Skill Align TASC Mathematics Methods - Foundation Level 3 Practice Pack 0 Section A - 2026 Edition
Current for 2026 | Free-online Pack 0 v2.0 booklet 1 of 2 | 80 marks | 80 minutes | calculators prohibited.
- Paper
- Section A - Calculator Prohibited
- Reading
- Included in the shared 15-minute preparation time
- Writing
- 80 minutes
- Assessment
- 80 marks
Section A conditions: calculators are prohibited. Supply the current TASC Mathematics Methods - Foundation Level 3 information sheet with this booklet. Internet access and external communication are not permitted. No public PDF or reference-sheet download is supplied with Pack 0.
Part 1 - Criterion 4: Algebra (16 marks)
Answer all 4 questions in this part without a calculator. One-mark questions do not require working. For questions worth 2 or more marks, show relevant working. Give exact values where required unless a question states otherwise.
Question 1
1 markQuestion 2
3 marksQuestion 3
5 marksQuestion 4
7 marksPart 2 - Criterion 5: Linear, quadratic and cubic functions (16 marks)
Answer all 4 questions in this part without a calculator. One-mark questions do not require working. For questions worth 2 or more marks, show relevant working. Give exact values where required unless a question states otherwise.
Question 5
1 markQuestion 6
3 marksQuestion 7
5 marksQuestion 8
7 marksPart 3 - Criterion 6: Logarithmic, exponential and trigonometric functions (16 marks)
Answer all 4 questions in this part without a calculator. One-mark questions do not require working. For questions worth 2 or more marks, show relevant working. Give exact values where required unless a question states otherwise.
Question 9
1 markQuestion 10
3 marksQuestion 11
5 marksQuestion 12
7 marksPart 4 - Criterion 7: Differential calculus (16 marks)
Answer all 4 questions in this part without a calculator. One-mark questions do not require working. For questions worth 2 or more marks, show relevant working. Give exact values where required unless a question states otherwise.
Question 13
1 markQuestion 14
3 marksQuestion 15
5 marksQuestion 16
7 marksPart 5 - Criterion 8: Probability and statistics (16 marks)
Answer all 4 questions in this part without a calculator. One-mark questions do not require working. For questions worth 2 or more marks, show relevant working. Give exact values where required unless a question states otherwise.
Question 17
1 markQuestion 18
3 marksQuestion 19
5 marksQuestion 20
7 marksWorked Solutions And Marking Guide
General marking principles
- Award one mark for each independently observable checkpoint described for the question.
- Apply consequential marking when a correct later method uses an earlier incorrect value.
- Accept algebraically equivalent exact forms and mathematically valid alternative no-calculator methods with sufficient working.
- Do not award full marks for an unsupported answer when the question requires working or reasoning.
Question 1
Indicative answer: -31.
The discriminant is b²-4ac=3²-4(2)(5)=9-40=-31.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 2
Indicative answer: 10-5sqrt3.
Multiply numerator and denominator by the conjugate 2-sqrt3. The denominator becomes (2+sqrt3)(2-sqrt3)=1, so the result is 5(2-sqrt3)=10-5sqrt3.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 3
Indicative answer: 3x²-12x+5=3(x-2)²-7, and x=2pmsqrt5.
Factor 3 from the quadratic terms and complete the square: 3(x²-4x)+5=3((x-2)²-4)+5=3(x-2)²-7. The equation is then 3(x-2)²-7=8, so (x-2)²=5 and x=2pmsqrt5.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 4
Indicative answer: The solutions are x=-1,1,3, and the expanded equation is x³-3x²-x+3=0.
The substitution gives u³-4u=0, or u(u-2)(u+2)=0. Thus u=-2,0,2, and because x=u+1, x=-1,1,3. Expanding gives (x-1)³-4(x-1)=x³-3x²-x+3. Substitution of each listed value gives zero, consistent with the complete factorisation in u.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (7 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 5
Indicative answer: (0,-8).
At the y-axis, x=0, so h(0)=(-2)(1)(4)=-8.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 6
Indicative answer: The gradient is -2, the rule is f(x)=-2x+3, and the zero is x=frac32.
