Maths HL Paper 2

Healium

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

MyHair

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

Healium

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

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

LadyGaga!

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

pp_me

Answers?

straight_As

Leaving cert maths has gone to the shitter.

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

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

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

G_R

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

Couldnt have been better

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.

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

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!?

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

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

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.

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.

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.

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?!

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

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

Let me know if you need answers.

chaoticmess

mind_master wrote: »

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)....

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.

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

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)

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....

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.

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

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!

Eanna B

mind_master wrote: »

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....

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...

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.