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# MAT Prep Thread - 2nd November 2016

1. (Original post by KloppOClock)
..
find the lengths of AR, BR, BP, CP, CQ and AQ first

your aim is to do big triangle minus the three smaller outer ones
2. (Original post by DylanJ42)
find the lengths of AR, BR, BP, CP, CQ and AQ first

your aim is to do big triangle minus the three smaller outer ones
ok thank you
3. (Original post by Moogle679)
Awesome thank you, I really wasn't sure how it all worked as you mentioned it's really not specific at all. Does it show anywhere rough marks for longer questions?
I haven't gone through many papers yet but in the 07 solutions it shows how many marks each part of the longer questions is worth:
4. Hi,
Not really too sure how to do it, I've looked at the solutions but I still don't quite understand it... Can someone please explain it step by step? thanks!
physicsmaths

Also on Question 5 - MAT 2012, I got everything right except the last question where they ask what is (x 8k, y 8k)...
I have no idea how you go from (X8, Y8) = (16,0) to (X8k, Y8k) = (16^k, 0)....
5. Okay I am confused by this part of the question. From what I have understood, it is saying to prove the numbers in the bottom two rows and columns are the same in an n*n square if you apply R then C or C then R once. So in order to show this, you would have to show this holds for n= 1,2,3 and n>3 right? So why is this explanation good enough?

6. (Original post by KloppOClock)
Okay I am confused by this part of the question. From what I have understood, it is saying to prove the numbers in the bottom two rows and columns are the same in an n*n square if you apply R then C or C then R once. So in order to show this, you would have to show this holds for n= 1,2,3 and n>3 right? So why is this explanation good enough?

I have found some of their explanations to not be the easiest to understand or as detailed as you'd possibly like.

I understood it as this: R and C both only ever affect the top 2 rows or the left 2 columns. So therefore as long as n > 2 the right and bottom n-2 columns aren't affected by either R or C whichever order they are performed in. It's a tricky one to explain imo.
7. (Original post by Bruhh)
I have found some of their explanations to not be the easiest to understand or as detailed as you'd possibly like.

I understood it as this: R and C both only ever affect the top 2 rows or the left 2 columns. So therefore as long as n > 2 the right and bottom n-2 columns aren't affected by either R or C whichever order they are performed in. It's a tricky one to explain imo.
i understand that for n>3 but for n=3, r then c and c then r change the top left corner of the final 2x2 square, so wouldnt you have to prove for n=3 as well as just stating that statement for n>3
8. (Original post by KloppOClock)
i understand that for n>3 but for n=3, r then c and c then r change the top left corner of the final 2x2 square, so wouldnt you have to prove for n=3 as well as just stating that statement for n>3
You don't need to prove anything for the question, it just says explain which is why its quite ambiguous as to how much detail is needed in the answer.

Yeah the top left corner being changed is fine? It just has to be the same either way round you apply R and C, which it would be. If you look at the n=3 example they give, both the bottom row and right column remain unchanged whether you do it the way they've shown - R and then C or the other way C and then R.
9. (Original post by KloppOClock)
i understand that for n>3 but for n=3, r then c and c then r change the top left corner of the final 2x2 square, so wouldnt you have to prove for n=3 as well as just stating that statement for n>3
n=3 isn't a special case. It can be treated by the general argument. The effect of C, whether done first or second, is to do nothing on the right n-2 columns. So RC and CR both have the effect of just R on those columns.
10. (Original post by Bruhh)
You don't need to prove anything for the question, it just says explain which is why its quite ambiguous as to how much detail is needed in the answer.

Yeah the top left corner being changed is fine? It just has to be the same either way round you apply R and C, which it would be. If you look at the n=3 example they give, both the bottom row and right column remain unchanged whether you do it the way they've shown - R and then C or the other way C and then R.
well yeah the bottom row and right column dont change, but it says the last two, not one. when n>3 thats easy to show as the affected rows and columns dont crossover when you apply r then c. but with n=3. after you apply r then c or c then r, the top left square of the remaining 2x2 square has been transformed twice, which surely you would then need to show that this transformation is the same regardless of order as well as just saying the initial statement?
11. (Original post by KloppOClock)
well yeah the bottom row and right column dont change, but it says the last two, not one. when n>3 thats easy to show as the affected rows and columns dont crossover when you apply r then c. but with n=3. after you apply r then c or c then r, the top left square of the remaining 2x2 square has been transformed twice, which surely you would then need to show that this transformation is the same regardless of order as well as just saying the initial statement?
Well it says 'the right n-2 columns' so when n is say 4 this will be the right 2 columns, but when n is 3 the n-2 columns is only the first column. I think it's just bad wording.
12. (Original post by KloppOClock)
this is what im confused about
Spoiler:
Show
R and C effect the top row and left most column, you're showing it the other way round?
13. (Original post by Bruhh)
R and C effect the top row and left most column, you're showing it the other way round?
i deleted the post but i think youre right about the wording being all the rows and columns apart from the first two, makes a lot more sense
14. (Original post by KloppOClock)
i deleted the post but i think youre right about the wording being all the rows and columns apart from the first two, makes a lot more sense
15. Can someone help me with 1E on 2014? I keep getting 81?
I got two solutions of the first derivative as x=0,x=60deg so when i sub in 60 deg it wworks out as 81? I am so confused where I've gone wrong.
16. (Original post by Mystery.)
Can someone help me with 1E on 2014? I keep getting 81?
I got two solutions of the first derivative as x=0,x=60deg so when i sub in 60 deg it wworks out as 81? I am so confused where I've gone wrong.
Have you tried manipulating the expression?
17. (Original post by Injective)
Have you tried manipulating the expression?
Yeah I understand you can do that but I don't get why taking the derivative gives you the wrong answer..
18. (Original post by Mystery.)
Yeah I understand you can do that but I don't get why taking the derivative gives you the wrong answer..
Yes, I get 36 for x=pi/3 - giving us the largest value.

Edit: show your working, as I don't know where you went wrong.
19. (Original post by Mystery.)
Yeah I understand you can do that but I don't get why taking the derivative gives you the wrong answer..
You're putting 60 degrees in wrong - specifically I think you're not squaring sqrt(3)/2 to 3/4.
20. (Original post by Mystery.)
Yeah I understand you can do that but I don't get why taking the derivative gives you the wrong answer..
Post your working, so we can see where you went wrong.

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