Mr M's OCR (not OCR MEI) Further Pure 1 Answers June 2013

Something small, but do you think you'd lose marks for not putting brackets around arg z-3i ?


Cheers

For question 6)i) I put (Z-3i)=arg(pi/4) (so basicly put the arg on the wrong side) - did the same thing again in my inequalities for ii). How many marks do you think I would lose?
Thanks for the answers

I don't think you will get anything for 6i but I might be wrong.

On 6ii you will get 2 marks for the arg z inequality

And no to the FP3 answers.

Hi Mr M, for the last question I only worked out the first line of cofactors and when I did my inverse I forgot to use the sign matrix so I ended up with different numerators but had the a+3. How many marks would I lose?

Also for 6 I put arg (z-3i) but I didn't specify the angle and I also put C=z-3i but didn't specify the length 3. How many marks would I lose?

alloww, i put 1/2n(n+1)(2n^2-1)..................anyone no what that would get me??

Mr M,

for question 6 I wrote it as arg L =( pi/4) + 3i

how many marks will I lose?

Mr M, do you know how the 90 UMS grade boundary is worked out? If say the A boundary is 63 and B boundary is 57 say (like Jan 2013)

I know, usually, there is a ' cap' whereby you double the distance between the A and B boundary. But in this case, that would be 75/72, which isn't possible,

In cases like this, is 90UMS just calculated half way between the A boundary and max mark? so 67/68 in the above case?

I'm not sure - does it really matter?

Not entirely, I only ask because if the A/80 UMS boundary is quite high for this paper (has been 65 in the past), it could mean 90 UMS is near on full marks. Thanks anyway!

Mr M's OCR (not OCR MEI) Further Pure 1 Answers June 2013


1. $a=\sqrt 3$

$|z|=2\sqrt{3}$

$z^*-3=-i\sqrt3$ (6 marks)


2. (i) $\begin{pmatrix} 7 & 23 \end{pmatrix}$

(ii) $\begin{pmatrix} 6 & -15 \\4& -10 \end{pmatrix}$

Singular because the determinant = 0 (5 marks)


3. $2\sqrt5+3i$ and $-2\sqrt5-3i$ (6 marks)


4. Proof (6 marks)


5. $n^3(n+1)$ (6 marks)


6. (i) arg (z - 3i) = $\frac{\pi}{4}$

$|z-3i|=3$

(ii) $|z-3i| \leq 3$ and $\frac{\pi}{4} \leq$ arg (z - 3i) $\leq \frac{\pi}{2}$ (3 marks)


7 (i) $\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}$ (2 marks)

(ii) $\begin{pmatrix} 1 & 0 \\0& -1 \end{pmatrix}$ (2 marks)

(iii) $\begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}$ (2 marks)

(iv) Reflection in the line y = x (2 marks)


8. $\frac{3k-11}{k}$ (6 marks)


9. (i) Show (2 marks)

(ii) Show (6 marks)


10. (i) $a=-3$ (5 marks)

(ii) $\frac{1}{a+3} \begin{pmatrix} 1 & -1 & 1 \\7 & a-4 & 1-2a \\-11 & 8-a & 3a-2 \end{pmatrix}$

$x=\frac{2}{a+3}$

$y=\frac{2-4a}{a+3}$

$z=\frac{7a-1}{a+3}$ (7 marks)

Q6i) i put |c-3i|=3 is that ok? Will i drop any marks

In question 7 is it fine to leave the rotation matrix in trig form?

Hi. I think I got everything right other than two silly mistakes:

For Q6 ii) I only did the circle inequality, for the other one I put arg (z-3i) > (greater than or same as) pi/4 but forgot pi/2 completely.

The other error was on Q9) ii), I forgot to times by 1/3 so i couldn't prove it.

How many marks will I loose Mr M?