O-LevelAdditional MathematicsDifferentiation and integration4037/22 · Oct/Nov 2019 · Q57 Marks
At the point where x = 1 on the curve y = k/(x+1)², the normal has a gradient of 1/3.
- (i)Find the value of the constant k.[4]
- (ii)Using your value of k, find the equation of the tangent to the curve at x = 2.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- B1 dy/dx = -2k/(x+1)³ oe Unsimplified
- M1 Gradient of normal = -(x+1)³/2k or Gradient of tangent = -3
- M1 8/2k = -(-3) or 2k/8 = -3 or 8/(2k) = 1/3 at x = 1 or equate gradient of tangent to -3 at x = 1
- A1 k = 12
- B1 FT x = 2 → dy/dx = -2k/27 or their -2k/27
- B1 FT y = 4/3 or their k/9
- B1 isw y - 4/3 = -8/9 (x-2) or y = -8/9 x + 28/9
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