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Maths HL Paper 2

  • 14-06-2010 11:11am
    #1
    Registered Users, Registered Users 2 Posts: 1,586 ✭✭✭


    ....beautiful. Ran out of time,but such a lovely paper. Hopefully get my required c3 after the disaster of paper 1


«13456

Comments

  • Registered Users, Registered Users 2 Posts: 46 MyHair


    Completely agree, nothing in that paper I could't get done.
    It was awesome! Paper 1 was an absolute nightmare through >_<


  • Registered Users, Registered Users 2 Posts: 1,586 ✭✭✭Healium


    Paper 1 was a disaster for me,too.... Just need a C3.... Fingers crossed


  • Closed Accounts Posts: 834 ✭✭✭Reillyman


    Easiest Paper 2 I ever done! Can't believe how easy it was, I'd say I dropped a max of 30 marks! Hope the marking scheme won't be cruel now to make up for the easy paper! Need a C3 for my course, fcuking love it!:D:D:D


  • Closed Accounts Posts: 140 ✭✭LadyGaga!


    Need a C3 as well *joins the bandwagon of fingers crossed*


  • Registered Users, Registered Users 2 Posts: 197 ✭✭pp_me


    Answers?


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  • Closed Accounts Posts: 193 ✭✭straight_As


    Leaving cert maths has gone to the shitter.

    That paper doesn't compare to the early 2000's.


  • Closed Accounts Posts: 315 ✭✭Making It Bad


    lovely stuff, marking scheme will be the balls though or else there's gonna be a load more A's than usual


  • Closed Accounts Posts: 175 ✭✭Blerdiii


    paper one was certainly easier!
    i definately should have done vectors instead of circle but as long as i only llost 20/30 marks i can still get my a1!
    nice paper though :)


  • Moderators, Education Moderators Posts: 5,028 Mod ✭✭✭✭G_R


    awesome paper, couldnt have been better!!!!!!!!!!!!!!!!!!!
    I'm so happy, it makes up for my average p1 performance.

    Couldnt have been better


  • Banned (with Prison Access) Posts: 58 ✭✭echoindia756


    it's hard to tell how good/bad I did but I dno. we'll soon find out in about 8 weeks or so!
    Ya know on section B? part (a), what did ye get for the max area and how did ye get it?

    Also part B of the circle, part 2 of that, i could not get it.


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  • Closed Accounts Posts: 311 ✭✭H2student


    Very ironic paper for me. After a better-than-expected performance of paper I, I was convinced that paper II was going to be very hard.

    I wasn't bothered about studying Q8 for P II, in fact I only looked at it for 5 mins at 6pm last night and ran off. Before that I spent 6-7 hours studying the rest of the paper(good for my standards).

    Today, I went in and went to Q 8 straight away to see how bad it is. "FACEPALM" Maclaurin series as a part C. - No way I can do that.

    "off to Q2" I went "Wtf" a few times. done part A and gave up on part b & c after a few minutes. Off to probability, did it and again "wtf@"part c about the bench thing. Also, couldn't finish off the proof despite looking over it the night before. Statistics was okay...

    Then the circle, spent 20 mins on part C doing all sorts of things before realising that the circle can be read as a midpoint....

    Went to the line, not too sure about part C.

    At this point of the paper I said "goodbye A, B". Here's the ironic part, I managed to do Q8 perfectly (I think) o_O. Then with 15 mins left, I managed to finish off Q2 b,c. & finish most of Q 3c! Still no chance of getting an A from my estimation but B is very likely now :D


  • Closed Accounts Posts: 56 ✭✭johndoe91


    was a very easy paper.... got everythiing up to the option....screwed up the area though as soon as i walked out igot it! :(

    plus couldnt convert the maclaurin series from cosx to sin2x!?

    can you still get an A1 if i got everything else right!?


  • Registered Users, Registered Users 2 Posts: 814 ✭✭✭JerCotter7


    H2student wrote: »
    Very ironic paper for me. After a better-than-expected performance of paper I, I was convinced that paper II was going to be very hard.

    I wasn't bothered about studying Q8 for P II, in fact I only looked at it for 5 mins at 6pm last night and ran off. Before that I spent 6-7 hours studying the rest of the paper(good for my standards).

    Today, I went in and went to Q 8 straight away to see how bad it is. "FACEPALM" Maclaurin series as a part C. - No way I can do that.

