Skill Align HSC Mathematics Advanced - Free Online Pack 0
Full-length Mathematics Advanced showcase paper
- Paper
- HSC Paper Showcase
- Reading
- 10 minutes
- Writing
- 3 hours
- Assessment
- 100 marks
NESA-approved calculators may be used. The Mathematics Advanced, Mathematics Extension 1 and Mathematics 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. Each question is worth 1 mark.
Question 1
1 mark- y=3
- y=-2
- y=1
- x=3
Question 2
1 mark- 1
- 2
- frac12
- ln2
Question 3
1 mark- 0.5
- 0.25
- 0.75
- 1
Question 4
1 mark- 1081.60
- 1080.00
- 1040.00
- 1160.00
Question 5
1 mark- 2
- 8
- -2
- 17
Question 6
1 mark- x=fracpi3,frac(2pi)3
- x=fracpi6,frac(5pi)6
- x=fracpi3,frac(5pi)3
- x=frac(2pi)3,frac(4pi)3
Question 7
1 mark- 9
- 4
- 5
- 13
Question 8
1 mark- frac5(14)
- ((25) / (64))
- ((10) / (21))
- frac38
Question 9
1 mark- 2.25
- 1.5
- 3.0
- 120
Question 10
1 mark- 38
- 50
- 62
- 144
Section II
Attempt Questions 11-31. Allow about 2 hours and 45 minutes for this section. Show sufficient reasoning and working to support each answer.
Question 11
4 marksQuestion 12
4 marksQuestion 13
4 marksQuestion 14
5 marksQuestion 15
4 marksQuestion 16
5 marksQuestion 17
4 marksQuestion 18
4 marksQuestion 19
5 marksQuestion 20
5 marksQuestion 21
5 marksQuestion 22
4 marksQuestion 23
5 marksQuestion 24
4 marksQuestion 25
4 marksQuestion 26
4 marksQuestion 27
4 marksQuestion 28
4 marksQuestion 29
4 marksQuestion 30
4 marksQuestion 31
4 marksWorked Solutions And Marking Guide
Section I Question 1
Answer: y=3
As xtoinfty, e^(-x)to0, so f(x)to3.
Section I Question 2
Answer: 1
Since f'(x)=2x / (x²+1), substituting x=1 gives 1.
Section I Question 3
Answer: 0.5
The density is symmetric about x=0.5, so half the probability lies on each side.
Section I Question 4
Answer: 1081.60
Compute 1000(1.04)²=1081.60.
Section I Question 5
Answer: 2
Standardise using (68-60) / 4=2.
Section I Question 6
Answer: x=fracpi3,frac(2pi)3
Sine is positive in quadrants I and II with reference angle pi / 3.
Section I Question 7
Answer: 9
The accumulated change is int_0²(3x²-2x),dx=4, so F(2)=5+4=9.
Section I Question 8
Answer: frac5(14)
Multiply 5 / 8 by 4 / 7 to obtain 20 / 56=5 / 14.
Section I Question 9
Answer: 2.25
Two periods multiply the population by (1.5)²=2.25.
Section I Question 10
Answer: 62
One standard deviation above the mean is 50+12=62.
Section II Question 11
(a) f(g(x))=2x²-1, with x=pm2.
Substitute g(x) into f; solve 2x²-1=7.
(b) [-1,infty).
Use x^2ge0; identify the minimum value 2(0)-1=-1.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Substitute g(x) into f.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Solve 2x²-1=7.; Obtains the correct final result f(g(x))=2x²-1, with x=pm2.
Part b (1 mark)
Use x^2ge0.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Identify the minimum value 2(0)-1=-1.; Completes the required reasoning and reaches [-1,infty).
Section II Question 12
(a) 5ln2 hours.
Set 800e^(-0.2t)=400; solve e^(-0.2t)=1 / 2 using logarithms.
(b) 5ln8 hours.
