Maths HL Paper 2

EuropeanSon

Normally I'd say you would get the marks. If the paper was well answered, however, which most people seem to think it was, they will look for any way to take marks off, and that is one possible way.

Making It Bad

EuropeanSon wrote: »

Normally I'd say you would get the marks. If the paper was well answered, however, which most people seem to think it was, they will look for any way to take marks off, and that is one possible way.

This is true I guess.

OConallain

Q5 part3, get the area of the triangle (1/2)(B)(H) an then equate that to the area formula prove in part (i).

OConallain

tracker-man wrote: »

Question 8 anyone?

Area of the rectangle?

maximum are for p was rt3 (which i fekn forgot to prove by geting the 2nd derivitive) then subbing that back into the expression the area was 12rt3...or so i got!

OConallain

tracker-man wrote: »

Question 8 anyone?

Area of the rectangle?

JerCotter7 wrote: »

CB isn't A.

yes it is. vectors of same lenght distance and direct are the same therefor A = CB

BL1993

foxpoint1 wrote: »

i stated,if pythagaros therom is correct then the hypontenuse squared is equal to the others squared.
by looking at the values given it was pretty obvious which one was A.

Does not matter how obvious it is, you still have to prove it.

Crochur

Ok so I worked out my marks based off the answers given here and trying to guess at the blunders and I'm at 93% exactly, 558 marks. Most of the things i got wrong were part bs so I'm just wondering if everyone thought the test was easy and the marking scheme is going to be changed accordingly, do you reckon I could lose my A1?? I really hope I don't because I worked SO hard at maths brought that up from a C2 in the mocks

drumlover22

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.

You got it right, colinear means they are on the same line, so if one can be written in terms of the other, they are colinear

EuropeanSon

Crochur wrote: »

Ok so I worked out my marks based off the answers given here and trying to guess at the blunders and I'm at 93% exactly, 558 marks. Most of the things i got wrong were part bs so I'm just wondering if everyone thought the test was easy and the marking scheme is going to be changed accordingly, do you reckon I could lose my A1?? I really hope I don't because I worked SO hard at maths brought that up from a C2 in the mocks

Firstly, it depends which answers you were going off. Some of the people who posted full solutions had several answers wrong (I'm talking mainly about the paper 2 solutions). So you may have gotten more (or less) right than you thought.

Secondly, the general feeling seems to be that the paper was easy enough, which will mean a tough marking scheme. However, 93%, if that's accurate, gives a reasonable margin for error, of 18 marks overall between the two papers. Blunders are unlikely to be hugely exaggerated, so I'd say you'll probably get the A1, assuming the 93% figure is accurate.

**Timbuk2**

Some people have posted up awkward ways of proving that A was the longest side. The simplest way, as I see it, is examining the sides

[latex]a = p^2 + q^2, b=p^2-q^2, c=2pq[/latex]
We can instantly know that [latex]p^2-q^2[/latex] is not the longest side.

Therefore, we examine whether a is longer than c, or c is longer than a
Is a > c?
[latex]p^2 + q^2 > 2pq[/latex]
[latex]p^2 - 2pq + q^2 > 0[/latex]
[latex](p-q)^2 > 0[/latex] which is true

Garykelleher

Anybody not do Question 8 Section B? The advanced probability question couldn't have been easier.

JerCotter7

OConallain wrote: »

yes it is. vectors of same lenght distance and direct are the same therefor A = CB

The length is the same. So it's |A| = |CB|.

What your trying to say is going from the origin along CB is equal to A.

JerCotter7

**Timbuk2** wrote: »

Some people have posted up awkward ways of proving that A was the longest side. The simplest way, as I see it, is examining the sides

[latex]a = p^2 + q^2, b=p^2-q^2, c=2pq[/latex]
We can instantly know that [latex]p^2-q^2[/latex] is not the longest side.

Therefore, we examine whether a is longer than c, or c is longer than a
Is a > c?
[latex]p^2 + q^2 > 2pq[/latex]
[latex]p^2 - 2pq + q^2 > 0[/latex]
[latex](p-q)^2 > 0[/latex] which is true

But what if p or q are negative? Then it could be the longest side? Personally I just think doing the theorem and getting it right shows that it was the longest side. If that was all section c was then you might get a blunder for not showing. But since it was made up of 3 parts I doubt they will for that one part.

OConallain

JerCotter7 wrote: »

But what if p or q are negative? Then it could be the longest side? Personally I just think doing the theorem and getting it right shows that it was the longest side. If that was all section c was then you might get a blunder for not showing. But since it was made up of 3 parts I doubt they will for that one part.

its said in the question that p and q are natural. this could also have been proved using the cosine rule. cos of the angle will be equal to a fraction with zero as the denominator therefore the angle will be 90degrees

EuropeanSon

**Timbuk2** wrote: »

Some people have posted up awkward ways of proving that A was the longest side. The simplest way, as I see it, is examining the sides

[latex]a = p^2 + q^2, b=p^2-q^2, c=2pq[/latex]
We can instantly know that [latex]p^2-q^2[/latex] is not the longest side.

