A-LevelMathematicsDifferentiation9709/33 · May/June 2018 · Q89 Marks
The equation of a curve is 2x³ – y³ – 3xy² = 2a³, where a is a non-zero constant.
- (i)Show that dy/dx = (2x²-y²)/(y²+2xy)[4]
- (ii)Find the coordinates of the two points on the curve at which the tangent is parallel to the y-axis.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- B1 State or imply 3y²(dy/dx) as derivative of y³
- B1 State or imply 3y² + 6xy (dy/dx) as derivative of 3xy²
- M1 Equate derivative of LHS to zero and solve for dy/dx
- A1 Obtain the given answer
- M1* Equate denominator to zero and solve for y
- A1 Obtain y = 0 and x = a
- M1(dep*) Obtain y = ax and substitute in curve equation to find x or to find y
- A1 Obtain x = -a
- A1 Obtain y = 2a
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