    "off to Q2" I went "Wtf" a few times. done part A and gave up on part b & c after a few minutes. Off to probability, did it and again "wtf@"part c about the bench thing. Also, couldn't finish off the proof despite looking over it the night before. Statistics was okay...

    Then the circle, spent 20 mins on part C doing all sorts of things before realising that the circle can be read as a midpoint....

    Went to the line, not too sure about part C.

    At this point of the paper I said "goodbye A, B". Here's the ironic part, I managed to do Q8 perfectly (I think) o_O. Then with 15 mins left, I managed to finish off Q2 b,c. & finish most of Q 3c! Still no chance of getting an A from my estimation but B is very likely now :D

    No you can't? It was (6,2) I think not (3,1)


  • Closed Accounts Posts: 177 ✭✭chaoticmess


    LadyGaga! wrote: »
    Need a C3 as well *joins the bandwagon of fingers crossed*

    Yep me too!
    There seems to be a lot of crossed fingers!

    Overall I thought the paper was pretty reasonable, a couple of iffy bits and I didn't get EVERYTHING out, I only answered part A of question 8 (eek!) but I think it went ok... well enough for my C3 anyway. :)


  • Closed Accounts Posts: 311 ✭✭H2student


    JerCotter7 wrote: »
    No you can't? It was (6,2) I think not (3,1)

    ah crap, I guess I just tricked myself into thinking that :P. What was the radius? I think I got something like root 6. Doesn't matter I guess, hope I get more than attempt marks.


  • Registered Users, Registered Users 2 Posts: 814 ✭✭✭JerCotter7


    H2student wrote: »
    ah crap, I guess I just tricked myself into thinking that :P. What was the radius? I think I got something like root 6. Doesn't matter I guess, hope I get more than attempt marks.

    Can't remember the slope right now. You had to get the slope of y=2x then perpendicular to that line at (2,4) draw a line and then perpendicular to that line again down to (4,-2). Then the distance formula from either of those points to the midpoint. Can't do it now though because I gotta go to Irish orals. Sorry.


  • Closed Accounts Posts: 48 Killio9


    i think we all got lucky at how easy the paper was this year! that paper2 was f**king beautiful! apart from 2 [C] where they asked about them being collinear! and 4 [c] part 3 was fair tough aswell! that was about it though?!


  • Users Awaiting Email Confirmation Posts: 114 ✭✭UglyFuc


    i thought the paper was grand for the mostpart

    1a) (x-3)^2+(x+4)^2 = 17

    bi) (4,5) rad +=3
    bii) distance from center to the line should be 3. use the perp. distance got 17 so put down +17 and -17

    c) g=-6 f=-2 c=-28

    2a) -i-3j

    bi) modulas v is root (1 + k). u.v was -2+k
    bii) ****er wouldnt work out for me ended up with k = root 5 and k = -2

    c) attempted it, got the same answer for part i and ii, so one of them could be right

    3a)a was 8 i think

    bi) (0,k/5) and (-k/4,0)
    bii)k was +or- 20

    ci) equation was x(m-1)+y(m+1) + c
    cii) got m to be -1

    4a) 30,150
    4b 0, 120, 180, 300

    ci) was handy. since the height of the 3 smaller triangles is the radius, half the base by the radius (1/2 ar) = the area of each add the 3 together and factorise
    cii) if its right angled, pythagorus theroum will come out as 0=0. so get a^2 = b^2 +c^2 and multiply them out
    iii) no idea

    5a) tan2x formula. and sub in tan x wherever it goes
    ci) square them out, change them around with formula from tables

    8a let u equal to logex, let dv=dx
    dudx= 1/x v=x


  • Closed Accounts Posts: 16 mind_master


    Tell me i'm cocky if you will, but i don't think i dropped a single mark in that exam. Did question 1-5 in Section A and Q8 in Section B. Question 7(b) was tough i thought. So between both papers i'm looking for a 99% average :D

    Let me know if you need answers.


  • Closed Accounts Posts: 177 ✭✭chaoticmess


    Let me know if you need answers.

    If you have time to post answers for question 2 c (ii) and questions 4 and 5 that'd be fantastic! Oh and 6 (c).... :)


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  • Closed Accounts Posts: 311 ✭✭H2student


    When they say prove it's collinear, I somehow got r=2c, and I said as R can be written in terms of C, they're collinear. Not sure if it's right though.


