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A-LevelMathematicsIntegration9709/32 · May/June 2024 · Q69 Marks

The diagram shows the curve y = xe^(-ax), where a is a positive constant, and its maximum point M. [Figure 6.1]

  1. (a)Find the exact coordinates of M.[4]
  2. (b)Find the exact value of ∫[0 to 2/a] xe^(-ax) dx.[5]

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Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

  1. [*M1] Use correct product rule. Or equivalent. Condone incorrect chain rule. M0 if a value is used for a (not equivalent work).
  2. [A1] Obtain correct derivative. E.g. dy/dx = -axe^(-ax) + e^(-ax).
  3. [DM1] Equate derivative to zero and solve for x.
  4. [A1] Obtain x = 1/a, y = 1/(ae). ISW. Or exact equivalent.
  5. [*M1] Use integration by parts to obtain pxe^(-ax) + q∫e^(-ax) dx. Condone sign error in parts formula and omission of dx. M0 if a value is used for a (not equivalent work).
  6. [A1] Obtain - (1/a)xe^(-ax) + (1/a)∫e^(-ax) dx. OE.
  7. [A1] Complete integration to obtain - (1/a)xe^(-ax) - (1/a²)e^(-ax). OE.
  8. [DM1] Correct use of limits 0 and 2/a in an expression of the form rxe^(-ax) + se^(-ax). ((-2/a²e^(-2)) - (-1/a²e^(-2)) + 0 + (1/a²))
  9. [A1] Obtain (1/(a²))(1 - 3e^(-2)). ISW. Or simplified 2-term equivalent, e.g. (e²-3)/(a²e²).

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About this question

Topic
Integration, A-Level Mathematics (syllabus 9709). Structured question, 9 marks.
Source
A-Level Mathematics, May/June 2024 Paper 3 Variant 2, question 6.

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