4755 MEI FP1 OCR 14th May 2015

Aph

22

Complex numbers :sogood:

and asymptotes this is the math exam heeding to infinity in complexity so good luck :woo:

Reply 1

Dan205

8

Gulp...
Was quite looking forward to this, but after M2 today anything could come up... :/

(edited 8 years ago)

Good luck all!

OP

Original post by Dan205

Gulp...
Was quite looking forward to this, but after M2 today anything could come up... :/

Today will be better... I hope at least.

Reply 4

Dan205

8

Original post by Aph

Today will be better... I hope at least.

Well it can't be any worse :wink:

you doing any last minute revision?

Posted from TSR Mobile

Reply 5

Aph

OP

22

Original post by Dan205

Well it can't be any worse :wink:

you doing any last minute revision?

Posted from TSR Mobile

dont jinx it!!!!!

nope, I tend to find that that does more harm then good. you shouldn't revise within the hour before your test but just relax ready.

Reply 6

Dan205

8

What did we think?

Posted from TSR Mobile

Reply 7

Aph

OP

22

Original post by Dan205

What did we think?

Posted from TSR Mobile

liked it, actually loved it :biggrin:

Reply 8

chriswhitehurst

Could not get the proof by induction question after getting them all right on every past paper :frown:

second to last question on the transformation matrix I think I got wrong (put it into my graphical calculator). Also did people get only real answers for solving the cubic equation or???

Reply 9

Aph

OP

22

Original post by chriswhitehurst

Could not get the proof by induction question after getting them all right on every past paper :frown:

second to last question on the transformation matrix I think I got wrong (put it into my graphical calculator). Also did people get only real answers for solving the cubic equation or???

the p,q,r cubic? i got p=-7, q=3 and I've forgotten r
I thought the proof by induction was fairly simple :s-smilie:

Reply 10

chriswhitehurst

Original post by Aph

the p,q,r cubic? i got p=-7, q=3 and I've forgotten r
I thought the proof by induction was fairly simple :s-smilie:

Yeah that one with alpha, alpha squared etc. When it put it into a form of z's and asked to find the roots, I've just realised you take alpha and its complex conjugate as two roots... damn. It should have been I just didn't get it straight away and paniced :/

Reply 11

danbowmcfc

Well that went very well. r was 123 I think and the proof by induction was very nice. Part ii of the summation question was the only really ugly question. And maybe question 8

(edited 8 years ago)

Reply 12

Aph

OP

22

Original post by chriswhitehurst

Yeah that one with alpha, alpha squared etc. When it put it into a form of z's and asked to find the roots, I've just realised you take alpha and its complex conjugate as two roots... damn. It should have been I just didn't get it straight away and paniced :/

oh the second one. you also had -3 a root and 1 was a root in the last part there. but panicking doesnt help :redface:

Reply 13

chriswhitehurst

Original post by Aph

oh the second one. you also had -3 a root and 1 was a root in the last part there. but panicking doesnt help :redface:

so yeah i got that wrong.... cant help it :/ are you Y13 resitting?

Reply 14

Aph

OP

22

Original post by chriswhitehurst

so yeah i got that wrong.... cant help it :/ are you Y13 resitting?

no I'm year 12.

Reply 15

ComputerMaths97

17

I did really well but read value of center of circle wrong (was (z-(-1-j)) i read it as (z-(-1+j)) for some reason :frown:

)so will lose 2-3 marks :frown:

So screwed now rrg anyway! :

Here are answers I can remember

1) x=5/18
y=1/27

proof by induction was to get 3^k=5/2

k=1/3 for sigma question

Area of A''B''C'' was 192 square units

roots were 5+/-4j and -3

There were two questions similar (answers of p q and r) that have confused me but here's all I can remember :
r=20 p=-11 q=12 r=123 may be some mix ups but yeah from above point I know it got them right xD

z=1 for the long quartic = cubic as quartic = z(cubic) so z=cubic/cubic = 1 (had loads of mates try argue that, I'm right lol xD)

(MT)^-1 = 1/48( 0 8) something like that, i got like 0, 1/6 , -1/8 , 1/12 so yeah xD
(-6 4)

Can't remember much more! Ask me!