  • Closed Accounts Posts: 48 Killio9


    H2student wrote: »
    When they say prove it's collinear, I somehow got r=2c, and I said as R can be written in terms of C, they're collinear. Not sure if it's right though.

    i ended up going along a long winding road and just gave up and whatever i had there just but a tick and "true therefore o,r,c are colinear" lol


  • Closed Accounts Posts: 16 mind_master


    Q2 c(i) q= c + (1/2)a

    (c)(ii) aq= c - (1/2)a

    This is because aq = q-a

    aq= c + (1/2)a - a
    = c - (1/2)a


    Q4 (a) 30 and 150

    (b) x = 0,120,240,360

    (c) (i) If you draw two more radii to where the circle touches the triangle, you will see the big triangle split up into 3 smaller ones. One with side a, one with b, and one with c. Each of these has a height of r. So to get the area use (1/2)(base)(height) and simplify which gives you the answer.

    (ii) use pythagoras to prove a^2=b^2+c^2

    For this one i'm not sure if you have to prove first that a is the longest side. But its easily done by letting a>b and letting a>c which gives two true statements.

    (iii) For this you draw a right-angled triangle with a circle inside and use the method from (i) to get the area. Then get the area using (1/2)(base)(height) and equal them. You get (r=pq-q^2) and as you are told in part (ii) that p,q are natural numbers this means that (pq-q^2) while be a whole number..



    Q5 (a) simple really , use tan 2A formula from log tables and fill in.

    (b) (i) Get Cos A and Cos C seperately and add
    (ii) use cos(A+B) rule from log tables. Find sin A by using cos A and drawing a right angles triangle to find sin A. Same for Sin C. Sub in answers.

    (c) (i) Straightforward
    (ii) use result from (i) using 4x=A and x=B
    Eventually you get tan3x=(1/root 3)

    Solve for x and you get 10,70,130,190,250,310



    Q6 (c) Note: did not do this in exam, possibility of being wrong here.

    All six seated together
    > 6!x6! = 518400
    No two seated together
    > 6! = 720

    (518400/720) = 720
    (very possibly wrong, not sure)


  • Closed Accounts Posts: 56 ✭✭johndoe91


    H2student wrote: »
    When they say prove it's collinear, I somehow got r=2c, and I said as R can be written in terms of C, they're collinear. Not sure if it's right though.

    well you got the right idea but i think you had to show that vector OC was the same as vector CR! r=2c but technically that can be true when r is parallel to c! but i'd say you got most of the marks....


  • Closed Accounts Posts: 16 mind_master


    H2student wrote: »
    When they say prove it's collinear, I somehow got r=2c, and I said as R can be written in terms of C, they're collinear. Not sure if it's right though.


    Thats what i did. But i also said OC is the midpoint of OR therefore they are collinear which is the same thing really. Thats the only part of the exam i'm a bit iffy about.


  • Closed Accounts Posts: 16 mind_master


    UglyFuc wrote: »
    i thought the paper was grand for the mostpart

    1a) (x-3)^2+(x+4)^2 = 17

    bi) (4,5) rad +=3
    bii) distance from center to the line should be 3. use the perp. distance got 17 so put down +17 and -17

    c) g=-6 f=-2 c=-28

    2a) -i-3j

    bi) modulas v is root (1 + k). u.v was -2+k
    bii) ****er wouldnt work out for me ended up with k = root 5 and k = -2

    c) attempted it, got the same answer for part i and ii, so one of them could be right

    3a)a was 8 i think

    bi) (0,k/5) and (-k/4,0)
    bii)k was +or- 20

    ci) equation was x(m-1)+y(m+1) + c
    cii) got m to be -1

    4a) 30,150
    4b 0, 120, 180, 300

    ci) was handy. since the height of the 3 smaller triangles is the radius, half the base by the radius (1/2 ar) = the area of each add the 3 together and factorise
    cii) if its right angled, pythagorus theroum will come out as 0=0. so get a^2 = b^2 +c^2 and multiply them out
    iii) no idea

    5a) tan2x formula. and sub in tan x wherever it goes
    ci) square them out, change them around with formula from tables

    8a let u equal to logex, let dv=dx
    dudx= 1/x v=x

    The bolded are answers i got different

    Q1 (c) i got c=+20

    Q2 (b)(i) modulus is root(1+k^2)
    (ii) k=(1/3) or k=-3


    Q3(c)(ii) You get m= + or - 1, but -1 cannot be the answer as the slope of f(L) is [-(m-1)/(m+1)] therefore using m=-1 would give you 0 on the bottom line and you cannot divide by zero. Therefore the only answer is plus 1. I'm sure alot of people got tricked by this. In the question it says value(s) with the s in brackets meaning there isn't necessarily 2 solutions (well thats what i assumed it to mean). In any case m=-1 doesn't work.


