Maths HL Paper 2

chaoticmess · 14-06-2010 5:31pm

mind_master wrote: »

(c) (i) If you draw two more radii to where the circle touches the triangle, you will see the big triangle split up into 3 smaller ones. One with side a, one with b, and one with c. Each of these has a height of r. So to get the area use (1/2)(base)(height) and simplify which gives you the answer.

(iii) For this you draw a right-angled triangle with a circle inside and use the method from (i) to get the area. Then get the area using (1/2)(base)(height) and equal them. You get (r=pq-q^2) and as you are told in part (ii) that p,q are natural numbers this means that (pq-q^2) while be a whole number..

haha, I don't think I'd have worked those out in the exam somehow..!

mind_master wrote: »

(b) (i) Get Cos A and Cos C seperately and add
(ii) use cos(A+B) rule from log tables. Find sin A by using cos A and drawing a right angles triangle to find sin A. Same for Sin C. Sub in answers.

Ah, I just couldn't seem to find sinA and sinC... but I did the rest and wrote down the cos(A+B) part so hopefully will get a few attempt marks.....

mind_master wrote: »

Q6 (c) Note: did not do this in exam, possibility of being wrong here.

All six seated together
> 6!x6! = 518400
No two seated together
> 6! = 720

(518400/720) = 720
(very possibly wrong, not sure)

Hmm, I had 6!x6.... not sure how I worked that one out though.....

Thanks for posting up your answers, I got quite a few the same as those so hopefully they were right!

Oh and it was 2c (iii) not (ii)... oops sorry haha!

But that one seems to be under debate from a couple others too!

SantryRed · 14-06-2010 5:33pm

Q6 (c) I did arrangements.

The six people sitting beside each other was 6!6!.

And then not sitting beside each other was 6x5x5x4x4x3x3x2x2x1x1.

I put the answers over each other and got 6 times more likely. It worked out so perfectly I think it has to be right

blacklionboy · 14-06-2010 5:34pm

for the 4 c iii) i found the are of the triangle on two ways. Area= 0.5r(a+b+c) as proved earlier.
and area=0.5(Sin90)(p^2-q^2)(2pq)
equaled those and got r=p^2-pq or something like that which must be real numbers!!

blacklionboy · 14-06-2010 5:35pm

What did people get for 7b)???
I got
i) 7/20
ii)13/20
iii)53/120

Eanna B · 14-06-2010 5:36pm

mind_master wrote: »

Q6 (c) Note: did not do this in exam, possibility of being wrong here.

All six seated together
> 6!x6! = 518400
No two seated together
> 6! = 720

(518400/720) = 720
(very possibly wrong, not sure)

that's pretty close yeah... i only coped yesterday how to do these properly...

full seats=F, empty seats= _
seated together:
F F F F F F _ _ _ _ _ is one arrangement... the 6 people can be seated 6! ways

the block of people can be in 6 positions

_ F F F F F F _ _ _ _

_ _ F F F F F F _ _ _ etc...

so 6x6!

SantryRed · 14-06-2010 5:37pm

Eanna B wrote: »

that's pretty close yeah... i only coped yesterday how to do these properly...

full seats=F, empty seats= _
seated together:
F F F F F F _ _ _ _ _ is one arrangement... the 6 people can be seated 6! ways

the block of people can be in 6 positions

_ F F F F F F _ _ _ _

_ _ F F F F F F _ _ _ etc...

so 6x6!

But the people can change positions too!

pp_me · 14-06-2010 5:37pm

What did everyone get for 1 c (ii)

i got (x-3)Sqd + (y-1)Sqd =10..

JerCotter7 · 14-06-2010 5:38pm

pp_me wrote: »

What did everyone get for 1 c (ii)

i got (x-3)Sqd + (y-1)Sqd =10..

It was x-6 and y-2

mind_master · 14-06-2010 5:39pm

julius21 wrote: »

i terribly regret not having attempted trig as 4,5 seemed to be more straightforwards than 6.

6) a: 20/120
b: proof
c: i thought this was similar to a birthday problem and i got all six together = 1/11^5. Then tried to see what ways could two persons sit together getting something like 154931/161051.... horrible attempt

7) a: 26x36x36x36 which is greater than 1 000 000
b: i) 7/20
ii) 1/40
iii) 5/24 (not reliable)

c: i) mean= 3a SD= (rt2)(a)
ii)mean = 3a+5 SD= 3(rt2)(a)

8) a: (x)lnx - x + C
b: i) (2p)(9-p^2)
ii) area = 12(rt3) p = rt3
c: i) didn't take them down
iii) 331/1440

could anyone tell me if they disagree or agree? Still puzzled about 6c...

