Skill Align TASC Mathematics Specialised Level 4 Practice Examination - Pack 0 - 2026 Edition
Mathematics Specialised Level 4 | Original Skill Align five-section practice examination covering Criteria 4-8 | 30 compulsory questions | 180 marks | 180 minutes working time.
- Paper
- Practice Examination
- Reading
- 15 minutes
- Writing
- 180 minutes
- Assessment
- 180 marks
The current TASC Mathematics Specialised Information Sheet may be used. TASC-approved calculators and their functions may be used throughout unless a question states otherwise. Internet access, AI tools, messaging and external communication are not permitted.
Section A - Criterion 4: Sequences and series - 36 marks
Answer all six compulsory questions. Show complete proof, exact reasoning and all required working. Suggested working time: 36 minutes.
Question 1
4 marksQuestion 2
6 marksQuestion 3
8 marksQuestion 4
4 marksQuestion 5
6 marksQuestion 6
8 marksSection B - Criterion 5: Matrices and linear algebra - 36 marks
Answer all six compulsory questions. Show row operations, transformation order and three-dimensional reasoning where required. Suggested working time: 36 minutes.
Question 7
4 marksQuestion 8
6 marksQuestion 9
8 marksQuestion 10
5 marksQuestion 11
5 marksQuestion 12
8 marksSection C - Criterion 6: Differential calculus, areas and volumes - 36 marks
Answer all six compulsory questions. Give exact values unless an approximation is requested and distinguish signed from geometric area. Suggested working time: 36 minutes.
Question 13
5 marksQuestion 14
5 marksQuestion 15
8 marksQuestion 16
4 marksQuestion 17
6 marksQuestion 18
8 marksSection D - Criterion 7: Integration techniques and differential equations - 36 marks
Answer all six compulsory questions. Show substitutions, transformed limits, constants of integration and domain restrictions. Suggested working time: 36 minutes.
Question 19
4 marksQuestion 20
6 marksQuestion 21
8 marksQuestion 22
6 marksQuestion 23
4 marksQuestion 24
8 marksSection E - Criterion 8: Complex numbers - 36 marks
Answer all six compulsory questions. State argument intervals and boundary conditions precisely and use exact complex-number notation. Suggested working time: 36 minutes.
Question 25
4 marksQuestion 26
6 marksQuestion 27
8 marksQuestion 28
4 marksQuestion 29
6 marksQuestion 30
8 marksSample Responses And Practice Marking Guidance
General marking principles
- Use the task-specific checkpoints to allocate marks and accept mathematically equivalent exact working. Apply the stated TASC fractional outcomes where they override whole-checkpoint awards.
- For a 1-mark item, a correct answer with or without working receives 1 mark; an incorrect answer with some incorrect working receives 0.5 mark.
- For a 2-mark item, relevant working is required for full marks; a correct answer without working is capped at 1.5 marks.
- For an item worth 3 marks or more, a correct answer without relevant working is capped at half marks.
- Apply consequential marking where later work correctly uses an earlier value without materially simplifying the later task.
Criterion Element coverage
Teacher reference only. Elements use the current TASC Mathematics Specialised Level 4 course and External Assessment Specifications.
| Question | Assessed Element |
|---|---|
| Q3(a) | sequences and their terms |
| Q3(b) | finite arithmetic and geometric series |
| Q3(c) | infinite geometric series and the condition for convergence |
| Q3(d) | formal convergence and divergence of sequences |
| Q6(a) | standard series summations |
| Q6(b) | mathematical induction |
| Q6(c) | the method of differences |
| Q6(d) | Maclaurin series |
| Q9(a) | matrix properties |
| Q9(b) | matrix equations and inverses |
| Q9(c) | Gauss-Jordan elimination, including symbolic or parameter systems |
| Q9(d) | geometric interpretation of systems as points, lines or planes |
| Q12(a) | lines and planes in three dimensions |
| Q12(b) | linear and composite transformations |
| Q12(c) | trigonometric addition and double-angle identities from transformations |
| Q12(d) | determinants and area scale factors |
| Q15(a) | derivatives, including inverse trigonometric, exponential and logarithmic functions |
| Q15(b) | tangents and normals for explicit and implicit relations |
| Q15(c) | stationary points and points of inflection |
| Q15(d) | concavity and curve analysis |
| Q18(a) | definite integrals, areas and volumes of revolution |
| Q18(b) | definite integrals, areas and volumes of revolution |
| Q18(c) | definite integrals, areas and volumes of revolution |
| Q18(d) | definite integrals, areas and volumes of revolution |
| Q21(a) | partial fractions |
| Q21(b) | partial fractions |
| Q21(c) | trigonometric and inverse-trigonometric integration |
| Q21(d) | integration by parts |
| Q24(a) | integration by substitution, including linear substitutions |
| Q24(b) | first-order differential equations, including separable and homogeneous forms |
| Q24(c) | first-order differential equations, including separable and homogeneous forms |
| Q24(d) | practical modelling with differential equations |
| Q27(a) | complex-number notation, equations, and real and imaginary parts |
| Q27(b) | Cartesian, polar and Euler forms |
| Q27(c) | De Moivre's theorem, roots and polynomial equations |
| Q27(d) | the conjugate-root theorem for real polynomials |
| Q30(a) | polynomial equations arising from geometric progressions |
| Q30(b) | polynomial equations arising from geometric progressions |
| Q30(c) | Argand loci and regions defined by multiple conditions |
| Q30(d) | Argand loci and regions defined by multiple conditions |
Section A Question 1
Indicative answer: ((1) / (r(r+2)))=frac12(frac1r-frac1(r+2)).
Let ((1) / (r(r+2)))=((A) / (r))+((B) / (r+2)). Solving A+B=0 and 2A=1 gives A=frac12, B=-frac12.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Introduces constants A and B in a partial-fraction decomposition.
- Forms 1=A(r+2)+Br.
- Solves A=frac12 and B=-frac12.
- States the correct difference frac12(frac1r-frac1(r+2)).
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches ((1) / (r(r+2)))=frac12(frac1r-frac1(r+2)). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Introduces constants A and B in a partial-fraction decomposition. Giving the final result without establishing: States the correct difference frac12(frac1r-frac1(r+2)).
Section A Question 2
Indicative answer: S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))), so lim_(ntoinfty)S_n=frac34.
Writing the terms in difference form cancels every interior reciprocal except 1, frac12, frac1(n+1) and frac1(n+2).
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Substitutes the difference from part (a) into the summation.
- Writes enough expanded terms to show the two-step cancellation pattern.
- Retains the uncancelled initial terms 1+frac12.
- Retains the uncancelled final terms -frac1(n+1)-frac1(n+2).
- Obtains S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))).
- Uses both remainder terms tending to zero to conclude the limit is frac34.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))), so lim_(ntoinfty)S_n=frac34. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Substitutes the difference from part (a) into the summation. Giving the final result without establishing: Uses both remainder terms tending to zero to conclude the limit is frac34.
Section A Question 3
(a) u_3=3 / 4.
Substitute n=3 into the explicit term.
(b) 93 / 16.
Use S_5=a(1-r⁵) / (1-r) with a=3 and r=1 / 2.
(c) It converges because |r|=1 / 2<1, and S_infty=6.
Apply the convergence condition before using the infinite geometric-series formula.
(d) For varepsilon>0, choose an integer N>varepsilon⁻¹-1. Then ngeq N gives |v_n|=1 / (n+1)<varepsilon, so v_nto0.
Choose an index from the epsilon inequality and verify the bound for all later terms.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- Obtains u_3=3 / 4 .
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches u_3=3 / 4. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains u_3=3 / 4 . Giving the final result without establishing: Obtains u_3=3 / 4 .
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Selects and substitutes into the finite geometric-series formula.
- Simplifies the exact sum to 93 / 16 .
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches 93 / 16. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Selects and substitutes into the finite geometric-series formula. Giving the final result without establishing: Simplifies the exact sum to 93 / 16 .
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Establishes |r|<1 and therefore convergence.
- Obtains S_infty=6 .
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches It converges because |r|=1 / 2<1, and S_infty=6. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Establishes |r|<1 and therefore convergence. Giving the final result without establishing: Obtains S_infty=6 .
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- States the quantified varepsilon-N definition.
- Chooses a valid integer threshold for the arbitrary positive epsilon.
- Verifies |v_n-0|<varepsilon and concludes convergence.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches For varepsilon>0, choose an integer N>varepsilon⁻¹-1. Then ngeq N gives |v_n|=1 / (n+1)<varepsilon, so v_nto0. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: States the quantified varepsilon-N definition. Giving the final result without establishing: Verifies |v_n-0|<varepsilon and concludes convergence.
Section A Question 4
Indicative answer: e^(-2x)approx1-2x+2x²-frac43x^3.
Substitute -2x into e^u=1+u+((u²) / (2!))+((u³) / (3!))+ × s.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Starts from the Maclaurin expansion of e^u.
- Substitutes u=-2x.
- Simplifies the quadratic coefficient to 2.
