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A-LevelMathematicsDifferentiation9709/22 · Feb/Mar 2022 · Q25 Marks

A curve has equation y = 7 + 4 ln(2x + 5). Find the equation of the tangent to the curve at the point (−2, 7), giving your answer in the form y = mx + c.

  1. (a)A curve has equation y = 7 + 4 ln(2x + 5). Find the equation of the tangent to the curve at the point (−2, 7), giving your answer in the form y = mx + c.[5]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

  1. [M1] Differentiate to obtain form k/(2x+5) [Any non-zero constant k]
  2. [A1] Obtain correct 8/(2x+5)
  3. [A1] Substitute x = -2 to obtain gradient 8
  4. [M1] Attempt equation of tangent through (-2, 7) with numerical gradient
  5. [A1] Obtain y = 8x + 23

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About this question

Topic
Differentiation, A-Level Mathematics (syllabus 9709). Structured question, 5 marks.
Source
A-Level Mathematics, Feb/Mar 2022 Paper 2 Variant 2, question 2.

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