Higher Level Maths Paper 2 Solutions

Calorimeterman

Seeing as both the other two threads have become polluted with ordinary level maths, I decided to post my higher level answers here. So, here goes:

1 a) x2+y2+6x-4y-4=0

(i) Very noice
(ii) k=6.5

c) (i) x2+y2-10x-2y+1=0
(ii) Also very noice

2 a) + or - 5

b) (i) t= -1
(ii) 135 degrees

c) (iii) k= 1/3 l=2/3

3 a) I dunno, didn't write it down, am sure I got it though...

b) (iii) a' 3,0
b' 17, -4
c' 24, -6
d' 8,5

C) Massively tipped to come up... I learnt it yesterday!!!

4 a) 63/65

b) noice

c) (sr)= 2 cosx

5 a) 24

b) (ii) 0,30,180

c) (i) Easy proof!
(i) Basic algebra

8 a) turned out to be 0, which is less than 1

B (ii) tricky I got 9root3 divided by 2 though

c) (i) and (ii) were fine... (iii) was tricky...

Please confirm my answers, I think I did really well, but I'm not sure...

Cokehead Mother

Calorimeterman wrote: »

b) (iii) a' 3,0
b' 17, -4
c' 24, -6
d' 8,5 I can't remember, but I think mine looked something like these

C) Massively tipped to come up... I learnt it yesterday!!!

4 a) 63/65 Same

b) noice

c) (sr)= 2 cosx same

5 a) 24 think i got the same

b) (ii) 0,30,180 same

c) (i) Easy proof!
(i) Basic algebra

8 a) turned out to be 0, which is less than 1 same

B (ii) tricky I got 9root3 divided by 2 thoughsame

c) (i) and (ii) were fine... (iii) was tricky...

Please confirm my answers, I think I did really well, but I'm not sure...

Hope yours are right.

Decerto

for vectors c part iii i got 2 and -4 anyone else?
also can anyone eexplain triangle thing in trig i got first part sr but i had no idea how to relate it to tan

Cokehead Mother

Decerto wrote: »

for vectors c part iii i got 2 and -4 anyone else?
also can anyone eexplain triangle thing in trig i got first part sr but i had no idea how to relate it to tan

Here's a crappy MS Paint solution...

mahamageehad

HL answers...well wot i got neways, so no guarantees!!

Q1- a: (x+3)²+(y-2)²=9
bii: k= 6.5
ci: x²+y²-10x-2y+1=0
cii: grr hadnt a clu, neva learnt JC theorems!!
Q2- a: +25 -25
bi: t=4
bii: 80degrees
ci: a + kac
cii: lbd-a-c
Q3- a: 3x+7y-9=0
biii: a'=(3,0), b'=(17,-4), c'=(24,-6), d'=(8,5)
Q5- a: 7.59 cm.....i no dis is totally wrong tho!!
bi: sin4xcos2x-cos4xsin2x
bii......totally 4got how 2 do it on da day!!
Q6- a: 66%
b: 1 over root3(2+root3)n + 2over root3(2- root3)n=o
ci: 30x over x³+15x²+74x+220
cii: x=7
Q7- ai: 126
aii: 70
bi: 1 over 270,725
bii: 6084 over 270,725
biii: 468 over 270,725
c: totally blanked
Q9- a: 256 over 625
bi: 5 over 36
bii: 91 over 216
biii: WTF??!
c: between 180 and 220, i used two tailed test...was it supposed 2 b 1 tailed???

Decerto

Cokehead Mother wrote: »

Here's a crappy MS Paint solution...

ffs i started out like tht and expanded sin a-b then just thought this is aint it:(

MnMs

Calorimeterman wrote: »

Seeing as both the other two threads have become polluted with ordinary level maths, I decided to post my higher level answers here. So, here goes:

1 a) x2+y2+6x-4y-4=0

(i) Very noice
(ii) k=6.5

c) (i) x2+y2-10x-2y+1=0
(ii) Also very noice

2 a) + or - 5

b) (i) t= -1
(ii) 135 degrees

c) (iii) k= 1/3 l=2/3

3 a) I dunno, didn't write it down, am sure I got it though...

b) (iii) a' 3,0
b' 17, -4
c' 24, -6
d' 8,5

C) Massively tipped to come up... I learnt it yesterday!!!

4 a) 63/65

b) noice

c) (sr)= 2 cosx

5 a) 24

b) (ii) 0,30,180 as far as i can remember there were way more answers for this question.

c) (i) Easy proof!
(i) Basic algebra

8 a) turned out to be 0, which is less than 1

B (ii) tricky I got 9root3 divided by 2 though

c) (i) and (ii) were fine... (iii) was tricky...

Please confirm my answers, I think I did really well, but I'm not sure...

.

mahamageehad

answers for Q2, 6,7,9 any1??

ian.f

Calorimeterman wrote: »

5b) (ii) 0,30,180

Ya left out 2 or 3 answers (can't remember) but those 2 are right so you should get most of the marks

jennyq

On that 5(b)(ii) one there's a couple more answers, as far as I can remember you had cos3x = 0 & then 3x = 90, 270, 450 then divide all by 3.

Really hoping that your 8(b)(ii) is right 'cos I got the same & after ages trying to work out part (i) :rolleyes:

Redisle

ian.f wrote: »

Ya left out 2 or 3 answers (can't remember) but those 2 are right so you should get most of the marks

Yeah I got 0,30,90,150,180

The rest of the answers in the op look to be the same as mine though. Hopefully we are both right

Btw I also got 9 root 3 over 2 for the 8(b) ii, so hopefully its right!