The gradient is (-5-7) / (4-(-2))=-12 / 6=-2. Using (-2,7) gives 7=4+b, so b=3. Solving -2x+3=0 gives x=frac32.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 7
Indicative answer: The rule is q(x)=-2(x+1)(x-5), its axis is x=2, its maximum point is (2,18), and its range is q(x)leq18.
Write q(x)=a(x+1)(x-5). The y-intercept gives 10=-5a, so a=-2. The axis lies midway between the zeros at x=2, and q(2)=-2(3)(-3)=18. Since the leading coefficient is negative, this is the maximum and the range follows.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 8
(a) The graph crosses at (-2,0) and touches at (1,0).
The factor x+2 has multiplicity one, so the graph crosses at x=-2. The factor x-1² has multiplicity two, so the graph touches at x=1.
(b) The y-intercept is (0,-2). As xto-infty, h(x)toinfty, and as xtoinfty, h(x)to-infty.
Substitution gives h(0)=-(2)(1)=-2. The leading term is -x³, so the graph rises to the left and falls to the right.
(c) A labelled cubic that rises on the left, crosses at -2,0, passes through 0,-2, touches the x-axis at 1,0, and falls on the right.
Plot the three intercepts, use the stated multiplicities at the two zeros, and connect them with the end behaviour of a negative cubic.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Part 8(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 8(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 8(c) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 9
Indicative answer: -2.
Since frac1(25)=5⁻², the logarithm is -2.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 10
Indicative answer: M(0)=480, M(12)=120, and the half-life is 6 time units.
At t=0, the exponent is 0, so M(0)=480. At t=12, the exponent is 2, giving M(12)=480(frac12)²=120. Increasing t by 6 increases the exponent by 1, so the quantity is multiplied by frac12; the half-life is therefore 6 time units.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 11
Indicative answer: The domain is x>1, and the solution is x=3.
Both logarithm arguments must be positive, giving x>1. Combining the logarithms gives log_2(x²-1)=3, so x²-1=8 and x=pm3. Only x=3 satisfies the domain restriction.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 12
Indicative answer: The amplitude is 3, the period is π, and the range is [-4,2]. The solutions are x=((π) / (6)),((5π) / (6)),((7π) / (6)),((11π) / (6)).
The vertical scale gives amplitude 3, and the input factor 2 gives period 2π / 2=π. Shifting down by 1 gives range [-4,2]. The equation becomes cos(2x)=frac12. For 0leq2xleq4π, the exact angles are 2x=((π) / (3)),((5π) / (3)),((7π) / (3)),((11π) / (3)), and division by 2 gives the four listed inputs.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (7 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 13
Indicative answer: ((dy) / (dx))=-9x^2.
The power rule gives ((dy) / (dx))=-9x²; the constant differentiates to zero.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 14
Indicative answer: f'(2)approx-2.5.
Use f'(2)approx((f(2.04)-f(1.96)) / (2.04-1.96))=((7.94-8.14) / (0.08))=-2.5.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 15
Indicative answer: a=1, b=4, and f'(-1)=7.
Differentiating gives f'(x)=3ax²+b. The two conditions give 3a+b=7 and 12a+b=16. Subtraction gives 9a=9, so a=1 and then b=4. Hence f'(x)=3x²+4, and f'(-1)=7.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 16
Indicative answer: v(t)=2t-6 m / s. The particle is at rest at t=3, moves in the negative direction for 0leq t<3, and in the positive direction for 3<tleq5. The total distance is 13 m.
Velocity is v(t)=s'(t)=2t-6, so rest occurs at t=3. The velocity is negative before 3 and positive after 3. The positions are s(0)=8, s(3)=-1 and s(5)=3, so the two travelled distances are 9 m and 4 m. Their sum is 13 m.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (7 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 17
Indicative answer: ((18) / (60))=0.30.
Experimental probability is relative frequency, so the estimate is 18 / 60=0.30.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 18
Indicative answer: There are 12 codes, and the probability is frac12.
There are 4 choices for the first digit and 3 for the second, giving 12 codes. An even code ends in 2 or 4; for each ending there are 3 possible first digits, so 6 codes are even and the probability is 6 / 12=frac12.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 19
Indicative answer: n(Wcup M)=37; 11 carry neither; P(Wcup M)=((37) / (48)); and P(Mmid W)=((13) / (29)).
