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A-LevelMathematicsTrigonometry9709/22 · Feb/Mar 2019 · Q711 Marks

The parametric equations of a curve are x = 2t - sin 2t, y = 5t + cos 2t, for 0 ≤ t ≤ ½π. At the point P on the curve, the gradient of the curve is 2.

  1. (i)Show that the value of the parameter at P satisfies the equation 2 sin 2t – 4 cos 2t = 1.[4]
  2. (ii)By first expressing 2 sin 2t – 4 cos 2t in the form R sin(2t – α), where R > 0 and 0 < α < ½π, find the coordinates of P. Give each coordinate correct to 3 significant figures.[7]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2019). The mark or mark type is shown in brackets.

  1. [B1] Obtain dx/dt = 2 – 2 cos 2t
  2. [B1] Obtain dy/dt = 5 – 2 sin 2t
  3. [M1] Equate attempt at dy/dx to 2 and rearrange
  4. [A1] Confirm equation 2 sin 2t – 4 cos 2t = 1
  5. [B1] State R = √20 or 4.47
  6. [M1] Use appropriate trigonometry to find α
  7. [A1] Obtain α = 1.107 with no errors seen
  8. [M1] Carry out correct method to find value of t
  9. [A1] Obtain t = 0.666
  10. [M1] Substitute value of t between 0 and ½π into expressions for x and y
  11. [A1] Obtain x = 0.361, y = 3.57

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About this question

Topic
Trigonometry, A-Level Mathematics (syllabus 9709). Structured question, 11 marks.
Source
A-Level Mathematics, Feb/Mar 2019 Paper 2 Variant 2, question 7.

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