I'm a little sweat and do all the past papers and I feel this years papers have been worse than the past papers I've done😶 hopefully the grade boundaries and 100ums will be a lot lower? Anyone agree?

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Reply 45

Roxanne18

Has anyone got an unofficial mark scheme for this ?

Reply 46

LaurenLovesMaths

[QUOTE=Physics Optimist;65801937]Thanks - this was the only one where I think I made a mistake. I got q as -7, -1 and 5. I didn't like 3 values and my Casio didn't agree with me :smile:

I think I have one sign error so a probable 1/5 marks?

How did you get that?! I got -1 and couldn't work out the others which meant I completely screwed up that question :frown:

ended up spending like 25 minutes on it too

Reply 47

LaurenLovesMaths

[QUOTE=Olsmarto;65805605]I'm a little sweat and do all the past papers and I feel this years papers have been worse than the past papers I've done😶 hopefully the grade boundaries and 100ums will be a lot lower? Anyone agree?

Posted from TSR Mobile

Other than the complex number question, I didn't do the questions any more difficult, though there were some 5/6 markers which would usually be split into smaller questions

Reply 48

Chickenslayer69

Original post by Olsmarto

Probably ~58 for an A. The paper was good except the higher mark ones which were much harder than usual, especially complex numbers question.

(edited 7 years ago)

Reply 49

Physics Optimist

Original post by LaurenLovesMaths

How did you get that?! I got -1 and couldn't work out the others which meant I completely screwed up that question :frown:

ended up spending like 25 minutes on it too

I subbed p+qi and p-2i into the equation, hence getting 4 simultaneous equations in p and q (2real and 2 imag). Two of them gave -1 for q but the others are prob incorrect. Never mind 😄

Reply 50

ollycostello

Original post by bheaton

Loved the matrix question anyone remember what they got for the last part of the question???

2root3 and 2

Reply 51

sam_97

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, and compare both sides, 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

(edited 7 years ago)

Reply 52

Roxanne18

Original post by sam_97

Did anyone get

(\frac{\sqrt 3}{2},2)

for the 6 mark matrices question?

We had to find the coordinates which were mapped onto

(0,-4)

after being transformed by

\mathbf{A^2}

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} . \begin{bmatrix} 4 & 0 \\[6pt] 0 & 1 \end{bmatrix} . \begin{bmatrix} x \\[6pt] y \end{bmatrix} = \begin{bmatrix} 0 \\[6pt] -4 \end{bmatrix}

\:\:\Rightarrow

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

If the multiply the new matrices together, then

2x - \frac{\sqrt 3}{2}y = 0

and

- 2 \sqrt 3x - \frac{1}{2}y = -4

.

By solving these two equations simultaneously, we get

x = \frac{\sqrt 3}{2}

and

y = 2

.

Hope someone recognises this!

Yay, I got this too!!!

Reply 53

bheaton

Original post by ollycostello

2root3 and 2

YESSS!!! I got that

Reply 54

sam_97

Original post by Roxanne18

Yay, I got this too!!!

Great! Hopefully it's correct

Reply 55

bheaton

Is anyone else taking MS2B on Tuesday ?

Reply 56

sam_97

Original post by bheaton

Is anyone else taking MS2B on Tuesday ?

I am. Here's the link to the MS2B exam thread http://www.thestudentroom.co.uk/showthread.php?t=4121015

Reply 57

LaurenLovesMaths

[QUOTE=Physics Optimist;65808229]I subbed p+qi and p-2i into the equation, hence getting 4 simultaneous equations in p and q (2real and 2 imag). Two of them gave -1 for q but the others are prob incorrect. Never mind 😄

That's what I did but I just got -1, and then through doing something weird I got -8i?! I don't know hahaha