- Simplifies the cubic coefficient to -frac43.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches e^(-2x)approx1-2x+2x²-frac43x^3. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Starts from the Maclaurin expansion of e^u. Giving the final result without establishing: Simplifies the cubic coefficient to -frac43.
Section A Question 5
Indicative answer: The polynomial gives 0.669333ldots. Since e^(-0.4)=0.670320ldots, the absolute error is about 0.000987, an underestimate of less than 0.001.
Substitution gives 1-0.4+0.08-0.010666ldots=0.669333ldots.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Substitutes x=0.2 into every term of the cubic polynomial.
- Evaluates the linear contribution as -0.4.
- Evaluates the quadratic contribution as 0.08.
- Evaluates the cubic contribution as -0.010666ldots.
- Obtains 0.669333ldots and compares it with 0.670320ldots.
- Calculates the absolute error and correctly identifies the approximation as an underestimate with error below 0.001.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches The polynomial gives 0.669333ldots. Since e^(-0.4)=0.670320ldots, the absolute error is about 0.000987, an underestimate of less than 0.001. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Substitutes x=0.2 into every term of the cubic polynomial. Giving the final result without establishing: Calculates the absolute error and correctly identifies the approximation as an underestimate with error below 0.001.
Section A Question 6
(a) 15.
Use the standard sum of the first five positive integers.
(b) The statement is true for n=1. Assuming it for n=k, adding 2k+1 gives k²+2k+1=(k+1)², so the result follows.
Verify the base case, assume the statement at k and add the next odd number.
(c) 4 / 5.
Write 1 / [r(r+1)]=1 / r-1 / (r+1) and cancel consecutive terms.
(d) 1+x+((x²) / (2))+((x³) / (6))+O(x⁴).
Use successive derivatives at zero and the Maclaurin coefficient formula.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- Obtains the exact sum 15 .
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches 15. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains the exact sum 15 . Giving the final result without establishing: Obtains the exact sum 15 .
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Verifies the base case 1=1^2.
- Uses the induction hypothesis and obtains (k+1)² before stating the conclusion.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches The statement is true for n=1. Assuming it for n=k, adding 2k+1 gives k²+2k+1=(k+1)², so the result follows. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Verifies the base case 1=1^2. Giving the final result without establishing: Uses the induction hypothesis and obtains (k+1)² before stating the conclusion.
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Decomposes the summand as 1 / r-1 / (r+1).
- Cancels the intermediate terms and obtains 4 / 5 .
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches 4 / 5. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Decomposes the summand as 1 / r-1 / (r+1). Giving the final result without establishing: Cancels the intermediate terms and obtains 4 / 5 .
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Obtains the constant and linear terms 1+x.
- Obtains the quadratic term x² / 2.
- Obtains the cubic term x³ / 6 with an appropriate remainder.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches 1+x+((x²) / (2))+((x³) / (6))+O(x⁴). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains the constant and linear terms 1+x. Giving the final result without establishing: Obtains the cubic term x³ / 6 with an appropriate remainder.
Section B Question 7
Indicative answer: [begin(array)(ccc|c)1&1&1&20&-3&1&-70&3&2&4end(array)].
Use R_2arrow R_2-2R_1 and R_3arrow R_3+R_1.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Writes the correct 3 × 4 augmented matrix.
- States R_2arrow R_2-2R_1.
- Obtains row 2 as [0,-3,1mid-7].
- Obtains row 3 as [0,3,2mid4] using R_3arrow R_3+R_1.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches [begin(array)(ccc|c)1&1&1&20&-3&1&-70&3&2&4end(array)]. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Writes the correct 3 × 4 augmented matrix. Giving the final result without establishing: Obtains row 3 as [0,3,2mid4] using R_3arrow R_3+R_1.
Section B Question 8
Indicative answer: The reduced matrix is [begin(array)(ccc|c)1&0&0&10&1&0&20&0&1&-1end(array)], so (x,y,z)=(1,2,-1).
Adding rows 2 and 3 gives 3z=-3, so z=-1. Back-substitution gives y=2 and x=1.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Eliminates y from row 3 to obtain an equation in z.
- Obtains z=-1.
- Uses row 2 to obtain y=2.
- Uses row 1 to obtain x=1.
- Presents the reduced row-echelon matrix or equivalent complete row reduction.
- States the unique ordered triple (1,2,-1).
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches The reduced matrix is [begin(array)(ccc|c)1&0&0&10&1&0&20&0&1&-1end(array)], so (x,y,z)=(1,2,-1). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Eliminates y from row 3 to obtain an equation in z. Giving the final result without establishing: States the unique ordered triple (1,2,-1).
Section B Question 9
(a) lambda-1.
Apply the two-by-two determinant rule.
(b) dfrac(1)(lambda-1)begin(bmatrix)lambda&-1-1&1end(bmatrix).
Use the adjugate and the non-zero determinant from part (a).
(c) Use R_2arrow R_2-R_1, giving (lambda-1)y=mu.
Apply the row operation to both coefficient and augmented entries.
(d) For lambdane1 the lines meet once. For lambda=1,mu=0 they coincide. For lambda=1,mune0 they are distinct and parallel.
Use the determinant and reduced equation, then translate rank and consistency into geometry.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- Obtains det A_lambda=lambda-1 .
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches lambda-1. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains det A_lambda=lambda-1 . Giving the final result without establishing: Obtains det A_lambda=lambda-1 .
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Uses the reciprocal factor 1 / (lambda-1).
- Forms the correct adjugate matrix.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches dfrac(1)(lambda-1)begin(bmatrix)lambda&-1-1&1end(bmatrix). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Uses the reciprocal factor 1 / (lambda-1). Giving the final result without establishing: Forms the correct adjugate matrix.
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Applies R_2arrow R_2-R_1 to the full augmented row.
- Obtains (lambda-1)y=mu .
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches Use R_2arrow R_2-R_1, giving (lambda-1)y=mu. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Applies R_2arrow R_2-R_1 to the full augmented row. Giving the final result without establishing: Obtains (lambda-1)y=mu .
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Identifies the unique intersection when the determinant is non-zero.
- Identifies the coincident-line case and infinitely many solutions.
- Identifies the distinct parallel-line case and no solution.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches For lambdane1 the lines meet once. For lambda=1,mu=0 they coincide. For lambda=1,mune0 they are distinct and parallel. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Identifies the unique intersection when the determinant is non-zero. Giving the final result without establishing: Identifies the distinct parallel-line case and no solution.
Section B Question 10
Indicative answer: M=RS=begin(bmatrix)0&-11&2end(bmatrix) and P'=(-2,5).
Because the shear occurs first, multiply R by S, then apply the product to the column vector for P.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Identifies the composite as RS, not SR.
- Forms the first row of RS correctly.
- Forms the second row of RS correctly.
- Multiplies Mbegin(bmatrix)12end(bmatrix).
- Obtains the image P'=(-2,5).
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches M=RS=begin(bmatrix)0&-11&2end(bmatrix) and P'=(-2,5). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Identifies the composite as RS, not SR. Giving the final result without establishing: Obtains the image P'=(-2,5).
Section B Question 11
Indicative answer: The image line is y=-3x, and the area scale factor is 1.
A point (t,t) maps to (-t,3t), so y=-3x. Also |det M|=1.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Parameterises the original line as (t,t).
- Maps (t,t) to (-t,3t).
- Eliminates t to obtain y=-3x.
- Calculates det M=1.
- Interprets |det M|=1 as the area scale factor.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches The image line is y=-3x, and the area scale factor is 1. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Parameterises the original line as (t,t). Giving the final result without establishing: Interprets |det M|=1 as the area scale factor.
Section B Question 12
(a) For example, (2,-1,2).
Read the coefficients of the Cartesian plane equation.
(b) begin(bmatrix)cosalphacosbeta-sinalphasinbeta&-(cosalphasinbeta+sinalphacosbeta)sinalphacosbeta+cosalphasinbeta&cosalphacosbeta-sinalphasinbetaend(bmatrix).
Multiply the rotation matrices in the stated order.
(c) cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta and sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta.
Match corresponding entries of the matrices.
(d) 1 square unit.
The original area is 1 and |det S|=1.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- States a non-zero scalar multiple of (2,-1,2) .
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches For example, (2,-1,2). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: States a non-zero scalar multiple of (2,-1,2) . Giving the final result without establishing: States a non-zero scalar multiple of (2,-1,2) .
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Obtains both diagonal entries correctly.
- Obtains the signed off-diagonal entries correctly.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches begin(bmatrix)cosalphacosbeta-sinalphasinbeta&-(cosalphasinbeta+sinalphacosbeta)sinalphacosbeta+cosalphasinbeta&cosalphacosbeta-sinalphasinbetaend(bmatrix). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains both diagonal entries correctly. Giving the final result without establishing: Obtains the signed off-diagonal entries correctly.
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- States the cosine addition identity.
- States the sine addition identity.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta and sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: States the cosine addition identity. Giving the final result without establishing: States the sine addition identity.
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Finds the original triangle area.
- Calculates |det S|=1.
- Obtains the unchanged image area.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches 1 square unit. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Finds the original triangle area. Giving the final result without establishing: Obtains the unchanged image area.
