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. A six-page Skill Align-authored Mathematics Advanced, Mathematics Extension 1 and Mathematics Extension 2 reference sheet is included at the back of the question paper.
Section I
Attempt Questions 1-10. Allow about 15 minutes for this section. Select the best answer for each question.
Question 1
1 mark- 40
- 80
- 120
- 160
Question 2
1 mark- frac1(sqrt3)
- ((sqrt3) / (2))
- 2sqrt3
- frac2(sqrt3)
Question 3
1 mark- -4
- 2
- -2
- 4
Question 4
1 mark- ((80) / (243))
- ((40) / (243))
- ((20) / (81))
- ((160) / (243))
Question 5
1 mark- -3
- 1
- -1
- 3
Question 6
1 mark- 2
- 4
- 6
- 8
Question 7
1 mark- -3
- -1
- 0
- 1
Question 8
1 mark- frac12tan⁻¹2
- tan⁻¹2
- 2tan⁻¹2
- frac14ln5
Question 9
1 mark- k²
- (k+1)²
- 2k+1
- 2k+2
Question 10
1 mark- [0,infty)
- (-infty,2]
- [2,infty)
- 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 marksQuestion 12
14 marksQuestion 13
16 marksQuestion 14
15 marksWorked 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=π / 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.
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.
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.
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.
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.
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.
Do not credit by itself: Uses integration bounds that do not match the stated interval or the chosen substitution.
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.
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.
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.
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.
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. Where this part explicitly reuses an earlier result, apply consequential error follow-through only to mathematically valid subsequent work.
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.
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.
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.
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.
Do not credit by itself: Uses an event, conditioning denominator or probability model that does not match the question.
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.
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.
Do not credit by itself: Uses an integer endpoint instead of the required half-unit continuity-corrected boundary. Uses an event, conditioning denominator or probability model that does not match the question.
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.
Do not credit by itself: Uses an integer endpoint instead of the required half-unit continuity-corrected boundary. Uses an event, conditioning denominator or probability model that does not match the question.
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.
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.
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.
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.
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.
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.
Do not credit by itself: Retains a value that is inconsistent with the stated domain, range or required branch.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| 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 | Q1 | 1 | ___ | Review the listed combinatorics 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. |
| Statistical Analysis | Q4, Q13(a), Q13(b), Q13(c), Q13(d), Q13(e), Q13(f) | 17 | ___ | Review the listed statistical analysis 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 | Q5, Q12(a), Q12(b), Q12(c), Q12(d), Q12(e), Q12(f) | 15 | ___ | Review the listed vectors questions, their worked solutions and the evidence-specific marking criteria. |