Maths HL Paper 2

EuropeanSon

Roro4Brit wrote: »

I did my LC in 2002 but I always like to give the Maths papers a stab every year and see how I get on! Have yet to fail one yet :P

I know, I know....weirdo or what. Anyhow this year I don't know anybody doing the LC - anyone any idea where I can get my hands on the papers online - preferably paper II?

Congrats to all and I hope the hard work paid off!

examinations.ie have an archive, they should be there.

EuropeanSon

Reillyman wrote: »

My last few lines went like this;

[latex]\frac{|3(4)+4(5)+k|}{\sqrt{3^2+4^2}} = 3[/latex]

[latex]|32+k|=15[/latex]

[latex]-32-k=15 [/latex]

[latex]k=-47[/latex]

or

[latex]32+k=15[/latex]

[latex]k=-17[/latex]

not sure if that's the correct metod of evaluating absolute values or not but that's what I did.

The answers are right, but surely you meant 32+k rather than 32k?

UglyFuc

Reillyman wrote: »

My last few lines went like this;

[latex]\frac{|3(4)+4(5)+k|}{\sqrt{3^2+4^2}} = 3[/latex]

[latex]|32+k|=15[/latex]

[latex]-32-k=15 [/latex]

[latex]k=-47[/latex]

or

[latex]32+k=15[/latex]

[latex]k=-17[/latex]

not sure if that's the correct metod of evaluating absolute values or not but that's what I did.

well thats the first time ive known the modulas on the perp. distance formula to be actually used. ragin! suppose il lose about 3 marks. method is right but an error

Reillyman

EuropeanSon wrote: »

The answers are right, but surely you meant 32+k rather than 32k?

Where?

EuropeanSon

Reillyman wrote: »

Where?

My mistake, it didn't show up correctly on my iTouch screen.

Trackmedown

EuropeanSon wrote: »

If it worked, it should be grand, unless it specifically stated to use trig. As for 6(c), that's the right answer.

u have just made my day!! woooo!!! the oxegen timetable has just ruined my day but i think thats for another thread... lol

student01

**Timbuk2** wrote: »

I agree though. You have to admit that studying maths is much more 'fun' than studying for other subjects. I'd much rather do 50 maths sums that have to learn off an essay (uhh writing and rewriting, regurgitating etc.)

FUN?! *sigh* I wish i had the that attitude towards maths!
Over the last 2years, i've probably spent more time at HL maths than all my other subjects put together!..And i still wont get an A or B
I also hate learning off essays..but thankfully i find languages easy so i just make up the essays on the day :P

**Timbuk2**

Reillyman wrote: »

My last few lines went like this;

[latex]\frac{|3(4)+4(5)+k|}{\sqrt{3^2+4^2}} = 3[/latex]

[latex]|32+k|=15[/latex]

[latex]-32-k=15 [/latex]

[latex]k=-47[/latex]

or

[latex]32+k=15[/latex]

[latex]k=-17[/latex]

not sure if that's the correct metod of evaluating absolute values or not but that's what I did.

I did the exact same, and got the same answers. If your ever in doubt about evaluation absolute value signs, by the way (it's not always obvious), just square both sides. You will get a quadratic, in k, which have solutions -17, -47.

student01 wrote: »

FUN?! *sigh* I wish i had the that attitude towards maths!
Over the last 2years, i've probably spent more time at HL maths than all my other subjects put together!..And i still wont get an A or B
I also hate learning off essays..but thankfully i find languages easy so i just make up the essays on the day :P

I envy you! I wish I found languages easy. It's a much more useful skill to have in real life than maths (Although I'd prefer maths anyday to languages, but that's because I'm better at maths than languages).

JerCotter7

**Timbuk2** wrote: »

I envy you! I wish I found languages easy. It's a much more useful skill to have in real life than maths (Although I'd prefer maths anyday to languages, but that's because I'm better at maths than languages).

Really? I would rather maths and logical thinking any day of the week. Suppose I just hate languages too much.

Pimptastic

Just out of interest... Are the people who do applied maths doing it in school or did you have to do it outside of school on your own?

**Timbuk2**

Pimptastic wrote: »

Just out of interest... Are the people who do applied maths doing it in school or did you have to do it outside of school on your own?

I do it outside school, but in a group with a grinds teacher. He runs Applied Maths classes - he's an excellent teacher. I think it would be quite difficult to do by yourself - it's an enjoyable but difficult subject.

Loucol

JerCotter7 wrote: »

Really? I would rather maths and logical thinking any day of the week. Suppose I just hate languages too much.

Yeah i agree much prefer being better at maths, much more useful problem solving and all that. But i also despise all languages esp english... so happy never have to write and english essay ever again in my life.