Section C Question 13
Indicative answer: The stationary points are (-2,-4), a local maximum, and (2,4), a local minimum.
Use f'(x)=1-frac4(x²) to locate the stationary points and f''(x)=frac8(x³) to classify them.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Differentiates to obtain f'(x)=1-frac4(x²).
- Solves f'(x)=0 to obtain x=-2 and x=2.
- Calculates the stationary-point coordinates (-2,-4) and (2,4).
- Obtains f''(x)=frac8(x³).
- Uses f''(-2)<0 and f''(2)>0 to classify the local maximum and local minimum.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches The stationary points are (-2,-4), a local maximum, and (2,4), a local minimum. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Differentiates to obtain f'(x)=1-frac4(x²). Giving the final result without establishing: Uses f''(-2)<0 and f''(2)>0 to classify the local maximum and local minimum.
Section C Question 14
Indicative answer: The asymptotes are x=0 and y=x. The range is (-infty,-4]cup[4,infty).
For a proposed value y, the equation y=x+frac4x is equivalent to x²-yx+4=0, which has a real solution exactly when y²-16ge0.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Identifies x=0 as the vertical asymptote.
- Uses f(x)-x=frac4xto0 as |x|toinfty to identify the oblique asymptote y=x.
- Rearranges y=x+frac4x to the quadratic x²-yx+4=0.
- Requires discriminant y²-16ge0 for real x.
- States the range (-infty,-4]cup[4,infty), including the endpoint values attained at the stationary points.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches The asymptotes are x=0 and y=x. The range is (-infty,-4]cup[4,infty). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Identifies x=0 as the vertical asymptote. Giving the final result without establishing: States the range (-infty,-4]cup[4,infty), including the endpoint values attained at the stationary points.
Section C Question 15
(a) f'(x)=1e^(1x).
Apply the relevant standard derivative and the chain rule where required.
(b) The tangent is y-1=-(x-1), and the normal is y-1=x-1.
Implicit differentiation gives 2x+y+(x+2y)y'=0, so y'=-1 at the point.
(c) x=pm1.
Differentiate, set the derivative to zero and retain every solution in the domain.
(d) There is a local maximum at x=-1, a local minimum at x=1, and an inflection at x=0, where g''(x)=6x changes sign.
Use the second derivative and verify a sign change of concavity at each proposed inflection.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- f'(x)=1e^(1x)
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches f'(x)=1e^(1x). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: f'(x)=1e^(1x). Giving the final result without establishing: f'(x)=1e^(1x).
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Uses implicit differentiation to obtain tangent slope -1.
- States both the tangent and perpendicular normal equations.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches The tangent is y-1=-(x-1), and the normal is y-1=x-1. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Uses implicit differentiation to obtain tangent slope -1. Giving the final result without establishing: States both the tangent and perpendicular normal equations.
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Obtains g'(x)=3(x²-1).
- Solves to obtain x=pm1.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches x=pm1. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains g'(x)=3(x²-1). Giving the final result without establishing: Solves to obtain x=pm1.
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Uses the second derivative to classify every stationary point.
- Solves the second-derivative condition for all possible inflection points.
- Verifies the required concavity change and states every inflection.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches There is a local maximum at x=-1, a local minimum at x=1, and an inflection at x=0, where g''(x)=6x changes sign. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Uses the second derivative to classify every stationary point. Giving the final result without establishing: Verifies the required concavity change and states every inflection.
Section C Question 16
Indicative answer: V=piint_0²(16-y⁴),dy.
The outer radius is 4 and the inner radius is y^2.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Identifies the outer radius as 4.
- Identifies the inner radius as y^2.
- Uses the washer area π(R²-r²).
- Uses the correct limits 0le yle2.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches V=piint_0²(16-y⁴),dy. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Identifies the outer radius as 4. Giving the final result without establishing: Uses the correct limits 0le yle2.
Section C Question 17
Indicative answer: V=((128π) / (5)) cubic units.
Integrate 16-y⁴ to obtain 16y-((y⁵) / (5)), then evaluate at 0 and 2.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Integrates 16 to 16y.
- Integrates y⁴ to ((y⁵) / (5)).
- Applies the upper limit y=2.
- Applies the lower limit y=0.
- Simplifies 32-((32) / (5)) to ((128) / (5)).
- States ((128π) / (5)) with cubic units.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches V=((128π) / (5)) cubic units. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Integrates 16 to 16y. Giving the final result without establishing: States ((128π) / (5)) with cubic units.
Section C Question 18
(a) int_0^2x(2-x),dx.
Use the non-negative curve as the height above the x-axis.
(b) ((4) / (3)) square units.
Find an antiderivative and apply both limits.
(c) piint_0²[x(2-x)]²,dx.
Use V=piint y²,dx and simplify the square before evaluating.
(d) ((16π) / (15)) cubic units.
Integrate the simplified disk-method integrand and apply both limits.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- States int_0^2x(2-x),dx.
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches int_0^2x(2-x),dx. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: States int_0^2x(2-x),dx. Giving the final result without establishing: States int_0^2x(2-x),dx.
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Integrates the area integrand correctly.
- Applies the limits and obtains ((4) / (3)).
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches ((4) / (3)) square units. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Integrates the area integrand correctly. Giving the final result without establishing: Applies the limits and obtains ((4) / (3)).
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Uses the disk method with the correct limits.
- Squares and simplifies the radius function correctly.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches piint_0²[x(2-x)]²,dx. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Uses the disk method with the correct limits. Giving the final result without establishing: Squares and simplifies the radius function correctly.
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Finds a correct antiderivative including the factor of π.
- Applies both limits without a sign or power error.
- Simplifies to the exact volume ((16π) / (15)).
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches ((16π) / (15)) cubic units. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Finds a correct antiderivative including the factor of π. Giving the final result without establishing: Simplifies to the exact volume ((16π) / (15)).
Section D Question 19
Indicative answer: ((5x+7) / ((x+1)(x+2)))=frac2(x+1)+frac3(x+2).
Equating coefficients gives A+B=5 and 2A+B=7.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Writes ((A) / (x+1))+((B) / (x+2)).
- Forms 5x+7=A(x+2)+B(x+1).
- Solves A=2.
- Solves B=3 and states the decomposition.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches ((5x+7) / ((x+1)(x+2)))=frac2(x+1)+frac3(x+2). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Writes ((A) / (x+1))+((B) / (x+2)). Giving the final result without establishing: Solves B=3 and states the decomposition.
Section D Question 20
Indicative answer: I=3ln3-ln2=ln(((27) / (2))).
An antiderivative is 2ln(x+1)+3ln(x+2).
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Integrates frac2(x+1) as 2ln|x+1|.
- Integrates frac3(x+2) as 3ln|x+2|.
- Evaluates the antiderivative at x=1.
- Evaluates the antiderivative at x=0.
- Simplifies to 3ln3-ln2.
- Combines logarithms to obtain ln(((27) / (2))).
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches I=3ln3-ln2=ln(((27) / (2))). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Integrates frac2(x+1) as 2ln|x+1|. Giving the final result without establishing: Combines logarithms to obtain ln(((27) / (2))).
Section D Question 21
(a) dfrac1x-dfrac1(x+1).
Combine the proposed simple fractions and equate coefficients.
(b) ln|x|-ln|x+1|+C.
Integrate each simple fraction and include the constant of integration.
(c) sin⁻¹(x / 2)+C.
Use the inverse-sine antiderivative supplied on the current information sheet.
(d) e^(2x)(x / 2-1 / 4)+C.
Take u=x and integrate the exponential for dv.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- Obtains dfrac1x-dfrac1(x+1) .
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches dfrac1x-dfrac1(x+1). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains dfrac1x-dfrac1(x+1) . Giving the final result without establishing: Obtains dfrac1x-dfrac1(x+1) .
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Integrates each simple fraction with the correct coefficient and absolute value.
- States ln|x|-ln|x+1|+C .
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches ln|x|-ln|x+1|+C. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Integrates each simple fraction with the correct coefficient and absolute value. Giving the final result without establishing: States ln|x|-ln|x+1|+C .
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Matches the integrand to the supplied inverse-sine form.
- Includes the constant and obtains the exact antiderivative.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches sin⁻¹(x / 2)+C. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Matches the integrand to the supplied inverse-sine form. Giving the final result without establishing: Includes the constant and obtains the exact antiderivative.
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Chooses u and v consistently.
- Applies the integration-by-parts formula with the correct sign.
- Simplifies to the stated antiderivative.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches e^(2x)(x / 2-1 / 4)+C. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Chooses u and v consistently. Giving the final result without establishing: Simplifies to the stated antiderivative.
Section D Question 22
Indicative answer: y=x(ln x+2), for x>0.
With y=vx, y'=v+xv'. Hence xv'=1, so v=ln x+C, and the initial condition gives C=2.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Sets y=vx.
- Differentiates to obtain y'=v+x((dv) / (dx)).
- Substitutes into the differential equation and simplifies to x((dv) / (dx))=1.
- Integrates to obtain v=ln x+C.
- Returns to y=x(ln x+C).
