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IB · MATH AA SL

Mathematics AA: Standard Level

Calculus — Topic 5

Name: ____________________Date: October 10, 2026
  1. 1.

    Let g(x) = x³ − 6x² + 9x. Find the coordinates of all stationary points and classify each as a local maximum or local minimum.

    [6 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates to obtain g′(x) = 3x² − 12x + 9. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Solves 3(x − 1)(x − 3) = 0, giving x = 1 and x = 3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates g(1) = 4 and g(3) = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Uses g″(x) = 6x − 12 or an equivalent sign test. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Classifies (1, 4) as a local maximum. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Work through this mathematical step: Classifies (3, 0) as a local minimum. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Coordinates require both x and y values; roots of g′ alone are not a complete answer.

    Marking points

    • Differentiates to obtain g′(x) = 3x² − 12x + 9.
    • Solves 3(x − 1)(x − 3) = 0, giving x = 1 and x = 3.
    • Calculates g(1) = 4 and g(3) = 0.
    • Uses g″(x) = 6x − 12 or an equivalent sign test.
    • Classifies (1, 4) as a local maximum.
    • Classifies (3, 0) as a local minimum.

    Examiner tip: Coordinates require both x and y values; roots of g′ alone are not a complete answer.

  2. 2.

    Marking analysis: A learner attempts the following task: “Let g(x) = x³ − 6x² + 9x. Find the coordinates of all stationary points and classify each as a local maximum or local minimum.” Their response addresses only this point: “Differentiates to obtain g′(x) = 3x² − 12x + 9.” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [6 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates to obtain g′(x) = 3x² − 12x + 9. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Solves 3(x − 1)(x − 3) = 0, giving x = 1 and x = 3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates g(1) = 4 and g(3) = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Uses g″(x) = 6x − 12 or an equivalent sign test. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Classifies (1, 4) as a local maximum. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Requirement 6: Identifies the missing requirement: Classifies (3, 0) as a local minimum. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    8. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates to obtain g′(x) = 3x² − 12x + 9.
    • Identifies the missing requirement: Solves 3(x − 1)(x − 3) = 0, giving x = 1 and x = 3.
    • Identifies the missing requirement: Calculates g(1) = 4 and g(3) = 0.
    • Identifies the missing requirement: Uses g″(x) = 6x − 12 or an equivalent sign test.
    • Identifies the missing requirement: Classifies (1, 4) as a local maximum.
    • Identifies the missing requirement: Classifies (3, 0) as a local minimum.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  3. 3.

    Evaluate the definite integral from 0 to 2 of (3x² + 2) dx. Interpret your answer as the signed area between the curve and the x-axis on this interval.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Build a supported judgement: identify the claim, use the question's evidence, consider a relevant limitation or alternative, and make the conclusion depend on that evidence. There may be more than one defensible answer.
    2. Work through this mathematical step: Finds an antiderivative x³ + 2x. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Substitutes the limits: (8 + 4) − 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Obtains a signed area of 12 square units. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Show the substituted upper and lower limits, even when the lower contribution is zero.

    Marking points

    • Finds an antiderivative x³ + 2x.
    • Substitutes the limits: (8 + 4) − 0.
    • Obtains a signed area of 12 square units.

    Examiner tip: Show the substituted upper and lower limits, even when the lower contribution is zero.

  4. 4.

    Marking analysis: A learner attempts the following task: “Evaluate the definite integral from 0 to 2 of (3x² + 2) dx. Interpret your answer as the signed area between the curve and the x-axis on this interval.” Their response addresses only this point: “Finds an antiderivative x³ + 2x.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Finds an antiderivative x³ + 2x. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Substitutes the limits: (8 + 4) − 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Obtains a signed area of 12 square units. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Finds an antiderivative x³ + 2x.
    • Identifies the missing requirement: Substitutes the limits: (8 + 4) − 0.
    • Identifies the missing requirement: Obtains a signed area of 12 square units.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  5. 5.

    Find the equation of the tangent to the curve y = x² − 3x + 5 at the point where x = 2.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates to obtain dy/dx = 2x − 3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Substitutes x = 2 to find the gradient = 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates the y-coordinate: y = 4 − 6 + 5 = 3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Uses y − y₁ = m(x − x₁) with point (2, 3) and gradient 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: States the tangent equation y = x + 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The tangent's gradient comes from the derivative evaluated at the given x, not the derivative expression itself.

    Marking points

    • Differentiates to obtain dy/dx = 2x − 3.
    • Substitutes x = 2 to find the gradient = 1.
    • Calculates the y-coordinate: y = 4 − 6 + 5 = 3.
    • Uses y − y₁ = m(x − x₁) with point (2, 3) and gradient 1.
    • States the tangent equation y = x + 1.

