Home

/

Exam Packs

/

HSC Mathematics Extension 1 Year 12 packs

/

Mathematics Extension 1 Free Online - Pack 0

/

HSC Paper Showcase

HSC Mathematics Extension 1 Free Online Pack 0 — HSC Paper Showcase

Read HSC Paper Showcase online for free, including every question, worked solution, marking note and diagnostic action. No public PDF download or checkout is provided.

HSC Year 12 Final Exam 2026 Edition - Pack 0 v1.0
HSC Paper Showcase is free to read in your browser. There is no public checkout or PDF download.

Exam-pack paper structure

This full-length showcase paper is available to read online.

HSC Paper Showcase

14 questions

70 marks

Estimated duration: Reading time 10 minutes; writing time 2 hours

Reading: 10 minutes · Writing: 2 hours

Read HSC Paper Showcase online

Skill Align

Skill Align HSC Mathematics Extension 1 - Free Online Pack 0

Full-length Mathematics Extension 1 showcase paper

Paper
HSC Paper Showcase
Reading
10 minutes
Writing
2 hours
Assessment
70 marks

NESA-approved calculators may be used. The Mathematics Advanced, Extension 1 and Extension 2 Reference Sheet is supplied separately as “Mathematics_Advanced_Extension_1_and_2_Reference_Sheet.pdf”. Resource: https://www.nsw.gov.au/education-and-training/nesa/curriculum/hsc-exam-papers/mathematics-advanced-extension-reference-sheet

Section I

Attempt Questions 1-10. Allow about 15 minutes for this section. Select the best answer for each question.

Question 1

1 mark
What is the coefficient of x³ in (1+2x)⁵?
  1. 40
  2. 80
  3. 120
  4. 160

Question 2

1 mark
If f(x)=sin⁻¹x, what is f'(1 / 2)?
  1. frac1(sqrt3)
  2. ((sqrt3) / (2))
  3. 2sqrt3
  4. frac2(sqrt3)

Question 3

1 mark
The polynomial P(x)=x³+ax-4 has factor x-2. What is a?
  1. -4
  2. 2
  3. -2
  4. 4

Question 4

1 mark
If Xsimoperatorname(Bin)(6,frac13), what is P(X=2)?
  1. ((80) / (243))
  2. ((40) / (243))
  3. ((20) / (81))
  4. ((160) / (243))

Question 5

1 mark
Vectors a=(1,1) and b=(1,lambda) are perpendicular. What is lambda?
  1. -3
  2. 1
  3. -1
  4. 3

Question 6

1 mark
How many solutions does tan(2x)=1 have for 0le x<2pi?
  1. 2
  2. 4
  3. 6
  4. 8

Question 7

1 mark
For ((dy) / (dx))=x-y, what is the gradient of the solution curve at (2,1)?
  1. -3
  2. -1
  3. 0
  4. 1

Question 8

1 mark
Evaluate int_0¹((1) / (1+4x²)),dx.
  1. frac12tan⁻¹2
  2. tan⁻¹2
  3. 2tan⁻¹2
  4. frac14ln5

Question 9

1 mark
In an induction proof for sum_(r=1)^(n)r²=frac(n(n+1)(2n+1))6, which term is added to the n=k case?
  1. (k+1)²
  2. 2k+1
  3. 2k+2

Question 10

1 mark
For f(x)=x²+2, xge0, what is the domain of f⁻¹?
  1. [0,infty)
  2. (-infty,2]
  3. [2,infty)
  4. mathbb R

Section II

Attempt Questions 11-14. Allow about 1 hour and 45 minutes for this section. Show relevant mathematical reasoning and calculations.

Question 11

15 marks
Let F(x)=x³-3x²-9x+27.
(a) 2 marks
Factorise F(x) completely.
(b) 2 marks
State the zeros of F, including multiplicities.
(c) 2 marks
Show that F'(x)=3(x-3)(x+1).
(d) 3 marks
Find both stationary points and classify them.
(e) 2 marks
Find the equation of the tangent to y=F(x) at x=0.
(f) 4 marks
Find the exact area enclosed by y=F(x) and the x-axis between its two distinct zeros.

Question 12

14 marks
Points A=(1,2), B=(7,5) and C=(4,-1) are given. Let d=overrightarrow(AB).
(a) 2 marks
Find d and |d|.
(b) 2 marks
Find the midpoint M of AB.
(c) 3 marks
The point P=A+td lies on y=x. Find t and P.
(d) 3 marks
Find the vector projection of overrightarrow(AC) onto d.
(e) 3 marks
Hence find the perpendicular distance from C to the line AB.
(f) 1 mark
Find the exact cosine of angle BAC.