Set 800e^(-0.2t)=100; take logarithms and solve for t.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Set 800e^(-0.2t)=400.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Solve e^(-0.2t)=1 / 2 using logarithms.; Obtains the correct final result 5ln2 hours.
Part b (1 mark)
Set 800e^(-0.2t)=100.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Take logarithms and solve for t.; Obtains the correct final result 5ln8 hours.
Section II Question 13
(a) k=frac16 and E(X)=frac43.
Use 6k=1; calculate 0(k)+1(2k)+2(3k).
(b) frac59.
Calculate E(X²)=7 / 3; subtract [E(X)]²=16 / 9.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Use 6k=1.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Calculate 0(k)+1(2k)+2(3k).; Obtains the correct final result k=frac16 and E(X)=frac43.
Part b (1 mark)
Calculate E(X²)=7 / 3.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Subtract [E(X)]²=16 / 9.; Obtains the correct final result frac59.
Section II Question 14
(a) f'(x)=e^(-x)(1-x), with stationary point (1,1 / e).
Apply the product rule; solve f'(x)=0; substitute x=1.
(b) The absolute maximum is 1 / e at x=1, and the tangent is y=1 / e.
Show f' changes from positive to negative at 1; compare with f(0)=0 and f(x)to0; use the zero gradient to write the horizontal tangent.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (3 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Apply the product rule.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Solve f'(x)=0.; Substitute x=1.; Obtains the correct final result f'(x)=e^(-x)(1-x), with stationary point (1,1 / e).
Part b (1 mark)
Show f' changes from positive to negative at 1.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Compare with f(0)=0 and f(x)to0.
Part b (1 mark)
Use the zero gradient to write the horizontal tangent.; Completes the required reasoning and reaches The absolute maximum is 1 / e at x=1, and the tangent is y=1 / e.
Section II Question 15
(a) V_n=5000(0.92)^(n-1), and V_6=3295.41.
Identify first term 5000 and ratio 0.92; evaluate 5000(0.92)^5.
(b) 2171.94.
Apply ten years of declining-balance depreciation: 5000(0.92)¹⁰approx2171.94.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Identify first term 5000 and ratio 0.92.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Evaluate 5000(0.92)^5.; Obtains the correct final result V_n=5000(0.92)^(n-1), and V_6=3295.41.
Part b (1 mark)
Apply ten years of declining-balance depreciation: 5000(0.92)¹⁰approx2171.94.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Obtains the correct final result 2171.94.
Section II Question 16
(a) Maximum 4 at x=2; area ((32) / (3)) square units.
Use the quadratic vertex at x=2; evaluate int_0⁴(4x-x²),dx.
(b) Average value frac83; x=2pm((2sqrt3) / (3)).
Divide the integral 32 / 3 by the interval length 4; solve x(4-x)=8 / 3; retain both solutions in [0,4].
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (3 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Use the quadratic vertex at x=2.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Evaluate int_0⁴(4x-x²),dx.; Obtains the correct final result Maximum 4 at x=2; area ((32) / (3)) square units.
Part b (1 mark)
Divide the integral 32 / 3 by the interval length 4.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Solve x(4-x)=8 / 3.
Part b (1 mark)
Retain both solutions in [0,4].; Obtains the correct final result Average value frac83; x=2pm((2sqrt3) / (3)).
Section II Question 17
(a) 0.1587.
Standardise 78 to obtain z=1; calculate 1-Phi(1).
(b) 79.692, approximately 79.7.
Use x=72+1.282(6); evaluate the percentile.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Standardise 78 to obtain z=1.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Calculate 1-Phi(1).; Obtains the correct final result 0.1587.
Part b (1 mark)
Use x=72+1.282(6).
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Evaluate the percentile.; Obtains the correct final result 79.692, approximately 79.7.
Section II Question 18
(a) Each additional kilometre is associated with 2.4 additional predicted minutes; the prediction is 39 minutes.
Interpret 2.4 using kilometres and minutes; substitute x=10.
(b) Residual 3 minutes; about 74% of delivery-time variation is explained by its linear association with route length.