    Examiner tip: The tangent's gradient comes from the derivative evaluated at the given x, not the derivative expression itself.

  6. 6.

    Marking analysis: A learner attempts the following task: “Find the equation of the tangent to the curve y = x² − 3x + 5 at the point where x = 2.” Their response addresses only this point: “Differentiates to obtain dy/dx = 2x − 3.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates to obtain dy/dx = 2x − 3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Substitutes x = 2 to find the gradient = 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates the y-coordinate: y = 4 − 6 + 5 = 3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Uses y − y₁ = m(x − x₁) with point (2, 3) and gradient 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: States the tangent equation y = x + 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates to obtain dy/dx = 2x − 3.
    • Identifies the missing requirement: Substitutes x = 2 to find the gradient = 1.
    • Identifies the missing requirement: Calculates the y-coordinate: y = 4 − 6 + 5 = 3.
    • Identifies the missing requirement: Uses y − y₁ = m(x − x₁) with point (2, 3) and gradient 1.
    • Identifies the missing requirement: States the tangent equation y = x + 1.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  7. 7.

    Differentiate y = (3x + 1)⁴ using the chain rule. Hence find the gradient of the curve at x = 0.

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Applies the chain rule: dy/dx = 4(3x + 1)³ × 3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Simplifies to dy/dx = 12(3x + 1)³. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Substitutes x = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Obtains gradient = 12. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Do not forget to multiply by the derivative of the inner function (here, 3) when applying the chain rule.

    Marking points

    • Applies the chain rule: dy/dx = 4(3x + 1)³ × 3.
    • Simplifies to dy/dx = 12(3x + 1)³.
    • Substitutes x = 0.
    • Obtains gradient = 12.

    Examiner tip: Do not forget to multiply by the derivative of the inner function (here, 3) when applying the chain rule.

  8. 8.

    Marking analysis: A learner attempts the following task: “Differentiate y = (3x + 1)⁴ using the chain rule. Hence find the gradient of the curve at x = 0.” Their response addresses only this point: “Applies the chain rule: dy/dx = 4(3x + 1)³ × 3.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Applies the chain rule: dy/dx = 4(3x + 1)³ × 3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Simplifies to dy/dx = 12(3x + 1)³. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Substitutes x = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains gradient = 12. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Applies the chain rule: dy/dx = 4(3x + 1)³ × 3.
    • Identifies the missing requirement: Simplifies to dy/dx = 12(3x + 1)³.
    • Identifies the missing requirement: Substitutes x = 0.
    • Identifies the missing requirement: Obtains gradient = 12.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  9. 9.

    Find the equation of the normal to the curve y = x³ at the point (1, 1).

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates to obtain dy/dx = 3x². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Substitutes x = 1 to find the tangent gradient = 3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Uses the normal gradient = −1/3, the negative reciprocal of the tangent gradient. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Uses y − 1 = −1/3(x − 1) with the given point. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: States the normal equation y = −x/3 + 4/3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient, not the same value.

    Marking points

    • Differentiates to obtain dy/dx = 3x².
    • Substitutes x = 1 to find the tangent gradient = 3.
    • Uses the normal gradient = −1/3, the negative reciprocal of the tangent gradient.
    • Uses y − 1 = −1/3(x − 1) with the given point.
    • States the normal equation y = −x/3 + 4/3.

    Examiner tip: The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent gradient, not the same value.

  10. 10.

    Marking analysis: A learner attempts the following task: “Find the equation of the normal to the curve y = x³ at the point (1, 1).” Their response addresses only this point: “Differentiates to obtain dy/dx = 3x².” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates to obtain dy/dx = 3x². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Substitutes x = 1 to find the tangent gradient = 3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Uses the normal gradient = −1/3, the negative reciprocal of the tangent gradient. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Uses y − 1 = −1/3(x − 1) with the given point. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: States the normal equation y = −x/3 + 4/3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates to obtain dy/dx = 3x².
    • Identifies the missing requirement: Substitutes x = 1 to find the tangent gradient = 3.
    • Identifies the missing requirement: Uses the normal gradient = −1/3, the negative reciprocal of the tangent gradient.
    • Identifies the missing requirement: Uses y − 1 = −1/3(x − 1) with the given point.
    • Identifies the missing requirement: States the normal equation y = −x/3 + 4/3.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  11. 11.

    A particle moves along a line so that its displacement is s(t) = t³ − 6t² + 9t metres, t ≥ 0 seconds. Find the particle's velocity function and determine when the particle is at rest.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates displacement to obtain v(t) = 3t² − 12t + 9. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Sets v(t) = 0 and factorises: 3(t − 1)(t − 3) = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States that the particle is at rest at t = 1 s and t = 3 s. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Velocity is the derivative of displacement; the particle is at rest exactly when velocity equals zero.

