Skill Align TASC Mathematics Methods Level 4 (MTM415117) Section B Calculator-Permitted Practice Booklet Pack 0 - 2026 Edition
Current for 2026 | Course code MTM415117. This original Skill Align booklet practises Section B only: 100 marks, approximately 100 minutes and calculator permitted. The separate 80-mark, calculator-prohibited Section A booklet is not included, so this resource is not a complete three-hour external-examination simulation.
- Paper
- Section B Calculator-Permitted Practice Booklet Showcase
- Preparation time
- Optional 15-minute preparation
- Suggested working time
- Approximately 100 minutes
- Assessment
- 100 marks
Section B conditions: TASC-approved calculators are permitted. The current TASC MTM415117 information sheet must accompany this booklet. Open the official information-sheet link displayed below and supply that sheet with this booklet. Internet access and external communication are not permitted during the practice session.
Part 1 - Functions and Graphs - Criterion 4 (20 marks)
Answer all four questions in this part. Question values vary from 3 to 8 marks; the value of every labelled subpart is printed beside that subpart. Show sufficient setup, working, reasoning and interpretation to support each answer.
Question 1
3 marksQuestion 2
4 marksQuestion 3
5 marksQuestion 4
8 marksPart 2 - Circular Functions - Criterion 5 (20 marks)
Answer all four questions in this part. Question values vary from 3 to 8 marks; the value of every labelled subpart is printed beside that subpart. Show sufficient setup, working, reasoning and interpretation to support each answer.
Question 5
3 marksQuestion 6
4 marksQuestion 7
5 marksQuestion 8
8 marksPart 3 - Differential Calculus - Criterion 6 (20 marks)
Answer all four questions in this part. Question values vary from 3 to 8 marks; the value of every labelled subpart is printed beside that subpart. Show sufficient setup, working, reasoning and interpretation to support each answer.
Question 9
3 marksQuestion 10
4 marksQuestion 11
5 marksQuestion 12
8 marksPart 4 - Integral Calculus - Criterion 7 (20 marks)
Answer all four questions in this part. Question values vary from 3 to 8 marks; the value of every labelled subpart is printed beside that subpart. Show sufficient setup, working, reasoning and interpretation to support each answer.
Question 13
3 marksQuestion 14
4 marksQuestion 15
5 marksQuestion 16
8 marksPart 5 - Statistics and Probability - Criterion 8 (20 marks)
Answer all four questions in this part. Question values vary from 3 to 8 marks; the value of every labelled subpart is printed beside that subpart. Show sufficient setup, working, reasoning and interpretation to support each answer.
Question 17
3 marksQuestion 18
4 marksQuestion 19
5 marksQuestion 20
8 marksWorked Solutions And Marking Guide
Question 1
(a) Domain xne-3; vertical asymptote x=-3.
The denominator is zero at x=-3, so that input is excluded and gives the vertical asymptote.
(b) Horizontal asymptote y=2.
The numerator and denominator have equal degree, so the asymptote is the ratio of leading coefficients.
(c) Range yne2.
Solving f(x)=2 gives -1=6, so the function never takes the value 2.
Mark allocation
- TASC-style partial-mark policy: one-mark items may receive 0.5 marks for valid working; items worth two or more marks receive consequential partial credit for correct later work after an earlier arithmetic slip.
- A correct answer without required working is capped at 1.5 marks for a two-mark item and at half marks for an item worth three or more marks.
- Accept algebraically equivalent exact forms and methods when sufficient working is shown. For a rounded answer, apply only the precision and acceptance interval stated for that subpart.
Detailed marking criteria
Part 1(a) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States the excluded input and vertical asymptote x=-3
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 1(b) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States the horizontal asymptote y=2
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 1(c) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States the range yne2
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 2
(a) Writes y=3e^(2x)-4; Isolates e^(2x)=(y+4) / 3
Let y=3e^(2x)-4. Then (y+4) / 3=e^(2x), so x=0.5ln((y+4) / 3). Swap x and y.
(b) Applies ln and divides by 2; Swaps variables correctly; States inverse domain x>-4 and range all real
Let y=3e^(2x)-4. Then (y+4) / 3=e^(2x), so x=0.5ln((y+4) / 3). Swap x and y.
Detailed marking criteria
Part 2(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Writes y=3e^(2x)-4
- Isolates e^(2x)=(y+4) / 3
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 2(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Applies ln and divides by 2
- Swaps variables correctly; States inverse domain x>-4 and range all real
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 3
(a) States mathematical domain t>-1; States contextual domain tgeq0
Set 4=12-4ln(t+1), so ln(t+1)=2 and t=e²-1=6.389... minutes. The logarithm gives mathematical domain t>-1, while elapsed time restricts the model to tgeq0.
(b) Obtains ln(t+1)=2; Solves t=e²-1; Rounds to 6.39 minutes
Set 4=12-4ln(t+1), so ln(t+1)=2 and t=e²-1=6.389... minutes. The logarithm gives mathematical domain t>-1, while elapsed time restricts the model to tgeq0.
Detailed marking criteria
Part 3(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States mathematical domain t>-1
- States contextual domain tgeq0
Acceptable alternatives: For the requested two-decimal answer, accept 6.385 leq t < 6.395 minutes; also accept the exact value e² - 1. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 3(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Obtains ln(t+1)=2
- Solves t=e²-1
- Rounds to 6.39 minutes
Acceptable alternatives: For the requested two-decimal answer, accept 6.385 leq t < 6.395 minutes; also accept the exact value e² - 1. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 4
(a) Reflect in the x-axis, stretch vertically by factor 2, translate 1 unit right and 5 units up.