Calculate observed minus predicted, 42-39=3; calculate r²=0.7396 and interpret in context.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Interpret 2.4 using kilometres and minutes.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Substitute x=10.; Completes the required reasoning and reaches Each additional kilometre is associated with 2.4 additional predicted minutes; the prediction is 39 minutes.
Part b (1 mark)
Calculate observed minus predicted, 42-39=3.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Calculate r²=0.7396 and interpret in context.; Completes the required reasoning and reaches Residual 3 minutes; about 74% of delivery-time variation is explained by its linear association with route length.
Section II Question 19
(a) angle A=25^circ, angle B=40^circ, and angle C=115^circ.
At A, compare bearings 065^circ and 090^circ. At B, compare bearings 270^circ and 310^circ. Subtract the two angles from 180^circ.
(b) ACapprox12.8 km.
Identifies that AC is opposite 40^circ and AB is opposite 115^circ; applies the sine rule ((AC) / (sin 40^circ))=((18) / (sin 115^circ)); evaluates and rounds to ACapprox12.8 km.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (3 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
At A, compare bearings 065^circ and 090^circ.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
At B, compare bearings 270^circ and 310^circ.; Subtract the two angles from 180^circ.; Obtains the correct final result angle A=25^circ, angle B=40^circ, and angle C=115^circ.
Part b (1 mark)
Identifies that AC is opposite 40^circ and AB is opposite 115^circ.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Applies the sine rule ((AC) / (sin 40^circ))=((18) / (sin 115^circ)).
Part b (1 mark)
Evaluates and rounds to ACapprox12.8 km.
Section II Question 20
(a) B_1=19,500; the 600 repayment exceeds the first month's 100 interest.
Substitute B_0=20000; compare 0.005B_0=100 with the repayment.
(b) B_n=120000-100000(1.005)^n.
Iterate the recurrence to identify the accumulated repayment series; sum the finite geometric series; simplify using 600 / 0.005=120000.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (3 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Substitute B_0=20000.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Compare 0.005B_0=100 with the repayment.; Completes the required reasoning and reaches B_1=19,500; the 600 repayment exceeds the first month's 100 interest.
Part b (1 mark)
Iterate the recurrence to identify the accumulated repayment series.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Sum the finite geometric series.
Part b (1 mark)
Simplify using 600 / 0.005=120000.; Completes the required reasoning and reaches B_n=120000-100000(1.005)^n.
Section II Question 21
(a) c=frac34.
Require total probability 1; evaluate int_0^2x(2-x),dx=4 / 3.
(b) F(x)=((3x²) / (4))-((x³) / (4)), and P(X>1)=frac12.
Integrate the density from 0 to x; evaluate F(1)=1 / 2; use P(X>1)=1-F(1).
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (3 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Require total probability 1.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Evaluate int_0^2x(2-x),dx=4 / 3.; Obtains the correct final result c=frac34.
Part b (1 mark)
Integrate the density from 0 to x.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Evaluate F(1)=1 / 2.
Part b (1 mark)
Use P(X>1)=1-F(1).; Obtains the correct final result F(x)=((3x²) / (4))-((x³) / (4)), and P(X>1)=frac12.
Section II Question 22
(a) Reflect in the x-axis, stretch vertically by factor 2, then translate 3 units right and 5 units up.
Read the multiplier -2; read the horizontal and vertical translations from vertex form.
(b) Vertex (3,5); range yle5.
Identify the vertex from the translated form; use the downward opening to state the maximum range.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Read the multiplier -2.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Read the horizontal and vertical translations from vertex form.; Obtains the correct final result Reflect in the x-axis, stretch vertically by factor 2, then translate 3 units right and 5 units up.
Part b (1 mark)
Identify the vertex from the translated form.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Use the downward opening to state the maximum range.; Obtains the correct final result Vertex (3,5); range yle5.
Section II Question 23
(a) S(r)=2pi r²+256pi / r.