    Marking points

    • Differentiates displacement to obtain v(t) = 3t² − 12t + 9.
    • Sets v(t) = 0 and factorises: 3(t − 1)(t − 3) = 0.
    • States that the particle is at rest at t = 1 s and t = 3 s.

    Examiner tip: Velocity is the derivative of displacement; the particle is at rest exactly when velocity equals zero.

  12. 12.

    Marking analysis: A learner attempts the following task: “A particle moves along a line so that its displacement is s(t) = t³ − 6t² + 9t metres, t ≥ 0 seconds. Find the particle's velocity function and determine when the particle is at rest.” Their response addresses only this point: “Differentiates displacement to obtain v(t) = 3t² − 12t + 9.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates displacement to obtain v(t) = 3t² − 12t + 9. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Sets v(t) = 0 and factorises: 3(t − 1)(t − 3) = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that the particle is at rest at t = 1 s and t = 3 s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates displacement to obtain v(t) = 3t² − 12t + 9.
    • Identifies the missing requirement: Sets v(t) = 0 and factorises: 3(t − 1)(t − 3) = 0.
    • Identifies the missing requirement: States that the particle is at rest at t = 1 s and t = 3 s.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  13. 13.

    Find ∫(4x³ − 6x + 2) dx.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Integrates each term using the power rule. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Obtains x⁴ − 3x² + 2x. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Adds the constant of integration + C. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: An indefinite integral always requires the constant of integration, C, for full marks.

    Marking points

    • Integrates each term using the power rule.
    • Obtains x⁴ − 3x² + 2x.
    • Adds the constant of integration + C.

    Examiner tip: An indefinite integral always requires the constant of integration, C, for full marks.

  14. 14.

    Marking analysis: A learner attempts the following task: “Find ∫(4x³ − 6x + 2) dx.” Their response addresses only this point: “Integrates each term using the power rule.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Integrates each term using the power rule. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Obtains x⁴ − 3x² + 2x. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Adds the constant of integration + C. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Integrates each term using the power rule.
    • Identifies the missing requirement: Obtains x⁴ − 3x² + 2x.
    • Identifies the missing requirement: Adds the constant of integration + C.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  15. 15.

    For the curve y = 12/x, x ≠ 0, find the equation of the tangent at x = 3.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates to obtain dy/dx = −12/x². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Evaluates the gradient at x = 3 as −4/3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Finds the point of contact (3, 4). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Gives y − 4 = (−4/3)(x − 3), or y = −4x/3 + 8. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Write 12/x as 12x⁻¹ before differentiating. Use the original function to find the point and the derivative to find the gradient.

    Marking points

    • Differentiates to obtain dy/dx = −12/x².
    • Evaluates the gradient at x = 3 as −4/3.
    • Finds the point of contact (3, 4).
    • Gives y − 4 = (−4/3)(x − 3), or y = −4x/3 + 8.

    Examiner tip: Write 12/x as 12x⁻¹ before differentiating. Use the original function to find the point and the derivative to find the gradient.

  16. 16.

    Marking analysis: A learner attempts the following task: “For the curve y = 12/x, x ≠ 0, find the equation of the tangent at x = 3.” Their response addresses only this point: “Differentiates to obtain dy/dx = −12/x².” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates to obtain dy/dx = −12/x². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Evaluates the gradient at x = 3 as −4/3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Finds the point of contact (3, 4). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Gives y − 4 = (−4/3)(x − 3), or y = −4x/3 + 8. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates to obtain dy/dx = −12/x².
    • Identifies the missing requirement: Evaluates the gradient at x = 3 as −4/3.
    • Identifies the missing requirement: Finds the point of contact (3, 4).
    • Identifies the missing requirement: Gives y − 4 = (−4/3)(x − 3), or y = −4x/3 + 8.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  17. 17.

    Determine the interval(s) over which the function h(x) = x³ − 3x is increasing.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates to obtain h′(x) = 3x² − 3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Sets h′(x) = 0 and solves 3(x − 1)(x + 1) = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Obtains critical values x = −1 and x = 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Tests the sign of h′(x) in each region (e.g. at x = −2, 0, 2). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: States h′ > 0 for x < −1 or x > 1. Also accept the maximal increasing intervals (−∞, −1] and [1, ∞), including the stationary endpoints. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A function increases where its derivative is positive; test one value from each interval created by the critical points.