Inclusion-exclusion gives n(Wcup M)=29+21-13=37. Therefore 48-37=11 hikers carry neither and P(Wcup M)=37 / 48. Of the 29 hikers who carry water, 13 also carry a map, so P(Mmid W)=13 / 29.
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 20
Indicative answer: There are 495 panels; 210 contain exactly two juniors; the probability of at least one junior is ((92) / (99)); and the final probability is frac8(33).
The total is binom(12)(4)=495. Exactly two juniors can be chosen in binom52binom72=210 ways. The complement of at least one junior is an all-senior panel, so the probability is (495-binom74) / 495=460 / 495=((92) / (99)). For the final condition, include the specified senior, exclude the specified junior and choose the remaining three people from the other ten: binom(10)(3)=120. Thus the probability is 120 / 495=frac8(33).
Mark allocation
- Retain exact values unless the prompt explicitly permits another form.
- Apply consequential marking to a valid later method after an earlier arithmetic or transcription error.
Detailed marking criteria
Integrated response (7 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any algebraically equivalent exact form. Accept another mathematically valid calculator-free method when sufficient working is shown.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Section A | Part 1 | Criterion 4 | Quadratic discriminant | Skill: evaluating a quadratic discriminant | Expected result: \(-31\). | Q1 | 1 | ___ | Rework Question 1 without a calculator: identify the three coefficients before substituting into the discriminant. |
| Section A | Part 1 | Criterion 4 | Rationalising a denominator | Skill: rationalising with a conjugate | Expected result: \(10-5\sqrt3\). | Q2 | 3 | ___ | Rework Question 2 without a calculator: multiply by the conjugate and simplify the difference-of-squares denominator. |
| Section A | Part 1 | Criterion 4 | Completing the square and solving | Skill: completing the square and solving an exact quadratic equation | Expected result: \(3x^2-12x+5=3(x-2)^2-7\), and \(x=2\pm\sqrt5\). | Q3 | 5 | ___ | Rework Question 3 without a calculator: factor the leading coefficient from the quadratic terms, complete the square and use the resulting form to isolate the squared expression. |
| Section A | Part 1 | Criterion 4 | Cubic equation by substitution | Skill: solving a shifted cubic through a temporary variable | Expected result: The solutions are \(x=-1,1,3\), and the expanded equation is \(x^3-3x^2-x+3=0\). | Q4 | 7 | ___ | Rework Question 4 without a calculator: reduce the shifted expression to a simple cubic, solve it completely, convert back and check the expanded form. |
| Section A | Part 2 | Criterion 5 | Cubic function value | Skill: finding a cubic y-intercept | Expected result: \((0,-8)\). | Q5 | 1 | ___ | Rework Question 5 without a calculator: substitute zero for the input. |
| Section A | Part 2 | Criterion 5 | Linear function from two values | Skill: constructing and using a linear function | Expected result: The gradient is \(-2\), the rule is \(f(x)=-2x+3\), and the zero is \(x=\frac32\). | Q6 | 3 | ___ | Rework Question 6 without a calculator: use the two supplied function values as points, then determine the intercept and solve for the zero. |
| Section A | Part 2 | Criterion 5 | Quadratic features | Skill: recovering and analysing a downward quadratic | Expected result: The rule is \(q(x)=-2(x+1)(x-5)\), its axis is \(x=2\), its maximum point is \((2,18)\), and its range is \(q(x)\leq18\). | Q7 | 5 | ___ | Rework Question 7 without a calculator: use the intercept form to determine the scale, then use symmetry and orientation for the maximum and range. |
| Section A | Part 2 | Criterion 5 | Factorised cubic graph | Skill: Interpret factor multiplicity on a cubic graph; Determine a cubic intercept and end behaviour; Sketch a factorised cubic from its defining features | Expected result: (a) The graph crosses at \((-2,0)\) and touches at \((1,0)\).; (b) The y-intercept is \((0,-2)\). As \(x\to-\infty\), \(h(x)\to\infty\), and as \(x\to\infty\), \(h(x)\to-\infty\).; (c) A labelled cubic that rises on the left, crosses at \(-2,0\), passes through \(0,-2\), touches the x-axis at \(1,0\), and falls on the right. | Q8 | 7 | ___ | Rework every labelled subpart of Question 8 using the worked solution and item-specific checkpoints. |
