Past paper practice · Mathematics Extension 1

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.

14 questions34 parts70 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

Solving an absolute-value inequality

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Functions & graphs

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An absolute value less than 5 means $-5<2x+3<5$ . Subtract 3 throughout, then divide by 2.

Answer

-4<x<1

Question 21 mark

Projecting one vector onto another

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Vectors

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The scalar product is $\mathbf u\cdot\mathbf v=1(2)+2(-3)=-4$ , while $|\mathbf v|^2=2^2+(-3)^2=13$ . Multiply $\mathbf v$ by their ratio.

Answer

-\dfrac4{13}(2\mathbf i-3\mathbf j)

Question 31 mark

Changing variables in an integral

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IntegrationTrigonometry

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With $x=5\sin\theta$ , we have $dx=5\cos\theta\,d\theta$ and $25-x^2=25\cos^2\theta$ . The endpoints become $-\pi/6$ and $\pi/6$ .
Thus $\dfrac{dx}{25-x^2}=\dfrac15\sec\theta\,d\theta$ ; the cosine is positive throughout these limits.

Answer

\dfrac15\displaystyle\int_{-\pi/6}^{\pi/6}\sec\theta\,d\theta

Question 41 mark

Bernoulli mean and variance

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ProbabilityBinomial distribution

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The success probability is $p=P(X=1)=2/3$ . A Bernoulli variable has mean $p$ and variance $p(1-p)$ .

Answer

E(X)=\dfrac23,\quad\operatorname{Var}(X)=\dfrac29

Question 51 mark

Counting trigonometric solutions

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Trigonometry

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Write $\sin x=\cos(\pi/2-x)$ , then use the sum-to-product identity:
$\cos5x+\sin x=2\cos(2x+\pi/4)\cos(3x-\pi/4).$
The first factor gives $x=\pi/8+k\pi/2$ , with $k=0,1,2,3$ . The second gives $x=\pi/4+k\pi/3$ , with $k=0,1,\ldots,5$ . None coincide, so there are $4+6$ distinct solutions.

Answer

D: 10 solutions.

Question 61 mark

Using odd symmetry in integration

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IntegrationInverse trigonometry

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The function $\tan^{-1}x$ is odd and $x^2$ is even, so their product is odd. An odd integrand has zero integral between opposite limits.

Answer

\displaystyle\int_{-a}^{a}x^2\tan^{-1}x\,dx=0

Question 71 mark

Reading a slope field

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Differential equations

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The slopes depend on the horizontal position only. They are zero on the $y$ -axis, positive on both sides, and steeper as $|x|$ grows. These are the features of $x^2$ .

Answer

\dfrac{dy}{dx}=x^2

Question 81 mark

Ordering collinear points

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Vectors

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Rearrange $\mathbf c=3\mathbf a-2\mathbf b$ to obtain $\mathbf a=(2/3)\mathbf b+(1/3)\mathbf c$ . Both weights are positive and add to 1, so A is inside the segment BC.

Answer

A: A lies between B and C.

Question 91 mark

Finding the angle between vectors

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VectorsTrigonometry

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Since $\mathbf a+\mathbf b=-\mathbf c$ , squaring the magnitudes gives $49=9+25+2(3)(5)\cos\theta$ . Hence $\cos\theta=1/2$ .

Answer

\theta=\pi/3

Question 101 mark

Second derivative of an inverse

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DifferentiationInverse functions

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From $f(g(x))=x$ , differentiate twice: $f'(g)g'=1$ and $f''(g)(g')^2+f'(g)g''=0$ .
At $x=1$ , $g(1)=3$ and $g'(1)=1/2$ . Therefore $4(1/2)^2+2g''(1)=0$ .

Answer

g''(1)=-\dfrac12

Section II Written responses

Question 11(a)2 marks

Finding a rational inverse

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Inverse functionsAlgebra

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Set $y=1-1/(x-2)$ . Then $1-y=1/(x-2)$ , so $x=2+1/(1-y)$ . Exchange the input and output.

Answer

f^{-1}(x)=2-\dfrac1{x-1},\quad x\ne1

Question 11(b)3 marks

Factoring a trigonometric equation

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Trigonometry

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Use $\sin2\theta=2\sin\theta\cos\theta$ to obtain $\sin\theta(2\cos\theta-1)=0$ .
The sine factor gives the endpoints 0 and $\pi$ ; the other factor gives $\cos\theta=1/2$ , hence $\theta=\pi/3$ in the required interval.