The coefficient -2 gives the reflection and vertical stretch; the nested and external constants give the translations.
(b) Point of inflection (1,5); y-intercept (0,7); x-intercept (1+√3(5 / 2),0).
The transformed centre is (1,5). Substitution gives f(0)=7, while f(x)=0 gives (x-1)³=5 / 2.
(c) A decreasing cubic through the labelled intercepts and point of inflection, rising to the left and falling to the right.
The negative leading cubic determines the end behaviour. The curve is smooth and changes concavity at (1,5).
Detailed marking criteria
Part 4(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Identifies the reflection and vertical stretch by factor 2
- Identifies translations 1 right and 5 up
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 4(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States the point of inflection (1,5)
- Finds the y-intercept (0,7)
- Finds the exact x-intercept
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 4(c) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Draws a decreasing cubic with correct end behaviour
- Labels the point of inflection and both intercepts
- Shows the concavity change at (1,5)
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 5
(a) Isolates sin x=√3 / 2
sin x=√3 / 2 has reference angle π / 3 and is positive in Quadrants I and II.
(b) Identifies reference angle π / 3; Selects Quadrant I; Selects Quadrant II; States π / 3 and 2π / 3
sin x=√3 / 2 has reference angle π / 3 and is positive in Quadrants I and II.
Detailed marking criteria
Part 5(a) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Isolates sin x=√3 / 2
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 5(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Identifies reference angle π / 3; Selects Quadrant I
- Selects Quadrant II; States π / 3 and 2π / 3
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 6
(a) States amplitude 3; Calculates period π
Write 2x-π / 2=2(x-π / 4). Thus amplitude is 3, period π, shift right π / 4 and vertical shift -1.
(b) Factors to identify shift right π / 4; States midline y=-1; States range [-4,2]
Write 2x-π / 2=2(x-π / 4). Thus amplitude is 3, period π, shift right π / 4 and vertical shift -1.
Detailed marking criteria
Part 6(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States amplitude 3
- Calculates period π
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 6(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Factors to identify shift right π / 4
- States midline y=-1; States range [-4,2]
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 7
(a) Recognises sin(π-x)=sin x; Recognises cos(2π-x)=cos x
Using supplementary and full-turn identities, sin(π-x)=sin x and cos(2π-x)=cos x.
(b) Uses correct signs; Combines the terms; States sin x+cos x
Using supplementary and full-turn identities, sin(π-x)=sin x and cos(2π-x)=cos x.
Detailed marking criteria
Part 7(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Recognises sin(π-x)=sin x
- Recognises cos(2π-x)=cos x
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 7(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Uses correct signs
- Combines the terms
- States sin x+cos x
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 8
(a) Range 3 m leq h leq 25 m; period 40 s.
The midline is 14 m, the amplitude is 11 m and the angular coefficient gives period 40 seconds.
(b) At t = 20 s, the height is 25 m.
Maximum height occurs when cos(π t / 20)=-1, first at t=20.
(c) The cycle passes through (0,3), (10,14), (20,25), (30,14) and (40,3).
Quarter-period points occur every 10 seconds, beginning at the minimum height.
Detailed marking criteria
Part 8(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States the range 3 m to 25 m
- States the period 40 s
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 8(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Finds the first maximum time 20 s
- States the maximum height 25 m
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 8(c) (4 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Uses axes t in seconds and h in metres
- Plots both minima and the maximum
- Plots the midline crossings at t=10 and t=30
- Draws a smooth cosine cycle through the labelled points
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 9
(a) States the product rule
The product rule gives 2xe^x+x²e^x=e^x(x²+2x). At x=1 the gradient is 3e.
(b) Obtains 2xe^x; Obtains x²e^x; Factorises the derivative correctly; Evaluates f'(1)=3e
The product rule gives 2xe^x+x²e^x=e^x(x²+2x). At x=1 the gradient is 3e.
Detailed marking criteria
Part 9(a) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States the product rule
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 9(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Obtains 2xe^x; Obtains x²e^x
- Factorises the derivative correctly; Evaluates f'(1)=3e
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 10
(a) Finds f'=(1-ln x) / x²; Solves ln x=1 to get x=e
f'=(1-ln x) / x². It is zero when ln x=1, so x=e. The numerator changes positive to negative, hence a maximum; f(e)=1 / e.
(b) Uses the derivative sign change; Classifies a maximum; States coordinate (e,1 / e)
f'=(1-ln x) / x². It is zero when ln x=1, so x=e. The numerator changes positive to negative, hence a maximum; f(e)=1 / e.
Detailed marking criteria
Part 10(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Finds f'=(1-ln x) / x²
- Solves ln x=1 to get x=e
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 10(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Uses the derivative sign change
- Classifies a maximum; States coordinate (e,1 / e)
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 11
(a) Identifies the outer power function; Obtains 5(3x²+1)⁴
The outer derivative is 5(3x²+1)⁴ and the inner derivative is 6x, giving 30x(3x²+1)⁴.
(b) Differentiates inner function to 6x; Multiplies the factors; States 30x(3x²+1)⁴
The outer derivative is 5(3x²+1)⁴ and the inner derivative is 6x, giving 30x(3x²+1)⁴.
