Maths HL Paper 2 solutions

Maybe_Memories

PLEASE DO NOT POST UNTIL Qs 1,4,5,6,7,8 HAVE BEEN UPLOADED. THANKS!

Right, this is all I had time to do. The paper absolutely wrecked my head. Probably cause I havent done a paper 2 in about a year. I'll do 2,3 and 11 tomorrow morning. Also, my pencil broke so I had to use pen for a lot of it, and I hate using pen, it causes me to make mistakes.. So yeah, currently scanning the solutions.

Maybe_Memories

Question 1

Maybe_Memories

Question 4
C done the standard way and a fancy way i cooked up using integration.

210 is also a possible solution to b I'm told

Maybe_Memories

Question 5

Maybe_Memories

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.

Swifty_N wrote: »

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.

Maybe_Memories

Question 7

EDIT
Last part in B should be [(.5)^4](4) squared, whatever that is.
1/16

Maybe_Memories

Question 8

NOTE: the denominator in B should be 4 - pi, not 4 + pi


Now you can post and point out all my mistakes

LilMissCiara

Maybe_Memories wrote: »

Question 7

I think your Q7b is wrong..!

iii) The possibility of the ball landing on N. Would it not be 1/5? as there are 5 endings, one of which is N.

iv) going by above, if the possibility of P is 1/5, then 2 balls on P should be 1/5*1/5 = 1/25.

Sorry I just want to know how I did!

Dongl

For Q6 part C iv) what I did was thus.

52C5 (so number of ways to choose 5 from 52)

[Ace][Ace][Ace][Ace][One of 52-4(aces) cards]

So there are 48 different ways of picking all four aces and one other card.

48 divided by 52C5 = 0.000018468

Two significant figures = 0.000018

LilMissCiara

Dongl wrote: »

For Q6 part C iv) what I did was thus.

52C5 (so number of ways to choose 5 from 52)

[Ace][Ace][Ace][Ace][One of 52-4(aces) cards]

So there are 48 different ways of picking all four aces and one other card.

48 divided by 52C5 = 0.000018468

Two significant figures = 0.000018

Me too..! Except I left them in scientific notation (1.85X10^-5)

Dongl

LilMissCiara wrote: »

I think your Q7b is wrong..!

iii) The possibility of the ball landing on N. Would it not be 1/5? as there are 5 endings, one of which is N.

iv) going by above, if the possibility of P is 1/5, then 2 balls on P should be 1/5*1/5 = 1/25.

Sorry I just want to know how I did!

I got this wrong in the exam but I reckon I've solved it since.
There is not an equal chance of it reaching any of the end points.
It's easier to elaborate based on the ii part.

There are 3 ways to get to H (as we see in part i)
Similarly there are three ways to get to J.
There is however only one way to get to G and one way to get to K.

So there are 8 ways to get from A to H/G/J/K

So P(H) is 3/8 not 1/4

Extend it further for N and P. I stupidly just multiplied my P(N) by itself to account for the second marble, assuming since it was a part B it would just feed into itself.

EDIT:Actually come to think of it, that was probably the most deceptively hard probability question I have come across...

Maybe_Memories

P(N) = 3/8

Sorry for confusion

LilMissCiara

OH FOR FúCK SAKE! :O :O

Why oh why didn't I see that in the exam..?! :O

Dongl

LilMissCiara wrote: »

OH FOR FúCK SAKE! :O :O

Why oh why didn't I see that in the exam..?! :O

Welcome to my thoughts on question 7 for the last 30 minutes. It's only marginally comforting to have figured it out :S

Still There are some marks for method to be had. Long as you have the probability princepals in mind you'll get something.

what.to.do

Maybe_Memories wrote: »

Question 8

That damn + C ....
I forgot it somewhere on Friday, swore I'd never do it again and WHAT DID I DO??

Whoops.

electrictrad

Once again, many thanks. . .you're putting minds at ease all over this country. . .

Q4 B possibly has 210 degrees as a 4th answer. . .

also, the neat way of proving q1 part c might be to get measure of r1,r2, and distance to centres. . .if r2's a tangent, they form 2 right-angled triangles. . .proving that the right angled triangle with the distance between the centres as the hyp should be sufficient. . .