I'm not going to comment on 6 or 7(b) as i'm truly unsure, but 7 (a) and (c) are correct.

As is everything else.

Eanna B · 14-06-2010 5:41pm

SantryRed wrote: »

But the people can change positions too!

the 6! part is the arrangement of people inside the block of people,
the x6 part is the block of people moving...

mind_master · 14-06-2010 5:42pm

SantryRed wrote: »

Q6 (c) I did arrangements.

The six people sitting beside each other was 6!6!.

And then not sitting beside each other was 6x5x5x4x4x3x3x2x2x1x1.

I put the answers over each other and got 6 times more likely. It worked out so perfectly I think it has to be right

That sounds pretty good actually, i checked with one of my friends and she got the same.

SantryRed · 14-06-2010 5:42pm

Eanna B wrote: »

the 6! part is the arrangement of people inside the block of people,
the x6 part is the block of people moving...

Yes but is it not 6! x 6! ?

Because the first 6 is for the block they move along. But the second 6! is for their different positions they can be in?

What answer did you get?

pp_me · 14-06-2010 5:43pm

JerCotter7 wrote: »

It was x-6 and y-2

Fcuk

karen-xxxxx · 14-06-2010 5:44pm

Anyone have answers for Question five part c? ii?

My circle was x^2 + y^2 -12x - 4y +20 = 0.

Same as anyone? :P

johnmcdnl · 14-06-2010 5:45pm

for the question 6c the one of the 11 people at the bar is it...

xSxSxSxSxSxSx so no 2 of them are seated together.... ie only 1 possible combination.....

while xxxxxxSSSSS, SxxxxxxSSSS, SSxxxxxxSSS, SSSxxxxxxSS, SSSSxxxxxxS, SSSSSxxxxxx is the combinations where there all together...

ie 6 times.....

x=person.. S=Seat

Eanna B · 14-06-2010 5:45pm

SantryRed wrote: »

Yes but is it not 6! x 6! ?

Because the first 6 is for the block they move along. But the second 6! is for their different positions they can be in?

What answer did you get?

my final answer was 6.6! over 6! = 6...

JerCotter7 · 14-06-2010 5:46pm

pp_me wrote: »

Fcuk

I made the same mistake as you until I asked for graph paper and realised it couldn't be a tangent at that angle unless I moved the centre.

mind_master · 14-06-2010 5:46pm

Eanna B wrote: »

the 6! part is the arrangement of people inside the block of people,
the x6 part is the block of people moving...

It is 6!x6!

example..

The 6 people are ABCDEF

So we can have
ABCDEF_ _ _ _ _
_ABCDEF_ _ _ _

keep going and you get 6 possbilities.
Put B first and keep going and you get 6
same for C-F
But A can be first with B and C swapped, so you can have the letters arranged in 6! possibilities.

But they can also be arranged in 6! ways with respect to position on chairs.

That makes alot more sense in my head.

mind_master · 14-06-2010 5:49pm

Eanna B wrote: »

the 6! part is the arrangement of people inside the block of people,
the x6 part is the block of people moving...

karen-xxxxx wrote: »

Anyone have answers for Question five part c? ii?

My circle was x^2 + y^2 -12x - 4y +20 = 0.

Same as anyone? :P

Bingo

I kept doing it wrong and realized my stupidity when the answer worked out so simply!

blacklionboy · 14-06-2010 5:49pm

UglyFuc wrote: »

i thought the paper was grand for the mostpart

1a) (x-3)^2+(x+4)^2 = 17

bi) (4,5) rad +=3
bii) distance from center to the line should be 3. use the perp. distance got 17 so put down +17 and -17

c) g=-6 f=-2 c=-28

2a) -i-3j

bi) modulas v is root (1 + k). u.v was -2+k
bii) ****er wouldnt work out for me ended up with k = root 5 and k = -2

c) attempted it, got the same answer for part i and ii, so one of them could be right

3a)a was 8 i think

bi) (0,k/5) and (-k/4,0)
bii)k was +or- 20

ci) equation was x(m-1)+y(m+1) + c
cii) got m to be -1

4a) 30,150
4b 0, 120, 180, 300

ci) was handy. since the height of the 3 smaller triangles is the radius, half the base by the radius (1/2 ar) = the area of each add the 3 together and factorise
cii) if its right angled, pythagorus theroum will come out as 0=0. so get a^2 = b^2 +c^2 and multiply them out
iii) no idea

5a) tan2x formula. and sub in tan x wherever it goes
ci) square them out, change them around with formula from tables

8a let u equal to logex, let dv=dx
dudx= 1/x v=x

karen-xxxxx wrote: »

Anyone have answers for Question five part c? ii?

