Maths HL Paper 2 solutions

RHunce

RHunce wrote: »

The way I (most foolishly) saw it was that to get to H or J on the 3rd line that you could only use 4 of the lines coming from the 2nd line to the 3rd line out of the 6 lines available. Not taking into account the other alternative options from part i) just making sure it was landing on H or J.

Seeing as only 2 lines go to H and 2 other lines go to J would that not mean out of the 6 lines available to go to the third line that 4 are being used to go to H or J therefore being 4/6 or 2/3?

anyone?

Maybe_Memories

RHunce wrote: »

anyone?

There's 3 ways of going to H and 3 to J

RHunce

I understand that but 2 options to J and H go through E so are they not only counted once?

There's 6 lines coming from the second row going to the third. 2 go to H and 2 go to J. 2 + 2 = 4 all over 6 = 2/3

HxGH

I'm boarderline failing/passing

Please God. I don't want to repeat!

Swifty_N

Maybe_Memories wrote: »

Question 6

NOTE: Part C, confused the crap out of me. Either I was doing it completely wrong or there was a misprint, because I dont see how you could get it to be 2 significant figures.

Please point out my mistake for a correct version.

Sorry to butt in but I was looking at the paper for nosiness sake.
I would do 6 part(c) using combinations.

(i) 5 diamonds so
13C5/52C5 = 1287/2598960 = 5.0E-4

(ii) all same suit so
5 diamonds OR 5 hearts OR 5 clubs OR 5 spades
(1287/2598960) times 4 = 5148/2598960 = 2.0E-3

(iii) ace, two, three, four and five of diamonds
Only one way to do this so
5C5/52C5 = 1/2598960 = 3.8E-7

(iv) four aces and any other card
(4C4 x 48C1)/52C5 = 48/2598960 = 1.8E-5

Pretty sure these are right. I have a degree in stats.
Hope they help.

zuluin

for (c)(i) can you say that the distace between centre of circles is less than the sum of the radii of the circles and so they intersect?

LilMissCiara

zuluin wrote: »

for (c)(i) can you say that the distace between centre of circles is less than the sum of the radii of the circles and so they intersect?

That shows they intersect but doesn't get the points of intersection!

PaddyP056

Maybe_Memories wrote: »

Question 6

NOTE: Part C, confused the crap out of me. Either I was doing it completely wrong or there was a misprint, because I dont see how you could get it to be 2 significant figures.

Please point out my mistake for a correct version.

u 4got for part (iii) to multiply by 5! as ace to 5 of diamonds can be arranged in that many ways

Maybe_Memories

zuluin wrote: »

for (c)(i) can you say that the distace between centre of circles is less than the sum of the radii of the circles and so they intersect?

It's fine for (i) as it never said to actually find the points.
You'd need to get them in (ii) though

zuluin

i know but for part (c)(i) it asked to only show that the circles intersect. for part(c)(ii) i went on to find point of intersections and then find tangents and all so will it be allright?

zuluin

ok thanks maybe memories

LilMissCiara

zuluin wrote: »

i know but for part (c)(i) it asked to only show that the circles intersect at 2 points. for part(c)(ii) i went on to find point of intersections and then find tangents and all so will it be allright?

Your way doesn't prove they intersect twice.

zuluin

if they intersect only once the distance between their centres would be equal to the sum of two radii. if they intersect twice the distace between their centres would be less than the sum of two radii. so i think it would prove that they intersect twice.

resistantdoor

Maybe_Memories wrote: »

There are 3 paths it can take to get to H and 3 to J,
so the probability of getting to H or J is 3(1/2)^3 + 3(1/2)^3
= 3/4

I was referring to 7 b IV
There are 4 possible paths to P
There are 16 total possible results.
Therefore P(one landing at P) = 1/4
Hence, P(both landing at P) = (1/4)*(1/4) = 1/16.
QED?

BrendaN_f

LilMissCiara wrote: »

Your way doesn't prove they intersect twice.

yes it does. a circle is always going to intersect twice, unless the distance between centres = the sum of radii

Maybe_Memories

resistantdoor wrote: »

I was referring to 7 b IV
There are 4 possible paths to P
There are 16 total possible results.
Therefore P(one landing at P) = 1/4
Hence, P(both landing at P) = (1/4)*(1/4) = 1/16.
QED?