Question 13

16 marks
Each component is defective independently with probability 0.05. Let X be the number of defective components in a batch of 200.
(a) 2 marks
State the distribution of X, and find its mean.
(b) 2 marks
Find operatorname(Var)(X).
(c) 3 marks
Write an exact expression for P(Xle1).
(d) 2 marks
Explain why a normal approximation is reasonable.
(e) 3 marks
Using P(Zle1.46)=0.9279, estimate P(Xge15) with a continuity correction.
(f) 4 marks
Using P(Zle2.33)=0.9901, find the smallest integer k for which the normal approximation gives P(Xge k)<0.01.

Question 14

15 marks
For xge0, a quantity y satisfies ((dy) / (dx))=(x+1)(6-y) and y=2 when x=0.
(a) 1 mark
State the equilibrium solution.
(b) 2 marks
Separate the variables.
(c) 3 marks
Integrate both sides.
(d) 3 marks
Use the initial condition to find y explicitly.
(e) 3 marks
Show that y is increasing and remains below 6 for xge0.
(f) 3 marks
Find the exact value of x when y=5.

HSC is administered by the NSW Education Standards Authority (NESA). Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by, or endorsed by NESA or the NSW Government.

Copyright (c) 2026 Skill Align. Free for personal, non-commercial online viewing at https://skillalign.au. You may share the Skill Align page link. Except as permitted by law or with Skill Align's prior written permission, the pack itself must not be resold, copied, redistributed, republished, automatically extracted, or uploaded to a question bank.

Worked Solutions And Marking Guide

Section I Question 1

Answer: 80

The coefficient is binom53 2³=10(8)=80.

Section I Question 2

Answer: frac2(sqrt3)

Use f'(x)=1 / √(1-x²), then substitute x=1 / 2.

Section I Question 3

Answer: -2

The factor theorem gives P(2)=8+2a-4=0, so a=-2.

Section I Question 4

Answer: ((80) / (243))

Evaluate binom62(1 / 3)²(2 / 3)⁴=80 / 243.

Section I Question 5

Answer: -1

Set a × b=1+lambda=0, giving lambda=-1.

Section I Question 6

Answer: 4

Since 2x=pi / 4+kpi, four values of x lie in the stated interval.

Section I Question 7

Answer: 1

Substitute x=2 and y=1 into the differential equation.

Section I Question 8

Answer: frac12tan⁻¹2

Use the antiderivative frac12tan⁻¹(2x) and apply the limits.

Section I Question 9

Answer: (k+1)²

The next partial sum contains exactly one new term, namely (k+1)^2.

Section I Question 10

Answer: [2,infty)

The inverse domain is the range of f, which is [2,infty).

Section II Question 11

(a) F(x)=(x-3)²(x+3).

Group the terms to obtain x²(x-3)-9(x-3); factor x-3; factor the resulting difference of squares.

(b) x=-3 is simple and x=3 is double.

Read the roots from the factorisation; identify the repeated factor (x-3)^2.

(c) F'(x)=3(x-3)(x+1).

Differentiate the expanded polynomial; factor 3(x²-2x-3).

(d) A local maximum at (-1,32) and a local minimum at (3,0).

Solve F'(x)=0; evaluate F(-1) and F(3); use the derivative sign changes to classify both points.

(e) y=-9x+27.

Evaluate F(0)=27; evaluate F'(0)=-9; use point-gradient form.

(f) 108 square units.

Use the factorisation to establish F(x)ge0 on [-3,3]; integrate x³-3x²-9x+27; apply the limits -3 and 3; obtain 108.

Mark allocation

  • Part a (2 marks): Group the terms to obtain x²(x-3)-9(x-3); factor x-3; factor the resulting difference of squares; States F(x)=(x-3)²(x+3).
  • Part b (2 marks): Read the roots from the factorisation; identify the repeated factor (x-3)²; States x=-3 is simple and x=3 is double.
  • Part c (2 marks): Differentiate the expanded polynomial; factor 3(x²-2x-3); States F'(x)=3(x-3)(x+1).
  • Part d (3 marks): Solve F'(x)=0; evaluate F(-1) and F(3); use the derivative sign changes to classify both points; States A local maximum at (-1,32) and a local minimum at (3,0).
  • Part e (2 marks): Evaluate F(0)=27; evaluate F'(0)=-9; use point-gradient form; States y=-9x+27.
  • Part f (4 marks): Use the factorisation to establish F(x)ge0 on [-3,3]; integrate x³-3x²-9x+27; apply the limits -3 and 3; obtain 108; States 108 square units.

