AS June 2019 Paper 1 Q6
6 A transformation T is represented by the matrix \(\mathbf{T}\) where \(\mathbf{T} = \begin{pmatrix} x^2 + 1 & -4 \\ 3 - 2x^2 & x^2 + 5 \end{pmatrix}\).
A quadrilateral \(Q\), whose area is 12 units, is transformed by T to \(Q'\).
Find the smallest possible value of the area of \(Q'\). [5]
| Scheme | Marks | AO |
|---|---|---|
| \((\Delta =) \begin{vmatrix} x^2 + 1 & -4 \\ 3 - 2x^2 & x^2 + 5 \end{vmatrix}\) \(= (x^2 + 1)(x^2 + 5) - (-4)(3 - 2x^2)\) | M1 | 1.1 |
| \(= x^4 - 2x^2 + 17\) | A1 | 1.1 |
| \(\Delta = (x^2 - 1)^2 + 16\) | M1* | 3.1a |
| \((x^2 = 1) \Rightarrow \Delta_{\min} = 16\) | A1 | 1.1 |
| So the smallest possible area is 192 (units) | A1ft (dep*) | 3.2a |
| [5] |
Notes
M1: Expanding the determinant of T. Condone sign error and/or one “\(x\)” rather than “\(x^2\)”
M1*: Attempt to find minimum value of quadratic in \(x^2\)
or from \(\dfrac{\mathrm{d}\Delta}{\mathrm{d}x} = 4x^3 - 4x = 0\)
A1ft: Their \(\Delta_{\min} \times 12\)