4MA0 Higher June 2016 Paper 4R Q20
20 The sketch shows the curve with equation \(y = x^2 + 4\) and the line with equation \(y = x + 10\)
The line cuts the curve at the points \(A\) and \(B\).
\(M\) is the midpoint of \(AB\).
Find the coordinates of \(M\).
Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| \(x^2 + 4 = x + 10\) | M1 |
| \(x^2 - x - 6\) (= 0) | M1 |
| \((x - 3)(x + 2)\) (= 0) | M1 |
| \(x = 3\), \(x = -2\) | A1 |
| \(x = 3, y = 13, x = -2, y = 8\) | M1 |
| (0.5, 10.5) | A1 |
| (6) |
Notes
M1: Equations equal to each other
M1: for reduction to 3 term quadratic
M1: Factorisation or correct use of quadratic formula
A1: Correct values for \(x\) dep on M2
M1: (\(y\)=)10 + 3 and (\(y\)=)10 − 2 or (\(y\) mid=) 10 + 0.5 dep on previous A1 awarded
A1: dep on previous A1 awarded
Alternative
| Scheme | Marks |
|---|---|
| \(x^2 + 4 = x + 10\) | M1 |
| \(x^2 - x - 6\) (= 0) | M1 |
| Sum of roots = 1 so midpoint has \(x\) coordinate 0.5 | M1 |
| A1 | |
| M1 | |
| (0.5, 10.5) | A1 |
Notes
M1: Equations equal to each other
M1: for reduction to 3 term quadratic
M1: for Sum of roots = 1 and midpoint has \(x\)-coordinate = sum of roots ÷2
A1: 0.5 dep on M2
M1: 0.5 + 10 dep on previous A1 awarded
A1: 10.5 dep on previous A1 awarded
Alternative
| Scheme | Marks |
|---|---|
| \(y = (y - 10)^2 + 4\) | M1 |
| \(y^2 - 21y + 104\) (= 0) | M1 |
| \((y - 8)(y - 13)\) (= 0) | M1 |
| \(y = 8\), \(y = 13\) | A1 |
| \(x = 3, y = 13, x = -2, y = 8\) | M1 |
| (0.5, 10.5) | A1 |
Notes
M1: Correct substitution of \(y - 10\) for \(x\)
M1: for reduction to 3 term quadratic
M1: Factorisation or correct use of quadratic formula
A1: Correct values for \(y\) dep on M2
M1: (\(x\)=)13 – 10 and (\(x\)=)8 – 10 or (\(x\) mid)=10.5 – 10 dep on previous A1 awarded
A1: dep on previous A1 awarded