AS June 2023 Paper 1 Q9
9.
Under \(T\), the point \(A(a, 2)\) and the point \(B(4, -a)\), where \(a\) is a constant, are transformed to the points \(A'\) and \(B'\) respectively.
Given that the distance \(A' B'\) is \(\sqrt{58}\), determine the possible values of \(a\). (5)| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix}k & -2 & 7\\ -3 & -5 & 2\\ k & k & 4\end{vmatrix} = k(-20 - 2k) + 2(-12 - 2k) + 7(-3k + 5k)\) or \(\begin{vmatrix}k & -2 & 7 & k & -2\\ -3 & -5 & 2 & -3 & -5\\ k & k & 4 & k & k\end{vmatrix} = k(-5)(4) - 2(2)(k) + 7(-3)(k)\) \(\qquad - 7(-5)(k) - k(2)(k) - (-2)(-3)(4)\) | M1 | 1.1b |
| \(-2k^2 - 10k - 24\ (= 0)\) isw | A1 | 1.2 |
| \(b^2 - 4ac = (10)^2 - 4(-2)(-24) = \ldots\) \(b^2 - 4ac = (5)^2 - 4(-1)(-12) = \ldots\) Or \(k^2 + 5k + 12 = 0 \Rightarrow (k + 2.5)^2 + 5.75 = 0 \Rightarrow (k + 2.5)^2 = -5.75\) \(-2k^2 - 10k - 24 = 0 \Rightarrow -2(k + 2.5)^2 - 11.5 = 0 \Rightarrow (k + 2.5)^2 = -5.75\) Or \(k^2 + 5k + 12 \Rightarrow (k + 2.5)^2 + 5.75 \Rightarrow (k + 2.5)^2 \geqslant 0\) or \(-2k^2 - 10k - 24 = 0 \Rightarrow -2(k + 2.5)^2 - 11.5 = 0 \Rightarrow -2(k + 2.5)^2 \leqslant 0\) Or \(\dfrac{\mathrm{d}\left(-2k^2 - 10k - 24\right)}{\mathrm{d}k} = -4k - 10 = 0 \Rightarrow k = -2.5 \Rightarrow \text{determinant} = -11.5\) Or \(k = \dfrac{10 \pm \sqrt{(-10)^2 - 4(-2)(-24)}}{2(-2)} = \dfrac{-5 \pm \sqrt{23}\mathrm{i}}{2}\) | M1 | 1.1b |
| \(b^2 - 4ac = -92 < 0\) therefore no real roots so non-singular \(b^2 - 4ac = -23 < 0\) therefore no real roots so non-singular Or Square of negative is not real therefore non-singular Or \((k + 2.5)^2 + 5.75 > 0\) therefore no real roots so non-singular \(-2(k + 2.5)^2 - 11.5 < 0\) therefore no real roots so non-singular Or As negative quadratic maximum value of determinant = −11.5 therefore no real roots so non-singular Or Imaginary roots therefore no real roots so non-singular | A1 | 2.4 |
| (4) |
Notes
(Corrected from the printed mark scheme: the completed-square, quadratic-formula and vertex lines are printed with \(-2k^2 - 10k - 25\), \(-2(k + 2.5)^2 - 12.5\), \(-4(-2)(-25)\), "determinant = −5.75" and "maximum value of determinant = −5.25". The determinant is \(-2k^2 - 10k - 24 = -2(k + 2.5)^2 - 11.5\), so these are typed with −24, −11.5 and a maximum value of −11.5.)
M1: Correct method to find the determinant, condone a single sign slip but not on second term must be +2 (…)
Note: May expand along any row or column.