Also, I think on Q6c significant figures means from the first non-zero number you arrive at, eg 0.00050. . .odd way of asking it. . .

Open to correction on this also, but I believe the answer for q7 b(iv) is 1/16. . .you added instead of multiplied. . .

resistantdoor

Question 4 part b:
Sinx = -1/2
X = 330 or 210

Question 7 part b -iv:
They BOTH need to land at N.
Therefore P(both landing at N) = (1/4)*(1/4) = 1/16

KarlD93

How did you get the tangets in Q1c by differentiating???

red_red_wine

I did it the fancy integration way too, although I was mad with myself when I came out for not seeing the easier, far shorter and more time-productive way to do it :P

Maybe_Memories

The circle is just a function, differentiate to get the slope, you have the points so you can throw everything into the equation of a line

resistantdoor

Did anyone get .000089 for Q6 part c - IV?

Also, did anyone else get 1/2 for Q7 part b - II?

Maybe_Memories

resistantdoor wrote: »

Question 4 part b:
Sinx = -1/2
X = 330 or 210

Question 7 part b -iv:
They BOTH need to land at N.
Therefore P(both landing at N) = (1/4)*(1/4) = 1/16

But probability for N is wrong, should have been 3/8
P(both on N) should be 9/64 I THINK.

Dongl

I think your Q7 part B is wrong mate.

Okay so theres 2x4x6x8 ways to go from A to L/M/N/P/Q =384

And theres 2 x 4 x 2 paths from N to A = 64

64 divided by 384 is one sixth.

Paths from P 2x3x3x2 = 36

36/384 = 1/12

P both at P is 1/12 by //12 = 1/144

KarlD93

In Q6c(iii), I think it should be 4/52*4/51*4/50*4/49*1/48?

Maybe_Memories

Dongl wrote: »

I think your Q7 part B is wrong mate.

Okay so theres 2x4x6x8 ways to go from A to L/M/N/P/Q =384

And theres 2 x 4 x 2 paths from N to A = 64

64 divided by 384 is one sixth.

Paths from P 2x3x3x2 = 36

36/384 = 1/12

P both at P is 1/12 by //12 = 1/144

It should be [(.5)^4](6) squared, whatever that is

Maybe_Memories

9/64 ^^^^

resistantdoor

Maybe_Memories wrote: »

But probability for N is wrong, should have been 3/8
P(both on N) should be 9/64 I THINK.

Sh*te. I should have posted P(both at P) not P(both at N).
Maybe I'm correct now?

RHunce

Dongl wrote: »

I got this wrong in the exam but I reckon I've solved it since.
There is not an equal chance of it reaching any of the end points.
It's easier to elaborate based on the ii part.

There are 3 ways to get to H (as we see in part i)
Similarly there are three ways to get to J.
There is however only one way to get to G and one way to get to K.

So there are 8 ways to get from A to H/G/J/K

So P(H) is 3/8 not 1/4

Extend it further for N and P. I stupidly just multiplied my P(N) by itself to account for the second marble, assuming since it was a part B it would just feed into itself.

EDIT:Actually come to think of it, that was probably the most deceptively hard probability question I have come across...

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?

Maybe_Memories

KarlD93 wrote: »

In Q6c(iii), I think it should be 4/52*4/51*4/50*4/49*1/48?

But theres only 1 of each of the ace, 2, 3, 4 and 5 of diamonds?

So you only have a 1/(number of cards left in deck) chance of getting that particular card

Mr. Maths

Dongl wrote: »

I think your Q7 part B is wrong mate.

Okay so theres 2x4x6x8 ways to go from A to L/M/N/P/Q =384

And theres 2 x 4 x 2 paths from N to A = 64

64 divided by 384 is one sixth.

Paths from P 2x3x3x2 = 36

36/384 = 1/12

P both at P is 1/12 by //12 = 1/144

Wrongwrongwrong
You're assuming that the marble can go to any place on the level below, when in fact it can only go to one of the two below it. The answers posted are right.

Maybe_Memories

resistantdoor wrote: »

Sh*te. I should have posted P(both at P) not P(both at N).
Maybe I'm correct now?

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

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...