A-LevelMathematicsDifferentiation9709/12 · May/June 2021 · Q1110 Marks
The gradient of a curve is given by dy/dx = 6(3x – 5)³ – kx², where k is a constant. The curve has a stationary point at (2, -3.5).
- (a)Find the value of k.[2]
- (b)Find the equation of the curve.[4]
- (c)Find d²y/dx².[2]
- (d)Determine the nature of the stationary point at (2, -3.5).[2]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [M1] At stationary point dy/dx = 0 so 6(3×2-5)²-k×2² = 0
- [A1] [k=] 3/2
- [*M1] [y=] 6/(4×3) (3x-5)⁴-k/3x³ [+c]
- [A1 FT] (1/2)(3x-5)⁴ - (k/3)x³ [+c]
- [DM1] -3.5 = (1/2)(3×2-5)⁴ - (3/2)/3×2³ +c [leading to -3.5 + c = -3.5]
- [A1] y=(1/2)(3x-5)⁴-(1/2)x³
- [B2,1,0 FT] [3x][18(3x-5)²][-2kx]
- [M1] [At x = 2] d²y/dx² = 54(3×2-5)²-4k or 48
- [A1] [>0] Minimum
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