Answer

\theta=0,\ \pi/3,\ \pi

Question 11(c)2 marks

Integrating a trigonometric product

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IntegrationTrigonometry

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The product-to-sum identity gives $\sin3x\cos x=\tfrac12(\sin4x+\sin2x)$ . Integrate each sine term, including its chain-rule divisor.

Answer

\displaystyle-\frac18\cos4x-\frac14\cos2x+C

Question 11(d)2 marks

Sketching a scaled inverse cosine

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Functions & graphsInverse trigonometry

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The input of inverse cosine must satisfy $-1\le2x\le1$ , so the domain is $[-1/2,1/2]$ . Scaling its output by $1/3$ gives the range $[0,\pi/3]$ .
Draw a smooth decreasing curve through $(-1/2,\pi/3)$ , $(0,\pi/6)$ and $(1/2,0)$ . Include both endpoints; the tangents become vertical there.

Solution sketch for Question 11(d); key values and curve behaviour are described in the steps above. — Original solution sketch; see the steps for key values.

Answer

Domain

[-1/2,1/2]

; range

[0,\pi/3]

Question 11(e)1 mark

Making vectors parallel

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Vectors

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The vector $(2,6)$ is twice $(1,3)$ . The same scale factor must apply to both components.

Answer

m=3

Question 11(f)2 marks

Using sums and products of roots

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PolynomialsAlgebra

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Put the fractions over a common denominator: $1/(\alpha\beta)+1/(\alpha\gamma)+1/(\beta\gamma)=(\alpha+\beta+\gamma)/(\alpha\beta\gamma)$ .
Vieta’s formulas give the sum $-6/2=-3$ and product $-(-1)/2=1/2$ .

Answer

-6

Question 11(g)3 marks

Integrating a squared cosine

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IntegrationTrigonometry

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Reduce the power using $\cos^2(3x)=(1+\cos6x)/2$ . An antiderivative is $x/2+\sin6x/12$ .
The sine terms vanish at both $x=\pi/6$ and $x=\pi/3$ . The remaining difference is $\pi/6-\pi/12$ .

Answer

\pi/12

Question 12(a)3 marks

Related rates in a sector

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Rates of changeDifferentiation

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The fixed area gives $\tfrac12r^2\theta=10$ . At $r=4$ , $\theta=20/16=5/4$ .
Differentiate with respect to time: $r\theta\,dr/dt+\tfrac12r^2\,d\theta/dt=0$ . Substituting $d\theta/dt=2$ gives $5\,dr/dt+16=0$ .

Answer

\dfrac{dr}{dt}=-3.2\text{ cm s}^{-1}

: the radius is decreasing.

Question 12(b)2 marks

Volume of revolution about the y-axis

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IntegrationVolumes of revolution

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Express the radius of a horizontal disc in terms of its height: $x=1/y$ . Its area is $\pi x^2=\pi/y^2$ .
$V=\pi\int_1^a y^{-2}\,dy=\pi[-1/y]_1^a.$

Answer

V=\pi(1-1/a)

cubic units.

Question 12(c)3 marks

Induction with factorials

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Mathematical induction

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Base case $n=1$ : the sum is 1 and $2!-1=1$ .
Assume $\sum_{j=1}^{k}j(j!)=(k+1)!-1$ . Adding the next term gives
$\sum_{j=1}^{k+1}j(j!)=(k+1)!-1+(k+1)(k+1)!=(k+2)!-1.$
This is precisely the required statement for $k+1$ . The base case and implication establish the identity for every positive integer.

Answer

\displaystyle\sum_{j=1}^{n}j(j!)=(n+1)!-1

for all integers

n\ge1

Question 12(d)3 marks

Separating a differential equation

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Differential equationsInverse trigonometry

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Write $dy/\sqrt{4-y^2}=dx$ . Integration gives $\sin^{-1}(y/2)=x+C$ .
At $(x,y)=(0,1)$ , $C=\pi/6$ . Take sine to obtain the solution on the increasing branch through the initial point.
The sine expression solves the original equation on $-2\pi/3\le x\le\pi/3$ ; on this interval its cosine factor is non-negative.

Answer

y=2\sin(x+\pi/6)

on the branch through

(0,1)

Question 12(e)(i)1 mark

Writing a trigonometric expression as one sine

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Trigonometry

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Expand $2\sin(x-\alpha)=2\cos\alpha\sin x-2\sin\alpha\cos x$ . Matching coefficients gives $\cos\alpha=\sqrt3/2$ and $\sin\alpha=1/2$ .

