Maths HL Paper 2 solutions

LilMissCiara

I could be wrong but for the vectors (Q2bii)

Q = 5(P perpendicular)

Which means Q is 5 times the size of P and is perpendicular to it?

Which means the angle should be 90 degrees?

Which means Cos(90) = 0.

Justifies the answers of some people I know.... And Root0 is 0..

Maybe_Memories

Doing Q2 now

Maybe_Memories

Due to tech issues I cant post the actual solutions, but they will be up tonight.

Anyway, answers.

2. (a). s = 6, t = -1

(b). (i). OQ = 20i + 15j

(ii). |<OQP| means the angle between OQ and QP,
and you get cosA = 5/root(26)

(c). (i) AB = AD - (1/2)BC
AC = AD + (1/2)BC

sparagon

i got -5 over root 26..

rosskind

I got root5/26, I think.

This may have been asked already, but in 6c, what was 'two significant places' about?? Isn't that 0.01 at most, surely the answers were much smaller than that unless I misread the question??

Maybe_Memories

rosskind wrote: »

I got root5/26, I think.

This may have been asked already, but in 6c, what was 'two significant places' about?? Isn't that 0.01 at most, surely the answers were much smaller than that unless I misread the question??

That's exactly what I thought, which is why my scanned solutions for that question aren't fully finished.

It turns out significant figures are not the same as decimal places.

MedMan101

LilMissCiara wrote: »

I could be wrong but for the vectors (Q2bii)

Q = 5(P perpendicular)

Which means Q is 5 times the size of P and is perpendicular to it?

Which means the angle should be 90 degrees?

Which means Cos(90) = 0.

Justifies the answers of some people I know.... And Root0 is 0..

ya thats the angle for QOP. the question was looking for OQP. I was doing it the first way at the start until I realised the mistake. Sneaky enough by them!

Maybe_Memories

Btw, people, add me on FB for the laugh. Link in my profile on this.
Just tell me your boards username first :P

MathsManiac

Just to clarify / amalgamate what several posters have said: you don't have to find the points of intersection for 1(c).

(i) If the distance d from one centre to the other is less than the sum of the radii and greater than their difference, then the circles intersect at two points.

(ii) If the tangent to one circle at the point(s) of intersection passes through the centre of the other circle, then the two centres and the point of contact from a right-angled triangle, which you can check for with Pythagoras' theorem, using only the two radii and d.

Kedo93

Hey thanks for the solutions, Memories :P Keep up the good work! You think you could get the answers for Q3 Up anytime soon? Would be delighted to see them

dougery

Maybe_Memories wrote: »

Question 8

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


Now you can post and point out all my mistakes

Thanks for posting up the solutions. My head was wrecked trying to think of those answers. That's the thing about a maths problem. It will niggle away in your head for days until you get the solution. I have one in there at the moment. Could you give me the solution to the project maths paper 2 HL Question 7 (c) and Q8.

dougery

thanks

SignTheContract

Swifty_N wrote: »

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.

I tried doing this a different way and the answer does not match and I am wondering why...here's how i did it and anyone who can spot my error feel free to point it out.

(4/52 * 3/51 * 2/50 * 1/49 * 48/48) * 5! = 4.4E-4

So the probably of 1st ace = 4/52, then only 3 left so 3/51 etc and the probably of the 5th card doesn't matter and these can be arranged in 5! different ways. What mistake am I making?

SignTheContract

Anyone...please help!!! It's driving me mad!

dougery

Soz. No good to you. Didn't do that paper. Just having a look.

SignTheContract

SignTheContract wrote: »

I tried doing this a different way and the answer does not match and I am wondering why...here's how i did it and anyone who can spot my error feel free to point it out.

(4/52 * 3/51 * 2/50 * 1/49 * 48/48) * 5! = 4.4E-4

So the probably of 1st ace = 4/52, then only 3 left so 3/51 etc and the probably of the 5th card doesn't matter and these can be arranged in 5! different ways. What mistake am I making?

I still can't see why I am getting a different answer from the combinations method

lostatsea

The best way to do these probability questions is to work out the probability of a particular event and then multiply your answer by the number of arrangements.
You are asked for the probability that the 5 cards drawn contain 4 aces.
Work out p(Ace of Diamonds, Ace of Hearts, Ace of Clubs, Ace of Spades, Any card)
= 1/52 * 1/51 * 1/50 * 1/49 * 48/48 * 5!

Now apply your method:
p(Ace, Ace, Ace, Ace, Any card)
= 4/52 * 3/51 * 2/50 * 1/49 * 48/48 * ?

I now need to multiply by the number of arrangements.
Think MISSISSIPPI: The number of arrangements of 11 letters with 4S, 4I and 2P is 11!/(4!4!2!). Changing letters that are alike does not give you a new arrangement so you divide by the number of ways you can arrange letters that are alike.

Therefore the number of arrangements of 4 Aces and 1 other card is 5!/4!
p(Ace, Ace, Ace, Ace, Any card)
= 4/52 * 3/51 * 2/50 * 1/49 * 48/48 * 5!/4!
which gives you the same answer as above.

Kedo93

Guys full solutions are up for both papers here. They're done out very well imo.
http://www.studentxpress.ie/papers.htm