The Student Room Group

FP1 OCR(not MEI) Jan 2011

Scroll to see replies

Reply 20
That was the easiest FP1 exam i've ever done
Original post by L100
That was the easiest FP1 exam i've ever done


So did you get inconsistent or infinite solutions for 9iii? :/
Original post by snowboardinsteve
So did you get inconsistent or infinite solutions for 9iii? :/

infinite.
Reply 23
Sat it today as well, didn't find it too demanding and was surprised that it didn't require you to find inverse matrices anywhere.

Places where I disagree:

Spoiler



I shaded between x = 4, y = -2 and y = x - 2



8.ii. q=2/3


I got 1/3

I may well be wrong though - I'm very careless sometimes :redface:

Edit: just seen other post giving the same thing. Spoiler'd :o:
(edited 13 years ago)
Original post by Goldfishy
Sat it today as well, didn't find it too demanding and was surprised that it didn't require you to find inverse matrices anywhere.

Places where I disagree:


I shaded between x = 4, y = -2 and y = x - 2



I got 1/3

I may well be wrong though - I'm very careless sometimes :redface:

If you're wrong, that'd make two of us... :p:
I agree.
I'm here!

Give me 10 mins to LaTeX it all.
Reply 26
can anyone remind me what was question 4 :s-smilie:? I don't really remember putting down 4, -4. . . .
Reply 27
Original post by Farhan.Hanif93
If you're wrong, that'd make two of us... :p:
I agree.


Fingers crossed then, I guess we'll know in 10mins anyway :tongue: How did you find the paper overall?
Original post by Goldfishy
Fingers crossed then, I guess we'll know in 10mins anyway :tongue: How did you find the paper overall?

It was a nice chilled start to my exam period, I will be surprised if I lose more than 1 or 2 marks. I took it EXTRA slowly to make sure that I didn't make any sign errors but it's still pretty likely that I've made a numerical error somewhere... :erm:
Yourself? Good luck, I'm sure you did fine anyways!
Reply 29
Original post by Mr M
I'm here!

Give me 10 mins to LaTeX it all.


yes mr m thank you :smile:
OCR FP1 (not MEI) answers Jan 2011


1. i) (79)\begin{pmatrix} 7 & 9 \end{pmatrix} (2 marks)

ii) (18)\begin{pmatrix} 18 \end{pmatrix} (2 marks)

iii) (12462)\begin{pmatrix} 12 & -4 \\6 & -2 \end{pmatrix} (3 marks)


2. i) 12+13i-12+13i (2 marks)

ii) 2714i37\frac{27-14i}{37} (4 marks)


3. Proof by induction (4 marks)


4. a=4a=4 and b=4b=-4 (6 marks)


5. A2\mathbf{A}^2 (3 marks)


6. i) a) Vertical line Re(z)=4Re(z) = 4 (2 marks)

b) Half line starting at 2i-2i making angle of π4\frac{\pi}{4} with real axis. (3 marks)

ii) Shade triangle to left of vertical line and between half line and the line Im (z) = -2 (a line parallel to real axis). (3 marks)


7. i) (1301)\begin{pmatrix} 1 & 3 \\0 & 1 \end{pmatrix} (2 marks)

ii) Enlargement centre (0, 0) scale factor 3\sqrt 3 (2 marks)

iii) a) Parallelogram vertices at (0, 0) (2, 0) (6, 2) (8, 2) (3 marks)

b) Determinant = 4 so the area scale factor = 4 (2 marks)


8. i) Show ... (4 marks)

ii) q=13q=\frac{1}{3} (5 marks)


9. i) a27a+6a^2-7a+6 (3 marks)

ii) a=1a=1 and a=6a=6 (3 marks)

iii) Infinite number of solutions (evidence of consistency needed) (3 marks)


10. i) Show ... (2 marks)

ii) 121(n+1)(n+2)\frac{1}{2} - \frac{1}{(n+1)(n+2)} (6 marks)

iii) Show ... (3 marks)


MISTAKE CORRECTED - SORRY TO THOSE I WORRIED - I KNEW WHAT I WAS TALKING ABOUT ANYWAY!
(edited 13 years ago)
Reply 31
Original post by Mr M
OCR FP1 (not MEI) answers Jan 2011


1. i) (79)\begin{pmatrix} 7 & 9 \end{pmatrix} (2 marks)

ii) (18)\begin{pmatrix} 18 \end{pmatrix} (2 marks)

iii) (12462)\begin{pmatrix} 12 & -4 \\6 & -2 \end{pmatrix} (3 marks)


2. i) 12+13i-12+13i (2 marks)

ii) 2714i37\frac{27-14i}{37} (4 marks)


3. Proof by induction (4 marks)


4. a=4a=4 and b=4b=-4 (6 marks)


5. A2\mathbf{A}^2 (3 marks)


6. i) a) Vertical line Re(z)=4Re(z) = 4 (2 marks)

b) Half line starting at 2i-2i making angle of π4\frac{\pi}{4} with real axis. (3 marks)

ii) Shade triangle to left of vertical line and between half line and real axis. (3 marks)