Answer

\sqrt3\sin x-\cos x=2\sin(x-\pi/6)

, so

\alpha=\pi/6

Question 12(e)(ii)2 marks

Solving with a phase shift

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Trigonometry

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Part (i) turns the equation into $2\sin(x-\pi/6)=1$ .
For $-\pi/6\le x-\pi/6\le11\pi/6$ , sine equals $1/2$ at $\pi/6$ and $5\pi/6$ . Add $\pi/6$ to each.

Answer

x=\pi/3,\ \pi

Question 13(a)3 marks

Choosing the correct solution branch

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Differential equations

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Separate and integrate: $y\,dy=5\,dx$ , so $y^2/2=5x+C$ . The initial value gives $C=8$ , hence $y^2=10x+16$ .
The initial value is negative, so choose the negative square root. Exclude the endpoint where $y=0$ , because the original differential equation divides by $y$ .

Answer

y=-\sqrt{10x+16},\quad x>-8/5

Question 13(b)2 marks

Circular arrangements with a restriction

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Combinatorics

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Fix one guest to remove rotational duplicates. The unrestricted count is $7!$ .
For the forbidden arrangements, combine the two specified guests into one block. The resulting seven objects have $6!$ circular arrangements, with 2 orders inside the block. Subtract these.

Answer

7!-2(6!)=3600

arrangements.

Question 13(c)4 marks

Angle between velocity and acceleration

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VectorsMotionDifferentiation

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Differentiate the position vector: $\mathbf v=(1,2t/9)$ and $\mathbf a=(0,2/9)$ .
The dot-product formula gives $\cos(\pi/4)=2t/\sqrt{81+4t^2}$ . The angle is acute, so $t>0$ .
Squaring yields $1/2=4t^2/(81+4t^2)$ , hence $4t^2=81$ . Keep the positive time.

Answer

t=9/2=4.5

time units.

Question 13(d)4 marks

Normal approximation to a binomial count

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Binomial distributionNormal distribution

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With replacement, $X\sim\operatorname{Bin}(100,p)$ . Since $100p=20$ , $p=0.2$ and $\sigma=\sqrt{100(0.2)(0.8)}=4$ .
The lower-tail probability is $1-0.0668=0.9332$ . The supplied normal table gives $P(Z\le1.5)=0.9332$ .
Using the standard normal approximation requested in the question, $(k-20)/4=1.5$ .

Answer

k=26

Question 13(e)(i)1 mark

Rearranging Pascal’s identity

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CombinatoricsBinomial coefficients

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Use Pascal’s identity with $n=m+1$ and $r=R+1$ : $\binom{m+1}{R+1}=\binom mR+\binom m{R+1}$ . Move the last term to the other side.

Answer

\displaystyle\binom mR=\binom{m+1}{R+1}-\binom m{R+1}

Question 13(e)(ii)2 marks

Telescoping a sum of binomial coefficients

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CombinatoricsBinomial coefficients

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Apply part (i) with $R=2000$ . Each term $\binom m{2000}$ becomes $\binom{m+1}{2001}-\binom m{2001}$ .
Summing for $m=2000,\ldots,2050$ , all intermediate terms cancel, leaving $\binom{2051}{2001}-\binom{2000}{2001}$ . The last term is zero because 2001 objects cannot be chosen from 2000.

Answer

\displaystyle\sum_{m=2000}^{2050}\binom m{2000}=\binom{2051}{2001}

Question 14(a)2 marks

Divisibility of a product of factors

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ProofNumber theory

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Among the positive factors of 60, only four are odd: 1, 3, 5 and 15. Any seven selected factors therefore include at least three even numbers, supplying more than the two factors of 2 needed.
Only six factors avoid a multiple of 3: 1, 2, 4, 5, 10 and 20. Likewise only six avoid a multiple of 5: 1, 2, 3, 4, 6 and 12. Seven selections must therefore supply both a factor of 3 and a factor of 5.
Thus the product is divisible by $2^2\times3\times5=60$ .

Answer

The product of any seven distinct positive factors of 60 is a multiple of 60.

Question 14(b)4 marks

Comparing horizontal projectiles

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Projectile motionMotion

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Let $OB=h$ , so $OA=kh$ . Both particles start with zero vertical speed. Their respective flight durations are $t_A=\sqrt{2kh/g}$ and $t_B=\sqrt{2h/g}$ .
They have the same horizontal range, so $Ut_A=Vt_B$ . Dividing gives $V/U=t_A/t_B=\sqrt{k}$ . The launch delay does not change either particle’s flight duration measured from its own launch.

