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?