Therefore, we examine whether a is longer than c, or c is longer than a
Is a > c?
[latex]p^2 + q^2 > 2pq[/latex]
[latex]p^2 - 2pq + q^2 > 0[/latex]
[latex](p-q)^2 > 0[/latex] which is true

Correct. That's how I did it.

JamesJB

EuropeanSon wrote: »

Correct. That's how I did it.

Seconded. It was the last part I did and I had to do it in a rather messy way, but I got it. I just wrote something like p(p-q) or whatever the last part was has to be a whole number since they are both integers, should suffice.

EDIT: I'm referring to the next part, sorry. Misread post. Need sleep actually...

**Timbuk2**

OConallain wrote: »

its said in the question that p and q are natural. this could also have been proved using the cosine rule. cos of the angle will be equal to a fraction with zero as the denominator therefore the angle will be 90degrees

Oh I never thought of using the Cosine Rule - that's genius

I just used Pythagoras, and finished it off with a 'As Pythagoras Theroem holds for the three sides, it is a right-angled triangle with a hypoteneuse of side A' - I think that will be ok.

hunii07

I thought maths paper 2 was pretty ok I got most of it out I think.....I hated maths ppr1 though a complete nightmare for me..

JamesJB

**Timbuk2** wrote: »

Oh I never thought of using the Cosine Rule - that's genius

I just used Pythagoras, and finished it off with a 'As Pythagoras Theroem holds for the three sides, it is a right-angled triangle with a hypoteneuse of side A' - I think that will be ok.

I was similar, except i first said 'longest side = A' and proved it with the p^2 + q^2 -2pq part or w/e it was > 0 always, therefore greater than p^2 - q^2 and 2pq. Then find area 2 ways and put them equal to each other. You get like p(p - q) or something after cancelling, so yeah you get a whole number :cool:

irish_man

OConallain wrote: »

maximum are for p was rt3 (which i fekn forgot to prove by geting the 2nd derivitive) then subbing that back into the expression the area was 12rt3...or so i got!

normally you don't lose marks for not proving it

muffinz

Looking at all your answers depresses me, coz as far as i remember i didnt get any of them
I thought i did well in paper 2 and all!!

I reckon i got a c3 overall, i hope so i dont want a D, but thats usually what i get in maths

JamesJB

irish_man wrote: »

normally you don't lose marks for not proving it

I'd believe that but they could make the marking sheme harsh to account for so many people feeling that the exam was the easiest in the past 6-7 years... Hopefully not!

hunii07

blacklionboy wrote: »

I got x^2+y^2= 20 but im sure its wrong!

I got the same I think.....:P

OConallain

EuropeanSon wrote: »

Correct. That's how I did it.

but dose that actully prove its a right angles triangle? dat jus prove dat p^2 + q^2 is greater than 2pq?!

EuropeanSon

OConallain wrote: »

but dose that actully prove its a right angles triangle? dat jus prove dat p^2 + q^2 is greater than 2pq?!

Yeah. I did that first to prove p^2 + q^2 was the longest side (the hypotenuse). Then I proved it was right angled with Pythagorus' theorum.

**Timbuk2**

OConallain wrote: »

but dose that actully prove its a right angles triangle? dat jus prove dat p^2 + q^2 is greater than 2pq?!

No, that's just to prove that the side A is the hypoteneuse. You then use Pythagoras Theorem, or the Cosine Rule, to prove that it is a right angled triangle.

Edit: Beaten by EuropeanSon :P

Areq

MathsNerd31 wrote: »

I've attached full solutions to both papers.

where did you get them from? :eek:

EuropeanSon

MathsNerd31 wrote: »

I've attached full solutions to both papers.

I appear to have gotten 100%, or very close to it. Nice.

I'd question your answer to Paper 1 Q4.(c)(iii), however. I had loads of time, so I worked it out manually by adding each term, and got 27430 or something similar, my memory is hazy.

EuropeanSon

MathsNerd31 wrote: »

The maths don't lie. My answer is right.

Perhaps. I don't think so, though.

Are you a teacher? A student?

BL1993

MathsNerd31 wrote: »

The maths don't lie. My answer is right.

Well your answer for 7.b. part 3 is wrong. It is done using:

p(A and B but not C) + P(A and C but not + P(C and B but not A) + P(A and b and c) where A is the novel, B is the drama and C is the poem. To justify myself, how can the prob. to study atleast 2 be smaller than the prob to study 3?

I get 19/24 btw.