    Marking points

    • Differentiates to obtain h′(x) = 3x² − 3.
    • Sets h′(x) = 0 and solves 3(x − 1)(x + 1) = 0.
    • Obtains critical values x = −1 and x = 1.
    • Tests the sign of h′(x) in each region (e.g. at x = −2, 0, 2).
    • States h′ > 0 for x < −1 or x > 1. Also accept the maximal increasing intervals (−∞, −1] and [1, ∞), including the stationary endpoints.

    Examiner tip: A function increases where its derivative is positive; test one value from each interval created by the critical points.

  18. 18.

    Marking analysis: A learner attempts the following task: “Determine the interval(s) over which the function h(x) = x³ − 3x is increasing.” Their response addresses only this point: “Differentiates to obtain h′(x) = 3x² − 3.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates to obtain h′(x) = 3x² − 3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Sets h′(x) = 0 and solves 3(x − 1)(x + 1) = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Obtains critical values x = −1 and x = 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Tests the sign of h′(x) in each region (e.g. at x = −2, 0, 2). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: States h′ > 0 for x < −1 or x > 1. Also accept the maximal increasing intervals (−∞, −1] and [1, ∞), including the stationary endpoints. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates to obtain h′(x) = 3x² − 3.
    • Identifies the missing requirement: Sets h′(x) = 0 and solves 3(x − 1)(x + 1) = 0.
    • Identifies the missing requirement: Obtains critical values x = −1 and x = 1.
    • Identifies the missing requirement: Tests the sign of h′(x) in each region (e.g. at x = −2, 0, 2).
    • Identifies the missing requirement: States h′ > 0 for x < −1 or x > 1. Also accept the maximal increasing intervals (−∞, −1] and [1, ∞), including the stationary endpoints.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  19. 19.

    Water flows into a tank at a rate given by R(t) = 6t − t² litres per minute, for 0 ≤ t ≤ 6. Find the total volume of water that flows into the tank during this time.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Recognises that total volume is the definite integral of the rate function from 0 to 6. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Finds the antiderivative 3t² − t³/3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Substitutes the limits t = 6 and t = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Obtains a total volume of 36 litres. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A rate function must be integrated, not evaluated at a single time, to find a total accumulated quantity.

    Marking points

    • Recognises that total volume is the definite integral of the rate function from 0 to 6.
    • Finds the antiderivative 3t² − t³/3.
    • Substitutes the limits t = 6 and t = 0.
    • Obtains a total volume of 36 litres.

    Examiner tip: A rate function must be integrated, not evaluated at a single time, to find a total accumulated quantity.

  20. 20.

    Marking analysis: A learner attempts the following task: “Water flows into a tank at a rate given by R(t) = 6t − t² litres per minute, for 0 ≤ t ≤ 6. Find the total volume of water that flows into the tank during this time.” Their response addresses only this point: “Recognises that total volume is the definite integral of the rate function from 0 to 6.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Recognises that total volume is the definite integral of the rate function from 0 to 6. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Finds the antiderivative 3t² − t³/3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Substitutes the limits t = 6 and t = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains a total volume of 36 litres. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Recognises that total volume is the definite integral of the rate function from 0 to 6.
    • Identifies the missing requirement: Finds the antiderivative 3t² − t³/3.
    • Identifies the missing requirement: Substitutes the limits t = 6 and t = 0.
    • Identifies the missing requirement: Obtains a total volume of 36 litres.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  21. 21.

    A farmer has 40 m of fencing to enclose a rectangular field, using an existing wall as one side (so fencing is only needed for the other three sides). Find the dimensions that maximise the enclosed area, and calculate this maximum area.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Sets up the constraint x + 2y = 40, where x is the side parallel to the wall and y are the two perpendicular sides, and expresses x = 40 − 2y. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Forms the area function A = xy = (40 − 2y)y = 40y − 2y². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Differentiates: dA/dy = 40 − 4y. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Sets dA/dy = 0 and solves to obtain y = 10. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Finds x = 20, and states the maximum area = 200 m². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Always reduce the problem to a single variable using the constraint before differentiating — trying to optimise a function of two variables directly with single-variable calculus does not work.

    Marking points

    • Sets up the constraint x + 2y = 40, where x is the side parallel to the wall and y are the two perpendicular sides, and expresses x = 40 − 2y.
    • Forms the area function A = xy = (40 − 2y)y = 40y − 2y².
    • Differentiates: dA/dy = 40 − 4y.
    • Sets dA/dy = 0 and solves to obtain y = 10.
    • Finds x = 20, and states the maximum area = 200 m².

    Examiner tip: Always reduce the problem to a single variable using the constraint before differentiating — trying to optimise a function of two variables directly with single-variable calculus does not work.

  22. 22.