My circle was x^2 + y^2 -12x - 4y +20 = 0.

Same as anyone? :P

I got x^2+y^2= 20 but im sure its wrong!

Making It Bad · 14-06-2010 5:50pm

How did you do 4(c)iii? First two parts were easy enough but couldn't get that out...

EuropeanSon · 14-06-2010 5:50pm

Easiest Paper II ever, after a piss easy paper I. Can't believe Q2(c), it was like an (a) part. Seriously doubt I got less than 100% overall, absolutely ****ing delighted.

Irish was awful though.

gant0 · 14-06-2010 5:51pm

Very easy paper.....my minimum c3 is looking very likely now

wonderfulname · 14-06-2010 5:51pm

Lovely paper!

bii)k was +or- 20

hmm i think i got +/-1..

did anyone do question 9?

EuropeanSon · 14-06-2010 5:52pm

mind_master wrote: »

It is 6!x6!

example..

The 6 people are ABCDEF

So we can have
ABCDEF_ _ _ _ _
_ABCDEF_ _ _ _

keep going and you get 6 possbilities.
Put B first and keep going and you get 6
same for C-F
But A can be first with B and C swapped, so you can have the letters arranged in 6! possibilities.

But they can also be arranged in 6! ways with respect to position on chairs.

That makes alot more sense in my head.

Incorrect. It's 6.6!/6!=6.

You are counting the rearrangement twice.

mind_master · 14-06-2010 5:53pm

Eanna B wrote: »

a slope with with zero on the bottom only means it's a vertical line... the other slope turned out to be 1 then(ie. 45deg) ... i think zero worked aswell because i think the formula is +/- Tan(theta) and something cancelled....

Ok fair point but when you are subbing it into the tan(theta) formula, you also have situations where you have m+1 on the bottom line, in that case you definitely cannot divide by zero. I'm not sure about the value(s) thing anymore but i dont think m=-1 works. But thinking back, 0 might work.

blacklionboy · 14-06-2010 5:54pm

Making It Bad wrote: »

How did you do 4(c)iii? First two parts were easy enough but couldn't get that out...

area=0.5r(a+b+c)=0.5(Sin90)bc

use p and q things for a,b and c

write r in terms of pq simplify and it p^2-pq which is natural

JerCotter7 · 14-06-2010 5:55pm

EuropeanSon wrote: »

Easiest Paper II ever, after a piss easy paper I. Can't believe Q2(c), it was like an (a) part. Seriously doubt I got less than 100% overall, absolutely ****ing delighted.

Irish was awful though.

Yea question 2c was simple. I went back to it afterwards to make sure that I hadn't done the wrong thing because it seemed too easy.

mind_master · 14-06-2010 5:57pm

EuropeanSon wrote: »

Incorrect. It's 6.6!/6!=6.

You are counting the rearrangement twice.

Hmmm just did the question out on paper with some thought and i think it is 6.6!, ah well, i didn't do the question in any case. I'm still getting my A1

Reillyman · 14-06-2010 5:57pm

Yes the 6(c) Q is definitely 6.

I drew out diagrams, complete with stick men:p Made it much easier.

There are 11 seats;

O O O O O O O O O O O

There are 6 people;

:):):):):)

P(E) = Favourable Outcomes

Total Outcomes

There is only 1 way they can all be apart;

I O I O I O I O I O I

Total Outcomes = 11C6
= 462

P(E) = 1/462

And there's 6 ways they can all be beside each other;

I I I I I I I O O O O O
O I I I I I I O O O O
O O I I I I I I O O O
O O O I I I I I I O O
O O O O I I I I I I O
O O O O O I I I I I I

P(E) = 6/462

Therefore

(1/462) : (6/462)

6 times as likely!

Maths HL Paper 2

Comments