    Q4 (b) i've posted my answer


  • Closed Accounts Posts: 177 ✭✭chaoticmess


    H2student wrote: »
    When they say prove it's collinear, I somehow got r=2c, and I said as R can be written in terms of C, they're collinear. Not sure if it's right though.

    Yeah I did the same thing as you.... but then realised that C could be written in terms of A and B, and they weren't collinear so that didn't really prove anything..... and then I just gave up! ;)


  • Closed Accounts Posts: 7 Eanna B


    Q3(c)(ii) You get m= + or - 1, but -1 cannot be the answer as the slope of f(L) is [-(m-1)/(m+1)] therefore using m=-1 would give you 0 on the bottom line and you cannot divide by zero. Therefore the only answer is plus 1. I'm sure alot of people got tricked by this. In the question it says value(s) with the s in brackets meaning there isn't necessarily 2 solutions (well thats what i assumed it to mean). In any case m=-1 doesn't work.

    a slope with with zero on the bottom only means it's a vertical line... the other slope turned out to be 1 then(ie. 45deg) ... i think zero worked aswell because i think the formula is +/- Tan(theta) and something cancelled....


  • Closed Accounts Posts: 1 julius21


    i terribly regret not having attempted trig as 4,5 seemed to be more straightforwards than 6.

    6) a: 20/120
    b: proof
    c: i thought this was similar to a birthday problem and i got all six together = 1/11^5. Then tried to see what ways could two persons sit together getting something like 154931/161051.... horrible attempt

    7) a: 26x36x36x36 which is greater than 1 000 000
    b: i) 7/20
    ii) 1/40
    iii) 5/24 (not reliable)

    c: i) mean= 3a SD= (rt2)(a)
    ii)mean = 3a+5 SD= 3(rt2)(a)

    8) a: (x)lnx - x + C
    b: i) (2p)(9-p^2)
    ii) area = 12(rt3) p = rt3
    c: i) didn't take them down
    iii) 331/1440

    could anyone tell me if they disagree or agree? Still puzzled about 6c...


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  • Registered Users, Registered Users 2 Posts: 814 ✭✭✭JerCotter7


    H2student wrote: »
    When they say prove it's collinear, I somehow got r=2c, and I said as R can be written in terms of C, they're collinear. Not sure if it's right though.

    That's the right answer. r=2c means they are on the same line.


  • Closed Accounts Posts: 177 ✭✭chaoticmess



    (c) (i) If you draw two more radii to where the circle touches the triangle, you will see the big triangle split up into 3 smaller ones. One with side a, one with b, and one with c. Each of these has a height of r. So to get the area use (1/2)(base)(height) and simplify which gives you the answer.

    (iii) For this you draw a right-angled triangle with a circle inside and use the method from (i) to get the area. Then get the area using (1/2)(base)(height) and equal them. You get (r=pq-q^2) and as you are told in part (ii) that p,q are natural numbers this means that (pq-q^2) while be a whole number..

    haha, I don't think I'd have worked those out in the exam somehow..! :(

    (b) (i) Get Cos A and Cos C seperately and add
    (ii) use cos(A+B) rule from log tables. Find sin A by using cos A and drawing a right angles triangle to find sin A. Same for Sin C. Sub in answers.

    Ah, I just couldn't seem to find sinA and sinC... but I did the rest and wrote down the cos(A+B) part so hopefully will get a few attempt marks.....

    Q6 (c) Note: did not do this in exam, possibility of being wrong here.

    All six seated together
    > 6!x6! = 518400
    No two seated together
    > 6! = 720

    (518400/720) = 720
    (very possibly wrong, not sure)

    Hmm, I had 6!x6.... not sure how I worked that one out though.....

    Thanks for posting up your answers, I got quite a few the same as those so hopefully they were right! :)
    Oh and it was 2c (iii) not (ii)... oops sorry haha! :) But that one seems to be under debate from a couple others too!


