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O-LevelAdditional MathematicsDifferentiation and integration4037/12 · Oct/Nov 2011 · Q1111 Marks

Answer only one of the following two alternatives. EITHER A curve has equation y = e⁻ˣ (Acos2x + Bsin2x). At the point (0, 4) on the curve, the gradient of the tangent is 6. OR A curve has equation y = ln(x²-1) / (x²-1), for x > 1.

  1. EITHER (i)Find the value of A.[1]
  2. EITHER (ii)Show that B = 5.[5]
  3. EITHER (iii)Find the value of x, where 0 < x < π/2 radians, for which y has a stationary value.[5]
  4. OR (i)Show that dy/dx = (kx(1 – ln(x²-1))) / (x² - 1)², where k is a constant to be found.[4]
  5. OR (ii)Hence find the approximate change in y when x increases from √5 to √5 + p, where p is small.[2]
  6. OR (iii)Find, in terms of e, the coordinates of the stationary point on the curve.[5]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2011). The mark or mark type is shown in brackets.

  1. [B1] B1 for A = 4
  2. [M1] M1 for differentiation of a product
  3. [B1] B1 for (-2Asin 2x + 2Bcos2x)e^(-x)
  4. [B1] B1 for -e^(-x) (Acos2x + Bsin2x)
  5. [M1] M1 for substitution to find B
  6. [A1] A1
  7. [M1] M1 for their dy/dx = 0
  8. [M1] M1 for attempt to simplify
  9. [M1] M1 for attempt to obtain tan 2x = p/q
  10. [M1] M1 for attempt to solve tan 2x = k
  11. [A1] A1 for x = 0.216
  12. [M1] M1 for differentiation of a quotient
  13. [B1] B1 for 2x/(x²-1)
  14. [A1] A1 for all else correct
  15. [A1] A1 for rearrangement to get k = 2
  16. [M1] M1 for substitution of x = √5 and correct method
  17. [√A1] √A1 on their k
  18. [M1] M1 for dy/dx = 0 and attempt to simplify
  19. [A1] A1 for ln(x²-1)=1
  20. [A1] A1 for x²-1=e or 2.72
  21. [A1] A1 for x = √(1+e)
  22. [A1] A1 for y = 1/e

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About this question

Topic
Differentiation and integration, O-Level Additional Mathematics (syllabus 4037). Structured question, 11 marks.
Source
O-Level Additional Mathematics, Oct/Nov 2011 Paper 1 Variant 2, question 11.

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