Reply 16

ComputerMaths97

17

I would've got 100% if I could've read properly :frown:

What do you guys think about grade boundaries?

Reply 17

Aph

OP

22

Original post by joe12345marc

I would've got 100% if I could've read properly :frown:

What do you guys think about grade boundaries?

Same if not higher then last year.

Reply 18

ComputerMaths97

17

Original post by Aph

Same if not higher then last year.

I'm so pissed off. I got everything right. Despite thinking it was a plus and not a minus :frown:

SO f*ing screwed rggggg

Reply 19

Jam Goodman

1

I'll try and write down what I remember from the exam:

I think the first half of the paper was rather straight forward. I dropped a minus sign on Q8 and had to rewrite it so I can remember it quite clearly.

Q? on induction)

u_1 = 3, u_{k+1} = ?

prove

u_k=\frac{3^k+5}{2}

Q? on sums) i)

\sum_{r=1}^n(2r-1)=2\left(\sum_{r=1}^n r \right) - n

Q? on roots of polynonials) i)

f(x) = 2x^3+px^2+qx+r (?), x=4 \text{is a root}, \sum \alpha = 6, \alpha\beta\gamma = 4(?)

Q? on geometry with complex number) plot: (i)

arg(z-(-1-j))=\frac{\pi}{4}

, (ii)

|z-(1+2j)|=2

and (iii)

|z-(1+2j)| \geq 2 \text{and} 0 \leq arg(z-(-1-j)) \leq \frac{\pi}{4}

I got a figure roughly like this http://bit.ly/1JLc65k though to be perfectly honest, I'm not entirely confident with it.

Unparseable latex formula:

=2\left(\frac{n}{2}(n+1)) - n =...=n^2

QED

ii)

\frac{\sum_{r=1}^n(2r-1)}{\sum_{r=n+1}^{2n}(2r-1)}= \frac{n^2}{\left(\sum_{r=1}^{2n}(2r-1) - \sum_{r=1}^{n}(2r-1)\right)}

= \frac{n^2}{(2n)^2-n^2}=\frac{1}{3} \therefore k=\frac{1}{3}

Q8) i)

\alpha =5+4j \Rightarrow \alpha^2=9+40j, \alpha^3=-115+236j

ii) Polynomial was

f(z) = z^3+qz^2+11z+r

\Rightarrow f(\alpha) = (\alpha)^3-+q(\alpha)^2+11(\alpha)+r = 0

\Rightarrow (r+9q-60)+(280+40q)j= 0

q,r\in\mathbb{R}\Rightarrow q=-7, r=123

\therefore f(z) = z^3-7z^2+11z+123

iii)

\beta = \alpha^*=5-4j \Rightarrow (z-\alpha)(z-\beta)=0

\Rightarrow z^2-10z+41=0

(z-(-3))(z^2-10z+41)=z^3-7z^2+11z+123 \therefore \gamma = -3

iv)

z^4+qz^3+11z^2+rz=z^3+qz^2+11z+r \Rightarrow (z-1)(z^3+qz^2+11z+r)=0

\Rightarrow (z-1)(z-\alpha)(z-\beta)(z-\gamma)=0

\therefore z=1,-3,(5+4j),(5-4j)

Q9) i)

A=(0,0), B=(0,2), C=(4,1)

ii)

M= \begin{pmatrix} 4 & 0 \\0 & 2 \end{pmatrix} \Rightarrow

M is an enlargement, scale factor 4 in the x direction, scale factor 2 in the y direction

iii) Find

(MT)^{-1} = \frac{1}{48}(?)

iv)

Area(ABC) = \frac{1}{2} (2) (4) = 4

det(MT) = 48 \therefore Area (A''B''C'')=192

Stupidly, I thought that B was (2,0) so got the final area wrong. D'oh! I also didn't simplify the fractions in Q1 though they can't possibly penalise me for that?... Ah well, apparently I'm an idiot. Who knew?

This is my first post so if I've written any errors, please let me know. Also, does anyone remember the equivalence relation from the induction question or the polynomial from the roots of equations question or how to embed images in a post. Thanks a lot!

(edited 8 years ago)

4755 MEI FP1 OCR 14th May 2015

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