- Uses y(1)=2 to obtain C=2 and states x>0.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches y=x(ln x+2), for x>0. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Sets y=vx. Giving the final result without establishing: Uses y(1)=2 to obtain C=2 and states x>0.
Section D Question 23
Indicative answer: y(e)=3e and .((dy) / (dx))|_(x=e)=4.
Substitute x=e into y=x(ln x+2), then use y'=1+((y) / (x)).
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Uses ln e=1.
- Obtains y(e)=e(1+2)=3e.
- Substitutes x=e and y=3e into 1+((y) / (x)).
- Obtains the gradient 4.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches y(e)=3e and .((dy) / (dx))|_(x=e)=4. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Uses ln e=1. Giving the final result without establishing: Obtains the gradient 4.
Section D Question 24
(a) u=2x+1, with dx=du / 2.
Use the inner linear expression.
(b) y=x²+x+2.
Integrate the function of x and apply the initial value.
(c) y=x(ln x+C).
Set v=y / x; then xv'=1.
(d) P(t)=100e^t and t=ln2.
Separate variables and apply the initial condition.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- States a consistent substitution and differential.
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches u=2x+1, with dx=du / 2. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: States a consistent substitution and differential. Giving the final result without establishing: States a consistent substitution and differential.
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Finds the general antiderivative.
- Uses the initial condition correctly.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches y=x²+x+2. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Finds the general antiderivative. Giving the final result without establishing: Uses the initial condition correctly.
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Makes the homogeneous substitution.
- Integrates and back-substitutes.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches y=x(ln x+C). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Makes the homogeneous substitution. Giving the final result without establishing: Integrates and back-substitutes.
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Obtains the exponential solution family.
- Applies the initial value.
- Solves the doubling equation.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches P(t)=100e^t and t=ln2. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains the exponential solution family. Giving the final result without establishing: Solves the doubling equation.
Section E Question 25
Indicative answer: z=2operatorname(cis)((π) / (4)),,2operatorname(cis)((3π) / (4)),,2operatorname(cis)((5π) / (4)),,2operatorname(cis)((7π) / (4)).
Write -16=16operatorname(cis)(π+2kpi), take fourth roots and use k=0,1,2,3.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Writes -16 with modulus 16 and argument π+2kpi.
- Takes the fourth root of the modulus to obtain 2.
- Uses angles ((π+2kpi) / (4)).
- Lists the four distinct roots for k=0,1,2,3.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches z=2operatorname(cis)((π) / (4)),,2operatorname(cis)((3π) / (4)),,2operatorname(cis)((5π) / (4)),,2operatorname(cis)((7π) / (4)). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Writes -16 with modulus 16 and argument π+2kpi. Giving the final result without establishing: Lists the four distinct roots for k=0,1,2,3.
Section E Question 26
Indicative answer: z⁴+16=(z²-2sqrt2,z+4)(z²+2sqrt2,z+4).
Pair each root with its complex conjugate. Each pair has product 4 and real-part sum pm2sqrt2.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Pairs the roots at angles ((π) / (4)) and ((7π) / (4)).
- Forms the factor z²-2sqrt2,z+4.
- Pairs the roots at angles ((3π) / (4)) and ((5π) / (4)).
- Forms the factor z²+2sqrt2,z+4.
- Multiplies or otherwise verifies cancellation of the z³ and z terms.
- States the complete real quadratic factorisation.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches z⁴+16=(z²-2sqrt2,z+4)(z²+2sqrt2,z+4). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Pairs the roots at angles ((π) / (4)) and ((7π) / (4)). Giving the final result without establishing: States the complete real quadratic factorisation.
Section E Question 27
(a) overline z=sqrt3-i.
Reverse only the imaginary sign.
(b) z=2e^(ipi / 6).
Find modulus 2 and the first-quadrant argument.
(c) z²=2+2sqrt3i.
Square the modulus and double the argument.
(d) The paired root is sqrt3-i and the factor is x²-2sqrt3x+4.
Apply the conjugate-root theorem and expand the pair.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- States overline z=sqrt3-i .
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches overline z=sqrt3-i. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: States overline z=sqrt3-i . Giving the final result without establishing: States overline z=sqrt3-i .
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Finds the modulus.
- Finds the principal argument and states the Euler form.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches z=2e^(ipi / 6). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Finds the modulus. Giving the final result without establishing: Finds the principal argument and states the Euler form.
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Squares the modulus and doubles the argument.
- Converts the result back to Cartesian form.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches z²=2+2sqrt3i. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Squares the modulus and doubles the argument. Giving the final result without establishing: Converts the result back to Cartesian form.
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Identifies the conjugate root.
- Uses the sum and product of the roots.
- Obtains the real monic factor.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches The paired root is sqrt3-i and the factor is x²-2sqrt3x+4. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Identifies the conjugate root. Giving the final result without establishing: Obtains the real monic factor.
Section E Question 28
Indicative answer: The boundaries are the included circle |z-(1+i)|=2 and the included rays operatorname(Arg)z=0 and operatorname(Arg)z=((π) / (3)), restricted to their common intersection; z=0 is excluded because its argument is undefined.
The non-strict inequalities include their boundaries, but the principal argument is not defined at the origin.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.
Mark-by-mark checkpoints:
- Identifies the circle centre 1+i.
- Identifies the circle radius 2.
- Identifies the two boundary rays at angles 0 and ((π) / (3)).
- States that the circle and rays are included but the origin is excluded because operatorname(Arg)0 is undefined.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches The boundaries are the included circle |z-(1+i)|=2 and the included rays operatorname(Arg)z=0 and operatorname(Arg)z=((π) / (3)), restricted to their common intersection; z=0 is excluded because its argument is undefined. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Identifies the circle centre 1+i. Giving the final result without establishing: States that the circle and rays are included but the origin is excluded because operatorname(Arg)0 is undefined.
Section E Question 29
Indicative answer: The region is the part of the closed disc lying in the closed sector 0lethetale((π) / (3)), excluding the origin. The points 2+i and 3+i belong to R; 1+3i does not.
Check both the distance from 1+i and the principal argument for each point.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Sketches the disc with centre 1+i and radius 2.
- Restricts the sketch to the sector between angles 0 and ((π) / (3)).
- Checks 2+i against both conditions and includes it.
- Checks 1+3i and excludes it because its argument exceeds ((π) / (3)).
- Checks 3+i as lying on the circle and within the sector.
- Uses solid boundaries and excludes the origin explicitly.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches The region is the part of the closed disc lying in the closed sector 0lethetale((π) / (3)), excluding the origin. The points 2+i and 3+i belong to R; 1+3i does not. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Sketches the disc with centre 1+i and radius 2. Giving the final result without establishing: Uses solid boundaries and excludes the origin explicitly.
Section E Question 30
(a) z³=1, with zne1.
Multiply the polynomial equation by z minus one and exclude the cancelled value.
(b) z=operatorname(cis)(2π / 3) or z=operatorname(cis)(4π / 3).
List the cube roots of unity and remove the real root one.
(c) (x-1)²+y^2leq4 and ygeq0.
Translate the modulus to a disk and the imaginary-part condition to a half-plane.
(d) It is the closed upper half of the disk centred at (1,0) with radius 2. Both boundary pieces are included, and they meet at (-1,0) and (3,0).
Intersect the closed disk with the upper half-plane and solve y equals zero on the circle.
Detailed marking criteria
Part a (1 mark)
Apply the current TASC one-mark-item outcomes. A correct answer with or without working receives 1 mark. An incorrect answer with some incorrect working receives 0.5 mark.
Mark-by-mark checkpoints:
- Obtains z³=1 while excluding z=1 .
Authority-specific partial-mark outcomes:
- 0.5 marks: Incorrect answer with some incorrect working.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 1 checkpoints and reaches z³=1, with zne1. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Obtains z³=1 while excluding z=1 . Giving the final result without establishing: Obtains z³=1 while excluding z=1 .
Part b (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Lists all three cube roots of unity before applying the exclusion.
- States exactly the two non-real solutions without duplication.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches z=operatorname(cis)(2π / 3) or z=operatorname(cis)(4π / 3). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Lists all three cube roots of unity before applying the exclusion. Giving the final result without establishing: States exactly the two non-real solutions without duplication.
Part c (2 marks)
Award one mark for each numbered checkpoint below, to a maximum of 2. A correct answer without relevant working is capped at 1.5 marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Converts the modulus condition to the correct disk inequality.
- Converts the imaginary-part condition to the correct half-plane.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 2 checkpoints and reaches (x-1)²+y^2leq4 and ygeq0. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Converts the modulus condition to the correct disk inequality. Giving the final result without establishing: Converts the imaginary-part condition to the correct half-plane.
Part d (3 marks)
Award one mark for each numbered checkpoint below, to a maximum of 3. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.
Mark-by-mark checkpoints:
- Identifies the correct upper half-disk.
- States that the arc and diameter are included.
- Obtains both diameter endpoints.
Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 3 checkpoints and reaches It is the closed upper half of the disk centred at (1,0) with radius 2. Both boundary pieces are included, and they meet at (-1,0) and (3,0). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.