Detailed marking criteria
Part 11(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Identifies the outer power function
- Obtains 5(3x²+1)⁴
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 11(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Differentiates inner function to 6x
- Multiplies the factors
- States 30x(3x²+1)⁴
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 12
(a) S'(r)=4π r-dfrac(600π)(r²).
Differentiate the power and reciprocal terms separately.
(b) r=√3(150) cm, so r=5.31 cm to two decimal places.
Solving S'(r)=0 gives 4π r³=600π, hence r³=150.
(c) S''(r)=4π+1200π / r³>0, so the stationary point is a minimum; h=300 / r²=10.63 cm.
The second derivative is positive for r>0. The volume relation gives h=300 / r².
Detailed marking criteria
Part 12(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Differentiates 2πr² to 4πr
- Differentiates 600π / r to -600π / r²
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 12(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Sets the derivative equal to zero
- Obtains the exact radius √3(150) cm
- Rounds to 5.31 cm
Acceptable alternatives: For two decimal places, accept 5.305 leq r < 5.315 cm; also accept the exact value √3(150) cm. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 12(c) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Obtains a positive second derivative for r>0
- Uses h=300 / r²
- States the minimum classification and h=10.63 cm
Acceptable alternatives: For two decimal places, accept 10.625 leq h < 10.635 cm. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 13
(a) Integrates 6x² to 2x³
Integrating term by term gives 2x³-2x²+3x+C.
(b) Integrates -4x to -2x²; Integrates 3 to 3x; Includes the constant of integration; States the complete antiderivative
Integrating term by term gives 2x³-2x²+3x+C.
Detailed marking criteria
Part 13(a) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Integrates 6x² to 2x³
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 13(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Integrates -4x to -2x²; Integrates 3 to 3x
- Includes the constant of integration; States the complete antiderivative
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 14
(a) Identifies the area integral from 0 to 2; Finds antiderivative 4x-x³ / 3
The function is nonnegative. Integral from 0 to 2 of (4-x²) dx = [4x-x³ / 3]_0²=8-8 / 3=16 / 3.
(b) Applies both limits; Simplifies 8-8 / 3; States 16 / 3 square units
The function is nonnegative. Integral from 0 to 2 of (4-x²) dx = [4x-x³ / 3]_0²=8-8 / 3=16 / 3.
Detailed marking criteria
Part 14(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Identifies the area integral from 0 to 2
- Finds antiderivative 4x-x³ / 3
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 14(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Applies both limits
- Simplifies 8-8 / 3; States 16 / 3 square units
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 15
(a) Finds intersections x=0,2; Identifies upper curve 2x
Intersections satisfy x²=2x, so x=0,2. On this interval 2x is above x². Integral 0 to 2 of (2x-x²) dx=4 / 3.
(b) Forms integral of 2x-x²; Evaluates the definite integral; States 4 / 3 square units
Intersections satisfy x²=2x, so x=0,2. On this interval 2x is above x². Integral 0 to 2 of (2x-x²) dx=4 / 3.
Detailed marking criteria
Part 15(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Finds intersections x=0,2
- Identifies upper curve 2x
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 15(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Forms integral of 2x-x²
- Evaluates the definite integral
- States 4 / 3 square units
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 16
(a) A(x)=x³-4x.
Integrate 3t²-4 and use the lower limit zero.
(b) A(2)=0; the positive and negative signed areas on the interval cancel.
Substitution gives 8-8=0, which is net accumulation rather than total geometric area.
(c) The graph crosses at x=2 / sqrt3. A decreases on 0<x<2 / sqrt3 and increases on 2 / sqrt3<x<2.
The sign of the integrand is the sign of A′, so it determines whether the accumulation function falls or rises.
Detailed marking criteria
Part 16(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Finds the antiderivative t³-4t
- Applies the limits to obtain A(x)=x³-4x
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 16(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Evaluates A(2)=0
- Explains that positive and negative signed areas cancel
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 16(c) (4 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Draws the upward-opening curve on the stated interval
- Labels the zero x=2 / √3
- Identifies the interval where A decreases
- Identifies the interval where A increases
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 17
(a) q=0.40.
The probabilities must sum to one, so q=1-0.15-0.25-0.20=0.40.
(b) E(X)=1.65. Over many repetitions, the mean value of X approaches 1.65.
E(X)=0(0.15)+1(0.25)+2(0.40)+3(0.20)=1.65.
Detailed marking criteria
Part 17(a) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Uses total probability one to obtain q=0.40
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 17(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Calculates E(X)=1.65
- Interprets 1.65 as the long-run mean
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 18
(a) np=6 and np(1-p)=4.2.
Use the binomial mean and variance formulas.
(b) p=0.30 and n=20.
Dividing the variance equation by the mean equation gives 1-p=0.7, so p=0.3 and n=6 / 0.3=20.
Detailed marking criteria
Part 18(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- States np=6
- States np(1-p)=4.2
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 18(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Determines p=0.30
- Determines n=20
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 19
(a) mu=60.
The two percentiles are equally distant from the mean, so mu=(52+68) / 2=60.
(b) sigma=6.24 to two decimal places; 60+1.2816(6.24)approx68.
The distance 68-60=8 equals 1.2816sigma, so sigma=8 / 1.2816=6.242ldots.
Detailed marking criteria
Part 19(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Recognises the percentile symmetry
- Calculates the mean as 60
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 19(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Forms 8=1.2816 sigma
- Calculates sigma=6.24 to two decimal places
- Substitutes to verify the 90th percentile
Acceptable alternatives: For two decimal places, accept 6.235 leq sigma < 6.245. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 20
(a) The parameter is p; the statistic is hat p=138 / 240.