Use h=128 / r²; substitute into S=2pi r²+2pi rh.
(b) r=4 cm and h=8 cm.
Solve S'(r)=4pi r-256pi / r²=0; obtain r=4 and hence h=8; use S''(4)>0 to justify the minimum.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (3 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Use h=128 / r^2.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Substitute into S=2pi r²+2pi rh.; Completes the required reasoning and reaches S(r)=2pi r²+256pi / r.
Part b (1 mark)
Solve S'(r)=4pi r-256pi / r²=0.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Obtain r=4 and hence h=8.
Part b (1 mark)
Use S''(4)>0 to justify the minimum.; Completes the required reasoning and reaches r=4 cm and h=8 cm.
Section II Question 24
(a) Mean 59; standard deviation 9.
Transform the mean using 1.5(42)-4; multiply the standard deviation by |1.5|.
(b) The transformed value is 71, and (71-59) / 9=4 / 3.
Calculate y=1.5(50)-4=71; standardise using the transformed mean and standard deviation.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Transform the mean using 1.5(42)-4.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Multiply the standard deviation by |1.5|.; Obtains the correct final result Mean 59; standard deviation 9.
Part b (1 mark)
Calculate y=1.5(50)-4=71.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Standardise using the transformed mean and standard deviation.; Completes the required reasoning and reaches The transformed value is 71, and (71-59) / 9=4 / 3.
Section II Question 25
(a) f(x)=2x³-6x²+2x+5.
Integrate f'(x); use f(0)=5 to determine the constant.
(b) x=1pm((sqrt6) / (3)).
Set 6x²-12x+2=0; apply the quadratic formula and simplify.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Integrate f'(x).
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Use f(0)=5 to determine the constant.; Obtains the correct final result f(x)=2x³-6x²+2x+5.
Part b (1 mark)
Set 6x²-12x+2=0.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Apply the quadratic formula and simplify.; Obtains the correct final result x=1pm((sqrt6) / (3)).
Section II Question 26
(a) A local maximum at (-1,2) and a local minimum at (1,-2).
Solve f'(x)=3x²-3=0; substitute the inputs and use f''(x)=6x.
(b) frac94 square units.
Factor to identify the zeros 0 and sqrt3; integrate 3x-x³ over the interval.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Solve f'(x)=3x²-3=0.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Substitute the inputs and use f''(x)=6x.; Completes the required reasoning and reaches A local maximum at (-1,2) and a local minimum at (1,-2).
Part b (1 mark)
Factor to identify the zeros 0 and sqrt3.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Integrate 3x-x³ over the interval.; Obtains the correct final result frac94 square units.
Section II Question 27
(a) Direction changes at t=1,3; s(1)=4 and s(3)=0.
Differentiate and factor v(t)=3(t-1)(t-3); verify sign changes and substitute the two times into s.
(b) 12 metres.
Evaluate s(0)=0 and s(4)=4; add the absolute changes 0to4, 4to0 and 0to4.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Differentiate and factor v(t)=3(t-1)(t-3).
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Verify sign changes and substitute the two times into s.; Obtains the correct final result Direction changes at t=1,3; s(1)=4 and s(3)=0.
Part b (1 mark)
Evaluate s(0)=0 and s(4)=4.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Add the absolute changes 0to4, 4to0 and 0to4.; Obtains the correct final result 12 metres.
Section II Question 28
(a) 300((1.004²⁴-1) / (0.004)).
Recognise the deposits as a geometric series; use the future-value annuity sum.
(b) 300((1.004²⁴-1) / (0.004))-7200.
Calculate total contributions 24(300)=7200; subtract them from the accumulated balance.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Recognise the deposits as a geometric series.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Use the future-value annuity sum.; Obtains the correct final result 300((1.004²⁴-1) / (0.004)).
Part b (1 mark)
Calculate total contributions 24(300)=7200.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Subtract them from the accumulated balance.; Obtains the correct final result 300((1.004²⁴-1) / (0.004))-7200.
