Review Maths P2 Higher

DiscoMouse

Ah, yeah, I meant 4. GOD, I CAN'T DO ANYTHING RIGHT! GAH!!

And yeah, I did what you did, minus the clever part where you worked out the first question using it.

ladyspud

meh don't really know how i got on weird.... it was hard but i was happy enuf i think!! I hope :rolleyes:

Gambra

ok, it went a lot better for me than Paper I (which I do believe I failed)

Q1 + 2 : Ignored both, simply could not get the circle question to work
Q 3: Forgot my proofs in (c)
Q 4: couldn't get (b)(ii), again forgot how to do it
Q 5: Found simple, got the graph, the asymptotes, and I guessed the (iv) part! But got it right!! Easiest question I've ever done
Q 6: Got everything but veryfying the second difference equation
Q 7:(c) in this was absolute ****. I'd no idea what was going on. Seemed yoiu had to apply a sequence formula to is

I'd say a C overall, just hope it'll rescue my Paper I from the fail..

Zoodlebop

I got the right answer by saying that and equilateral triangle in a circle has sides which bisect the radius, then pythang. to find half the radius and the ans. by 2 to find radius. It worked in the next question too. D'ye think that is logically ok?

sariane

Q 7c) in this was absolute ****. I'd no idea what was going on. Seemed yoiu had to apply a sequence formula to it

I know i couldn't do it!! i got like root (n squared over 3)???? how was it done??

Zoodlebop

Also , did anyone find 4 (b) a bit odd, i didn't get it till 1 min before the end, in a mad rush. Was the answer30 and 240 degrees, or something close?

Zoodlebop

I think Raphael is right, it was meant to weed out people who had learned methods, as supposed to understanding the material. People like me

joxer05

I just want to check what other people got. I got:

1.(a) (x-1)sqrd+(y+1)sqrd=8
(b)(ii) p=0, p=-12/35
(c)(i)I substituted in x into S and found bsqrd-4ac less than 0, implies no real roots, implies K doesnt intersect S
(c)(ii) (2,1)

2.(a) 3i-j
(b)(i) pq= 6i-8j, pr=4i+3j
(ii) s= 7i-j
(iii) Angle=45 degrees
(c) got as far as dot product= length b sqrd - length of c sqrd. did you have to say that length b = length c, implies dot product= 0?

3.(b) q(8,0) , r(0,-12)
(c)(ii) Angle=54 degrees

4.(a) A= 60 or 300 degrees
(b)(i) 2Cos(2x+30)Sin(x+30)
(ii) x E (30, 120, 150, 330)
(c)(i) r=k/root3
(ii) Area Circle= Pi K sqrd/3, Area Sector= Pi k sqrd/6

5.(a) WTF!!
(iv) wrote nothing
(b)(i) 272m
(ii) 110,000m sqrd

8.(b)(i) a(0,2-4m) , b(4-2/m,0)
(ii) no time
(c)(i) -1/3 less than x less than 1/3
(ii) r= 1/4, no time to finish :mad:

Shelga

I'm not too happy with paper 2. After paper 1 I was happy and hoped that Paper 2 would go the same way. After I saw Q5 I was completely thrown, but I had to do it because I have no clue about probability. Since when do they put graphs and asymptotes and all that jazz in a trig question???

Really really wanted an A, doesn't look too likely now. Ah well, I'm still happy enough.

sariane

could someone run through the answers they got for 6 and 7 and how they did 7c??? i can't do it!!!

whassupp2

6ai- 165
6aii- dunno
6bi- Un= 3(1)/\n + 5(1/6)/\n
6ci- 1/729000000
6cii- 1/2.13

7a- 120
7aii- 30
7bi- 324632
7bii- couldnt do
7biii- couldnt do
7biv- couldnt do

Got different answer fo 7c.
i got root 2n/\2 over 3

There the answers i got. dunno if they're right. some of them sound very strange

Faerie

joxer05 wrote:

I just want to check what other people got. I got:

1.(a) (x-1)sqrd+(y+1)sqrd=8
(b)(ii) p=0, p=-12/35
(c)(i)I substituted in x into S and found bsqrd-4ac less than 0, implies no real roots, implies K doesnt intersect S
(c)(ii) (2,1)

2.(a) 3i-j
(b)(i) pq= 6i-8j, pr=4i+3j
(ii) s= 7i-j
(iii) Angle=45 degrees
(c) got as far as dot product= length b sqrd - length of c sqrd. did you have to say that length b = length c, implies dot product= 0?