    Marking analysis: A learner attempts the following task: “A farmer has 40 m of fencing to enclose a rectangular field, using an existing wall as one side (so fencing is only needed for the other three sides). Find the dimensions that maximise the enclosed area, and calculate this maximum area.” Their response addresses only this point: “Sets up the constraint x + 2y = 40, where x is the side parallel to the wall and y are the two perpendicular sides, and expresses x = 40 − 2y.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Sets up the constraint x + 2y = 40, where x is the side parallel to the wall and y are the two perpendicular sides, and expresses x = 40 − 2y. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Forms the area function A = xy = (40 − 2y)y = 40y − 2y². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Differentiates: dA/dy = 40 − 4y. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Sets dA/dy = 0 and solves to obtain y = 10. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Finds x = 20, and states the maximum area = 200 m². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Sets up the constraint x + 2y = 40, where x is the side parallel to the wall and y are the two perpendicular sides, and expresses x = 40 − 2y.
    • Identifies the missing requirement: Forms the area function A = xy = (40 − 2y)y = 40y − 2y².
    • Identifies the missing requirement: Differentiates: dA/dy = 40 − 4y.
    • Identifies the missing requirement: Sets dA/dy = 0 and solves to obtain y = 10.
    • Identifies the missing requirement: Finds x = 20, and states the maximum area = 200 m².

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  23. 23.

    A particle's displacement is given by s(t) = t³ − 9t² + 24t, for t ≥ 0 (metres, seconds). Find the particle's acceleration function, and determine the value of t at which the particle's acceleration is zero.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates displacement twice: v(t) = 3t² − 18t + 24, then a(t) = 6t − 18. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Sets a(t) = 0: 6t − 18 = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Solves to obtain t = 3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: States that the particle's acceleration is zero at t = 3 s. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Acceleration is the derivative of velocity, which is itself the derivative of displacement — differentiating displacement twice always gives acceleration directly.

    Marking points

    • Differentiates displacement twice: v(t) = 3t² − 18t + 24, then a(t) = 6t − 18.
    • Sets a(t) = 0: 6t − 18 = 0.
    • Solves to obtain t = 3.
    • States that the particle's acceleration is zero at t = 3 s.

    Examiner tip: Acceleration is the derivative of velocity, which is itself the derivative of displacement — differentiating displacement twice always gives acceleration directly.

  24. 24.

    Marking analysis: A learner attempts the following task: “A particle's displacement is given by s(t) = t³ − 9t² + 24t, for t ≥ 0 (metres, seconds). Find the particle's acceleration function, and determine the value of t at which the particle's acceleration is zero.” Their response addresses only this point: “Differentiates displacement twice: v(t) = 3t² − 18t + 24, then a(t) = 6t − 18.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates displacement twice: v(t) = 3t² − 18t + 24, then a(t) = 6t − 18. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Sets a(t) = 0: 6t − 18 = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Solves to obtain t = 3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: States that the particle's acceleration is zero at t = 3 s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates displacement twice: v(t) = 3t² − 18t + 24, then a(t) = 6t − 18.
    • Identifies the missing requirement: Sets a(t) = 0: 6t − 18 = 0.
    • Identifies the missing requirement: Solves to obtain t = 3.
    • Identifies the missing requirement: States that the particle's acceleration is zero at t = 3 s.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  25. 25.

    Find the area of the region enclosed between the curves y = x² and y = 2x.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Finds the points of intersection by solving x² = 2x: x = 0 and x = 2. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Determines that 2x ≥ x² on the interval [0, 2]. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Sets up the integral ∫₀² (2x − x²) dx. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Integrates to obtain [x² − x³/3]₀². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Obtains area = 4/3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The area between two curves is always found by integrating (upper curve − lower curve), not either curve alone — always check which curve is on top across the interval first.

    Marking points

    • Finds the points of intersection by solving x² = 2x: x = 0 and x = 2.
    • Determines that 2x ≥ x² on the interval [0, 2].
    • Sets up the integral ∫₀² (2x − x²) dx.
    • Integrates to obtain [x² − x³/3]₀².
    • Obtains area = 4/3.

    Examiner tip: The area between two curves is always found by integrating (upper curve − lower curve), not either curve alone — always check which curve is on top across the interval first.

  26. 26.

    Marking analysis: A learner attempts the following task: “Find the area of the region enclosed between the curves y = x² and y = 2x.” Their response addresses only this point: “Finds the points of intersection by solving x² = 2x: x = 0 and x = 2.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Finds the points of intersection by solving x² = 2x: x = 0 and x = 2. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Determines that 2x ≥ x² on the interval [0, 2]. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Sets up the integral ∫₀² (2x − x²) dx. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Integrates to obtain [x² − x³/3]₀². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Obtains area = 4/3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Finds the points of intersection by solving x² = 2x: x = 0 and x = 2.
    • Identifies the missing requirement: Determines that 2x ≥ x² on the interval [0, 2].
    • Identifies the missing requirement: Sets up the integral ∫₀² (2x − x²) dx.
    • Identifies the missing requirement: Integrates to obtain [x² − x³/3]₀².
    • Identifies the missing requirement: Obtains area = 4/3.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  27. 27.