The unknown population proportion is estimated by the observed sample proportion.
(b) hat p=0.575; nhat p=138 and n(1-hat p)=102, both greater than 10.
The observed success and failure counts are both sufficiently large.
(c) operatorname(SE)(hat p)=0.0319.
operatorname(SE)(hat p)=√(0.575(0.425) / 240)=0.03191ldots.
(d) The interval is (0.512,0.638). It is a plausible range for the population proportion of voters who support the proposal.
0.575pm1.96(0.03191ldots)=(0.51245ldots,0.63755ldots).
Detailed marking criteria
Part 20(a) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Identifies p and the sample statistic hat p
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 20(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Calculates hat p=0.575
- Verifies both observed counts exceed 10
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 20(c) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Uses the estimated standard-error formula
- Rounds the standard error to 0.0319
Acceptable alternatives: For four decimal places, accept 0.03185 leq SE < 0.03195. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 20(d) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Mark-by-mark checkpoints:
- Calculates the margin of error
- Draws and labels the interval (0.512,0.638)
- Interprets the interval for the population proportion
Acceptable alternatives: For three-decimal endpoints, accept 0.5115 leq lower < 0.5125 and 0.6375 leq upper < 0.6385. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 1(a) | Skill: State the domain and vertical asymptote | Expected result: Domain x\(\ne\)−3; vertical asymptote x=−3. | Mark-by-mark evidence: States the excluded input and vertical asymptote x=−3 | Q1(a) | 1 | ___ | Rework Question 1(a): State the domain and vertical asymptote. Then verify these item-specific checkpoints: States the excluded input and vertical asymptote x=-3 |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 1(b) | Skill: State the horizontal asymptote | Expected result: Horizontal asymptote y=2. | Mark-by-mark evidence: States the horizontal asymptote y=2 | Q1(b) | 1 | ___ | Rework Question 1(b): State the horizontal asymptote. Then verify these item-specific checkpoints: States the horizontal asymptote y=2 |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 1(c) | Skill: State the range | Expected result: Range y\(\ne\)2. | Mark-by-mark evidence: States the range y\(\ne\)2 | Q1(c) | 1 | ___ | Rework Question 1(c): State the range. Then verify these item-specific checkpoints: States the range yne2 |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 2(a) | Skill: Let \(y=3e^{2x}-4\) and isolate \(e^{2x}\) | Expected result: Writes y=\(3e^{2x}\)-4; Isolates \(e^{2x}\)=(y+4)/3 | Mark-by-mark evidence: Writes y=\(3e^{2x}\)-4; Isolates \(e^{2x}\)=(y+4)/3 | Q2(a) | 2 | ___ | Rework Question 2(a): Let y=3e^(2x)-4 and isolate e^(2x). Then verify these item-specific checkpoints: Writes y=3e^(2x)-4; Isolates e^(2x)=(y+4) / 3 |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 2(b) | Skill: Take natural logarithms, find \(f^{-1}(x)\), and state its domain and range | Expected result: Applies ln and divides by 2; Swaps variables correctly; States inverse domain x>-4 and range all real | Mark-by-mark evidence: Applies ln and divides by 2; Swaps variables correctly; States inverse domain x>-4 and range all real | Q2(b) | 2 | ___ | Rework Question 2(b): Take natural logarithms, find f⁻¹(x), and state its domain and range. Then verify these item-specific checkpoints: Applies ln and divides by 2; Swaps variables correctly; States inverse domain x>-4 and range all real |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 3(a) | Skill: State the mathematical domain of \(L(t)\) and the contextual domain for elapsed time | Expected result: States mathematical domain t>-1; States contextual domain t\(\geq\)0 | Mark-by-mark evidence: States mathematical domain t>-1; States contextual domain t\(\geq\)0 | Q3(a) | 2 | ___ | Rework Question 3(a): State the mathematical domain of L(t) and the contextual domain for elapsed time. Then verify these item-specific checkpoints: States mathematical domain t>-1; States contextual domain tgeq0 |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 3(b) | Skill: Solve \(L(t)=4\). Give the exact time and the time in minutes correct to two decimal places | Expected result: Obtains \(\ln\left(t+1\right)\)=2; Solves t=\(e^{2}\)-1; Rounds to 6.39 minutes | Mark-by-mark evidence: Obtains \(\ln\left(t+1\right)\)=2; Solves t=\(e^{2}\)-1; Rounds to 6.39 minutes | Q3(b) | 3 | ___ | Rework Question 3(b): Solve L(t)=4. Give the exact time and the time in minutes correct to two decimal places. Then verify these item-specific checkpoints: Obtains ln(t+1)=2; Solves t=e²-1; Rounds to 6.39 minutes |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 4(a) | Skill: Describe the transformations that map \(y=x^3\) to \(y=f(x)\) | Expected result: Reflect in the x-axis, stretch vertically by factor 2, translate 1 unit right and 5 units up. | Mark-by-mark evidence: Identifies the reflection and vertical stretch by factor 2; Identifies translations 1 right and 5 up | Q4(a) | 2 | ___ | Rework Question 4(a): Describe the transformations that map y=x³ to y=f(x). Then verify these item-specific checkpoints: Identifies the reflection and vertical stretch by factor 2; Identifies translations 1 right and 5 up |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 4(b) | Skill: Find the point of inflection and both axis intercepts. Give the x-intercept exactly | Expected result: Point of inflection \((1,5)\); y-intercept \((0,7)\); x-intercept \(\left(1+\sqrt[3]{5/2},0\right)\). | Mark-by-mark evidence: States the point of inflection (1,5); Finds the y-intercept (0,7); Finds the exact x-intercept | Q4(b) | 3 | ___ | Rework Question 4(b): Find the point of inflection and both axis intercepts. Give the x-intercept exactly. Then verify these item-specific checkpoints: States the point of inflection (1,5); Finds the y-intercept (0,7); Finds the exact x-intercept |