3.(b) q(8,0) , r(0,-12)
(c)(ii) Angle=54 degrees

4.(a) A= 60 or 300 degrees
(b)(i) 2Cos(2x+30)Sin(x+30)
(ii) x E (30, 120, 150, 330)
(c)(i) r=k/root3
(ii) Area Circle= Pi K sqrd/3, Area Sector= Pi k sqrd/6

5.(a) WTF!!
(iv) wrote nothing
(b)(i) 272m
(ii) 110,000m sqrd

8.(b)(i) a(0,2-4m) , b(4-2/m,0)
(ii) no time
(c)(i) -1/3 less than x less than 1/3
(ii) r= 1/4, no time to finish :mad:

I got pretty much the same answers as you!
Except for part 1 if vectors, I have i-3j written on my exam paper....I just hope I wrote it correctly in my exam!! It would be such a stupid mistake to lose marks for!

Nichololas

whassupp2 wrote:

6ai- 165
6aii- dunno
6bi- Un= 3(1)/\n + 5(1/6)/\n
6ci- 1/729000000
6cii- 1/2.13

Can you explain how you got 1/729000000 (or (1/30) to the 6)?

I got (1/30) to the 5, reasoning being;

1 (the first day) x 1/30 x 1/30 x 1/30 x 1/30 x 1/30 x 1/30 (probability of each subsequent birthday being on the same day) x 30 (for each day of the month)

Seems right to me but it's niggling me since people got 1/729000000. If i'm wrong I think it's the x30 for each day of the month, although I seem to remember an example where you did the same thing.

Nichololas

Double post - should pay attention.

Attractive Nun

joxer05 wrote:

I just want to check what other people got. I got:

1.(a) (x-1)sqrd+(y+1)sqrd=8 same as me, iirc
(b)(ii) p=0, p=-12/35 same as me, iirc
(c)(i)I substituted in x into S and found bsqrd-4ac less than 0, implies no real roots, implies K doesnt intersect S same as me, iirc
(c)(ii) (2,1) couldn't figure out

2.(a) 3i-j same as me, iirc
(b)(i) pq= 6i-8j, pr=4i+3j same as me, iirc
(ii) s= 7i-j same as me, iirc
(iii) Angle=45 degrees same as me, iirc
(c) got as far as dot product= length b sqrd - length of c sqrd. did you have to say that length b = length c, implies dot product= 0? I think I actually got that far, didn't realise b=c...hmm

3.(b) q(8,0) , r(0,-12) didn't do Q3
(c)(ii) Angle=54 degrees

4.(a) A= 60 or 300 degrees same as me, iirc
(b)(i) 2Cos(2x+30)Sin(x+30) same as me, iirc
(ii) x E (30, 120, 150, 330) ****ed this up
(c)(i) r=k/root3 same as me, iirc
(ii) Area Circle= Pi K sqrd/3, Area Sector= Pi k sqrd/6 same as me, iirc

5.(a) WTF!!Aye, me too. Think I got (i) and (ii) though
(iv) wrote nothing I wrote something clearly incorrect
(b)(i) 272m same as me, iirc
(ii) 110,000m sqrd can't remember really, but maybe

8.(b)(i) a(0,2-4m) , b(4-2/m,0) same as me, iirc
(ii) no time I think I got m=1/2
(c)(i) -1/3 less than x less than 1/3 same as me, iirc
(ii) r= 1/4, no time to finish :mad:yeah I got that and finished it off as for part (i)

That's me.

DonaldDuck

joxer05 wrote:

I just want to check what other people got. I got:

1.(a) (x-1)sqrd+(y+1)sqrd=8
(b)(ii) p=0, p=-12/35
(c)(i)I substituted in x into S and found bsqrd-4ac less than 0, implies no real roots, implies K doesnt intersect S
(c)(ii) (2,1)

2.(a) 3i-j
(b)(i) pq= 6i-8j, pr=4i+3j
(ii) s= 7i-j
(iii) Angle=45 degrees
(c) got as far as dot product= length b sqrd - length of c sqrd. did you have to say that length b = length c, implies dot product= 0?