| Section A | Part 3 | Criterion 6 | Exact logarithm | Skill: evaluating an exact logarithm | Expected result: \(-2\). | Q9 | 1 | ___ | Rework Question 9 without a calculator: rewrite the argument as a power of the base. |
| Section A | Part 3 | Criterion 6 | Exponential decay model | Skill: interpreting an exact exponential decay model | Expected result: \(M(0)=480\), \(M(12)=120\), and the half-life is \(6\) time units. | Q10 | 3 | ___ | Rework Question 10 without a calculator: evaluate the model at the two supplied times and read the halving interval from the exponent. |
| Section A | Part 3 | Criterion 6 | Logarithmic equation | Skill: solving a logarithmic equation with its domain | Expected result: The domain is \(x>1\), and the solution is \(x=3\). | Q11 | 5 | ___ | Rework Question 11 without a calculator: state the argument restrictions first, combine the logarithms, solve the resulting equation and reject excluded values. |
| Section A | Part 3 | Criterion 6 | Transformed cosine function | Skill: analysing and solving a transformed cosine function | Expected result: The amplitude is \(3\), the period is \(\pi\), and the range is \([-4,2]\). The solutions are \(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\). | Q12 | 7 | ___ | Rework Question 12 without a calculator: read the graph parameters, isolate the cosine value and use exact unit-circle angles over the transformed interval. |
| Section A | Part 4 | Criterion 7 | Cubic power-rule differentiation | Skill: differentiating a cubic term | Expected result: \(\frac{dy}{dx}=-9x^2\). | Q13 | 1 | ___ | Rework Question 13 without a calculator: apply the power rule to the cubic term and differentiate the constant as zero. |
| Section A | Part 4 | Criterion 7 | Numerical derivative estimate | Skill: estimating a derivative symmetrically | Expected result: \(f'(2)\approx-2.5\). | Q14 | 3 | ___ | Rework Question 14 without a calculator: divide the change across equally spaced inputs by the full input interval. |
| Section A | Part 4 | Criterion 7 | Parameters from derivative values | Skill: recovering parameters from derivative values | Expected result: \(a=1\), \(b=4\), and \(f'(-1)=7\). | Q15 | 5 | ___ | Rework Question 15 without a calculator: differentiate first, solve the two parameter equations and evaluate the completed derivative. |
| Section A | Part 4 | Criterion 7 | Motion from position | Skill: analysing one-dimensional motion from a quadratic position model | Expected result: \(v(t)=2t-6\) m/s. The particle is at rest at \(t=3\), moves in the negative direction for \(0\leq t<3\), and in the positive direction for \(3<t\leq5\). The total distance is \(13\) m. | Q16 | 7 | ___ | Rework Question 16 without a calculator: differentiate for velocity, use its sign for direction and split the distance calculation at the rest time. |
| Section A | Part 5 | Criterion 8 | Experimental probability | Skill: calculating an experimental probability | Expected result: \(\frac{18}{60}=0.30\). | Q17 | 1 | ___ | Rework Question 17 without a calculator: divide the observed event frequency by the total number of trials. |
| Section A | Part 5 | Criterion 8 | Finite sample space | Skill: counting a finite ordered sample space | Expected result: There are \(12\) codes, and the probability is \(\frac12\). | Q18 | 3 | ___ | Rework Question 18 without a calculator: count all ordered codes, then count those with an even final digit. |
| Section A | Part 5 | Criterion 8 | Set and conditional probabilities | Skill: using set counts for union, complement and conditional probability | Expected result: \(n(W\cup M)=37\); \(11\) carry neither; \(P(W\cup M)=\frac{37}{48}\); and \(P(M\mid W)=\frac{13}{29}\). | Q19 | 5 | ___ | Rework Question 19 without a calculator: apply inclusion-exclusion to the two set counts, use the group total for the complement and condition on the water-carrying group. |
| Section A | Part 5 | Criterion 8 | Combination probabilities | Skill: using combinations for several panel events | Expected result: There are \(495\) panels; \(210\) contain exactly two juniors; the probability of at least one junior is \(\frac{92}{99}\); and the final probability is \(\frac8{33}\). | Q20 | 7 | ___ | Rework Question 20 without a calculator: establish the common sample-space count, then use direct products, a complement and fixed-member counting. |