Do not credit by itself: Omitting or contradicting: Identifies the correct upper half-disk. Giving the final result without establishing: Obtains both diameter endpoints.
Diagnostic Checklist
| Topic | Questions | Maximum marks | Score (0.5-mark increments) | Action |
|---|---|---|---|---|
| Section A | Criterion 4 | Sequences and series | Decompose a rational term for telescoping | Expected result: \(\frac{1}{r(r+2)}=\frac12\left(\frac1r-\frac1{r+2}\right)\). | Required evidence: Introduces constants \(A\) and \(B\) in a partial-fraction decomposition.; Forms \(1=A(r+2)+Br\).; Solves \(A=\frac12\) and \(B=-\frac12\).; States the correct difference \(\frac12(\frac1r-\frac1{r+2})\). | Q1 | 4 | ___ / 4 | Q1: reproduce this process without the solution: Introduces constants A and B in a partial-fraction decomposition; then Forms 1=A(r+2)+Br; then Solves A=frac12 and B=-frac12; then States the correct difference frac12(frac1r-frac1(r+2)). |
| Section A | Criterion 4 | Sequences and series | Telescope a finite series and justify its limit | Expected result: \(S_n=\frac34-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}\), so \(\lim_{n\to\infty}S_n=\frac34\). | Required evidence: Substitutes the difference from part (a) into the summation.; Writes enough expanded terms to show the two-step cancellation pattern.; Retains the uncancelled initial terms \(1+\frac12\).; Retains the uncancelled final terms \(-\frac1{n+1}-\frac1{n+2}\).; Obtains \(S_n=\frac34-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}\).; Uses both remainder terms tending to zero to conclude the limit is \(\frac34\). | Q2 | 6 | ___ / 6 | Q2: reproduce this process without the solution: Substitutes the difference from part (a) into the summation; then Writes enough expanded terms to show the two-step cancellation pattern; then Retains the uncancelled initial terms 1+frac12; then Retains the uncancelled final terms -frac1(n+1)-frac1(n+2); then Obtains S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))); then Uses both remainder terms tending to zero to conclude the limit is frac34. |
| Criterion 4 - Sequences and series | explicit term evaluation | Expected result: \(u_3=3/4\). | Required evidence: Obtains \(u_3=3/4\) . | Q3(a) | 1 | ___ / 1 | Q3(a): reproduce this process without the solution: Obtains u_3=3 / 4 . |
| Criterion 4 - Sequences and series | finite geometric series | Expected result: \(93/16\). | Required evidence: Selects and substitutes into the finite geometric-series formula.; Simplifies the exact sum to \(93/16\) . | Q3(b) | 2 | ___ / 2 | Q3(b): reproduce this process without the solution: Selects and substitutes into the finite geometric-series formula; then Simplifies the exact sum to 93 / 16 . |
| Criterion 4 - Sequences and series | infinite geometric-series convergence | Expected result: It converges because \(|r|=1/2<1\), and \(S_\infty=6\). | Required evidence: Establishes \(|r|<1\) and therefore convergence.; Obtains \(S_\infty=6\) . | Q3(c) | 2 | ___ / 2 | Q3(c): reproduce this process without the solution: Establishes |r|<1 and therefore convergence; then Obtains S_infty=6 . |
| Criterion 4 - Sequences and series | epsilon proof of convergence to zero | Expected result: For \(\varepsilon>0\), choose an integer \(N>\varepsilon^{-1}-1\). Then \(n\geq N\) gives \(|v_n|=1/(n+1)<\varepsilon\), so \(v_n\to0\). | Required evidence: States the quantified \(\varepsilon\)-\(N\) definition.; Chooses a valid integer threshold for the arbitrary positive epsilon.; Verifies \(|v_n-0|<\varepsilon\) and concludes convergence. | Q3(d) | 3 | ___ / 3 | Q3(d): reproduce this process without the solution: States the quantified varepsilon-N definition; then Chooses a valid integer threshold for the arbitrary positive epsilon; then Verifies |v_n-0|<varepsilon and concludes convergence. |
| Section A | Criterion 4 | Sequences and series | Derive a cubic Maclaurin polynomial | Expected result: \(e^{-2x}\approx1-2x+2x^2-\frac43x^3\). | Required evidence: Starts from the Maclaurin expansion of \(e^u\).; Substitutes \(u=-2x\).; Simplifies the quadratic coefficient to \(2\).; Simplifies the cubic coefficient to \(-\frac43\). | Q4 | 4 | ___ / 4 | Q4: reproduce this process without the solution: Starts from the Maclaurin expansion of e^u; then Substitutes u=-2x; then Simplifies the quadratic coefficient to 2; then Simplifies the cubic coefficient to -frac43. |
| Section A | Criterion 4 | Sequences and series | Use and assess a Maclaurin approximation | Expected result: The polynomial gives \(0.669333\ldots\). Since \(e^{-0.4}=0.670320\ldots\), the absolute error is about \(0.000987\), an underestimate of less than \(0.001\). | Required evidence: Substitutes \(x=0.2\) into every term of the cubic polynomial.; Evaluates the linear contribution as \(-0.4\).; Evaluates the quadratic contribution as \(0.08\).; Evaluates the cubic contribution as \(-0.010666\ldots\).; Obtains \(0.669333\ldots\) and compares it with \(0.670320\ldots\).; Calculates the absolute error and correctly identifies the approximation as an underestimate with error below \(0.001\). | Q5 | 6 | ___ / 6 | Q5: reproduce this process without the solution: Substitutes x=0.2 into every term of the cubic polynomial; then Evaluates the linear contribution as -0.4; then Evaluates the quadratic contribution as 0.08; then Evaluates the cubic contribution as -0.010666ldots; then Obtains 0.669333ldots and compares it with 0.670320ldots; then Calculates the absolute error and correctly identifies the approximation as an underestimate with error below 0.001. |
| Criterion 4 - Sequences and series | standard consecutive-integer summation | Expected result: \(15\). | Required evidence: Obtains the exact sum \(15\) . | Q6(a) | 1 | ___ / 1 | Q6(a): reproduce this process without the solution: Obtains the exact sum 15 . |
| Criterion 4 - Sequences and series | induction for the odd-number square identity | Expected result: The statement is true for \(n=1\). Assuming it for \(n=k\), adding \(2k+1\) gives \(k^2+2k+1=(k+1)^2\), so the result follows. | Required evidence: Verifies the base case \(1=1^2\).; Uses the induction hypothesis and obtains \((k+1)^2\) before stating the conclusion. | Q6(b) | 2 | ___ / 2 | Q6(b): reproduce this process without the solution: Verifies the base case 1=1²; then Uses the induction hypothesis and obtains (k+1)² before stating the conclusion. |
| Criterion 4 - Sequences and series | adjacent reciprocal telescoping | Expected result: \(4/5\). | Required evidence: Decomposes the summand as \(1/r-1/(r+1)\).; Cancels the intermediate terms and obtains \(4/5\) . | Q6(c) | 2 | ___ / 2 | Q6(c): reproduce this process without the solution: Decomposes the summand as 1 / r-1 / (r+1); then Cancels the intermediate terms and obtains 4 / 5 . |
| Criterion 4 - Sequences and series | direct exponential Maclaurin expansion | Expected result: \(1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)\). | Required evidence: Obtains the constant and linear terms \(1+x\).; Obtains the quadratic term \(x^2/2\).; Obtains the cubic term \(x^3/6\) with an appropriate remainder. | Q6(d) | 3 | ___ / 3 | Q6(d): reproduce this process without the solution: Obtains the constant and linear terms 1+x; then Obtains the quadratic term x² / 2; then Obtains the cubic term x³ / 6 with an appropriate remainder. |
| Section B | Criterion 5 | Matrices and linear algebra | Set up and begin Gauss-Jordan reduction | Expected result: \(\left[\begin{array}{ccc|c}1&1&1&2\\0&-3&1&-7\\0&3&2&4\end{array}\right]\). | Required evidence: Writes the correct \(3\times4\) augmented matrix.; States \(R_2\leftarrow R_2-2R_1\).; Obtains row 2 as \([0,-3,1\mid-7]\).; Obtains row 3 as \([0,3,2\mid4]\) using \(R_3\leftarrow R_3+R_1\). | Q7 | 4 | ___ / 4 | Q7: reproduce this process without the solution: Writes the correct 3 × 4 augmented matrix; then States R_2arrow R_2-2R_1; then Obtains row 2 as [0,-3,1mid-7]; then Obtains row 3 as [0,3,2mid4] using R_3arrow R_3+R_1. |
| Section B | Criterion 5 | Matrices and linear algebra | Complete Gauss-Jordan reduction and interpret a unique solution | Expected result: The reduced matrix is \(\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&2\\0&0&1&-1\end{array}\right]\), so \((x,y,z)=(1,2,-1)\). | Required evidence: Eliminates \(y\) from row 3 to obtain an equation in \(z\).; Obtains \(z=-1\).; Uses row 2 to obtain \(y=2\).; Uses row 1 to obtain \(x=1\).; Presents the reduced row-echelon matrix or equivalent complete row reduction.; States the unique ordered triple \((1,2,-1)\). | Q8 | 6 | ___ / 6 | Q8: reproduce this process without the solution: Eliminates y from row 3 to obtain an equation in z; then Obtains z=-1; then Uses row 2 to obtain y=2; then Uses row 1 to obtain x=1; then Presents the reduced row-echelon matrix or equivalent complete row reduction; then States the unique ordered triple (1,2,-1). |
