A-LevelMathematicsDifferentiation9709/22 · May/June 2023 · Q79 Marks
A curve has parametric equations x = (2t + 3)/(t + 2), y = t² + at + 1, where a is a constant. It is given that, at the point P on the curve, the gradient is 1.
- (a)Show that the value of t at P satisfies the equation 2t³ + (a + 8)t² + (4a + 8)t + 4a − 1 = 0.[4]
- (b)It is given that (t + 1) is a factor of 2t³ + (a + 8)t² + (4a + 8)t + 4a – 1. Find the value of a.[2]
- (c)Hence show that P is the only point on the curve at which the gradient is 1.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- 7(a)
- [M1*] Attempt use of quotient rule (or equivalent) to find dx/dt
- [A1] Obtain 1/(t+2)² or (unsimplified) equivalent.
- [DM1] Equate dy/dx to 1 Must be using their dy/dx
- [A1] Obtain (2t + a)(t+2)² =1 and expand to confirm given result
- 7(b)
- [M1] Substitute t = −1, equate to zero and attempt solution for a
- [A1] Obtain a = 3
- 7(c)
- [M1] Divide their cubic by t +1 at least as far as the x term
- [A1] Obtain 2t² + 9t +11
- [A1] Calculate discriminant, obtain -7 and conclude no further value of t
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 →