A-LevelMathematicsDifferentiation9709/12 · May/June 2022 · Q97 Marks
The equation of a curve is y = 3x + 1 − 4(3x + 1)½ for x > -1/3.
- (a)Find dy/dx and d²y/dx².[3]
- (b)Find the coordinates of the stationary point of the curve and determine its nature.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [B1 B1] dy/dx = 3 - 6(3x+1)^(-1/2)
- [B1] d²y/dx² = 9(3x+1)^(-3/2)
- [M1] dy/dx = 0 leading to 3 - 6(3x+1)^(-1/2) = 0
- [A1] (3x+1)^(1/2) = 2 leading to x = 1
- [A1] y = -4 [coordinates (1, -4)]
- [A1] d²y/dx² = 9(3(1)+1)^(-3/2) = 9/8 or > 0 so minimum
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →