SOLUTIONS Maths HL Paper 1

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!

Brokenghost

Hi guys - here's a question. In question 2, section b part 2 we are asked to create a new quadratic with the new roots a + 1/a and b +a/b...

The original quadratic was X squared - 2x + 5 = meaning that the original roots were 1+2i and 1-2i. Taking these figures, if we then calculate a + 1/a and b +1/b we get 2+2.5i and 2-2.5i (I think!). Taking these and using the formula X squared - x(sum of roots) + (product of roots) results in the quadratic X squared - 4x + 3/2 - is this an acceptable answer?

Cheers

sarahmocks

You wont lose any, there both equally valid ways of writing it. [/QUOTE]

Thanks! was annoyed that i'd forgotten something that simple!

thepikminman

Thanks you so much!

Tankosaur

There's an error in your solution for 2 ( b) part (ii)

you've written 5 + 1/5 -6/5 which is correct but the next lines is wrong.
You say 5 + 1 when it should be 5 - 1.

The answer the the quadratic is 20 ( 5 x 4) not 30 ( 5 x 6)

K_1

Tankosaur wrote: »

There's an error in your solution for 2 ( b) part (ii)

you've written 5 + 1/5 -6/5 which is correct but the next lines is wrong.
You say 5 + 1 when it should be 5 - 1.

The answer the the quadratic is 20 ( 5 x 4) not 30 ( 5 x 6)

I thought that too.


Another thing I noticed, in q1, c ii, you said t<0, but does t/= -1 as well as that would result in 2 of the roots being the same, and they are distinct?

I said t<0 and t /= -1, if I'm wrong will I lose much?!

Osric

K_1 wrote: »

I thought that too.


Another thing I noticed, in q1, c ii, you said t<0, but does t/= -1 as well as that would result in 2 of the roots being the same, and they are distinct?

I said t<0 and t /= -1, if I'm wrong will I lose much?!

No, I'm pretty sure you're correct - there's an exception to the value of t
Even if you're wrong (unlikely), it'll be -1 mark, for a slip

Didn't notice the exception during the exam, though

Tankosaur

K_1 wrote: »

I thought that too.


Another thing I noticed, in q1, c ii, you said t<0, but does t/= -1 as well as that would result in 2 of the roots being the same, and they are distinct?

I said t<0 and t /= -1, if I'm wrong will I lose much?!

That's been pointed out to me before but I still can't see it.

I'm not saying it's wrong I just don't know why you reject -1 when x^2 = -t/k^2

Osric

Tankosaur wrote: »

That's been pointed out to me before but I still can't see it.

I'm not saying it's wrong I just don't know why you reject -1 when x^2 = -t/k^2

If t=-1, then for the given factor (kx-t), when you isolate x you get t/k as the root, so putting in -1 as t gives -1/k as root. For the other factor, isolating x gives root(-t/k^2). If t=-1 here too, then x= root(--1/k^2), which can be plus or minus 1/k. One of those, the -1/k is the same as the given root. But since you must have 3 distinct roots, you can't have that value of t, giving you 2 equal roots

K_1

Osric wrote: »

If t=-1, then for the given factor (kx-t), when you isolate x you get t/k as the root, so putting in -1 as t gives -1/k as root. For the other factor, isolating x gives root(-t/k^2). If t=-1 here too, then x= root(--1/k^2), which can be plus or minus 1/k. One of those, the -1/k is the same as the given root. But since you must have 3 distinct roots, you can't have that value of t, giving you 2 equal roots

Great. Only problem now is a made an absolutely retarded mistake in the b part of that q, so there goes my a1

legend44

For Question 2 C (i) and (ii), I proved that it was true for 1 and 2 and that it would be therefore true for any no. greater than 1 as 2 + any constant is going to be > 2.
Is this any use? No i guess