SOLUTIONS Maths HL Paper 1

Maybe_Memories

Solutions to 1,2,6,7,8. The rest coming tomorrow.

**Disclaimer: I'm not a maths teacher or in any way involved in the marking. I'm simply going into 3rd year maths at Trinity College, in my LC got A2 in maths and physics and A1 in Applied maths. There may possibly be mistakes in my solutions. I'm not a machine. If you spot something, please say so.
They will not appear on your screen and you will have to click a link to view/download them, so no one is forcing to to view them. You do so at your own discretion.

Maybe_Memories

And the rest.

DepoProvera

Maybe_Memories wrote: »

Solutions to 1,2,6,7,8. The rest coming tomorrow.

**Disclaimer: I'm not a maths teacher or in any way involved in the marking. I'm simply going into 3rd year maths at Trinity College, in my LC got A2 in maths and physics and A1 in Applied maths. There may possibly be mistakes in my solutions. I'm not a machine. If you spot something, please say so.
They will not appear on your screen and you will have to click a link to view/download them, so no one is forcing to to view them. You do so at your own discretion.

EDIT: DERP I'm stupid.

Maybe_Memories

DepoProvera wrote: »

Forgot to check answers in 2a) 1/2 isn't a possible solution

From what I can see it is. LHS is root 4 and RHS is -2 which is correct.

Wesc.

For 8 (a) the integral of cosx is +sinx I think

Raeone

Any chance ya want to do the same for the ordinary level??

lemonz

Maybe_Memories wrote: »

From what I can see it is. LHS is root 4 and RHS is -2 which is correct.

Thanks for these, really reassuring. Would you lose marks for getting the two solutions and then rejecting the 1/2 afterwards or would they penalize it? :rolleyes:

Mista

Hmm... for 1c(i), can you prove x = t/k is a root, therefore kx - t is a factor?

And in 2c(ii), can you sustitute in -x?

Maybe_Memories

Wesc. wrote: »

For 8 (a) the integral of cosx is +sinx I think

Yes it is indeed.

This also means in b the answer should be (pi-4)/16

Apologies!

DepoProvera

ahhahaha I'm a ****ing retard...! So sorry! Well I guess that question is out the window so

Dazaq

Could u do 2c(iii)by squaring both sides

Maybe_Memories

Raeone wrote: »

Any chance ya want to do the same for the ordinary level??

Yeah I'll do it over the weekend!

lemonz wrote: »

Thanks for these, really reassuring. Would you lose marks for getting the two solutions and then rejecting the 1/2 afterwards or would they penalize it? :rolleyes:

No idea, like I said, not a teacher/involved with marking.

Mista wrote: »

Hmm... for 1c(i), can you prove x = t/k is a root, therefore kx - t is a factor?

And in 2c(ii), can you sustitute in -x?

I'm afraid not, since in part i you don't know that the thing is equal to zero.

I'm not really sure about your second question tbh.

Wesc.

Awh man I feel bad now :P Y'know for your integration (c) (ii)? I think you might of made a tiny error in that in calculating the area under the curve, you accidentily omitted the "2" where: 2ka = 2kroot(k)... you just said krootk.


Thanks for these though much appreciated!

cocopopsxx

For 1 c (i) I said x=t/k then sub it in the equation and got 0=0. Would that be correct as well?

sarahmocks

Maybe_Memories wrote: »

Solutions to 1,2,6,7,8. The rest coming tomorrow.

**Disclaimer: I'm not a maths teacher or in any way involved in the marking. I'm simply going into 3rd year maths at Trinity College, in my LC got A2 in maths and physics and A1 in Applied maths. There may possibly be mistakes in my solutions. I'm not a machine. If you spot something, please say so.
They will not appear on your screen and you will have to click a link to view/download them, so no one is forcing to to view them. You do so at your own discretion.

For question 2b part two, i didnt multiply across by 5 at the very end, will i lose many marks?

leaveiton

Thank you, this is really appreciated! You're very kind to do this

Maybe_Memories

Dazaq wrote: »

Could u do 2c(iii)by squaring both sides

Yup!

emmamurphy233

There are definitely two mistakes there. One in the alpha/beta question and in 8b)ii you didn't multiply properly! I made the same mistake in the exam, good thing I noticed it! Thanks! I reckon I got 245 marks. Kind of disappointed but what can ya do!

Maybe_Memories

Mista wrote: »

Hmm... for 1c(i), can you prove x = t/k is a root, therefore kx - t is a factor?

Actually scratch what I last said if you show that x=t/k makes the whole thing 0 then yes you can say its a root and hence kx-t is a factor!

cocopopsxx wrote: »

For 1 c (i) I said x=t/k then sub it in the equation and got 0=0. Would that be correct as well?

