AQA AS FP1 Further Pure 1 15th June 2016 [Exam Discussion Thread]

[QUOTE=Physics Optimist;65813665]

Same here, I think you'd get a fair few method marks for subbing w & w* into the equation because there was a lot of simplifying and multiplying out brackets there aha

Is anyone else taking MS2B on Tuesday ?

Did anyone get $(\frac{\sqrt 3}{2},2)$ for the 6 mark matrices question?

We had to find the coordinates of the point which is mapped onto $(0,-4)$, after being transformed by $\mathbf{A^2}$ and then reflected in the line $x - \sqrt{3}y = 0$. So, using the information from earlier question parts, this is how I arrived at my answer (sorry about the LaTex):

$\begin{bmatrix} \frac{1}{2} & -\frac{\sqrt 3}{2} \\[6pt] - \frac{\sqrt 3}{2} & -\frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} 4 & 0 \\[6pt] 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} x \\[6pt] y \end{bmatrix} = \begin{bmatrix} 0 \\[6pt] -4 \end{bmatrix} \: \therefore \: \begin{bmatrix} 2 & -\frac{\sqrt 3}{2} \\[8pt] - 2 \sqrt 3 & -\frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} x \\[6pt] y \end{bmatrix} = \begin{bmatrix} 0 \\[6pt] -4 \end{bmatrix}$

If we multiply the two matrices together, we get $2x - \frac{\sqrt 3}{2}y = 0$ and $- 2 \sqrt 3x - \frac{1}{2}y = -4$.

By solving these equations simultaneously, we can then show that $x = \frac{\sqrt 3}{2}$ and $y = 2$.

I thought the question was "Find the coordinates of this point AFTER the stretch occurs then the reflection" not the other way round, other people seem to be confused on this too

No you had to find the coordinates of P before

Did anyone get $(\frac{\sqrt 3}{2},2)$ for the 6 mark matrices question?

We had to find the coordinates of the point which is mapped onto $(0,-4)$, after being transformed by $\mathbf{A^2}$ and then reflected in the line $x - \sqrt{3}y = 0$. So, using the information from earlier question parts, this is how I arrived at my answer (sorry about the LaTex):

$\begin{bmatrix} \frac{1}{2} & -\frac{\sqrt 3}{2} \\[6pt] - \frac{\sqrt 3}{2} & -\frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} 4 & 0 \\[6pt] 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} x \\[6pt] y \end{bmatrix} = \begin{bmatrix} 0 \\[6pt] -4 \end{bmatrix} \: \therefore \: \begin{bmatrix} 2 & -\frac{\sqrt 3}{2} \\[8pt] - 2 \sqrt 3 & -\frac{1}{2} \end{bmatrix} \cdot \begin{bmatrix} x \\[6pt] y \end{bmatrix} = \begin{bmatrix} 0 \\[6pt] -4 \end{bmatrix}$

If we multiply the two matrices together, we get $2x - \frac{\sqrt 3}{2}y = 0$ and $- 2 \sqrt 3x - \frac{1}{2}y = -4$.

By solving these equations simultaneously, we can then show that $x = \frac{\sqrt 3}{2}$ and $y = 2$.

I thought the question was "Find the coordinates of this point AFTER the stretch occurs then the reflection" not the other way round, other people seem to be confused on this too

I got this aswell

Yeah man