| Section B | Part 1 | Criterion 4 | Functions and Graphs | Criterion 4 | Question 4(c) | Skill: Sketch the graph, labelling the point of inflection and intercepts, and show the correct end behaviour | Expected result: A decreasing cubic through the labelled intercepts and point of inflection, rising to the left and falling to the right. | Mark-by-mark evidence: Draws a decreasing cubic with correct end behaviour; Labels the point of inflection and both intercepts; Shows the concavity change at (1,5) | Q4(c) | 3 | ___ | Rework Question 4(c): Sketch the graph, labelling the point of inflection and intercepts, and show the correct end behaviour. Then verify these item-specific checkpoints: Draws a decreasing cubic with correct end behaviour; Labels the point of inflection and both intercepts; Shows the concavity change at (1,5) |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 5(a) | Skill: Isolate \(\sin x\) | Expected result: Isolates \(\sin x\)=\(\sqrt{3}\)/2 | Mark-by-mark evidence: Isolates \(\sin x\)=\(\sqrt{3}\)/2 | Q5(a) | 1 | ___ | Rework Question 5(a): Isolate sin x. Then verify these item-specific checkpoints: Isolates sin x=√3 / 2 |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 5(b) | Skill: State the exact reference angle and find every solution for \(0\leq x<2\pi\) | Expected result: Identifies reference angle \(\pi\)/3; Selects Quadrant I; Selects Quadrant II; States \(\pi\)/3 and 2\(\pi\)/3 | Mark-by-mark evidence: Identifies reference angle \(\pi\)/3; Selects Quadrant I; Selects Quadrant II; States \(\pi\)/3 and 2\(\pi\)/3 | Q5(b) | 2 | ___ | Rework Question 5(b): State the exact reference angle and find every solution for 0leq x<2π. Then verify these item-specific checkpoints: Identifies reference angle π / 3; Selects Quadrant I; Selects Quadrant II; States π / 3 and 2π / 3 |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 6(a) | Skill: For \(y=3\cos(2x-\pi/2)-1\), state the amplitude and period | Expected result: States amplitude 3; Calculates period \(\pi\) | Mark-by-mark evidence: States amplitude 3; Calculates period \(\pi\) | Q6(a) | 2 | ___ | Rework Question 6(a): For y=3cos(2x-π / 2)-1, state the amplitude and period. Then verify these item-specific checkpoints: States amplitude 3; Calculates period π |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 6(b) | Skill: State the phase shift, midline and range | Expected result: Factors to identify shift right \(\pi\)/4; States midline y=-1; States range [-4,2] | Mark-by-mark evidence: Factors to identify shift right \(\pi\)/4; States midline y=-1; States range [-4,2] | Q6(b) | 2 | ___ | Rework Question 6(b): State the phase shift, midline and range. Then verify these item-specific checkpoints: Factors to identify shift right π / 4; States midline y=-1; States range [-4,2] |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 7(a) | Skill: Simplify \(\sin(\pi-x)\) and \(\cos(2\pi-x)\) separately | Expected result: Recognises \(\sin\left(\pi-x\right)\)=\(\sin x\); Recognises \(\cos\left(2\pi-x\right)\)=\(\cos x\) | Mark-by-mark evidence: Recognises \(\sin\left(\pi-x\right)\)=\(\sin x\); Recognises \(\cos\left(2\pi-x\right)\)=\(\cos x\) | Q7(a) | 2 | ___ | Rework Question 7(a): Simplify sin(π-x) and cos(2π-x) separately. Then verify these item-specific checkpoints: Recognises sin(π-x)=sin x; Recognises cos(2π-x)=cos x |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 7(b) | Skill: Hence simplify the complete expression | Expected result: Uses correct signs; Combines the terms; States \(\sin x\)+\(\cos x\) | Mark-by-mark evidence: Uses correct signs; Combines the terms; States \(\sin x\)+\(\cos x\) | Q7(b) | 3 | ___ | Rework Question 7(b): Hence simplify the complete expression. Then verify these item-specific checkpoints: Uses correct signs; Combines the terms; States sin x+cos x |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 8(a) | Skill: State the range and period of the model, including units | Expected result: Range 3 m \(\leq\) h \(\leq\) 25 m; period 40 s. | Mark-by-mark evidence: States the range 3 m to 25 m; States the period 40 s | Q8(a) | 2 | ___ | Rework Question 8(a): State the range and period of the model, including units. Then verify these item-specific checkpoints: States the range 3 m to 25 m; States the period 40 s |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 8(b) | Skill: Find the first time the seat reaches its maximum height and state that height | Expected result: At t = 20 s, the height is 25 m. | Mark-by-mark evidence: Finds the first maximum time 20 s; States the maximum height 25 m | Q8(b) | 2 | ___ | Rework Question 8(b): Find the first time the seat reaches its maximum height and state that height. Then verify these item-specific checkpoints: Finds the first maximum time 20 s; States the maximum height 25 m |