See above!

sarahmocks wrote: »

For question 2b part two, i didnt multiply across by 5 at the very end, will i lose many marks?

You wont lose any, there both equally valid ways of writing it.

finality

Maybe_Memories wrote: »

I'm afraid not, since in part i you don't know that the thing is equal to zero.

But if it's a root it will equal zero when subbed in. I'd say that's a valid method, I did it in the exam and it gave 0, though I also did long division just in case.

edit: you beat me to it :P

Mista

Maybe_Memories wrote: »

Actually scratch what I last said if you show that x=t/k makes the whole thing 0 then yes you can say its a root and hence kx-t is a factor!

Yaay! Cheers

RedTexan

Where did the two come from in the last integration question after integrating?

cocopopsxx

Maybe_Memories wrote: »

Actually scratch what I last said if you show that x=t/k makes the whole thing 0 then yes you can say its a root and hence kx-t is a factor!



See above!



You wont lose any, there both equally valid ways of writing it.

Okay, yay, I was worried about it. Thank you.

And sorry bothering you again and but you know in 1 (b) (ii) where we had to find the cubic equation? I did it exactly the same way as you did and got the same answer but a few people in the other thread got a different answer so I was a bit worried but this will be an acceptable way ya?

Thanks a million for the solutions, appreciate your effort!

Maybe_Memories

RedTexan wrote: »

Where did the two come from in the last integration question after integrating?

The two tangents are symmetric so the area under both of them is equal to the 2(area under one of them).

cocopopsxx wrote: »

Okay, yay, I was worried about it. Thank you.

And sorry bothering you again and but you know in 1 (b) (ii) where we had to find the cubic equation? I did it exactly the same way as you did and got the same answer but a few people in the other thread got a different answer so I was a bit worried but this will be an acceptable way ya?

Thanks a million for the solutions, appreciate your effort!

I honestly can't think of any other possible way of doing it, but this way is definitely right.

cocopopsxx

Maybe_Memories wrote: »

I honestly can't think of any other possible way of doing it, but this way is definitely right.

Okay, thank you. (: Feeling much better after the solutions, waiting for Q5 now. Thanks a ton!

Dazaq

in the alpha and beta question,i do believe u made mistake wen getting the product of the 2 new roots ,the answer should be 4 not 6

RedTexan

Maybe_Memories wrote: »

The two tangents are symmetric so the area under both of them is equal to the 2(area under one of them).



I honestly can't think of any other possible way of doing it, but this way is definitely right.

In such a case should your limits not be a and zero and not a and minus a? I think that might be a mistake? You subtracted the 3k root k instead of 2k root k. I think the correct answer is 2k root k over 3

Dicksboro_man

in 1bii what i did for my new roots was:

(x+2)-1= x+1

(x-1) -1= x-2

(x-3)-1= x-4

Am i wrong yeah? :-/

Maybe_Memories

Dazaq wrote: »

in the alpha and beta question,i do believe u made mistake wen getting the product of the 2 new roots ,the answer should be 4 not 6

Yeah I know it's been mentioned already.

RedTexan wrote: »

In such a case should your limits not be a and zero and not a and minus a? I think that might be a mistake?

No, because the area under f(x) is still to be taken between -a and a, then you subtract the area under the tangents which is 2(area under one of them).

You had me scared for a minute.

RedTexan

No actually you dropped the 2 from infront of 2ka once you substituted for root k

emmamurphy233

I have a question about 8c)ii. I got the area by getting the curves area between a and -a, then subtracted the areas of the two equal right angled triangles that can be made if you draw a line down from the points of intersection to the x-axis. Is this a valid method?

Epsi

Solution for B part ii) question 2 is incorrect. 5 + 1/5 - 6/5 should be 4 , not 6.

Spattersonox

It's good to see what i was supposed to do but bad to see the FCUKING ridiculous mistakes i made :L

Maybe_Memories

RedTexan wrote: »

No actually you dropped the 2 from infront of 2ka once you substituted for root k

The area under the curve is 2/3a^3/2 + krootk
Under 2 tangents its 2krootk

So under curve - under tangents gives you 2/3a^3/2 - krootk

Epsi

my apologies , maybe memories ,I see its been pointed out already.

Maybe_Memories

emmamurphy233 wrote: »

I have a question about 8c)ii. I got the area by getting the curves area between a and -a, then subtracted the areas of the two equal right angled triangles that can be made if you draw a line down from the points of intersection to the x-axis. Is this a valid method?

Not fully sure what you mean?