Answer

\dfrac VU=\sqrt{k}

, which depends only on

k

Question 14(c)3 marks

Perpendicular clock hands

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VectorsTrigonometry

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Perpendicular vectors have zero dot product. Dividing that product by 2 and using $\cos(A-B)=\cos A\cos B+\sin A\sin B$ gives $\cos(11\pi t/360)=0$ .
The first two positive angles are $\pi/2$ and $3\pi/2$ . Therefore the times are $180/11$ and $540/11$ minutes after midnight.

Answer

t=16.364

minutes and

t=49.091

minutes.

Question 14(d)3 marks

Differentiating an inverse-trigonometric composite

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DifferentiationInverse trigonometry

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The chain rule gives $f'(x)=-\cos x/\sqrt{1-\sin^2x}=-\cos x/|\cos x|$ . The absolute value is essential.
Cosine is positive on $(0,\pi/2)$ and negative on $(\pi/2,\pi)$ . At $\pi/2$ , the graph has a corner, so no derivative exists.

Answer

f'(x)=\begin{cases}-1,&0<x<\pi/2,\\1,&\pi/2<x<\pi.\end{cases}

Undefined at

x=\pi/2

Question 14(e)3 marks

Combining roots with a tangent-sum identity

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PolynomialsTrigonometry

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Let the roots be $p=\tan\alpha$ , $q=\tan\beta$ , $r=\tan\gamma$ . Vieta gives $p+q+r=-b$ , $pq+pr+qr=c$ , and $pqr=1-b-c$ .
The three-angle tangent identity now gives $\tan(\alpha+\beta+\gamma)=\dfrac{p+q+r-pqr}{1-pq-pr-qr}=\dfrac{-b-(1-b-c)}{1-c}=-1$ . The assumption $c\ne1$ allows this division.
The angle sums are $-\pi/4+n\pi$ . The first positive one occurs at $n=1$ .

Answer

\alpha+\beta+\gamma=3\pi/4

is the smallest positive value.

2025 HSC
worked solutions

Choose your topics

Section I Multiple choice

Solving an absolute-value inequality

Projecting one vector onto another

Changing variables in an integral

Bernoulli mean and variance

Counting trigonometric solutions

Using odd symmetry in integration

Reading a slope field

Ordering collinear points

Finding the angle between vectors

Second derivative of an inverse

Section II Written responses

Finding a rational inverse

Factoring a trigonometric equation

Integrating a trigonometric product

Sketching a scaled inverse cosine

Making vectors parallel

Using sums and products of roots

Integrating a squared cosine

Related rates in a sector

Volume of revolution about the y-axis

Induction with factorials

Separating a differential equation

Writing a trigonometric expression as one sine

Solving with a phase shift

Choosing the correct solution branch

Circular arrangements with a restriction

Angle between velocity and acceleration

Normal approximation to a binomial count

Rearranging Pascal’s identity

Telescoping a sum of binomial coefficients

Divisibility of a product of factors

Comparing horizontal projectiles

Perpendicular clock hands

Differentiating an inverse-trigonometric composite

Combining roots with a tangent-sum identity

No matching question parts

2025 HSCworked solutions

Choose your topics

Section I Multiple choice

Solving an absolute-value inequality

Projecting one vector onto another

Changing variables in an integral

Bernoulli mean and variance

Counting trigonometric solutions

Using odd symmetry in integration

Reading a slope field

Ordering collinear points

Finding the angle between vectors

Second derivative of an inverse

Section II Written responses

Finding a rational inverse

Factoring a trigonometric equation

Integrating a trigonometric product

Sketching a scaled inverse cosine

Making vectors parallel

Using sums and products of roots

Integrating a squared cosine

Related rates in a sector

Volume of revolution about the y-axis

Induction with factorials

Separating a differential equation

Writing a trigonometric expression as one sine

Solving with a phase shift

Choosing the correct solution branch

Circular arrangements with a restriction

Angle between velocity and acceleration

Normal approximation to a binomial count

Rearranging Pascal’s identity

Telescoping a sum of binomial coefficients

Divisibility of a product of factors

Comparing horizontal projectiles

Perpendicular clock hands

Differentiating an inverse-trigonometric composite

Combining roots with a tangent-sum identity

No matching question parts

2025 HSC
worked solutions