| Section B | Part 2 | Criterion 5 | Circular Functions | Criterion 5 | Question 8(c) | Skill: Sketch one full cycle on the stated interval. Label the minimum, maximum, midline crossings and axes with units | Expected result: The cycle passes through (0,3), (10,14), (20,25), (30,14) and (40,3). | Mark-by-mark evidence: Uses axes t in seconds and h in metres; Plots both minima and the maximum; Plots the midline crossings at t=10 and t=30; Draws a smooth cosine cycle through the labelled points | Q8(c) | 4 | ___ | Rework Question 8(c): Sketch one full cycle on the stated interval. Label the minimum, maximum, midline crossings and axes with units. Then verify these item-specific checkpoints: Uses axes t in seconds and h in metres; Plots both minima and the maximum; Plots the midline crossings at t=10 and t=30; Draws a smooth cosine cycle through the labelled points |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 9(a) | Skill: Write the product-rule setup for differentiating \(f(x)=x^2e^x\) | Expected result: States the product rule | Mark-by-mark evidence: States the product rule | Q9(a) | 1 | ___ | Rework Question 9(a): Write the product-rule setup for differentiating f(x)=x^2e^x. Then verify these item-specific checkpoints: States the product rule |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 9(b) | Skill: Differentiate both factors, simplify \(f'(x)\), and evaluate \(f'(1)\) | Expected result: Obtains 2xe^x; Obtains \(x^{2}\)e^x; Factorises the derivative correctly; Evaluates f'(1)=3e | Mark-by-mark evidence: Obtains 2xe^x; Obtains \(x^{2}\)e^x; Factorises the derivative correctly; Evaluates f'(1)=3e | Q9(b) | 2 | ___ | Rework Question 9(b): Differentiate both factors, simplify f'(x), and evaluate f'(1). Then verify these item-specific checkpoints: Obtains 2xe^x; Obtains x²e^x; Factorises the derivative correctly; Evaluates f'(1)=3e |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 10(a) | Skill: Differentiate \(f(x)=\ln(x)/x\) and determine the stationary x-coordinate | Expected result: Finds f'=(1-\(\ln x\))/\(x^{2}\); Solves \(\ln x\)=1 to get x=e | Mark-by-mark evidence: Finds f'=(1-\(\ln x\))/\(x^{2}\); Solves \(\ln x\)=1 to get x=e | Q10(a) | 2 | ___ | Rework Question 10(a): Differentiate f(x)=ln(x) / x and determine the stationary x-coordinate. Then verify these item-specific checkpoints: Finds f'=(1-ln x) / x²; Solves ln x=1 to get x=e |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 10(b) | Skill: Use derivative signs to classify the stationary point and state its coordinates | Expected result: Uses the derivative sign change; Classifies a maximum; States coordinate (e,1/e) | Mark-by-mark evidence: Uses the derivative sign change; Classifies a maximum; States coordinate (e,1/e) | Q10(b) | 2 | ___ | Rework Question 10(b): Use derivative signs to classify the stationary point and state its coordinates. Then verify these item-specific checkpoints: Uses the derivative sign change; Classifies a maximum; States coordinate (e,1 / e) |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 11(a) | Skill: Identify the outer power function and differentiate it while retaining the inner expression | Expected result: Identifies the outer power function; Obtains 5\((3x^2+1)^{4}\) | Mark-by-mark evidence: Identifies the outer power function; Obtains 5\((3x^2+1)^{4}\) | Q11(a) | 2 | ___ | Rework Question 11(a): Identify the outer power function and differentiate it while retaining the inner expression. Then verify these item-specific checkpoints: Identifies the outer power function; Obtains 5(3x²+1)⁴ |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 11(b) | Skill: Differentiate the inner function, apply the chain rule, and state the simplified derivative | Expected result: Differentiates inner function to 6x; Multiplies the factors; States 30x\((3x^2+1)^{4}\) | Mark-by-mark evidence: Differentiates inner function to 6x; Multiplies the factors; States 30x\((3x^2+1)^{4}\) | Q11(b) | 3 | ___ | Rework Question 11(b): Differentiate the inner function, apply the chain rule, and state the simplified derivative. Then verify these item-specific checkpoints: Differentiates inner function to 6x; Multiplies the factors; States 30x(3x²+1)⁴ |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 12(a) | Skill: Find \(S'(r)\) | Expected result: \(S'(r)=4\pi r-\dfrac{600\pi}{r^2}\). | Mark-by-mark evidence: Differentiates 2πr² to 4πr; Differentiates 600\(\pi\)/r to −600\(\pi\)/r² | Q12(a) | 2 | ___ | Rework Question 12(a): Find S'(r). Then verify these item-specific checkpoints: Differentiates 2πr² to 4πr; Differentiates 600π / r to -600π / r² |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 12(b) | Skill: Find the stationary radius and give it correct to two decimal places | Expected result: \(r=\sqrt[3]{150}\text{ cm}\), so \(r=5.31\text{ cm}\) to two decimal places. | Mark-by-mark evidence: Sets the derivative equal to zero; Obtains the exact radius \(\sqrt[3]{150}\) cm; Rounds to 5.31 cm | Q12(b) | 3 | ___ | Rework Question 12(b): Find the stationary radius and give it correct to two decimal places. Then verify these item-specific checkpoints: Sets the derivative equal to zero; Obtains the exact radius √3(150) cm; Rounds to 5.31 cm |