| Criterion 5 - Matrices and linear algebra | parameter determinant and singular value | Expected result: \(\lambda-1\). | Required evidence: Obtains \(\det A_\lambda=\lambda-1\) . | Q9(a) | 1 | ___ / 1 | Q9(a): reproduce this process without the solution: Obtains det A_lambda=lambda-1 . |
| Criterion 5 - Matrices and linear algebra | symbolic inverse of a parameter matrix | Expected result: \(\dfrac{1}{\lambda-1}\begin{bmatrix}\lambda&-1\\-1&1\end{bmatrix}\). | Required evidence: Uses the reciprocal factor \(1/(\lambda-1)\).; Forms the correct adjugate matrix. | Q9(b) | 2 | ___ / 2 | Q9(b): reproduce this process without the solution: Uses the reciprocal factor 1 / (lambda-1); then Forms the correct adjugate matrix. |
| Criterion 5 - Matrices and linear algebra | single symbolic elimination in a two-line system | Expected result: Use \(R_2\leftarrow R_2-R_1\), giving \((\lambda-1)y=\mu\). | Required evidence: Applies \(R_2\leftarrow R_2-R_1\) to the full augmented row.; Obtains \((\lambda-1)y=\mu\) . | Q9(c) | 2 | ___ / 2 | Q9(c): reproduce this process without the solution: Applies R_2arrow R_2-R_1 to the full augmented row; then Obtains (lambda-1)y=mu . |
| Criterion 5 - Matrices and linear algebra | point-coincident-parallel classification for two lines | Expected result: For \(\lambda\ne1\) the lines meet once. For \(\lambda=1,\mu=0\) they coincide. For \(\lambda=1,\mu\ne0\) they are distinct and parallel. | Required evidence: Identifies the unique intersection when the determinant is non-zero.; Identifies the coincident-line case and infinitely many solutions.; Identifies the distinct parallel-line case and no solution. | Q9(d) | 3 | ___ / 3 | Q9(d): reproduce this process without the solution: Identifies the unique intersection when the determinant is non-zero; then Identifies the coincident-line case and infinitely many solutions; then Identifies the distinct parallel-line case and no solution. |
| Section B | Criterion 5 | Matrices and linear algebra | Compose transformations in the correct order | Expected result: \(M=RS=\begin{bmatrix}0&-1\\1&2\end{bmatrix}\) and \(P'=(-2,5)\). | Required evidence: Identifies the composite as \(RS\), not \(SR\).; Forms the first row of \(RS\) correctly.; Forms the second row of \(RS\) correctly.; Multiplies \(M\begin{bmatrix}1\\2\end{bmatrix}\).; Obtains the image \(P'=(-2,5)\). | Q10 | 5 | ___ / 5 | Q10: reproduce this process without the solution: Identifies the composite as RS, not SR; then Forms the first row of RS correctly; then Forms the second row of RS correctly; then Multiplies Mbegin(bmatrix)12end(bmatrix); then Obtains the image P'=(-2,5). |
| Section B | Criterion 5 | Matrices and linear algebra | Map a curve and interpret a determinant | Expected result: The image line is \(y=-3x\), and the area scale factor is \(1\). | Required evidence: Parameterises the original line as \((t,t)\).; Maps \((t,t)\) to \((-t,3t)\).; Eliminates \(t\) to obtain \(y=-3x\).; Calculates \(\det M=1\).; Interprets \(|\det M|=1\) as the area scale factor. | Q11 | 5 | ___ / 5 | Q11: reproduce this process without the solution: Parameterises the original line as (t,t); then Maps (t,t) to (-t,3t); then Eliminates t to obtain y=-3x; then Calculates det M=1; then Interprets |det M|=1 as the area scale factor. |
| Criterion 5 - Matrices and linear algebra | normal vector of a plane | Expected result: For example, \((2,-1,2)\). | Required evidence: States a non-zero scalar multiple of \((2,-1,2)\) . | Q12(a) | 1 | ___ / 1 | Q12(a): reproduce this process without the solution: States a non-zero scalar multiple of (2,-1,2) . |
| Criterion 5 - Matrices and linear algebra | composition of two rotations | Expected result: \(\begin{bmatrix}\cos\alpha\cos\beta-\sin\alpha\sin\beta&-(\cos\alpha\sin\beta+\sin\alpha\cos\beta)\\\sin\alpha\cos\beta+\cos\alpha\sin\beta&\cos\alpha\cos\beta-\sin\alpha\sin\beta\end{bmatrix}\). | Required evidence: Obtains both diagonal entries correctly.; Obtains the signed off-diagonal entries correctly. | Q12(b) | 2 | ___ / 2 | Q12(b): reproduce this process without the solution: Obtains both diagonal entries correctly; then Obtains the signed off-diagonal entries correctly. |
| Criterion 5 - Matrices and linear algebra | addition identities from rotations | Expected result: \(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\) and \(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\). | Required evidence: States the cosine addition identity.; States the sine addition identity. | Q12(c) | 2 | ___ / 2 | Q12(c): reproduce this process without the solution: States the cosine addition identity; then States the sine addition identity. |
| Criterion 5 - Matrices and linear algebra | area preserved by a shear | Expected result: 1 square unit. | Required evidence: Finds the original triangle area.; Calculates \(|\det S|=1\).; Obtains the unchanged image area. | Q12(d) | 3 | ___ / 3 | Q12(d): reproduce this process without the solution: Finds the original triangle area; then Calculates |det S|=1; then Obtains the unchanged image area. |
| Section C | Criterion 6 | Differential calculus, areas and volumes | Analyse stationary points of a rational function | Expected result: The stationary points are \((-2,-4)\), a local maximum, and \((2,4)\), a local minimum. | Required evidence: Differentiates to obtain \(f'(x)=1-\frac4{x^2}\).; Solves \(f'(x)=0\) to obtain \(x=-2\) and \(x=2\).; Calculates the stationary-point coordinates \((-2,-4)\) and \((2,4)\).; Obtains \(f''(x)=\frac8{x^3}\).; Uses \(f''(-2)<0\) and \(f''(2)>0\) to classify the local maximum and local minimum. | Q13 | 5 | ___ / 5 | Q13: reproduce this process without the solution: Differentiates to obtain f'(x)=1-frac4(x²); then Solves f'(x)=0 to obtain x=-2 and x=2; then Calculates the stationary-point coordinates (-2,-4) and (2,4); then Obtains f''(x)=frac8(x³); then Uses f''(-2)<0 and f''(2)>0 to classify the local maximum and local minimum. |
| Section C | Criterion 6 | Differential calculus, areas and volumes | Determine asymptotes and exact range | Expected result: The asymptotes are \(x=0\) and \(y=x\). The range is \((-\infty,-4]\cup[4,\infty)\). | Required evidence: Identifies \(x=0\) as the vertical asymptote.; Uses \(f(x)-x=\frac4x\to0\) as \(|x|\to\infty\) to identify the oblique asymptote \(y=x\).; Rearranges \(y=x+\frac4x\) to the quadratic \(x^2-yx+4=0\).; Requires discriminant \(y^2-16\ge0\) for real \(x\).; States the range \((-\infty,-4]\cup[4,\infty)\), including the endpoint values attained at the stationary points. | Q14 | 5 | ___ / 5 | Q14: reproduce this process without the solution: Identifies x=0 as the vertical asymptote; then Uses f(x)-x=frac4xto0 as |x|toinfty to identify the oblique asymptote y=x; then Rearranges y=x+frac4x to the quadratic x²-yx+4=0; then Requires discriminant y²-16ge0 for real x; then States the range (-infty,-4]cup[4,infty), including the endpoint values attained at the stationary points. |
| Criterion 6 - Differential calculus, areas and volumes | derivatives of advanced functions | Expected result: \(f'(x)=1e^{1x}\). | Required evidence: \(f'(x)=1e^{1x}\) | Q15(a) | 1 | ___ / 1 | Q15(a): reproduce this process without the solution: f'(x)=1e^(1x). |
| Criterion 6 - Differential calculus, areas and volumes | implicit tangent and normal | Expected result: The tangent is \(y-1=-(x-1)\), and the normal is \(y-1=x-1\). | Required evidence: Uses implicit differentiation to obtain tangent slope -1.; States both the tangent and perpendicular normal equations. | Q15(b) | 2 | ___ / 2 | Q15(b): reproduce this process without the solution: Uses implicit differentiation to obtain tangent slope -1; then States both the tangent and perpendicular normal equations. |
| Criterion 6 - Differential calculus, areas and volumes | stationary points | Expected result: \(x=\pm1\). | Required evidence: Obtains \(g'(x)=3(x^2-1)\).; Solves to obtain \(x=\pm1\). | Q15(c) | 2 | ___ / 2 | Q15(c): reproduce this process without the solution: Obtains g'(x)=3(x²-1); then Solves to obtain x=pm1. |