EuropeanSon

**Timbuk2** wrote: »

I did the exact same, and got the same answers. If your ever in doubt about evaluation absolute value signs, by the way (it's not always obvious), just square both sides. You will get a quadratic, in k, which have solutions -17, -47.



I envy you! I wish I found languages easy. It's a much more useful skill to have in real life than maths (Although I'd prefer maths anyday to languages, but that's because I'm better at maths than languages).

I could write quite a long, indignant answer to this post, but I won't. Suffice to say that without maths, most modern technology would not exist.

EuropeanSon

Pimptastic wrote: »

Just out of interest... Are the people who do applied maths doing it in school or did you have to do it outside of school on your own?

In school. We got 4 classes per week, we did it while others did LCVP.

Messi 10

Reillyman wrote: »

My last few lines went like this;

[latex]\frac{|3(4)+4(5)+k|}{\sqrt{3^2+4^2}} = 3[/latex]

[latex]|32+k|=15[/latex]

[latex]-32-k=15 [/latex]

[latex]k=-47[/latex]

or

[latex]32+k=15[/latex]

[latex]k=-17[/latex]

not sure if that's the correct metod of evaluating absolute values or not but that's what I did.

I got those answers aswell but I did it slightly differently. I also had

[latex]\frac{|3(4)+4(5)+k|}{\sqrt{3^2+4^2}} = 3[/latex]

But then I said the square root of 25 = +/-5

Then I cross multiplied and got k + 32 = +/-15

so K= +/-15 -32

So taking +15 gives -17 and taking -15 gives -47.

I presume both our methods are fine snce we got the correct answers.

timsnewbridge

1 (a) x^2 +y^2 -6x +8y +8=0
(b) (i) (4,5) r=3
(ii) -17,-47
(c) x^2 +y^2 -12x -4y +20=0
2 (a) -i-3j
(b)(i) u.v = k-2 , [v] = square root[k^2 + 1]
(ii) 1/3 , -3
(c) (i) c + 1/2 a
(ii) c - 1/2 a
(iii) worked out that r= 2c
3(a) 8
(b) (i) (0,k/5) (-k/4,0)
(ii) +/- 20
(c) (i) (m-1)x'+(m+1)y' +2c
(ii) +/- 1
(checked with friend, he said only one of these was valid) think it was +1
4(a) 30 or 150
(b) 0, 120, 240, 360
(c) (i)by gettin the area of OBA, OAC, OBC by using 1/2 (base)(height)
add together.
(ii) a^2 = b^2 + c^2
(iii) 5 mins left, didnt get it done
5 (a) use formula, works out
(b) (i) use cosine rule like they say, works out
(i) use formula, works out
(c) (i) square out, use formula, particualry sin^2a + cos^2a = 1, works out
(ii) use 2+2(cos)(A-B)
let A = 4x and B + x, put = to 2+2(s.root)(3)sin3x
works out.
(10, 70, 130, 190, 250,310)
6.(a) 5/33
(b) ya just write out proof
(c) didnt get this myself but my friend says six times.

7 (a) i got 1413720 initially and any other possibilities means its still > than 1 mill (i think this might be wrong though )
(b) (i) 7/20
(ii) 1/40
(iii) got this wrong, forgot what my friend said it was
(c) (i) 3a, a(s.root)2
(ii) 9a+5, 3a(s.root)2
8.(a) x(log(b,e)x -1) +c
(b) (i) 18p-2p^3
(ii) 12(s.root)3
(c) (i) ya.... im not writing these out there in tha book
(ii) ya worked out.
(ii) 331/1440

reckon i got an a1 in both papers in other news im ****ed for irish ord paper 2 (n) hope these answers help, there right unless stated!

Reillyman

Any chance you could write it out properly? Awful hard to read like that.

timsnewbridge

Reillyman wrote: »

Any chance you could write it out properly? Awful hard to read like that.

sorry lad that took like 5 mins and im not sure how to do it properly! what ya having touble with? (s.root) means square root of whatever is after it, x^3 means like x cubed, x^2 wud be x sqaured.

JamieByrne

Anyone else in the country do groups as der option?:D

Paper 2 was just as easy as paper 1 so i definitely got my A1!!!:D:D

Bigrob

timsnewbridge wrote: »

