OCR MEI AS Further Mathematics FP1 13/05/2013 PM

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Reply 60

Jackb1234

Original post by 123formyabc

Hard paper i reckon below 58 for an a. What lines did everyone get for the last question

Think I got y = x - 1 for one of them and y = -x for the other

Reply 61

Aaqil

Guys what did you get in the matrix questions because I was kind of baffled except the first and the last parts. I think the (ii) I got p'(2,-2), (iii) y=-x, (iv) y =6

Reply 62

abdcefghi

Original post by 123formyabc

Hard paper i reckon below 58 for an a. What lines did everyone get for the last question

I think I put y=6 for one answer then something like y=-x for another one, was in the last minute so not exactly reliable lol

Reply 63

abdcefghi

Original post by Aaqil

Guys what did you get in the matrix questions because I was kind of baffled except the first and the last parts. I think the (ii) I got p'(2,-2), (iii) y=-x, (iv) y =6

What was the original p coordinate can you remember?

Reply 64

UKBrah

Original post by Jackb1234

Think I got y = x - 1 for one of them and y = -x for the other

For the last one I got y = x, since I kept getting the collumn matrix (y y). I got y = -x for iii) but y =x for vi.

Ill post the question if any1 wants to check:

ill do rest in a second yea

(edited 10 years ago)

IMG_4653.jpg514.1KB

Reply 65

UKBrah

Original post by Aaqil

Guys what did you get in the matrix questions because I was kind of baffled except the first and the last parts. I think the (ii) I got p'(2,-2), (iii) y=-x, (iv) y =6

Yeah man thats all correct, checking the paper currently looks good.

Nah wait, p(-2,2) not what you wrote.

Reply 66

123formyabc

Great i put something similar. But.. It sounds like i messed up on he argand diagram i got 3 - j, will he error get carried forward? So ill only lose like a 2 marks for it?

Reply 67

UKBrah

Original post by Law-Hopeful

Can anyone remember what they got for question 6?
The question was x^3-5x^2+3x-6=0 has roots... Find the cubic with roots (a/3)+1, (B/3)+1 and (y/3)+1.

I got 9w^3-42w^2+60w-29=0?

Just did it and i got 9w^3 -42w^2 +60w -29 =0 after simplfying. :cool:

Reply 68

abdcefghi

Original post by UKBrah

Just did it and i got 9w^3 -42w^2 +60w -29 =0 after simplfying. :cool:

Thats definitely the right answer checked it on the calculator

Reply 69

UKBrah

Original post by abdcefghi

Thats definitely the right answer checked it on the calculator

haha thats what I did, i found alpha of the original equation and did a/3 +1 and it did indeed work.

Reply 70

username1107833

Original post by UKBrah

For the last one I got y = x, since I kept getting the collumn matrix (y y). I got y = -x for iii) but y =x for vi.

Ill post the question if any1 wants to check:

ill do rest in a second yea

omg yeah could you please post a few pictures of the whole paper. Thought it was vile. Didn't manage to prove by induction, fudged it up half way through...If grade boundaries are like a record low I may have got an A

Reply 71

UKBrah

Original post by theCreator

I got you man, sending from phone

Reply 72

abdcefghi

Can someone upload pictures of the whole paper please would be much appreciated :biggrin:

Reply 73

Political Cake

Far as I was concerned, it seemed a rather middle-of-the-road paper in terms of difficulty, maybe slightly more difficult than average, albeit MUCH nicer than the one this January.

I struggled to plot the argand diagram. As far as I remember, it asked you to plot z₁, z₂, z₁-z₂, and z₁+z₂.

\left | {z_{2}} \right | = 5

arg(z_{2}) = \frac{\pi}{4}

Using them both I got:

z_{2}​ = \frac{5\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}j

The roots question seemed a bit out of the way, I remember having an x³ coefficient of

9x^{3}

and the others were fairly high.

(edited 10 years ago)

Reply 74

UKBrah

Here lads

Can someone please find what the equation of the line is in the last part of 9?

Is it y =x? im dying to know

(edited 10 years ago)

Original post by UKBrah

Here lads

Can someone please find what the equation of the line is in the last part of 9?

Is it y =x? im dying to know

I got y=-x

Reply 76

UKBrah

Original post by Political Cake

\left | {z_{2}} \right | = 5

arg(z_{2}) = \frac{\pi}{4}

Using them both I got:

z_{2}​ = \frac{5\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}j

The roots question seemed a bit out of the way, I remember having an x³ coefficient of

9x^{3}

and the others were fairly high.

Yep I agree with your answer for z2.

The diagram was really cluttered with tons of surds. I think I had like 3 complex numbers in the positive x quadrant and 1 somewhere else.