| Criterion 6 - Differential calculus, areas and volumes | classification, inflection and concavity | Expected result: There is a local maximum at \(x=-1\), a local minimum at \(x=1\), and an inflection at \(x=0\), where \(g''(x)=6x\) changes sign. | Required evidence: Uses the second derivative to classify every stationary point.; Solves the second-derivative condition for all possible inflection points.; Verifies the required concavity change and states every inflection. | Q15(d) | 3 | ___ / 3 | Q15(d): reproduce this process without the solution: Uses the second derivative to classify every stationary point; then Solves the second-derivative condition for all possible inflection points; then Verifies the required concavity change and states every inflection. |
| Section C | Criterion 6 | Differential calculus, areas and volumes | Model a y-axis volume with washers | Expected result: \(V=\pi\int_0^2(16-y^4)\,dy\). | Required evidence: Identifies the outer radius as \(4\).; Identifies the inner radius as \(y^2\).; Uses the washer area \(\pi(R^2-r^2)\).; Uses the correct limits \(0\le y\le2\). | Q16 | 4 | ___ / 4 | Q16: reproduce this process without the solution: Identifies the outer radius as 4; then Identifies the inner radius as y²; then Uses the washer area π(R²-r²); then Uses the correct limits 0le yle2. |
| Section C | Criterion 6 | Differential calculus, areas and volumes | Evaluate an exact washer integral | Expected result: \(V=\frac{128\pi}{5}\) cubic units. | Required evidence: Integrates \(16\) to \(16y\).; Integrates \(y^4\) to \(\frac{y^5}{5}\).; Applies the upper limit \(y=2\).; Applies the lower limit \(y=0\).; Simplifies \(32-\frac{32}{5}\) to \(\frac{128}{5}\).; States \(\frac{128\pi}{5}\) with cubic units. | Q17 | 6 | ___ / 6 | Q17: reproduce this process without the solution: Integrates 16 to 16y; then Integrates y⁴ to ((y⁵) / (5)); then Applies the upper limit y=2; then Applies the lower limit y=0; then Simplifies 32-((32) / (5)) to ((128) / (5)); then States ((128π) / (5)) with cubic units. |
| Criterion 6 - Differential calculus, areas and volumes | area integral setup | Expected result: \(\int_0^2x(2-x)\,dx\). | Required evidence: States \(\int_0^2x(2-x)\,dx\). | Q18(a) | 1 | ___ / 1 | Q18(a): reproduce this process without the solution: States int_0^2x(2-x),dx. |
| Criterion 6 - Differential calculus, areas and volumes | evaluate a definite area | Expected result: \(\frac{4}{3}\) square units. | Required evidence: Integrates the area integrand correctly.; Applies the limits and obtains \(\frac{4}{3}\). | Q18(b) | 2 | ___ / 2 | Q18(b): reproduce this process without the solution: Integrates the area integrand correctly; then Applies the limits and obtains ((4) / (3)). |
| Criterion 6 - Differential calculus, areas and volumes | volume-of-revolution setup | Expected result: \(\pi\int_0^2[x(2-x)]^2\,dx\). | Required evidence: Uses the disk method with the correct limits.; Squares and simplifies the radius function correctly. | Q18(c) | 2 | ___ / 2 | Q18(c): reproduce this process without the solution: Uses the disk method with the correct limits; then Squares and simplifies the radius function correctly. |
| Criterion 6 - Differential calculus, areas and volumes | evaluate a volume of revolution | Expected result: \(\frac{16\pi}{15}\) cubic units. | Required evidence: Finds a correct antiderivative including the factor of pi.; Applies both limits without a sign or power error.; Simplifies to the exact volume \(\frac{16\pi}{15}\). | Q18(d) | 3 | ___ / 3 | Q18(d): reproduce this process without the solution: Finds a correct antiderivative including the factor of π; then Applies both limits without a sign or power error; then Simplifies to the exact volume ((16π) / (15)). |
| Section D | Criterion 7 | Integration techniques and differential equations | Resolve distinct linear factors | Expected result: \(\frac{5x+7}{(x+1)(x+2)}=\frac2{x+1}+\frac3{x+2}\). | Required evidence: Writes \(\frac{A}{x+1}+\frac{B}{x+2}\).; Forms \(5x+7=A(x+2)+B(x+1)\).; Solves \(A=2\).; Solves \(B=3\) and states the decomposition. | Q19 | 4 | ___ / 4 | Q19: reproduce this process without the solution: Writes ((A) / (x+1))+((B) / (x+2)); then Forms 5x+7=A(x+2)+B(x+1); then Solves A=2; then Solves B=3 and states the decomposition. |
| Section D | Criterion 7 | Integration techniques and differential equations | Integrate partial fractions with exact limits | Expected result: \(I=3\ln3-\ln2=\ln\left(\frac{27}{2}\right)\). | Required evidence: Integrates \(\frac2{x+1}\) as \(2\ln|x+1|\).; Integrates \(\frac3{x+2}\) as \(3\ln|x+2|\).; Evaluates the antiderivative at \(x=1\).; Evaluates the antiderivative at \(x=0\).; Simplifies to \(3\ln3-\ln2\).; Combines logarithms to obtain \(\ln(\frac{27}{2})\). | Q20 | 6 | ___ / 6 | Q20: reproduce this process without the solution: Integrates frac2(x+1) as 2ln|x+1|; then Integrates frac3(x+2) as 3ln|x+2|; then Evaluates the antiderivative at x=1; then Evaluates the antiderivative at x=0; then Simplifies to 3ln3-ln2; then Combines logarithms to obtain ln(((27) / (2))). |
| Criterion 7 - Integration techniques and differential equations | pack 0 partial-fraction decomposition | Expected result: \(\dfrac1x-\dfrac1{x+1}\). | Required evidence: Obtains \(\dfrac1x-\dfrac1{x+1}\) . | Q21(a) | 1 | ___ / 1 | Q21(a): reproduce this process without the solution: Obtains dfrac1x-dfrac1(x+1) . |
| Criterion 7 - Integration techniques and differential equations | pack 0 rational integration | Expected result: \(\ln|x|-\ln|x+1|+C\). | Required evidence: Integrates each simple fraction with the correct coefficient and absolute value.; States \(\ln|x|-\ln|x+1|+C\) . | Q21(b) | 2 | ___ / 2 | Q21(b): reproduce this process without the solution: Integrates each simple fraction with the correct coefficient and absolute value; then States ln|x|-ln|x+1|+C . |
| Criterion 7 - Integration techniques and differential equations | inverse-sine antiderivative | Expected result: \(\sin^{-1}(x/2)+C\). | Required evidence: Matches the integrand to the supplied inverse-sine form.; Includes the constant and obtains the exact antiderivative. | Q21(c) | 2 | ___ / 2 | Q21(c): reproduce this process without the solution: Matches the integrand to the supplied inverse-sine form; then Includes the constant and obtains the exact antiderivative. |
| Criterion 7 - Integration techniques and differential equations | exponential integration by parts | Expected result: \(e^{2x}(x/2-1/4)+C\). | Required evidence: Chooses u and v consistently.; Applies the integration-by-parts formula with the correct sign.; Simplifies to the stated antiderivative. | Q21(d) | 3 | ___ / 3 | Q21(d): reproduce this process without the solution: Chooses u and v consistently; then Applies the integration-by-parts formula with the correct sign; then Simplifies to the stated antiderivative. |
| Section D | Criterion 7 | Integration techniques and differential equations | Solve a homogeneous differential equation by substitution | Expected result: \(y=x(\ln x+2)\), for \(x>0\). | Required evidence: Sets \(y=vx\).; Differentiates to obtain \(y'=v+x\frac{dv}{dx}\).; Substitutes into the differential equation and simplifies to \(x\frac{dv}{dx}=1\).; Integrates to obtain \(v=\ln x+C\).; Returns to \(y=x(\ln x+C)\).; Uses \(y(1)=2\) to obtain \(C=2\) and states \(x>0\). | Q22 | 6 | ___ / 6 | Q22: reproduce this process without the solution: Sets y=vx; then Differentiates to obtain y'=v+x((dv) / (dx)); then Substitutes into the differential equation and simplifies to x((dv) / (dx))=1; then Integrates to obtain v=ln x+C; then Returns to y=x(ln x+C); then Uses y(1)=2 to obtain C=2 and states x>0. |
| Section D | Criterion 7 | Integration techniques and differential equations | Evaluate a solution and its derivative | Expected result: \(y(e)=3e\) and \(\left.\frac{dy}{dx}\right|_{x=e}=4\). | Required evidence: Uses \(\ln e=1\).; Obtains \(y(e)=e(1+2)=3e\).; Substitutes \(x=e\) and \(y=3e\) into \(1+\frac{y}{x}\).; Obtains the gradient \(4\). | Q23 | 4 | ___ / 4 | Q23: reproduce this process without the solution: Uses ln e=1; then Obtains y(e)=e(1+2)=3e; then Substitutes x=e and y=3e into 1+((y) / (x)); then Obtains the gradient 4. |
| Criterion 7 - Integration techniques and differential equations | linear substitution | Expected result: \(u=2x+1\), with \(dx=du/2\). | Required evidence: States a consistent substitution and differential. | Q24(a) | 1 | ___ / 1 | Q24(a): reproduce this process without the solution: States a consistent substitution and differential. |