| Section B | Part 3 | Criterion 6 | Differential Calculus | Criterion 6 | Question 12(c) | Skill: Justify that this radius gives a minimum and find the corresponding cylinder height correct to two decimal places | Expected result: \(S''(r)=4\pi+1200\pi/r^3>0\), so the stationary point is a minimum; \(h=300/r^2=10.63\text{ cm}\). | Mark-by-mark evidence: Obtains a positive second derivative for r>0; Uses h=300/r²; States the minimum classification and h=10.63 cm | Q12(c) | 3 | ___ | Rework Question 12(c): Justify that this radius gives a minimum and find the corresponding cylinder height correct to two decimal places. Then verify these item-specific checkpoints: Obtains a positive second derivative for r>0; Uses h=300 / r²; States the minimum classification and h=10.63 cm |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 13(a) | Skill: Integrate the \(6x^2\) term | Expected result: Integrates \(6x^{2}\) to \(2x^{3}\) | Mark-by-mark evidence: Integrates \(6x^{2}\) to \(2x^{3}\) | Q13(a) | 1 | ___ | Rework Question 13(a): Integrate the 6x² term. Then verify these item-specific checkpoints: Integrates 6x² to 2x³ |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 13(b) | Skill: Integrate the \(-4x\) and constant terms, then state the complete general antiderivative including the integration constant | Expected result: Integrates -4x to -\(2x^{2}\); Integrates 3 to 3x; Includes the constant of integration; States the complete antiderivative | Mark-by-mark evidence: Integrates -4x to -\(2x^{2}\); Integrates 3 to 3x; Includes the constant of integration; States the complete antiderivative | Q13(b) | 2 | ___ | Rework Question 13(b): Integrate the -4x and constant terms, then state the complete general antiderivative including the integration constant. Then verify these item-specific checkpoints: Integrates -4x to -2x²; Integrates 3 to 3x; Includes the constant of integration; States the complete antiderivative |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 14(a) | Skill: Write the area integral and determine its antiderivative | Expected result: Identifies the area integral from 0 to 2; Finds antiderivative 4x-\(x^{3}\)/3 | Mark-by-mark evidence: Identifies the area integral from 0 to 2; Finds antiderivative 4x-\(x^{3}\)/3 | Q14(a) | 2 | ___ | Rework Question 14(a): Write the area integral and determine its antiderivative. Then verify these item-specific checkpoints: Identifies the area integral from 0 to 2; Finds antiderivative 4x-x³ / 3 |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 14(b) | Skill: Apply both limits, simplify the result, and state the exact area | Expected result: Applies both limits; Simplifies 8-8/3; States 16/3 square units | Mark-by-mark evidence: Applies both limits; Simplifies 8-8/3; States 16/3 square units | Q14(b) | 2 | ___ | Rework Question 14(b): Apply both limits, simplify the result, and state the exact area. Then verify these item-specific checkpoints: Applies both limits; Simplifies 8-8 / 3; States 16 / 3 square units |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 15(a) | Skill: Find the intersection points and identify the upper curve | Expected result: Finds intersections x=0,2; Identifies upper curve 2x | Mark-by-mark evidence: Finds intersections x=0,2; Identifies upper curve 2x | Q15(a) | 2 | ___ | Rework Question 15(a): Find the intersection points and identify the upper curve. Then verify these item-specific checkpoints: Finds intersections x=0,2; Identifies upper curve 2x |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 15(b) | Skill: Write and evaluate the area integral | Expected result: Forms integral of 2x-\(x^{2}\); Evaluates the definite integral; States 4/3 square units | Mark-by-mark evidence: Forms integral of 2x-\(x^{2}\); Evaluates the definite integral; States 4/3 square units | Q15(b) | 3 | ___ | Rework Question 15(b): Write and evaluate the area integral. Then verify these item-specific checkpoints: Forms integral of 2x-x²; Evaluates the definite integral; States 4 / 3 square units |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 16(a) | Skill: Find an exact expression for \(A(x)\) | Expected result: \(A(x)=x^3-4x\). | Mark-by-mark evidence: Finds the antiderivative t³−4t; Applies the limits to obtain A(x)=x³−4x | Q16(a) | 2 | ___ | Rework Question 16(a): Find an exact expression for A(x). Then verify these item-specific checkpoints: Finds the antiderivative t³-4t; Applies the limits to obtain A(x)=x³-4x |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 16(b) | Skill: Evaluate \(A(2)\) and interpret the value as signed area | Expected result: A(2)=0; the positive and negative signed areas on the interval cancel. | Mark-by-mark evidence: Evaluates A(2)=0; Explains that positive and negative signed areas cancel | Q16(b) | 2 | ___ | Rework Question 16(b): Evaluate A(2) and interpret the value as signed area. Then verify these item-specific checkpoints: Evaluates A(2)=0; Explains that positive and negative signed areas cancel |