  • Closed Accounts Posts: 13,224 ✭✭✭✭SantryRed


    Q6 (c) I did arrangements.

    The six people sitting beside each other was 6!6!.

    And then not sitting beside each other was 6x5x5x4x4x3x3x2x2x1x1.

    I put the answers over each other and got 6 times more likely. It worked out so perfectly I think it has to be right :D


  • Registered Users, Registered Users 2 Posts: 100 ✭✭blacklionboy


    for the 4 c iii) i found the are of the triangle on two ways. Area= 0.5r(a+b+c) as proved earlier.
    and area=0.5(Sin90)(p^2-q^2)(2pq)
    equaled those and got r=p^2-pq or something like that which must be real numbers!!


  • Registered Users, Registered Users 2 Posts: 100 ✭✭blacklionboy


    What did people get for 7b)???
    I got
    i) 7/20
    ii)13/20
    iii)53/120


  • Closed Accounts Posts: 7 Eanna B


    Q6 (c) Note: did not do this in exam, possibility of being wrong here.

    All six seated together
    > 6!x6! = 518400
    No two seated together
    > 6! = 720

    (518400/720) = 720
    (very possibly wrong, not sure)

    that's pretty close yeah... i only coped yesterday how to do these properly...

    full seats=F, empty seats= _
    seated together:
    F F F F F F _ _ _ _ _ is one arrangement... the 6 people can be seated 6! ways

    the block of people can be in 6 positions

    _ F F F F F F _ _ _ _

    _ _ F F F F F F _ _ _ etc...

    so 6x6!


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  • Closed Accounts Posts: 13,224 ✭✭✭✭SantryRed


    Eanna B wrote: »
    that's pretty close yeah... i only coped yesterday how to do these properly...

    full seats=F, empty seats= _
    seated together:
    F F F F F F _ _ _ _ _ is one arrangement... the 6 people can be seated 6! ways

    the block of people can be in 6 positions

    _ F F F F F F _ _ _ _

    _ _ F F F F F F _ _ _ etc...

    so 6x6!

    But the people can change positions too!


  • Registered Users, Registered Users 2 Posts: 197 ✭✭pp_me


    What did everyone get for 1 c (ii)

    i got (x-3)Sqd + (y-1)Sqd =10..


  • Registered Users, Registered Users 2 Posts: 814 ✭✭✭JerCotter7


    pp_me wrote: »
    What did everyone get for 1 c (ii)

    i got (x-3)Sqd + (y-1)Sqd =10..

    It was x-6 and y-2


  • Closed Accounts Posts: 16 mind_master


    julius21 wrote: »
    i terribly regret not having attempted trig as 4,5 seemed to be more straightforwards than 6.

    6) a: 20/120
    b: proof
    c: i thought this was similar to a birthday problem and i got all six together = 1/11^5. Then tried to see what ways could two persons sit together getting something like 154931/161051.... horrible attempt

    7) a: 26x36x36x36 which is greater than 1 000 000
    b: i) 7/20
    ii) 1/40
    iii) 5/24 (not reliable)

    c: i) mean= 3a SD= (rt2)(a)
    ii)mean = 3a+5 SD= 3(rt2)(a)

    8) a: (x)lnx - x + C
    b: i) (2p)(9-p^2)
    ii) area = 12(rt3) p = rt3
    c: i) didn't take them down
    iii) 331/1440

    could anyone tell me if they disagree or agree? Still puzzled about 6c...

    I'm not going to comment on 6 or 7(b) as i'm truly unsure, but 7 (a) and (c) are correct.

    As is everything else.


  • Closed Accounts Posts: 7 Eanna B


    SantryRed wrote: »
    But the people can change positions too!

    the 6! part is the arrangement of people inside the block of people,
    the x6 part is the block of people moving...


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  • Closed Accounts Posts: 16 mind_master


    SantryRed wrote: »
    Q6 (c) I did arrangements.

    The six people sitting beside each other was 6!6!.

    And then not sitting beside each other was 6x5x5x4x4x3x3x2x2x1x1.

    I put the answers over each other and got 6 times more likely. It worked out so perfectly I think it has to be right :D


    That sounds pretty good actually, i checked with one of my friends and she got the same.