Detailed marking criteria

Part a (2 marks)

Group the terms to obtain x²(x-3)-9(x-3); factor x-3; factor the resulting difference of squares; States F(x)=(x-3)²(x+3)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part b (2 marks)

Read the roots from the factorisation; identify the repeated factor (x-3)²; States x=-3 is simple and x=3 is double

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part c (2 marks)

Differentiate the expanded polynomial; factor 3(x²-2x-3); States F'(x)=3(x-3)(x+1)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part d (3 marks)

Solve F'(x)=0; evaluate F(-1) and F(3); use the derivative sign changes to classify both points; States A local maximum at (-1,32) and a local minimum at (3,0)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part e (2 marks)

Evaluate F(0)=27; evaluate F'(0)=-9; use point-gradient form; States y=-9x+27

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part f (4 marks)

Use the factorisation to establish F(x)ge0 on [-3,3]; integrate x³-3x²-9x+27; apply the limits -3 and 3; obtain 108; States 108 square units

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Section II Question 12

(a) d=(6,3) and |d|=3sqrt5.

Subtract the coordinates of A from B; calculate √(6²+3²)=3sqrt5.

(b) M=(4,frac72).

Average the corresponding coordinates of A and B; state M=(4,7 / 2).

(c) t=frac13 and P=(3,3).

Write P=(1+6t,2+3t); impose 2+3t=1+6t; solve t=1 / 3 and substitute.

(d) operatorname(proj)_(d)overrightarrow(AC)=(frac65,frac35).

Calculate overrightarrow(AC)=(3,-3); calculate overrightarrow(AC) × d=9 and d × d=45; multiply d by 1 / 5.

(e) ((9) / (sqrt5)) units.

Subtract the projection from overrightarrow(AC); obtain the perpendicular component (9 / 5,-18 / 5); calculate its magnitude 9 / sqrt5.

(f) cosangle BAC=frac1(√10).

Use d × overrightarrow(AC)=9; divide by (3sqrt5)(3sqrt2).

Mark allocation

  • Part a (2 marks): Subtract the coordinates of A from B; calculate √(6²+3²)=3sqrt5; States d=(6,3) and |d|=3sqrt5.
  • Part b (2 marks): Average the corresponding coordinates of A and B; state M=(4,7 / 2); States M=(4,frac72).
  • Part c (3 marks): Write P=(1+6t,2+3t); impose 2+3t=1+6t; solve t=1 / 3 and substitute; States t=frac13 and P=(3,3).
  • Part d (3 marks): Calculate overrightarrow(AC)=(3,-3); calculate overrightarrow(AC) × d=9 and d × d=45; multiply d by 1 / 5; States operatorname(proj)_(d)overrightarrow(AC)=(frac65,frac35).
  • Part e (3 marks): Subtract the projection from overrightarrow(AC); obtain the perpendicular component (9 / 5,-18 / 5); calculate its magnitude 9 / sqrt5; States ((9) / (sqrt5)) units.
  • Part f (1 marks): Use d × overrightarrow(AC)=9; divide by (3sqrt5)(3sqrt2); States cosangle BAC=frac1(√10).

Detailed marking criteria

Part a (2 marks)

Subtract the coordinates of A from B; calculate √(6²+3²)=3sqrt5; States d=(6,3) and |d|=3sqrt5

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part b (2 marks)

Average the corresponding coordinates of A and B; state M=(4,7 / 2); States M=(4,frac72)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part c (3 marks)

Write P=(1+6t,2+3t); impose 2+3t=1+6t; solve t=1 / 3 and substitute; States t=frac13 and P=(3,3)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part d (3 marks)

Calculate overrightarrow(AC)=(3,-3); calculate overrightarrow(AC) × d=9 and d × d=45; multiply d by 1 / 5; States operatorname(proj)_(d)overrightarrow(AC)=(frac65,frac35)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part e (3 marks)

Subtract the projection from overrightarrow(AC); obtain the perpendicular component (9 / 5,-18 / 5); calculate its magnitude 9 / sqrt5; States ((9) / (sqrt5)) units

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part f (1 mark)

Use d × overrightarrow(AC)=9; divide by (3sqrt5)(3sqrt2); States cosangle BAC=frac1(√10)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Section II Question 13

(a) Xsimoperatorname(Bin)(200,0.05), with mean 10.