| Section B | Part 4 | Criterion 7 | Integral Calculus | Criterion 7 | Question 16(c) | Skill: Sketch \(y=3x^2-4\) for \(0\leq x\leq2\). Label its zero and use the graph to explain where \(A\) decreases and increases | Expected result: The graph crosses at \(x=2/\sqrt3\). A decreases on \(0<x<2/\sqrt3\) and increases on \(2/\sqrt3<x<2\). | Mark-by-mark evidence: Draws the upward-opening curve on the stated interval; Labels the zero x=2/√3; Identifies the interval where A decreases; Identifies the interval where A increases | Q16(c) | 4 | ___ | Rework Question 16(c): Sketch y=3x²-4 for 0leq xleq2. Label its zero and use the graph to explain where A decreases and increases. Then verify these item-specific checkpoints: Draws the upward-opening curve on the stated interval; Labels the zero x=2 / √3; Identifies the interval where A decreases; Identifies the interval where A increases |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 17(a) | Skill: Calculate \(q\) | Expected result: \(q=0.40\). | Mark-by-mark evidence: Uses total probability one to obtain q=0.40 | Q17(a) | 1 | ___ | Rework Question 17(a): Calculate q. Then verify these item-specific checkpoints: Uses total probability one to obtain q=0.40 |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 17(b) | Skill: Calculate \(E(X)\) and interpret this value | Expected result: \(E(X)=1.65\). Over many repetitions, the mean value of \(X\) approaches \(1.65\). | Mark-by-mark evidence: Calculates \(E(X)\)=1.65; Interprets 1.65 as the long-run mean | Q17(b) | 2 | ___ | Rework Question 17(b): Calculate E(X) and interpret this value. Then verify these item-specific checkpoints: Calculates E(X)=1.65; Interprets 1.65 as the long-run mean |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 18(a) | Skill: Write two equations involving \(n\) and \(p\) | Expected result: \(np=6\) and \(np(1-p)=4.2\). | Mark-by-mark evidence: States np=6; States np(1-p)=4.2 | Q18(a) | 2 | ___ | Rework Question 18(a): Write two equations involving n and p. Then verify these item-specific checkpoints: States np=6; States np(1-p)=4.2 |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 18(b) | Skill: Determine \(p\) and \(n\) | Expected result: \(p=0.30\) and \(n=20\). | Mark-by-mark evidence: Determines p=0.30; Determines n=20 | Q18(b) | 2 | ___ | Rework Question 18(b): Determine p and n. Then verify these item-specific checkpoints: Determines p=0.30; Determines n=20 |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 19(a) | Skill: Use symmetry to determine the mean \(\mu\) | Expected result: \(\mu=60\). | Mark-by-mark evidence: Recognises the percentile symmetry; Calculates the mean as 60 | Q19(a) | 2 | ___ | Rework Question 19(a): Use symmetry to determine the mean mu. Then verify these item-specific checkpoints: Recognises the percentile symmetry; Calculates the mean as 60 |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 19(b) | Skill: Calculate the standard deviation \(\sigma\), correct to two decimal places, and verify the 90th percentile | Expected result: \(\sigma=6.24\) to two decimal places; \(60+1.2816(6.24)\approx68\). | Mark-by-mark evidence: Forms 8=1.2816 \(\sigma\); Calculates \(\sigma\)=6.24 to two decimal places; Substitutes to verify the 90th percentile | Q19(b) | 3 | ___ | Rework Question 19(b): Calculate the standard deviation sigma, correct to two decimal places, and verify the 90th percentile. Then verify these item-specific checkpoints: Forms 8=1.2816 sigma; Calculates sigma=6.24 to two decimal places; Substitutes to verify the 90th percentile |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 20(a) | Skill: Identify the population parameter and the corresponding sample statistic | Expected result: The parameter is \(p\); the statistic is \(\hat p=138/240\). | Mark-by-mark evidence: Identifies p and the sample statistic \(\hat p\) | Q20(a) | 1 | ___ | Rework Question 20(a): Identify the population parameter and the corresponding sample statistic. Then verify these item-specific checkpoints: Identifies p and the sample statistic hat p |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 20(b) | Skill: Calculate \(\hat p\) and show that the large-sample approximation conditions are satisfied | Expected result: \(\hat p=0.575\); \(n\hat p=138\) and \(n(1-\hat p)=102\), both greater than \(10\). | Mark-by-mark evidence: Calculates \(\hat p\)=0.575; Verifies both observed counts exceed 10 | Q20(b) | 2 | ___ | Rework Question 20(b): Calculate hat p and show that the large-sample approximation conditions are satisfied. Then verify these item-specific checkpoints: Calculates hat p=0.575; Verifies both observed counts exceed 10 |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 20(c) | Skill: Calculate the estimated standard error of \(\hat p\), correct to four decimal places | Expected result: \(\operatorname{SE}(\hat p)=0.0319\). | Mark-by-mark evidence: Uses the estimated standard-error formula; Rounds the standard error to 0.0319 | Q20(c) | 2 | ___ | Rework Question 20(c): Calculate the estimated standard error of hat p, correct to four decimal places. Then verify these item-specific checkpoints: Uses the estimated standard-error formula; Rounds the standard error to 0.0319 |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Criterion 8 | Question 20(d) | Skill: Construct an approximate 95% confidence interval for \(p\), show it on a labelled proportion number line, and interpret the interval in context | Expected result: The interval is \((0.512,0.638)\). It is a plausible range for the population proportion of voters who support the proposal. | Mark-by-mark evidence: Calculates the margin of error; Draws and labels the interval (0.512,0.638); Interprets the interval for the population proportion | Q20(d) | 3 | ___ | Rework Question 20(d): Construct an approximate 95% confidence interval for p, show it on a labelled proportion number line, and interpret the interval in context. Then verify these item-specific checkpoints: Calculates the margin of error; Draws and labels the interval (0.512,0.638); Interprets the interval for the population proportion |