    For the curve y = x³ − 3x² + 2, find the coordinates of the point of inflection, using the second derivative.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Finds y′ = 3x² − 6x and y″ = 6x − 6. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Sets y″ = 0: 6x − 6 = 0, giving x = 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Confirms this is a genuine point of inflection, since y″ changes sign (from negative to positive) as x passes through 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates y(1) = 1 − 3 + 2 = 0, giving the point of inflection (1, 0). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A point of inflection requires the second derivative to actually change sign, not just equal zero — always check the sign on either side of the candidate x-value to confirm a genuine inflection point.

    Marking points

    • Finds y′ = 3x² − 6x and y″ = 6x − 6.
    • Sets y″ = 0: 6x − 6 = 0, giving x = 1.
    • Confirms this is a genuine point of inflection, since y″ changes sign (from negative to positive) as x passes through 1.
    • Calculates y(1) = 1 − 3 + 2 = 0, giving the point of inflection (1, 0).

    Examiner tip: A point of inflection requires the second derivative to actually change sign, not just equal zero — always check the sign on either side of the candidate x-value to confirm a genuine inflection point.

  28. 28.

    Marking analysis: A learner attempts the following task: “For the curve y = x³ − 3x² + 2, find the coordinates of the point of inflection, using the second derivative.” Their response addresses only this point: “Finds y′ = 3x² − 6x and y″ = 6x − 6.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Finds y′ = 3x² − 6x and y″ = 6x − 6. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Sets y″ = 0: 6x − 6 = 0, giving x = 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Confirms this is a genuine point of inflection, since y″ changes sign (from negative to positive) as x passes through 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates y(1) = 1 − 3 + 2 = 0, giving the point of inflection (1, 0). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Finds y′ = 3x² − 6x and y″ = 6x − 6.
    • Identifies the missing requirement: Sets y″ = 0: 6x − 6 = 0, giving x = 1.
    • Identifies the missing requirement: Confirms this is a genuine point of inflection, since y″ changes sign (from negative to positive) as x passes through 1.
    • Identifies the missing requirement: Calculates y(1) = 1 − 3 + 2 = 0, giving the point of inflection (1, 0).

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  29. 29.

    The area of a circular oil spill is increasing at a rate of 8 m² s⁻¹. Find the rate at which the radius is increasing at the instant when the radius is 5 m. (A = πr²)

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Differentiates A = πr² with respect to t, using the chain rule: dA/dt = 2πr(dr/dt). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Substitutes dA/dt = 8 and r = 5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Solves 8 = 10π(dr/dt). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Obtains dr/dt ≈ 0.255 m s⁻¹. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Related rates problems always connect two quantities that are both changing with time through a shared geometric formula — differentiate that formula with respect to time first, then substitute the known instantaneous values.

    Marking points

    • Differentiates A = πr² with respect to t, using the chain rule: dA/dt = 2πr(dr/dt).
    • Substitutes dA/dt = 8 and r = 5.
    • Solves 8 = 10π(dr/dt).
    • Obtains dr/dt ≈ 0.255 m s⁻¹.

    Examiner tip: Related rates problems always connect two quantities that are both changing with time through a shared geometric formula — differentiate that formula with respect to time first, then substitute the known instantaneous values.

  30. 30.

    Marking analysis: A learner attempts the following task: “The area of a circular oil spill is increasing at a rate of 8 m² s⁻¹. Find the rate at which the radius is increasing at the instant when the radius is 5 m. (A = πr²)” Their response addresses only this point: “Differentiates A = πr² with respect to t, using the chain rule: dA/dt = 2πr(dr/dt).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Differentiates A = πr² with respect to t, using the chain rule: dA/dt = 2πr(dr/dt). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Substitutes dA/dt = 8 and r = 5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Solves 8 = 10π(dr/dt). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains dr/dt ≈ 0.255 m s⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Differentiates A = πr² with respect to t, using the chain rule: dA/dt = 2πr(dr/dt).
    • Identifies the missing requirement: Substitutes dA/dt = 8 and r = 5.
    • Identifies the missing requirement: Solves 8 = 10π(dr/dt).
    • Identifies the missing requirement: Obtains dr/dt ≈ 0.255 m s⁻¹.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  31. 31.