A1: Correct simplified determinant
M1: Either
- Finds the value of the discriminant or sufficient working seen to identify the sign e.g. 100 – 192
- Completes the square an rearranges so that \((k \pm a)^2 = -b\)
- Completes the square and states that \((k \pm a)^2 \geqslant 0\)
- Completes the square and states that \(-\alpha(k \pm a)^2 \leqslant 0\)
- Differentiates the determinant to find the coordinates of the vertex
- Use the quadratic formula to find the imaginary roots
A1: Correct solution only
Either
- Correct value for the discriminant (may be implied), concludes less than 0, therefore no real roots and non singular.
- Correct completing the square and conclude no real roots as square root of negative therefore non singular
- Correct completing the square and shows > 0 therefore no real roots and non singular.
- Correct completing the square and shows < 0 therefore no real roots and non singular.
- Correct coordinates of the vertex and negative quadratic therefore no real roots and non singular.
- Use the quadratics formula to find the correct imaginary roots therefore no real roots/value for \(k\) and non singular.
Note \(k = \dfrac{-5 \pm \sqrt{23}\mathrm{i}}{2}\) which is not real is M0 A0 unless uses the quadratic formula or completing the square to show where this has come from
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}2 & -1\\ -3 & 0\end{pmatrix}\begin{pmatrix}a & 4\\ 2 & -a\end{pmatrix} = \begin{pmatrix}\ldots & \ldots\\ \ldots & \ldots\end{pmatrix}\) can be done separately for each point | M1 | 3.1a |
| \(\begin{pmatrix}2a - 2 & 8 + a\\ -3a & -12\end{pmatrix}\) or \((2a - 2,\ -3a)\) and \((8 + a,\ -12)\) | A1 | 1.1b |
| \(\sqrt{\left[(2a - 2) - (8 + a)\right]^2 + \left[-3a - (-12)\right]^2} = \sqrt{58}\) or \(\overrightarrow{AB} = \begin{pmatrix}8 + a\\ -12\end{pmatrix} - \begin{pmatrix}2a - 2\\ -3a\end{pmatrix} = \begin{pmatrix}10 - a\\ -12 + 3a\end{pmatrix}\) or \(\overrightarrow{BA} = \begin{pmatrix}2a - 2\\ -3a\end{pmatrix} - \begin{pmatrix}8 + a\\ -12\end{pmatrix} = \begin{pmatrix}a - 10\\ 12 - 3a\end{pmatrix}\) \((a - 10)^2 + (12 - 3a)^2 = 58\) or \((10 - a)^2 + (3a - 12)^2 = 58\) leading to a 3TQ | M1 | 3.1a |
| \(10a^2 - 92a + 186 = 0\) | A1 | 1.1b |
| \(a = 3,\ \dfrac{31}{5}\) o.e. cso | A1 | 1.1b |
| (5) | ||
| (9 marks) |
Notes
M1: Uses matrix Q to find the coordinates of the points \(A'\) and \(B'\). Condone a sign slip.
A1: Correct coordinates for the points \(A'\) and \(B'\), they do not need to be labelled
M1: Finds the distance between their points \(A'\) and \(B'\) which must not be equal to \(A\) and \(B\), sets equal to \(\sqrt{58}\), forms a 3TQ.
A1: Correct 3TQ form correct coordinates
A1: Correct values cso
Misread: A common misread is 3 instead of – 3, the first 3 mark only can be scored using the misread rule
M1: \(\begin{pmatrix}2 & -1\\ 3 & 0\end{pmatrix}\begin{pmatrix}a & 4\\ 2 & -a\end{pmatrix} = \begin{pmatrix}\ldots & \ldots\\ \ldots & \ldots\end{pmatrix}\)
A1: \(\begin{pmatrix}2a - 2 & 8 + a\\ 3a & 12\end{pmatrix}\) or \((2a - 2,\ 3a)\) and \((8 + a,\ 12)\)
M1: \(\sqrt{\left[(2a - 2) - (8 + a)\right]^2 + \left[3a - 12\right]^2} = \sqrt{58}\)
A0, A0 This does lead to the correct answer but can score the first three marks only.