Correct.
Where did I keep getting 6 from..? :pac:

JamJamJamJam

That's an impressive way to prove 5bii [(tan2a)/(tan2b)], but is it not just easier to consider (tan2a)/(tan2b) = 1 and then simplify away like mad and prove that a + b = 90?

resistantdoor

Maybe_Memories wrote: »

Correct.
Where did I keep getting 6 from..? :pac:

No idea. All I know is probability is a bitch but quite nice this year.
Thanks anyway for all the input. You're solutions put my mind at ease that maybe just maybe I salvaged that elusive A1

Maybe_Memories

JamJamJamJam wrote: »

That's an impressive way to prove 5bii [(tan2a)/(tan2b)], but is it not just easier to consider (tan2a)/(tan2b) = 1 and then simplify away like mad and prove that a + b = 90?

This was something I considered when proving it.
What you're asking is an extremely debatable topic.

The question says "given a + b = 90, show that tan2a/tan2b = -1".

You're taking the fact that tan2a/tan2b = -1 is given and trying to show
a + b = 90, when in fact its never said that tan2a/tan2b = -1

In research level mathematics, I doubt you'd be allowed to do this.
But at LC level, you'd get away with it I'd say.

JamJamJamJam

Maybe_Memories wrote: »

This was something I considered when proving it.
What you're asking is an extremely debatable topic.

The question says "given a + b = 90, show that tan2a/tan2b = -1".

You're taking the fact that tan2a/tan2b = -1 is given and trying to show
a + b = 90, when in fact its never said that tan2a/tan2b = -1

In research level mathematics, I doubt you'd be allowed to do this.
But at LC level, you'd get away with it I'd say.

I was thinking something like that might be a slight issue. I guess you could solve the equation to give a + b = 90 and then write every step going backwards until you reach (tan2a)/(tan2b)! Unfortunately, that's a bit unrealistic without being given that tan2a/tan2b is equal to -1...

Shanee.

HxGH wrote: »

I'm boarderline failing/passing

Please God. I don't want to repeat!

Judging by some of these answers i failed..mother of god why didnt i do pass

Danieljl

For 5b(ii) I took a+b=90 and rearranged it to be a=90-b.
Then filled this into tan2a/tan2b=-1.
then it turned into tan2(90-b) on the top and thus tan(180-2b) which is one of those "well behaved angles" but would be in the sin quadrant so it became -tan2b and this cancelled with the bottom tan2b to give -1. Reckon thats an acceptable way to do it?

Maybe_Memories

Danieljl wrote: »

For 5b(ii) I took a+b=90 and rearranged it to be a=90-b.
Then filled this into tan2a/tan2b=-1.
then it turned into tan2(90-b) on the top and thus tan(180-2b) which is one of those "well behaved angles" but would be in the sin quadrant so it became -tan2b and this cancelled with the bottom tan2b to give -1. Reckon thats an acceptable way to do it?

Perfect

River Song

Danieljl wrote: »

For 5b(ii) I took a+b=90 and rearranged it to be a=90-b.
Then filled this into tan2a/tan2b=-1.
then it turned into tan2(90-b) on the top and thus tan(180-2b) which is one of those "well behaved angles" but would be in the sin quadrant so it became -tan2b and this cancelled with the bottom tan2b to give -1. Reckon thats an acceptable way to do it?

I hope so because that's exactly what I did, bar I split the tan(180-2b) into tan180-tan2b/1+tan180tanb.

Kedo93

BrendaN_f wrote: »

yes it does. a circle is always going to intersect twice, unless the distance between centres = the sum of radii

Well what if one circle was inside the other? In that case the sum of the radii would be larger than the distance between their centres and still they wouldn't technically be intersecting.

MedMan101

anyone remember what they got for Q2 (b)(ii)? it was the q about finding the angle between the two lines. Its killing me not knowing the answer.

Maybe could you do out the solution Maybe Memories in your own time for q2?

LilMissCiara

MedMan101 wrote: »

anyone remember what they got for Q2 (b)(ii)? it was the q about finding the angle between the two lines. Its killing me not knowing the answer.

Maybe could you do out the solution Maybe Memories in your own time for q2?

I got 1. So did another girl in my year (10 of us do higher) but another 2 got 0...

To put 1 into surd form I just wrote Root3/Root3 :P

River Song

LilMissCiara wrote: »

I got 1. So did another girl in my year (10 of us do higher) but another 2 got 0...

To put 1 into surd form I just wrote Root3/Root3 :P

:P Good thinking, I actually giggled at that :P

I got a "Math Error" ><

jumpguy

MedMan101 wrote: »

anyone remember what they got for Q2 (b)(ii)? it was the q about finding the angle between the two lines. Its killing me not knowing the answer.

Maybe could you do out the solution Maybe Memories in your own time for q2?

I got a strange surd as did my friend (but it was a surd - thank feck). Was something like 13root3/28 (note - that's definitely wrong, I can't remember what I got at all). You had to be careful and make sure both vectors were going toward/away from the point.

MedMan101

ya i got cosQ = -1/root26. anybody else get this?