Standard +0.3 This question requires finding where a normal line meets a parabola by using the perpendicular gradient relationship, then solving simultaneously. It involves multiple standard steps (finding gradient of normal, gradient of curve, solving quadratic, substituting) but all techniques are routine for P1 level with no novel insight required, making it slightly easier than average.
Gradient of normal is – 1/3 → gradient of tangent is 3 SOI
B1 B1 FT
dy/dx = 2x – 5 = 3
M1
Differentiate and set = their 3 (numerical).
x = 4
*A1
Sub x = 4 into line → y = 7 & sub their (4, 7) into curve
DM1
OR sub x = 4 into curve → y = k ‒ 4 and sub their(4, k ‒ 4) into line
OR other valid methods deriving a linear equation in k (e.g. equating curve
with either normal or tangent and sub x = 4).
Answer
Marks
Guidance
k = 11
A1
Total:
6
Question
Answer
Marks
Question 6:
6 | Gradient of normal is – 1/3 → gradient of tangent is 3 SOI | B1 B1 FT | FT from their gradient of normal.
dy/dx = 2x – 5 = 3 | M1 | Differentiate and set = their 3 (numerical).
x = 4 | *A1
Sub x = 4 into line → y = 7 & sub their (4, 7) into curve | DM1 | OR sub x = 4 into curve → y = k ‒ 4 and sub their(4, k ‒ 4) into line
OR other valid methods deriving a linear equation in k (e.g. equating curve
with either normal or tangent and sub x = 4).
k = 11 | A1
Total: | 6
Question | Answer | Marks | Guidance