| Criterion 7 - Integration techniques and differential equations | equation with derivative a function of x | Expected result: \(y=x^2+x+2\). | Required evidence: Finds the general antiderivative.; Uses the initial condition correctly. | Q24(b) | 2 | ___ / 2 | Q24(b): reproduce this process without the solution: Finds the general antiderivative; then Uses the initial condition correctly. |
| Criterion 7 - Integration techniques and differential equations | homogeneous differential equation | Expected result: \(y=x(\ln x+C)\). | Required evidence: Makes the homogeneous substitution.; Integrates and back-substitutes. | Q24(c) | 2 | ___ / 2 | Q24(c): reproduce this process without the solution: Makes the homogeneous substitution; then Integrates and back-substitutes. |
| Criterion 7 - Integration techniques and differential equations | exponential growth model | Expected result: \(P(t)=100e^t\) and \(t=\ln2\). | Required evidence: Obtains the exponential solution family.; Applies the initial value.; Solves the doubling equation. | Q24(d) | 3 | ___ / 3 | Q24(d): reproduce this process without the solution: Obtains the exponential solution family; then Applies the initial value; then Solves the doubling equation. |
| Section E | Criterion 8 | Complex numbers | Apply De Moivre's theorem to roots | Expected result: \(z=2\operatorname{cis}\frac{\pi}{4},\,2\operatorname{cis}\frac{3\pi}{4},\,2\operatorname{cis}\frac{5\pi}{4},\,2\operatorname{cis}\frac{7\pi}{4}\). | Required evidence: Writes \(-16\) with modulus \(16\) and argument \(\pi+2k\pi\).; Takes the fourth root of the modulus to obtain \(2\).; Uses angles \(\frac{\pi+2k\pi}{4}\).; Lists the four distinct roots for \(k=0,1,2,3\). | Q25 | 4 | ___ / 4 | Q25: reproduce this process without the solution: Writes -16 with modulus 16 and argument π+2kpi; then Takes the fourth root of the modulus to obtain 2; then Uses angles ((π+2kpi) / (4)); then Lists the four distinct roots for k=0,1,2,3. |
| Section E | Criterion 8 | Complex numbers | Use conjugate root pairs for real factorisation | Expected result: \(z^4+16=(z^2-2\sqrt2\,z+4)(z^2+2\sqrt2\,z+4)\). | Required evidence: Pairs the roots at angles \(\frac{\pi}{4}\) and \(\frac{7\pi}{4}\).; Forms the factor \(z^2-2\sqrt2\,z+4\).; Pairs the roots at angles \(\frac{3\pi}{4}\) and \(\frac{5\pi}{4}\).; Forms the factor \(z^2+2\sqrt2\,z+4\).; Multiplies or otherwise verifies cancellation of the \(z^3\) and \(z\) terms.; States the complete real quadratic factorisation. | Q26 | 6 | ___ / 6 | Q26: reproduce this process without the solution: Pairs the roots at angles ((π) / (4)) and ((7π) / (4)); then Forms the factor z²-2sqrt2,z+4; then Pairs the roots at angles ((3π) / (4)) and ((5π) / (4)); then Forms the factor z²+2sqrt2,z+4; then Multiplies or otherwise verifies cancellation of the z³ and z terms; then States the complete real quadratic factorisation. |
| Criterion 8 - Complex numbers | complex conjugate | Expected result: \(\overline z=\sqrt3-i\). | Required evidence: States \(\overline z=\sqrt3-i\) . | Q27(a) | 1 | ___ / 1 | Q27(a): reproduce this process without the solution: States overline z=sqrt3-i . |
| Criterion 8 - Complex numbers | Cartesian to Euler form | Expected result: \(z=2e^{i\pi/6}\). | Required evidence: Finds the modulus.; Finds the principal argument and states the Euler form. | Q27(b) | 2 | ___ / 2 | Q27(b): reproduce this process without the solution: Finds the modulus; then Finds the principal argument and states the Euler form. |
| Criterion 8 - Complex numbers | De Moivre power from Cartesian data | Expected result: \(z^2=2+2\sqrt3i\). | Required evidence: Squares the modulus and doubles the argument.; Converts the result back to Cartesian form. | Q27(c) | 2 | ___ / 2 | Q27(c): reproduce this process without the solution: Squares the modulus and doubles the argument; then Converts the result back to Cartesian form. |
| Criterion 8 - Complex numbers | conjugate-root quadratic factor | Expected result: The paired root is \(\sqrt3-i\) and the factor is \(x^2-2\sqrt3x+4\). | Required evidence: Identifies the conjugate root.; Uses the sum and product of the roots.; Obtains the real monic factor. | Q27(d) | 3 | ___ / 3 | Q27(d): reproduce this process without the solution: Identifies the conjugate root; then Uses the sum and product of the roots; then Obtains the real monic factor. |
| Section E | Criterion 8 | Complex numbers | Interpret closed modulus and argument boundaries | Expected result: The boundaries are the included circle \(|z-(1+i)|=2\) and the included rays \(\operatorname{Arg}z=0\) and \(\operatorname{Arg}z=\frac{\pi}{3}\), restricted to their common intersection; \(z=0\) is excluded because its argument is undefined. | Required evidence: Identifies the circle centre \(1+i\).; Identifies the circle radius \(2\).; Identifies the two boundary rays at angles \(0\) and \(\frac{\pi}{3}\).; States that the circle and rays are included but the origin is excluded because \(\operatorname{Arg}0\) is undefined. | Q28 | 4 | ___ / 4 | Q28: reproduce this process without the solution: Identifies the circle centre 1+i; then Identifies the circle radius 2; then Identifies the two boundary rays at angles 0 and ((π) / (3)); then States that the circle and rays are included but the origin is excluded because operatorname(Arg)0 is undefined. |
| Section E | Criterion 8 | Complex numbers | Test points against a combined Argand region | Expected result: The region is the part of the closed disc lying in the closed sector \(0\le\theta\le\frac{\pi}{3}\), excluding the origin. The points \(2+i\) and \(3+i\) belong to \(R\); \(1+3i\) does not. | Required evidence: Sketches the disc with centre \(1+i\) and radius \(2\).; Restricts the sketch to the sector between angles \(0\) and \(\frac{\pi}{3}\).; Checks \(2+i\) against both conditions and includes it.; Checks \(1+3i\) and excludes it because its argument exceeds \(\frac{\pi}{3}\).; Checks \(3+i\) as lying on the circle and within the sector.; Uses solid boundaries and excludes the origin explicitly. | Q29 | 6 | ___ / 6 | Q29: reproduce this process without the solution: Sketches the disc with centre 1+i and radius 2; then Restricts the sketch to the sector between angles 0 and ((π) / (3)); then Checks 2+i against both conditions and includes it; then Checks 1+3i and excludes it because its argument exceeds ((π) / (3)); then Checks 3+i as lying on the circle and within the sector; then Uses solid boundaries and excludes the origin explicitly. |
| Criterion 8 - Complex numbers | three-term geometric polynomial identity | Expected result: \(z^3=1\), with \(z\ne1\). | Required evidence: Obtains \(z^3=1\) while excluding \(z=1\) . | Q30(a) | 1 | ___ / 1 | Q30(a): reproduce this process without the solution: Obtains z³=1 while excluding z=1 . |
| Criterion 8 - Complex numbers | non-real cube roots from a geometric sum | Expected result: \(z=\operatorname{cis}(2\pi/3)\) or \(z=\operatorname{cis}(4\pi/3)\). | Required evidence: Lists all three cube roots of unity before applying the exclusion.; States exactly the two non-real solutions without duplication. | Q30(b) | 2 | ___ / 2 | Q30(b): reproduce this process without the solution: Lists all three cube roots of unity before applying the exclusion; then States exactly the two non-real solutions without duplication. |
| Criterion 8 - Complex numbers | Cartesian conversion of an offset half-disk | Expected result: \((x-1)^2+y^2\leq4\) and \(y\geq0\). | Required evidence: Converts the modulus condition to the correct disk inequality.; Converts the imaginary-part condition to the correct half-plane. | Q30(c) | 2 | ___ / 2 | Q30(c): reproduce this process without the solution: Converts the modulus condition to the correct disk inequality; then Converts the imaginary-part condition to the correct half-plane. |
| Criterion 8 - Complex numbers | closed offset semicircular region | Expected result: It is the closed upper half of the disk centred at \((1,0)\) with radius 2. Both boundary pieces are included, and they meet at \((-1,0)\) and \((3,0)\). | Required evidence: Identifies the correct upper half-disk.; States that the arc and diameter are included.; Obtains both diameter endpoints. | Q30(d) | 3 | ___ / 3 | Q30(d): reproduce this process without the solution: Identifies the correct upper half-disk; then States that the arc and diameter are included; then Obtains both diameter endpoints. |
Total score (record in 0.5-mark increments): _____ / 180