3.(b) q(8,0) , r(0,-12)
(c)(ii) Angle=54 degrees

4.(a) A= 60 or 300 degrees
(b)(i) 2Cos(2x+30)Sin(x+30)
(ii) x E (30, 120, 150, 330)
(c)(i) r=k/root3
(ii) Area Circle= Pi K sqrd/3, Area Sector= Pi k sqrd/6

5.(a) WTF!!
(iv) wrote nothing
(b)(i) 272m
(ii) 110,000m sqrd

8.(b)(i) a(0,2-4m) , b(4-2/m,0)
(ii) no time
(c)(i) -1/3 less than x less than 1/3
(ii) r= 1/4, no time to finish :mad:

For the vectors part C

I just took "ah" as "h-a" which was "a+b+c-a"
And bc as "c-b"

giving me (b+c)(c-b)
Bc-cb
=0

CrimE

DonaldDuck wrote:

For the vectors part C

I just took "ah" as "h-a" which was "a+b+c-a"
And bc as "c-b"

giving me (b+c)(c-b)
Bc-cb
=0

Yeah same here, hadn't a clue when I looked at it first. I'm pretty sure that proves it but I can't understand why they would say it was the orthocentre when it isn't needed to prove the question...

DonaldDuck

CrimE wrote:

Yeah same here, hadn't a clue when I looked at it first. I'm pretty sure that proves it but I can't understand why they would say it was the orthocentre when it isn't needed to prove the question...

Probably suppose to do it a more difficult way

For 3(b) I just took "2 to 0" =2 units,ratio = 3:! therefor 2+6= 8 giving me
(8,0) for first point,then "0 to -9" = -0 units,therefore -9-3=(0,-12) Im prtty sure it wasnt meant to be done that way

Attractive Nun

DonaldDuck wrote:

For the vectors part C

I just took "ah" as "h-a" which was "a+b+c-a"
And bc as "c-b"

giving me (b+c)(c-b)
Bc-cb
=0

Hmm, I got that but didn't realise it solved the question. But wouldn't bc-cb = bc+bc = 2bc /= 0? If not, then D'Oh!

EDIT: Oh I see.

sully-gormo

Enlil_Nick wrote:

Can you explain how you got 1/729000000 (or (1/30) to the 6)?

I got (1/30) to the 5, reasoning being;

1 (the first day) x 1/30 x 1/30 x 1/30 x 1/30 x 1/30 x 1/30 (probability of each subsequent birthday being on the same day) x 30 (for each day of the month)

Seems right to me but it's niggling me since people got 1/729000000. If i'm wrong I think it's the x30 for each day of the month, although I seem to remember an example where you did the same thing.

That paper was so tough.
For 6(c) would it be (1/30) to the power of 6 but then multiplied by 30!
(wish i thought of that in the exam)
What did anyone write for 5a(iv) the one about invTan(tank)=k

Shark-69

What did you all get for Q3b ,Q3c(ii)....Q5b(ii).......Q8c(ii)..?

DonaldDuck

sully-gormo wrote:

That paper was so tough.
For 6(c) would it be (1/30) to the power of 6 but then multiplied by 30!
(wish i thought of that in the exam)
What did anyone write for 5a(iv) the one about invTan(tank)=k

I wrote 0<tank<90

But it should have been -90<tank<90

Nichololas

sully-gormo wrote:

That paper was so tough.
For 6(c) would it be (1/30) to the power of 6 but then multiplied by 30!
(wish i thought of that in the exam)
What did anyone write for 5a(iv) the one about invTan(tank)=k

You sure it's 30! and not just 30?

Nichololas

Jesus I gotta stop hitting Submit thread.

Nichololas

Problem with my browser not updating.

Raphael

DonaldDuck wrote:

For the vectors part C

I just took "ah" as "h-a" which was "a+b+c-a"
And bc as "c-b"

giving me (b+c)(c-b)
Bc-cb
=0

actually you get

C.C + C.B - B.C - B.B
|C|-|B|

And because o is the circumcentre, |C| = |B|, so |C|-|B| = 0

casanova_kid

I thought it was a load of crap, what was going on with question 5(A), i don't know. I got some mad answers all around the place though.

sully-gormo

Enlil_Nick wrote:

You sure it's 30! and not just 30?

i dunno; just saw it there
I spent my last 5 minutes just staring at that question and not having a clue about what to do

According to Radio 1 theres a lot of controversy about the paper. Apparently the chief examiner was trying to get students to think independently(according to his report). Govt want more ppl to honours maths but less and less are doing it and a ccunt of a paper like this dosnt help. Id say the probability and that trig gfraph will be marked easy enough

caesar

I found it vey tough, i f*cked up alot of bits and the only reason was because of the answers i was getting i wasn't confident that they were right and i wasted time trying to figure out where i went wrong when i probably didn't. For that probability one i got 1/729000000 and i was sure i was wrong. For the person that was wondering how to do it. I did this.

1 x 1/30 x 1/30 x 1/30 x 1/30 x 1/30 x 1/30= 1/729000000


The one was for the first student as it didn't matter what day they were born and it was 1 in 30 chances of the other six having they same birthday.

helles belles

i cried, i needed an a in that paper i think i got a c