Newton’s law gives $v\,dv/ds=10-0.1v^2$ .
Let $y=v^2$ . Then $dy/ds+0.2y=20$ , with $y(0)=0$ , so $v^2=100(1-e^{-s/5})$ .
At the ground, $s=5\ln(676/100)$ , hence $v^2=100(1-100/676)=14400/169$ .

Answer

v=\frac{120}{13}\approx9.23\text{ m s}^{-1}

Question 13(e)(i)2 marks

Irrationality by contradiction

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Suppose $\sqrt6=p/q$ in lowest positive integer terms. Then $p^2=6q^2$ , so $p$ is even; write $p=2r$ .
Then $2r^2=3q^2$ , which makes $q$ even too. This contradicts $p$ and $q$ having no common factor.

Answer

\sqrt6

is irrational.

Question 13(e)(ii)2 marks

Using a known irrational number

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Suppose $r=\sqrt3-\sqrt2$ is rational. Squaring gives $r^2=5-2\sqrt6$ , hence $\sqrt6=(5-r^2)/2$ would be rational.
This contradicts part (i).

Answer

\sqrt3-\sqrt2

is irrational.

Question 14(a)4 marks

Power failure followed by projectile motion

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VectorsMechanics

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The squared distance to the transmitter is $(3-t/4)^2+2^2+(t/4)^2$ . Setting it to 9 gives $t^2-12t+32=0$ , so the helicopter first enters the 3 m boundary at $t=4$ s.
At that instant its height is 52 m and its upward velocity is $1/4\text{ m s}^{-1}$ . If $\tau$ is time after power loss, height is $52+\tau/4-5\tau^2$ .
Setting the height to zero gives $20\tau^2-\tau-208=0$ . The positive root is $(1+129)/40$ .

Answer

\tau=\frac{13}{4}=3.25\text{ s}

Question 14(b)3 marks

Volume by discs and integration by parts

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IntegrationVolumes

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Squaring $y=x\sqrt{e^x}$ gives $y^2=x^2e^x$ . The disc method gives $V=\pi\int_0^3x^2e^x\,dx$ .
Integrating by parts twice gives the primitive $e^x(x^2-2x+2)$ . Evaluating at 3 and 0 yields $5e^3-2$ .

Answer

V=\pi(5e^3-2)

cubic units.

Question 14(c)(i)1 mark

Perpendicular vector condition

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Vectors

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With direction vector $\mathbf d=(6,2,-3)$ , use $\overrightarrow{BA}\cdot\mathbf d=0$ : $(4-3b,5-4b,2-4b)\cdot(6,2,-3)=28-14b=0$ .

Answer

b=2

Question 14(c)(ii)2 marks

Two points making a given angle

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Vectors

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Now $B=(6,8,8)$ , so $|AB|=\sqrt{4+9+36}=7$ . Triangle ABC is right-angled at A. With angle C equal to $\pi/4$ , it has $|AC|=|AB|=7$ .
The line direction $(6,2,-3)$ also has length 7. Thus the two choices are $C=A\pm(6,2,-3)$ .

Answer

\overrightarrow{OC}=(10,7,-1)

(-2,3,5)

Question 14(d)(i)2 marks

Separating a velocity equation

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MechanicsDifferential equations

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The equation of motion used here is $dv/dt=-(k+v^2)$ . Separating and integrating from initial velocity u to velocity v gives $t=-\int_u^v ds/(k+s^2)$ .
Use $\int ds/(k+s^2)=k^{-1/2}\tan^{-1}(s/\sqrt k)$ , since $k>0$ .

Answer

t=\frac1{\sqrt k}\left[\tan^{-1}\left(\frac u{\sqrt k}\right)-\tan^{-1}\left(\frac v{\sqrt k}\right)\right]

Question 14(d)(ii)3 marks

Testing the direction of motion

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MechanicsInequalities

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Substitute the given time in part (i): $\tan^{-1}(v/\sqrt k)=\tan^{-1}(u/\sqrt k)-\pi/4$ .
Since $u>0$ , the right side lies between $-\pi/4$ and $\pi/4$ . Therefore $v<0$ exactly when $\tan^{-1}(u/\sqrt k)<\pi/4$ . Monotonicity of inverse tangent gives $u<\sqrt k$ .

