June 2024 Paper 1 Q7
7 The point \(A\) has coordinates (1, 7), and the point \(B\) has coordinates (\(h\), 10).
Find the value of \(h\). [2]
Find the coordinates of the point \(C\). [2]
Determine the set of possible values of \(k\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{10 - 7}{h - 1} = 2\) | M1 | 1.1 |
| \(h = \tfrac{5}{2}\) | A1 | 1.1 |
| [2] |
Notes
M1: Equate attempt at gradient of line to 2
Must be attempting to use \(\dfrac{y_2 - y_1}{x_2 - x_1}\), with consistent order in numerator and denominator; allow one sign slip
Could also use informal methods
A1: Obtain \(h = \tfrac{5}{2}\) oe
| Scheme | Marks | AO |
|---|---|---|
| (4, 13) | B1FT | 1.1 |
| Obtain correct \(y\) coordinate of 13 | B1 | 1.1 |
| [2] |
Notes
B1FT: Obtain correct \(x\) coordinate for \(C\), following their \(h\) ie \(2h - 1\)
Must be clear that their \(2h - 1\) is the \(x\) coordinate
BOD if brackets omitted
B1: Must be clear that 13 is the \(y\) coordinate
BOD if brackets omitted
SC B1 if both values correct but given as a vector not a coordinate
| Scheme | Marks | AO |
|---|---|---|
| \(y - 7 = 2(x - 1)\) | M1 | 3.1a |
| \(x^2 - 4x + k = 2x + 5\) | M1 | 1.1 |
| \(x^2 - 6x + (k - 5) = 0\) | A1 | 1.1 |
| \(b^2 - 4ac = 36 - 4(k - 5)\) | M1* | 3.1a |
| Two points of intersection so \(b^2 - 4ac \gt 0\) \(36 - 4(k - 5) \gt 0\) | M1d* | 1.1 |
| \(56 - 4k \gt 0\) \(k \lt 14\) | A1 | 1.1 |
| [6] |
Notes
M1: Attempt equation of line through \(A\)
Allow one sign slip, but M0 if \(x\) and \(y\) coordinates transposed
Could use their \(B\) or \(C\) instead, but must still be using gradient of 2
M1: Equate line and curve
A1: Obtain correct quadratic, with like terms collected
Condone no ‘= 0’
M1*: Attempt discriminant of 3 term quadratic, resulting from equating line and curve
Correct discriminant for their quadratic
Condone any inequality or equality, or no, sign for this mark
M0 if using just \(x^2 - 4x + k\)
If the discriminant is initially embedded in the quadratic formula, then M1 is only awarded when it is considered in isolation
M1d*: Use \(b^2 - 4ac \gt 0\)
Inequality sign could be implied by final answer
M0 if incorrect inequality sign, including \(b^2 - 4ac \geqslant 0\)
A1: Obtain \(k \lt 14\)
Alt method for final 3 marks
| Scheme | Marks |
|---|---|
| \((x - 3)^2 - 9 + k - 5\) | M1* |
| \(k - 14 \lt 0\) | M1d* |
| \(k \lt 14\) | A1 |
M1*: Attempt completed square form
Correct expression for their quadratic
Allow unsimplified
M1d*: Set constant term \(\lt 0\)
M0 if constant term \(\leqslant 0\)
May see more informal method to determine inequality sign
A1: Obtain \(k \lt 14\)
Alt method (using differentiation to find point of intersection)
| Scheme | Marks |
|---|---|
| \(2x - 4 = 2\) \(x = 3\) | M1 |
| \(y - 7 = 2(x - 1)\) | M1 |
| \(y = 11\) \(9 - 12 + k = 11\) | M1 |
| \(k = 14\) | A1 |
| One point of intersection when \(k = 14\). It is a positive quadratic so translate in negative \(y\) direction for two points of intersection | M1 |
| \(k \lt 14\) | A1 |
M1: Differentiate equation of curve, equate to 2 and attempt \(x\)
M1: Attempt equation of line through \(A\)
Could use their \(B\) or \(C\) instead
M1: Attempt \(y\) value from line, and use their (3, 11) in equation of curve to attempt \(k\)
A1: Obtain \(k = 14\)
M1: Clear method to determine inequality sign
Could be algebraic or a sketch
Could be implied by final answer
A1: Obtain \(k \lt 14\)