A-LevelMathematicsDifferentiation9709/12 · May/June 2024 · Q1010 Marks

The equation of a curve is y = (5−2x)³/² + 5 for x < 5/2.

(a)A point P is moving along the curve in such a way that the y-coordinate of point P is decreasing at 5 units per second. Find the rate at which the x-coordinate of point P is increasing when y = 32.[4]
(b)Point A on the curve has y-coordinate 32. Point B on the curve is such that the gradient of the curve at B is −3. Find the equation of the perpendicular bisector of AB. Give your answer in the form ax+by+c = 0, where a, b and c are integers.[6]

Mark scheme

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

[B1] [x=]-2
[M1*] dy/dx = k(5-2x)¹⁄² = -2 × 3/2 (5-2x)¹⁄² Differentiating to get k(5-2x)¹⁄² only.
[DM1] dy/dt = dy/dx × dx/dt leading to -5 = -9 × dx/dt Correct statement linking their numerical expression for dy/dx with dx/dt and ± 5.
[A1] 5/9 or 0.556 = AWRT
[M1] k(5-2x)¹⁄² = -3 Equating their dy/dx of the form k(5-2x)¹⁄² to -3.
[A1] [B is] (2,6)
[M1*] Gradient AB m₁ = (32-6)/(-2-2) , gradient of perpendicular = -1/m₁ = -1/(-26/4) Clear use of difference in y co-ordinates / difference in x co-ordinates for points A and B, condone inconsistent order, and using m₁m₂=-1.
[M1*] Mid point is ( (-2+2)/2, (6+32)/2 ) = (0,19) Finding the midpoint of AB using A and B. If incorrect values used then all working must be clear. For A, y must be 32.
[DM1] y-19 = 2/13 (x-0) Finding the equation of the perpendicular bisector using their midpoint and their perpendicular gradient.
[A1] 2x-13y + 247=0 or ± integer multiples of this.

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