    For f(x) = x^2 e^(-x), with real x, find the stationary points and classify each using the sign of the derivative.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Apply the product rule. The exponential factor never changes sign, so the sign chart depends only on x(2 - x).
    2. The signs on (-infinity,0), (0,2), (2,infinity) are negative, positive, negative. Evaluate the original function, not the derivative, for the coordinates.

    Marking points

    • f'(x) = e^(-x)(2x - x^2) = e^(-x)x(2 - x).
    • Stationary inputs are x = 0 and x = 2 since e^(-x) is positive.
    • The corresponding points are (0,0) and (2,4/e^2).
    • Derivative changes from negative to positive at 0: local minimum.
    • Derivative changes from positive to negative at 2: local maximum.

    Examiner tip: A derivative zero locates a candidate; its sign change supplies the classification.

  32. 32.

    Marking analysis: A learner attempts the following task: “For f(x) = x^2 e^(-x), with real x, find the stationary points and classify each using the sign of the derivative.” Their response addresses only this point: “f'(x) = e^(-x)(2x - x^2) = e^(-x)x(2 - x).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: f'(x) = e^(-x)(2x - x^2) = e^(-x)x(2 - x). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Stationary inputs are x = 0 and x = 2 since e^(-x) is positive. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: The corresponding points are (0,0) and (2,4/e^2). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Derivative changes from negative to positive at 0: local minimum. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Derivative changes from positive to negative at 2: local maximum. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: f'(x) = e^(-x)(2x - x^2) = e^(-x)x(2 - x).
    • Identifies the missing requirement: Stationary inputs are x = 0 and x = 2 since e^(-x) is positive.
    • Identifies the missing requirement: The corresponding points are (0,0) and (2,4/e^2).
    • Identifies the missing requirement: Derivative changes from negative to positive at 0: local minimum.
    • Identifies the missing requirement: Derivative changes from positive to negative at 2: local maximum.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  33. 33.

    Find the intersections of y = 2x and y = x^2, then calculate the exact area enclosed between the curves.

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. The intersection inputs give integration limits. Testing x = 1 establishes which curve is upper.
    2. Integrate upper minus lower so the enclosed area is positive; substitute both limits into the antiderivative.

    Marking points

    • x^2 = 2x gives x = 0 and x = 2, at (0,0) and (2,4).
    • On 0 < x < 2, the line is above the parabola.
    • Area = integral from 0 to 2 of (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2.
    • Area = 4/3 square units.

    Examiner tip: Integrating one curve alone gives area to the axis, not the area between the curves.

  34. 34.

    Marking analysis: A learner attempts the following task: “Find the intersections of y = 2x and y = x^2, then calculate the exact area enclosed between the curves.” Their response addresses only this point: “x^2 = 2x gives x = 0 and x = 2, at (0,0) and (2,4).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: x^2 = 2x gives x = 0 and x = 2, at (0,0) and (2,4). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: On 0 < x < 2, the line is above the parabola. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Area = integral from 0 to 2 of (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Area = 4/3 square units. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: x^2 = 2x gives x = 0 and x = 2, at (0,0) and (2,4).
    • Identifies the missing requirement: On 0 < x < 2, the line is above the parabola.
    • Identifies the missing requirement: Area = integral from 0 to 2 of (2x - x^2) dx = [x^2 - x^3/3] from 0 to 2.
    • Identifies the missing requirement: Area = 4/3 square units.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  35. 35.

    Find the tangent to y = ln x at x = e. Prove that ln x <= x/e for every x > 0, stating when equality holds.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. A tangent equation alone does not prove a bound. Subtract the curve from the line and study the difference over its whole domain.
    2. The difference decreases to zero and then increases. This proves that no positive input puts the logarithm above the tangent.

    Marking points

    • At x = e, y = 1 and dy/dx = 1/e.
    • Tangent: y - 1 = (x - e)/e, or y = x/e.
    • Set g(x) = x/e - ln x; g'(x) = 1/e - 1/x.
    • g' is negative on (0,e) and positive on (e,infinity), so g has a global minimum at e.
    • g(e) = 0, hence g(x) >= 0 with equality only at x = e.

    Examiner tip: A local second-derivative check alone is not a global inequality proof.

  36. 36.