7. i) (1301)\begin{pmatrix} 1 & 3 \\0 & 1 \end{pmatrix} (2 marks)

ii) Enlargement centre (0, 0) scale factor 3\sqrt 3 (2 marks)

iii) a) Parallelogram vertices at (0, 0) (2, 0) (6, 2) (8, 2) (3 marks)

b) Determinant = 4 so the area scale factor = 4 (2 marks)


8. i) Show ... (4 marks)

ii) q=13q=\frac{1}{3} (5 marks)


9. i) a27a+6a^2-7a+6 (3 marks)

ii) a=1a=1 and a=6a=6 (3 marks)

iii) Infinite number of solutions (evidence of consistency needed) (3 marks)


10. i) Show ... (2 marks)

ii) 121(n+1)(n+2)\frac{1}{2} - \frac{1}{(n+1)(n+2)} (6 marks)

iii) Show ... (3 marks)


okay why is 6)ii) between line x=4 and the real axis?
:frown:
Original post by mrppaulo
okay why is 6)ii) between line x=4 and the real axis?
:frown:


Because it also had to satisfy the other inequality which had the condition the argument was greater than or equal to zero.
Original post by Mr M
OCR FP1 (not MEI) answers Jan 2011


1. i) (79)\begin{pmatrix} 7 & 9 \end{pmatrix} (2 marks)

ii) (18)\begin{pmatrix} 18 \end{pmatrix} (2 marks)

iii) (12462)\begin{pmatrix} 12 & -4 \\6 & -2 \end{pmatrix} (3 marks)


2. i) 12+13i-12+13i (2 marks)

ii) 2714i37\frac{27-14i}{37} (4 marks)


3. Proof by induction (4 marks)


4. a=4a=4 and b=4b=-4 (6 marks)


5. A2\mathbf{A}^2 (3 marks)


6. i) a) Vertical line Re(z)=4Re(z) = 4 (2 marks)

b) Half line starting at 2i-2i making angle of π4\frac{\pi}{4} with real axis. (3 marks)

ii) Shade triangle to left of vertical line and between half line and real axis. (3 marks)


7. i) (1301)\begin{pmatrix} 1 & 3 \\0 & 1 \end{pmatrix} (2 marks)

ii) Enlargement centre (0, 0) scale factor 3\sqrt 3 (2 marks)

iii) a) Parallelogram vertices at (0, 0) (2, 0) (6, 2) (8, 2) (3 marks)

b) Determinant = 4 so the area scale factor = 4 (2 marks)


8. i) Show ... (4 marks)

ii) q=13q=\frac{1}{3} (5 marks)


9. i) a27a+6a^2-7a+6 (3 marks)

ii) a=1a=1 and a=6a=6 (3 marks)

iii) Infinite number of solutions (evidence of consistency needed) (3 marks)


10. i) Show ... (2 marks)

ii) 121(n+1)(n+2)\frac{1}{2} - \frac{1}{(n+1)(n+2)} (6 marks)

iii) Show ... (3 marks)

I agree with all of those. For question 2ii) however, it did ask for the answer in the form x+yi.

EDIT: Although, the shaded triangle should the one enclosed by the half line, the line Im(z)=2Im(z) = -2 (as opposed to the real axis) and the line Re(z)=4Re(z)=4 surely?
(edited 13 years ago)
Reply 34
Original post by Mr M
Because it also had to satisfy the other inequality which had the condition the argument was greater than or equal to zero.


i thought if the argument was from that point between 0 and pi then you take -2i as the centre point and 0radians is a sort of imaginary horizontal line from that point :s-smilie:
Reply 35
full marks fuk yeah
Reply 36
hey, what was question 4 o_O? I still can't remember it xD sorry ~
Reply 37
for 7i) what was the question?
i thought it was what makes (1,0) go to (4,0) :/
Original post by Mr M
OCR FP1 (not MEI) answers Jan 2011
6. ii) Shade triangle to left of vertical line and between half line and real axis. (3 marks)

9. iii) Infinite number of solutions (evidence of consistency needed) (3 marks)


Damn. Why does the shading stop at the real axis? i thought it would be left of x = 2 and above y = -2 and underneath the line y = x - 2

Got inconsistent, but i see why i am wrong now :frown:
Reply 39
Original post by Farhan.Hanif93
It was a nice chilled start to my exam period, I will be surprised if I lose more than 1 or 2 marks. I took it EXTRA slowly to make sure that I didn't make any sign errors but it's still pretty likely that I've made a numerical error somewhere... :erm:
Yourself? Good luck, I'm sure you did fine anyways!


I had to wait for Mr M's solutions before I could say I found it pretty easy :redface:. Yep - I was just worried about a small slip-up too. Looks like you have full marks too :grin: Btw, did you qualify for BMO2 by any chance?
Original post by Mr M
x


I'm with the others on this point that it's bounded by y = -2 rather than the real axis.
(edited 13 years ago)

Quick Reply

Latest

Trending

Trending