Answer

0<u<\sqrt k

, using the relation in part (i), with force in the fixed negative direction.

Question 15(a)4 marks

Induction for a reciprocal sum

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ProofInduction

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Let $S_n=\sum_{r=n+1}^{2n}1/r$ . For $n=2$ , $S_2=1/3+1/4=14/24>13/24$ .
Assume $S_k>13/24$ for an integer $k\ge2$ . Comparing consecutive sums, $S_{k+1}-S_k=\frac1{2k+1}+\frac1{2k+2}-\frac1{k+1}=\frac1{(2k+1)(2k+2)}>0$ .
Thus $S_{k+1}>S_k>13/24$ , completing the induction.

Answer

\displaystyle\sum_{r=n+1}^{2n}\frac1r>\frac{13}{24}

for all integers

n>1

Question 15(b)4 marks

Reflecting a point in a line

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Vectors

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The diagram shows $PM$ perpendicular to the line. Write $M=A+\lambda(B-A)=(1+\lambda,-1+\lambda,2+\lambda)$ , using direction $(1,1,1)$ .
Perpendicularity gives $(P-M)\cdot(1,1,1)=0$ , so $(3-\lambda)-\lambda+(3-\lambda)=0$ . Hence $\lambda=2$ and $M=(3,1,4)$ .
Because M is the midpoint, $Q=2M-P$ .

Answer

Q=(2,3,3)

Question 15(c)3 marks

Length of a complex spiral

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Complex numbersSequences & series

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The first step runs from $z_1=8$ to $z_2=6i$ , so its length is $\sqrt{8^2+6^2}=10$ .
Each subsequent difference is multiplied by $3i/4$ , so its length is multiplied by $3/4$ . The total is the convergent geometric sum $10/(1-3/4)$ .

Answer

40

Question 15(d)(i)2 marks

A point on a translated unit circle

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Complex numbers

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The radius is 1, so $z_1-w=\cos(-2\pi/3)+i\sin(-2\pi/3)=-1/2-i\sqrt3/2$ . Add $w=2+i$ .

Answer

z_1=\frac32+i\left(1-\frac{\sqrt3}{2}\right)

Question 15(d)(ii)2 marks

Tangency from the origin

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Complex numbersVectors

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Write $z_2=x+iy$ . The circle gives $(x-2)^2+(y-1)^2=1$ . A radius is perpendicular to its tangent, so $(x,y)\cdot(x-2,y-1)=0$ .
Subtracting these equations gives $2x+y=4$ . Substitution produces $(x-2)(5x-6)=0$ . The point shown above and to the left of the centre is $(6/5,8/5)$ .

Answer

For the point shown,

z_2=\frac65+\frac85i

. The other tangent point is

2+0i

Question 16(a)4 marks

Unit-circle roots from an imaginary part

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Complex numbersTrigonometry

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Put $z=e^{i\theta}$ , with $0\le\theta<2\pi$ . The imaginary-part condition becomes $\sin3\theta+\sin2\theta+\sin\theta=0$ .
Using $\sin3\theta+\sin\theta=2\sin2\theta\cos\theta$ , factor as $\sin2\theta(2\cos\theta+1)=0$ .
The first factor gives $\theta=0,\pi/2,\pi,3\pi/2$ ; the second gives $\theta=2\pi/3,4\pi/3$ .

Answer

z\in\left\{1,-1,i,-i,\frac{-1+i\sqrt3}{2},\frac{-1-i\sqrt3}{2}\right\}

Question 16(b)(i)1 mark

Subtracting two related integrals

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IntegrationProof

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Combine the integrands over the common denominator: $\frac1{(1+x^2)^n}-\frac1{(1+x^2)^{n-1}}=\frac{1-(1+x^2)}{(1+x^2)^n}$ . Integrating both sides from 0 to 1 gives the required relation.