Section II Question 29
(a) theta=fracpi3; sector area ((50pi) / (3))text( cm)²; segment area ((50pi) / (3))-25sqrt3text( cm)^2.
Use 10=2(10)sin(theta / 2) to obtain theta=pi / 3; calculate sector area frac12r^2theta; calculate triangle area frac12r^2sintheta=25sqrt3; subtract triangle area from sector area.
Mark allocation
- Part a (4 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Use 10=2(10)sin(theta / 2) to obtain theta=pi / 3.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Calculate sector area frac12r^2theta.
Part a (1 mark)
Calculate triangle area frac12r^2sintheta=25sqrt3.
Part a (1 mark)
Subtract triangle area from sector area.; Obtains the correct final result theta=fracpi3; sector area ((50pi) / (3))text( cm)²; segment area ((50pi) / (3))-25sqrt3text( cm)^2.
Section II Question 30
(a) a=(3 / 2)^(1 / 3), so P(t)=1200(3 / 2)^(t / 3).
Use 1200a³=1800; solve a³=3 / 2 and substitute into the model.
(b) ((3ln2) / (ln(3 / 2))).
Set (3 / 2)^(t / 3)=2; take logarithms and solve for t.
Mark allocation
- Part a (2 marks): award one mark for each distinct criterion listed below.
- Part b (2 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Use 1200a³=1800.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Solve a³=3 / 2 and substitute into the model.; Obtains the correct final result a=(3 / 2)^(1 / 3), so P(t)=1200(3 / 2)^(t / 3).
Part b (1 mark)
Set (3 / 2)^(t / 3)=2.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part b (1 mark)
Take logarithms and solve for t.; Obtains the correct final result ((3ln2) / (ln(3 / 2))).
Section II Question 31
(a) 8 square units.
Factor x(x-2)(x+2) to locate the three intercepts; use odd symmetry to double the area on [0,2]; integrate 4x-x³ from 0 to 2; evaluate 2(4)=8.
Mark allocation
- Part a (4 marks): award one mark for each distinct criterion listed below.
Detailed marking criteria
Part a (1 mark)
Factor x(x-2)(x+2) to locate the three intercepts.
Acceptable alternatives: Accept an equivalent valid method or expression that demonstrates this mathematical decision.
Part a (1 mark)
Use odd symmetry to double the area on [0,2].
Part a (1 mark)
Integrate 4x-x³ from 0 to 2.
Part a (1 mark)
Evaluate 2(4)=8.; Obtains the correct final result 8 square units.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Calculus and Applications | Q2, Q7, Q14(a), Q14(b), Q16(a), Q16(b), Q23(a), Q23(b), Q25(a), Q25(b), Q26(a), Q26(b), Q27(a), Q27(b), Q31(a) | 33 | ___ | Review the listed calculus and applications items, their worked solutions and the evidence-specific marking criteria. |
| Functions, Graphs and Modelling | Q1, Q9, Q11(a), Q11(b), Q12(a), Q12(b), Q22(a), Q22(b), Q30(a), Q30(b) | 18 | ___ | Review the listed functions, graphs and modelling items, their worked solutions and the evidence-specific marking criteria. |
| Probability and Statistics | Q3, Q5, Q8, Q10, Q13(a), Q13(b), Q17(a), Q17(b), Q18(a), Q18(b), Q21(a), Q21(b), Q24(a), Q24(b) | 25 | ___ | Review the listed probability and statistics items, their worked solutions and the evidence-specific marking criteria. |
| Sequences and Financial Mathematics | Q4, Q15(a), Q15(b), Q20(a), Q20(b), Q28(a), Q28(b) | 14 | ___ | Review the listed sequences and financial mathematics items, their worked solutions and the evidence-specific marking criteria. |
| Trigonometry and Geometry | Q6, Q19(a), Q19(b), Q29(a) | 10 | ___ | Review the listed trigonometry and geometry items, their worked solutions and the evidence-specific marking criteria. |