Identify n=200 and p=0.05; calculate np=10.

(b) operatorname(Var)(X)=9.5.

Use np(1-p); calculate 200(0.05)(0.95)=9.5.

(c) 0.95²⁰⁰+10(0.95)¹⁹⁹.

Write P(X=0)=0.95²⁰⁰; write P(X=1)=200(0.05)(0.95)¹⁹⁹=10(0.95)¹⁹⁹; add the two probabilities.

(d) Both np=10 and n(1-p)=190 are at least 10.

Calculate both expected category counts; compare each with the applicable minimum of 10.

(e) P(Xge15)approx0.0721.

Approximate by Ysim N(10,9.5); use the boundary 14.5; standardise to z=(14.5-10) / √9.5approx1.46; take 1-0.9279=0.0721.

(f) k=18.

Use the corrected boundary k-0.5; require (k-0.5-10) / √9.5>2.33; obtain k>10.5+2.33sqrt(9.5)approx17.68; select the smallest integer k=18.

Mark allocation

  • Part a (2 marks): Identify n=200 and p=0.05; calculate np=10; States Xsimoperatorname(Bin)(200,0.05), with mean 10.
  • Part b (2 marks): Use np(1-p); calculate 200(0.05)(0.95)=9.5; States operatorname(Var)(X)=9.5.
  • Part c (3 marks): Write P(X=0)=0.95²⁰⁰; write P(X=1)=200(0.05)(0.95)¹⁹⁹=10(0.95)¹⁹⁹; add the two probabilities; States 0.95²⁰⁰+10(0.95)¹⁹⁹.
  • Part d (2 marks): Calculate both expected category counts; compare each with the applicable minimum of 10; States Both np=10 and n(1-p)=190 are at least 10.
  • Part e (3 marks): Approximate by Ysim N(10,9.5); use the boundary 14.5; standardise to z=(14.5-10) / √9.5approx1.46; take 1-0.9279=0.0721; States P(Xge15)approx0.0721.
  • Part f (4 marks): Use the corrected boundary k-0.5; require (k-0.5-10) / √9.5>2.33; obtain k>10.5+2.33sqrt(9.5)approx17.68; select the smallest integer k=18; States k=18.

Detailed marking criteria

Part a (2 marks)

Identify n=200 and p=0.05; calculate np=10; States Xsimoperatorname(Bin)(200,0.05), with mean 10

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part b (2 marks)

Use np(1-p); calculate 200(0.05)(0.95)=9.5; States operatorname(Var)(X)=9.5

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part c (3 marks)

Write P(X=0)=0.95²⁰⁰; write P(X=1)=200(0.05)(0.95)¹⁹⁹=10(0.95)¹⁹⁹; add the two probabilities; States 0.95²⁰⁰+10(0.95)¹⁹⁹

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part d (2 marks)

Calculate both expected category counts; compare each with the applicable minimum of 10; States Both np=10 and n(1-p)=190 are at least 10

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part e (3 marks)

Approximate by Ysim N(10,9.5); use the boundary 14.5; standardise to z=(14.5-10) / √9.5approx1.46; take 1-0.9279=0.0721; States P(Xge15)approx0.0721

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part f (4 marks)

Use the corrected boundary k-0.5; require (k-0.5-10) / √9.5>2.33; obtain k>10.5+2.33sqrt(9.5)approx17.68; select the smallest integer k=18; States k=18

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Section II Question 14

(a) y=6.

Set dy / dx=0; since x+1>0, solve 6-y=0.

(b) ((dy) / (6-y))=(x+1),dx.

Divide by 6-y; multiply by dx.

(c) -ln|6-y|=((x²) / (2))+x+C.

Integrate 1 / (6-y) with the negative sign; integrate x+1; include an arbitrary constant.

(d) y=6-4e^(-x² / 2-x).

Substitute x=0,y=2 to determine the constant; exponentiate the logarithmic equation; select 6-y>0 from the initial value.

(e) For xge0, 2le y<6 and dy / dx>0.

Use 0<e^(-x² / 2-x)le1 to bound y; infer 6-y>0; combine with x+1>0 in the differential equation.

(f) x=-1+√(1+2ln4).

Set 6-4e^(-x² / 2-x)=5; obtain x² / 2+x=ln4; solve the quadratic and retain the non-negative root.