Answer

I_n-I_{n-1}=\displaystyle\int_0^1\frac{-x^2}{(1+x^2)^n}\,dx

Question 16(b)(ii)3 marks

Evaluating a related integral

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Integration

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First, $I_1=[\tan^{-1}x]_0^1=\pi/4$ . To use part (i), differentiate $x/(1+x^2)$ .
Since $\frac{d}{dx}\frac{x}{1+x^2}=\frac{1-x^2}{(1+x^2)^2}$ , its integral is $I_2+\int_0^1\frac{-x^2}{(1+x^2)^2}\,dx=I_2+(I_2-I_1)$ , using part (i).
The boundary value is $[x/(1+x^2)]_0^1=1/2$ , so $2I_2-I_1=1/2$ .

Answer

I_2=\frac14+\frac\pi8

Question 16(c)(i)2 marks

Magnitude of a parametrised circle

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VectorsProof

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Expand the squared norm using the dot product: $|\mathbf v|^2=\cos^2\theta|\mathbf a|^2+\sin^2\theta|\mathbf b|^2+2\sin\theta\cos\theta(\mathbf a\cdot\mathbf b)$ .
A and B lie on the unit sphere, so their position vectors have length 1; they are perpendicular, so the cross term vanishes.

Answer

|\mathbf v(\theta)|^2=\cos^2\theta+\sin^2\theta=1

Question 16(c)(ii)2 marks

Intersection of two great circles

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Vectors

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Every point of $C_1$ satisfies $y=2x$ , while every point of $C_2$ satisfies $z=y$ . Thus an intersection has coordinates $(x,2x,2x)$ .
Both circles lie on the unit sphere, so $x^2+4x^2+4x^2=1$ , giving $x=\pm1/3$ . Each of these unit vectors lies in both circle planes.

Answer

\left(\frac13,\frac23,\frac23\right)

and

\left(-\frac13,-\frac23,-\frac23\right)

Question 16(c)(iii)4 marks

Constructing an orthonormal circle basis

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Vectors

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Take $\mathbf a=(0,1,0)$ . The projection of $\overrightarrow{OP}$ onto this unit vector is $(0,1/\sqrt6,0)$ . Subtract it to obtain $\mathbf r=(\sqrt3/2,0,-1/(2\sqrt3))$ .
This remainder has length $\sqrt{5/6}$ . Normalising gives $\mathbf b=(3/\sqrt{10},0,-1/\sqrt{10})$ , perpendicular to $\mathbf a$ .
The vectors span the plane through O, A and P. Both are unit vectors, so the supplied circle formula applies; P occurs when $\cos\theta=1/\sqrt6$ and $\sin\theta=\sqrt{5/6}$ .

Answer

\mathbf v(\theta)=\cos\theta\,(0,1,0)+\frac{\sin\theta}{\sqrt{10}}(3,0,-1),\quad0\le\theta<2\pi

2026 HSCworked solutions

Choose your topics

Section I Multiple choice

Conjugate roots

Vector magnitude

Equation of a sphere

Intersecting complex regions

Argument of a quotient

Negating quantifiers

Partial fractions and primitives

Complex triangle area

Comparing vector lines

Counterexample to an inflection claim

Section II Written responses

Writing a contrapositive

Adding complex numbers

Conjugate multiplication

Angle between unit vectors

Square roots of a complex number

Product with the conjugate

Completing the square in an integral

Modulus of a power

Principal argument of a power

Exponential form of a trigonometric identity

Integrating a fourth power of sine

Constructing a real polynomial

Applying the arithmetic-geometric mean inequality

Recovering displacement from acceleration

Exponential substitution

Difference of squares pattern

Distance in one oscillation

Tangent half-angle substitution

Vertical motion with quadratic resistance

Irrationality by contradiction

Using a known irrational number

Power failure followed by projectile motion

Volume by discs and integration by parts

Perpendicular vector condition

Two points making a given angle

Separating a velocity equation

Testing the direction of motion

Induction for a reciprocal sum

Reflecting a point in a line

Length of a complex spiral

A point on a translated unit circle

Tangency from the origin

Unit-circle roots from an imaginary part

Subtracting two related integrals

Evaluating a related integral

Magnitude of a parametrised circle

Intersection of two great circles

Constructing an orthonormal circle basis

No matching question parts

2026 HSC
worked solutions