    Marking analysis: A learner attempts the following task: “Find the tangent to y = ln x at x = e. Prove that ln x <= x/e for every x > 0, stating when equality holds.” Their response addresses only this point: “At x = e, y = 1 and dy/dx = 1/e.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: At x = e, y = 1 and dy/dx = 1/e. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Tangent: y - 1 = (x - e)/e, or y = x/e. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Set g(x) = x/e - ln x; g'(x) = 1/e - 1/x. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: g' is negative on (0,e) and positive on (e,infinity), so g has a global minimum at e. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: g(e) = 0, hence g(x) >= 0 with equality only at x = e. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: At x = e, y = 1 and dy/dx = 1/e.
    • Identifies the missing requirement: Tangent: y - 1 = (x - e)/e, or y = x/e.
    • Identifies the missing requirement: Set g(x) = x/e - ln x; g'(x) = 1/e - 1/x.
    • Identifies the missing requirement: g' is negative on (0,e) and positive on (e,infinity), so g has a global minimum at e.
    • Identifies the missing requirement: g(e) = 0, hence g(x) >= 0 with equality only at x = e.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  37. 37.

    f(x) = x^3 - 3kx, where k > 0. The vertical separation between its local maximum and local minimum is 32. Determine k.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Use the stationary-point condition to express both inputs in terms of k before substituting into the cubic.
    2. At k = 4 the points are (-2,16) and (2,-16), confirming a vertical gap of 32 rather than a horizontal gap.

    Marking points

    • f'(x) = 3x^2 - 3k, so stationary inputs are +/-sqrt(k).
    • f''(x) = 6x identifies -sqrt(k) as the maximum and +sqrt(k) as the minimum.
    • Their values are 2k^(3/2) and -2k^(3/2).
    • Vertical separation = 4k^(3/2) = 32.
    • k^(3/2) = 8, so k = 4.

    Examiner tip: Vertical separation uses function values; subtracting x-coordinates answers a different question.

  38. 38.

    Marking analysis: A learner attempts the following task: “f(x) = x^3 - 3kx, where k > 0. The vertical separation between its local maximum and local minimum is 32. Determine k.” Their response addresses only this point: “f'(x) = 3x^2 - 3k, so stationary inputs are +/-sqrt(k).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: f'(x) = 3x^2 - 3k, so stationary inputs are +/-sqrt(k). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: f''(x) = 6x identifies -sqrt(k) as the maximum and +sqrt(k) as the minimum. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Their values are 2k^(3/2) and -2k^(3/2). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Vertical separation = 4k^(3/2) = 32. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: k^(3/2) = 8, so k = 4. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: f'(x) = 3x^2 - 3k, so stationary inputs are +/-sqrt(k).
    • Identifies the missing requirement: f''(x) = 6x identifies -sqrt(k) as the maximum and +sqrt(k) as the minimum.
    • Identifies the missing requirement: Their values are 2k^(3/2) and -2k^(3/2).
    • Identifies the missing requirement: Vertical separation = 4k^(3/2) = 32.
    • Identifies the missing requirement: k^(3/2) = 8, so k = 4.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  39. 39.

    A temperature is modelled by T(t) = 20 + 60e^(-0.15t), where t is minutes after cooling begins. Find when T first reaches 35 degrees, the initial rate of temperature change, and whether the model reaches 20 degrees at a finite time.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Subtract the surrounding temperature before taking logarithms. The model is strictly decreasing for t >= 0, so this crossing is unique.
    2. The negative initial derivative means cooling. An exponential is always positive, so it approaches its ambient limit without exactly attaining it at finite time.

    Marking points

    • 35 = 20 + 60e^(-0.15t) gives e^(-0.15t) = 1/4.
    • t = ln(4)/0.15 = 9.24 minutes to three significant figures.
    • T'(t) = -9e^(-0.15t).
    • T'(0) = -9 degrees per minute.
    • T tends to 20 but remains above 20 for every finite t >= 0.

    Examiner tip: The horizontal asymptote is a limiting temperature, not an attained final reading in this model.

  40. 40.

    Marking analysis: A learner attempts the following task: “A temperature is modelled by T(t) = 20 + 60e^(-0.15t), where t is minutes after cooling begins. Find when T first reaches 35 degrees, the initial rate of temperature change, and whether the model reaches 20 degrees at a finite time.” Their response addresses only this point: “35 = 20 + 60e^(-0.15t) gives e^(-0.15t) = 1/4.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: 35 = 20 + 60e^(-0.15t) gives e^(-0.15t) = 1/4. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: t = ln(4)/0.15 = 9.24 minutes to three significant figures. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: T'(t) = -9e^(-0.15t). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: T'(0) = -9 degrees per minute. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: T tends to 20 but remains above 20 for every finite t >= 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: 35 = 20 + 60e^(-0.15t) gives e^(-0.15t) = 1/4.
    • Identifies the missing requirement: t = ln(4)/0.15 = 9.24 minutes to three significant figures.
    • Identifies the missing requirement: T'(t) = -9e^(-0.15t).
    • Identifies the missing requirement: T'(0) = -9 degrees per minute.
    • Identifies the missing requirement: T tends to 20 but remains above 20 for every finite t >= 0.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.