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- 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=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
| 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 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. |