  • Closed Accounts Posts: 13,224 ✭✭✭✭SantryRed


    Eanna B wrote: »
    the 6! part is the arrangement of people inside the block of people,
    the x6 part is the block of people moving...

    Yes but is it not 6! x 6! ?

    Because the first 6 is for the block they move along. But the second 6! is for their different positions they can be in?

    What answer did you get?


  • Registered Users, Registered Users 2 Posts: 197 ✭✭pp_me


    JerCotter7 wrote: »
    It was x-6 and y-2
    Fcuk


  • Closed Accounts Posts: 24 karen-xxxxx


    Anyone have answers for Question five part c? ii?

    My circle was x^2 + y^2 -12x - 4y +20 = 0.

    Same as anyone? :P


  • Registered Users, Registered Users 2 Posts: 3,739 ✭✭✭johnmcdnl


    for the question 6c the one of the 11 people at the bar is it...

    xSxSxSxSxSxSx so no 2 of them are seated together.... ie only 1 possible combination.....

    while xxxxxxSSSSS, SxxxxxxSSSS, SSxxxxxxSSS, SSSxxxxxxSS, SSSSxxxxxxS, SSSSSxxxxxx is the combinations where there all together...

    ie 6 times.....

    x=person.. S=Seat


  • Closed Accounts Posts: 7 Eanna B


    SantryRed wrote: »
    Yes but is it not 6! x 6! ?

    Because the first 6 is for the block they move along. But the second 6! is for their different positions they can be in?

    What answer did you get?

    my final answer was 6.6! over 6! = 6...


  • Registered Users, Registered Users 2 Posts: 814 ✭✭✭JerCotter7


    pp_me wrote: »
    Fcuk

    I made the same mistake as you until I asked for graph paper and realised it couldn't be a tangent at that angle unless I moved the centre.


  • Closed Accounts Posts: 16 mind_master


    Eanna B wrote: »
    the 6! part is the arrangement of people inside the block of people,
    the x6 part is the block of people moving...

    It is 6!x6!

    example..

    The 6 people are ABCDEF

    So we can have
    ABCDEF_ _ _ _ _
    _ABCDEF_ _ _ _

    keep going and you get 6 possbilities.
    Put B first and keep going and you get 6
    same for C-F
    But A can be first with B and C swapped, so you can have the letters arranged in 6! possibilities.

    But they can also be arranged in 6! ways with respect to position on chairs.

    That makes alot more sense in my head.


  • Closed Accounts Posts: 16 mind_master


    Eanna B wrote: »
    the 6! part is the arrangement of people inside the block of people,
    the x6 part is the block of people moving...
    Anyone have answers for Question five part c? ii?

    My circle was x^2 + y^2 -12x - 4y +20 = 0.

    Same as anyone? :P

    Bingo :D

    I kept doing it wrong and realized my stupidity when the answer worked out so simply!


  • Registered Users, Registered Users 2 Posts: 100 ✭✭blacklionboy


    UglyFuc wrote: »
    i thought the paper was grand for the mostpart

    1a) (x-3)^2+(x+4)^2 = 17

    bi) (4,5) rad +=3
    bii) distance from center to the line should be 3. use the perp. distance got 17 so put down +17 and -17

    c) g=-6 f=-2 c=-28

    2a) -i-3j

    bi) modulas v is root (1 + k). u.v was -2+k
    bii) ****er wouldnt work out for me ended up with k = root 5 and k = -2

    c) attempted it, got the same answer for part i and ii, so one of them could be right

    3a)a was 8 i think

    bi) (0,k/5) and (-k/4,0)
    bii)k was +or- 20

    ci) equation was x(m-1)+y(m+1) + c
    cii) got m to be -1

    4a) 30,150
    4b 0, 120, 180, 300

    ci) was handy. since the height of the 3 smaller triangles is the radius, half the base by the radius (1/2 ar) = the area of each add the 3 together and factorise
    cii) if its right angled, pythagorus theroum will come out as 0=0. so get a^2 = b^2 +c^2 and multiply them out
    iii) no idea

    5a) tan2x formula. and sub in tan x wherever it goes
    ci) square them out, change them around with formula from tables

    8a let u equal to logex, let dv=dx
    dudx= 1/x v=x
    Anyone have answers for Question five part c? ii?

    My circle was x^2 + y^2 -12x - 4y +20 = 0.

    Same as anyone? :P
    I got x^2+y^2= 20 but im sure its wrong!


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