Mark allocation

  • Part a (1 marks): Set dy / dx=0; since x+1>0, solve 6-y=0; States y=6.
  • Part b (2 marks): Divide by 6-y; multiply by dx; States ((dy) / (6-y))=(x+1),dx.
  • Part c (3 marks): Integrate 1 / (6-y) with the negative sign; integrate x+1; include an arbitrary constant; States -ln|6-y|=((x²) / (2))+x+C.
  • Part d (3 marks): Substitute x=0,y=2 to determine the constant; exponentiate the logarithmic equation; select 6-y>0 from the initial value; States y=6-4e^(-x² / 2-x).
  • Part e (3 marks): Use 0<e^(-x² / 2-x)le1 to bound y; infer 6-y>0; combine with x+1>0 in the differential equation; States For xge0, 2le y<6 and dy / dx>0.
  • Part f (3 marks): Set 6-4e^(-x² / 2-x)=5; obtain x² / 2+x=ln4; solve the quadratic and retain the non-negative root; States x=-1+√(1+2ln4).

Detailed marking criteria

Part a (1 mark)

Set dy / dx=0; since x+1>0, solve 6-y=0; States y=6

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part b (2 marks)

Divide by 6-y; multiply by dx; States ((dy) / (6-y))=(x+1),dx

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part c (3 marks)

Integrate 1 / (6-y) with the negative sign; integrate x+1; include an arbitrary constant; States -ln|6-y|=((x²) / (2))+x+C

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part d (3 marks)

Substitute x=0,y=2 to determine the constant; exponentiate the logarithmic equation; select 6-y>0 from the initial value; States y=6-4e^(-x² / 2-x)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part e (3 marks)

Use 0<e^(-x² / 2-x)le1 to bound y; infer 6-y>0; combine with x+1>0 in the differential equation; States For xge0, 2le y<6 and dy / dx>0

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Part f (3 marks)

Set 6-4e^(-x² / 2-x)=5; obtain x² / 2+x=ln4; solve the quadratic and retain the non-negative root; States x=-1+√(1+2ln4)

Acceptable alternatives: Accept an algebraically equivalent method with the same mathematical evidence.; Apply consequential error follow-through where later work is valid from an earlier stated value.

Do not credit by itself: Unsupported answer, missing required reasoning, or an invalid branch, bound, sign or continuity correction.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Calculus and Mathematical Modelling Q7, Q8, Q11(a), Q11(b), Q11(c), Q11(d), Q11(e), Q11(f), Q14(a), Q14(b), Q14(c), Q14(d), Q14(e), Q14(f) 32 ___ Review the listed calculus and mathematical modelling questions, their worked solutions and the evidence-specific marking criteria.
Combinatorics and Statistical Analysis Q1, Q4, Q13(a), Q13(b), Q13(c), Q13(d), Q13(e), Q13(f) 18 ___ Review the listed combinatorics and statistical analysis questions, their worked solutions and the evidence-specific marking criteria.
Functions and Polynomials Q3, Q10 2 ___ Review the listed functions and polynomials questions, their worked solutions and the evidence-specific marking criteria.
Proof Q9 1 ___ Review the listed proof questions, their worked solutions and the evidence-specific marking criteria.
Trigonometric Functions Q2, Q6 2 ___ Review the listed trigonometric functions questions, their worked solutions and the evidence-specific marking criteria.
Vectors and Motion Q5, Q12(a), Q12(b), Q12(c), Q12(d), Q12(e), Q12(f) 15 ___ Review the listed vectors and motion questions, their worked solutions and the evidence-specific marking criteria.

What is included

HSC Paper Showcase questions (70 marks)

Worked solutions and marking guidance shown online

Diagnostic checklist shown online

Free browser viewing with no checkout

No PDF or downloadable file

Independent practice resource

HSC is administered by the NSW Education Standards Authority (NESA). Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by, or endorsed by NESA or the NSW Government.

Each exam pack is listed with a pack label so parents do not buy the same pack twice. Future packs will use the next label for that state or curriculum.

Related Online Practice and curriculum

Pack 0 is a free online resource. These links open the related subscription practice, curriculum coverage, and free public sample questions.

Questions about this exam pack

What is included in Mathematics Extension 1 Free Online - Pack 0?

Pack 0 includes one full-length showcase paper, worked solutions, marking guidance and diagnostic checklists, all shown online.

Is Pack 0 really free?

Yes. Pack 0 can be read online without checkout or a monthly subscription.

Can I download Pack 0 as a PDF?

No. Pack 0 is intentionally online-only and no downloadable PDF is provided.

Are these official assessment authority examination questions?

No. The questions are original Skill Align material. Skill Align is independent and is not affiliated with, authorised by, sponsored by, approved by, or endorsed by any state assessment authority.