RedTexan

Maybe_Memories wrote: »

The area under the curve is 2/3a^3/2 + krootk
Under 2 tangents its 2krootk

So under curve - under tangents gives you 2/3a^3/2 - krootk

the area under the curve is 2/3 k^3/2 + 2kroot(k) it's in your solution, you just dropped the two by accident

Osric

Maybe_Memories

In your solution to 8cii, you have the area under the curve to 2/3a^3+2ka.
Then from that you replace the a with rootK, but you're getting 2/3k^3/2+krootK. Where did the 2 (in 2ka) go? I think it should be 2krootK. Then, when you take from that the area of the two triangle (2krootK), you'll get 2/3k^3/2

Adolescenteen

emmamurphy233 wrote: »

I have a question about 8c)ii. I got the area by getting the curves area between a and -a, then subtracted the areas of the two equal right angled triangles that can be made if you draw a line down from the points of intersection to the x-axis. Is this a valid method?

I did the same as yourself. Had to rush it a bit because I did an extra question, but hopeful I got some marks for it.

Epsi

Don't know if it's been pointed out , in question 8 b ii) integrating -cos4x gives you -Sin4x/4. 1/4 x 1/2 should also be 1/8. Making the final answer ( pi - 2 )/16 .

amymak

Thanks a million for putting these up, it really reassures me to see them. I think I got all my methods right, just the odd slip or blunder in a couple of places.

Maybe_Memories

RedTexan wrote: »

the area under the curve is 2/3 k^3/2 + 2kroot(k) it's in your solution, you just dropped the two by accident

Osric wrote: »

Maybe_Memories

In your solution to 8cii, you have the area under the curve to 2/3a^3+2ka.
Then from that you replace the a with rootK, but you're getting 2/3k^3/2+krootK. Where did the 2 (in 2ka) go? I think it should be 2krootK. Then, when you take from that the area of the two triangle (2krootK), you'll get 2/3k^3/2

Sorry, I get what you mean, you're both correct. Sorry.

earwax_man

Oh my God, I'm so much happier now; a B in maths IS possible :') THANK YOU!

Maybe_Memories

As you can all see the methods are all correct, however there are a few slips/blunders. Nothing major, I'd say if I handed in these solutions I would've lost about 15 marks max. Still in A1 land.
I know it can be frustrating since you all want to know how well you did, but please bare in mind I'm a human, not a machine, so please excuse the little mistakes.

If you find any more though please say so!

Adolescenteen

Maybe_Memories wrote: »

As you can all see the methods are all correct, however there are a few slips/blunders. Nothing major, I'd say if I handed in these solutions I would've lost about 15 marks max. Still in A1 land.
I know it can be frustrating since you all want to know how well you did, but please bare in mind I'm a human, not a machine, so please excuse the little mistakes.

If you find any more though please say so!

By the way, thank you putting these up. It is very much appreciated.

homolumo

emmamurphy233 wrote: »

I have a question about 8c)ii. I got the area by getting the curves area between a and -a, then subtracted the areas of the two equal right angled triangles that can be made if you draw a line down from the points of intersection to the x-axis. Is this a valid method?

As far as I can see that is what the OP did too.

Tankosaur

lemonz wrote: »

Thanks for these, really reassuring. Would you lose marks for getting the two solutions and then rejecting the 1/2 afterwards or would they penalize it? :rolleyes:

Technically the root of 4 has to be 2.

Don't know why but it's only when you have x^2 =4
then x= -2 or x=2

finality

Tankosaur wrote: »

Technically the root of 4 has to be 2.

Don't know why but it's only when you have x^2 =4
then x= -2 or x=2

Fairly sure this is correct. Sucks, because I didn't cancel 1/2 in the exam

tacofries

Maybe_Memories wrote: »

Solutions to 1,2,6,7,8. The rest coming tomorrow.

**Disclaimer: I'm not a maths teacher or in any way involved in the marking. I'm simply going into 3rd year maths at Trinity College, in my LC got A2 in maths and physics and A1 in Applied maths. There may possibly be mistakes in my solutions. I'm not a machine. If you spot something, please say so.
They will not appear on your screen and you will have to click a link to view/download them, so no one is forcing to to view them. You do so at your own discretion.

*Q2. b. ii. -the 4th line should be equal to 4 not 6??
Is 6. a. ii. definitely right? I messed pretty stupid if it is! i thought "a" is always equal to 1

I know your in college, but what did you think of the standard of the paper!? it wasnt the worst but there was tricky parts all the same,. i say i got a b3 partly due to bad question picking

Chavways

I'm desperately trying to ignore this.Had a look at 1a and found out I got the a value wrong.I'm afraid to read anymore.

RedTexan

For question 6 (c) (ii) was f'(x)=8cos^2(2x)+8cos(2x)+4? And for getting this far and making an equation with y=cos(2x) but going no further how many marks do ye reckon I'd get? I know it's anybody's guess, but just for a small bit of peace of mind!