1 (a) x^2 +y^2 -6x +8y +8=0
(b) (i) (4,5) r=3
(ii) -17,-47
(c) x^2 +y^2 -12x -4y +20=0
2 (a) -i-3j
(b)(i) u.v = k-2 , [v] = square root[k^2 + 1]
(ii) 1/3 , -3
(c) (i) c + 1/2 a
(ii) c - 1/2 a
(iii) worked out that r= 2c
3(a) 8
(b) (i) (0,k/5) (-k/4,0)
(ii) +/- 20
(c) (i) (m-1)x'+(m+1)y' +2c
(ii) +/- 1
(checked with friend, he said only one of these was valid) think it was +1
4(a) 30 or 150
(b) 0, 120, 240, 360
(c) (i)by gettin the area of OBA, OAC, OBC by using 1/2 (base)(height)
add together.
(ii) a^2 = b^2 + c^2
(iii) 5 mins left, didnt get it done
5 (a) use formula, works out
(b) (i) use cosine rule like they say, works out
(i) use formula, works out
(c) (i) square out, use formula, particualry sin^2a + cos^2a = 1, works out
(ii) use 2+2(cos)(A-B)
let A = 4x and B + x, put = to 2+2(s.root)(3)sin3x
works out.
(10, 70, 130, 190, 250,310)
6.(a) 5/33
(b) ya just write out proof
(c) didnt get this myself but my friend says six times.

7 (a) i got 1413720 initially and any other possibilities means its still > than 1 mill (i think this might be wrong though )
(b) (i) 7/20
(ii) 1/40
(iii) got this wrong, forgot what my friend said it was
(c) (i) 3a, a(s.root)2
(ii) 9a+5, 3a(s.root)2
8.(a) x(log(b,e)x -1) +c
(b) (i) 18p-2p^3
(ii) 12(s.root)3
(c) (i) ya.... im not writing these out there in tha book
(ii) ya worked out.
(ii) 331/1440

reckon i got an a1 in both papers in other news im ****ed for irish ord paper 2 (n) hope these answers help, there right unless stated!

How much are you blundered for leaving out a solution in 5 c)ii) ?

foxpoint1

howd ya get 6A to be that? i got 1/6

EuropeanSon

You are right. Several of his answers are incorrect, 6(a) was 4/10 x 5/12 = 1/6

BL1993

Just to let you guys know, for q.4, part (ii), you had to first prove a was the longest side before substituting values for a^2=b^2+c^2 and for part (iii), you had to find r in terms of p and q and prove that it was not a fraction as the question stated that p and q were natural numbers, ie whole. That was done by letting the area you found in part (i) = 0.5bc as b and c are not the longest sides in the right angled triangle in part (ii). Sub in values for b, a and c and voila, you get a q(p-q) for r which is a whole number.

Making It Bad

Bigrob wrote: »

How much are you blundered for leaving out a solution in 5 c)ii) ?

3 marks in other years

BL1993 wrote: »

Just to let you guys know, for q.4, part (ii), you had to first prove a was the longest side before substituting values for a^2=b^2+c^2

Why, are you sure? If pythagoras' therom holds true does that not allow you to immediately conclude it's a right angled triangle?

BL1993

Making It Bad wrote: »

Why, are you sure? If pythagoras' therom holds true does that not allow you to immediately conclude it's a right angled triangle?

Because you cannot just assume a was the biggest side. :P Since when are you allowed to make assumptions in maths? (bar induction). I do not see any reason as to why this question would be an exception. It does not state anywhere in the question that a was the biggest side.

However, I have a feeling in my gut that the marking scheme will not require people to prove a was the biggest side. However, at the same time, I would not be surprised if it was required.

JerCotter7

BL1993 wrote: »

Because you cannot just assume a was the biggest side. :P Since when are you allowed to make assumptions in maths? (bar induction). I do not see any reason as to why this question would be an exception. It does not state anywhere in the question that a was the biggest side.

However, I have a feeling in my gut that the marking scheme will not require people to prove a was the biggest side. However, at the same time, I would not be surprised if it was required.

Yes but when you get the right answer that proves that a is the longest side so it then makes your assumption right.

Making It Bad

BL1993 wrote: »

Because you cannot just assume a was the biggest side. :P Since when are you allowed to make assumptions in maths? (bar induction). I do not see any reason as to why this question would be an exception. It does not state anywhere in the question that a was the biggest side.

However, I have a feeling in my gut that the marking scheme will not require people to prove a was the biggest side. However, at the same time, I would not be surprised if it was required.

Well trial and error is an acceptable way of solving certain questions, I actually tried c^2=b^2+a^2 before I realised that would never work. I can't see why they wouldn't give you full marks. By proving a^2=b^2+c^2 you show it's the biggest side anyway, as it proves it's the hypotenuse!

EuropeanSon

JerCotter7 wrote: »

Yes but when you get the right answer that proves that a is the longest side so it then makes your assumption right.

I'd tend to agree with BL1993. Unless you show or state that A is the longest side or hypotenuse explicitly I'd expect you'll lose 3 marks.

Making It Bad

EuropeanSon wrote: »

I'd tend to agree with BL1993. Unless you show or state that A is the longest side or hypotenuse explicitly I'd expect you'll lose 3 marks.

Well my final statement was something like "therefore Pythagoras' therom holds true for the triangle => the triangle is right angled", hopefully this is not ambiguous and I won't lose marks.

foxpoint1

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.