ShogunWarrior

Yay I was delighted to see 9 root(3)/2 that's what I got. I was sure that y = 3root(3) - root(3)x/2 could not be right but it did work out as x=3 and seems pretty good.

That would make question eight my best! (50/50 hopefully)

Decerto

wat were peoples answers for 2c(iii) i got like 2 and -4

Qbritgohm

wat were the answers for Q9?? i just chanced it...for the last part asking what Annes chances of winning a game with two ppl in it i just said 1 in 2?!

fivetwenty

More or less the same as the OP thankfully - somone agreeing with me is just what I needed - However for 2(b) I got 151 degrees or something? The Cos inverse one.

Nanaki

For question 8 (b) what did you do after tidying the derivitave?
Couldn't work out how to bring side of length 6 into it >_<

for 8 (c) it was a nice question, the trick was getting d/dx(cos^2x) using the chain rule.
The higher derivitaves were a bit tricky, but the hint on the paper made it a bit easier.
Was delighted with the maclaurin!
Rest of the paper was hit and miss, vectors were great, circle was maybe half marks.
The line was alright, thought the transformations were a bit long winded, but not difficult. Only proof I didn't get was perp. dist.

According to someone on the radio (I was told) 4(c)(ii) was a ridiculous question!

Intend on having a go at the papers again over the summer =P

ShogunWarrior

For 8 (c) (ii) what I did is considering you have cos(x) = 1 + (x^2)/2 + (x^3)/8 just square boths sides so:

cos(x) = 1 + (x^2)/2 + (x^3)/8
( cos(x) )^2 = ( 1 + (x^2)/2 + (x^3)/8 )^2
( cos(x) )^2 = ( 1 + (x^2)/2 + (x^3)/8 )( 1 + (x^2)/2 + (x^3)/8 )

Simplify and the first three non-zero terms are the ones given in the question, also part c (iii) gave the same answer with this series as the calculator.

As for the min/max part I worked with the small triangle in the bottom right. The base must be (6-x)/2 and the bottom-right angle is 60 degrees so use:

sin(60)=O/H

sin(60) = y/((6-x)/2)
sin(60) = 2y/6-x
etc.. simplify

Slashman X

I changed Cos^2x into 1/2(1+Cos2x) then had the actual yolk as:

1/2(1+1-4x^2/2....)

Worked Out,
got 0.9605 or something for iii, anyone confirm?

ShogunWarrior

Yup, 0.9605

Slashman X

Sweet, anyone else think Question 3 was a piece of urine

Cokehead Mother

Slashman X wrote: »

Sweet, anyone else think Question 3 was a piece of urine

It was awful. Same as Paper One's question 3. Just boring, easy (a) and (b) parts and a proof. I really do hate these sort of questions that seem as though about 5 minutes of thought was put into them.

Timee

the infinite geometric series of Q9 (b) (iii).. i believe the answer is 6/11

you start with: p(anne getting it 1st time) + p(anne getting it 2nd time) + p(anne getting it 3rd time) ....

and you soon get a value for a and r, and use the formula.
hope that helps.

Timee

As for Q1(c)(ii), i showed that [oa] and [ob] are perpendicular, (forming a 90 degree angle) by finding both their slopes. Then the whole 'angle at the centre twice the angle on the same arc' rule showed the 45degrees bit.

orangetictac

Subtended arc therom

Katerrrs

Timee wrote: »

As for Q1(c)(ii), i showed that [oa] and [ob] are perpendicular, (forming a 90 degree angle) by finding both their slopes. Then the whole 'angle at the centre twice the angle on the same arc' rule showed the 45degrees bit.

Yeah, that's exactly what i did

Sugarplum_fairy

Timee wrote: »

As for Q1(c)(ii), i showed that [oa] and [ob] are perpendicular, (forming a 90 degree angle) by finding both their slopes. Then the whole 'angle at the centre twice the angle on the same arc' rule showed the 45degrees bit.

i joined the radii to a and b, got the distance between them and used the consine rule to show the angle at the centre was 90degrees, think that's ok?
for 8 cii i ended up going back and doing out the cosx^2 from the start usuing macclaurine, took me ages, think it worked out tho, i always seemed to make things difficult for myself...

Timee

Yeah that sounds just as good. I hope theyre nice to 'alternative methods' for this paper because i went a different route for the Prove Tan 22.5 = route2 - 1, I Used the Tan2A formula (i missed the hence..but the fact they said 'or otherwise', it should be good enough).

Heres wondering if 2(c) in paper one will still be given 20marks after so few people got it out

lemur option

mahamageehad wrote: »

HL answers...well wot i got neways, so no guarantees!!

Q6- a: 66%
b: 1 over root3(2+root3)n + 2over root3(2- root3)n=o
ci: 30x over x³+15x²+74x+220
cii: x=7
Q7- ai: 126
aii: 70
bi: 1 over 270,725
bii: 6084 over 270,725
biii: 468 over 270,725
c: totally blanked
QUOTE]

hi. I'm a math grind and I was asked to do the probability answers out for my student. These are nearly ok but not quite. (I didn't do out the difference equation)

Q6 ci is 30x over (x+6)(x+5)(x+4)= x³+15x²+74x+120 (think that was a type-o on your part)

Q7
aii is 7c4= 35 (not 8C4=70) you take both french and german out of the subject list meaning there are only 7 subjects to choose from.

biii is 198/270725
what you did is 4C2 and 13C2. what you were suposed to spot was that if there are three clubs and two aces then one card must be the ace of clubs.
thus answer is 1.3(3C1 aces). 12C2= 198

if anyone actually wants the answer to c. reply i have it done out but it's long.

Challenged

The Paper 2 questions and answers are up on:

www.studentxpress.ie/papers.htm