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# MAT Prep Thread - 2nd November 2016 watch

1. (Original post by m0.4444)
i got that one wrong. But i think its 2 - sqrt2
Ikr i made such a ****ing silly mistake. I'm kicking myself
2. Is that not the answer to B
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3. (Original post by m0.4444)
But how many inequalities where there with c and 0.
For the last bit of 2 I had no idea how to approach it i was goona do wat you did but i didnt have time so i came up with some like contradiction saying 2c-1=92 did anyone get anything like this. And when are the results published
4. (Original post by louisforrest)
Is that not the answer to B
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I got sqrt(2) - 1 I think but I can't be sure
5. That graph was a right?
6. Any predictions for boundaries?? easier or harder than last year??
7. What did people put for the penultimate multiple choice question (about ax+by) and why? I put "max(a, b)" but it was a slight guess - I know you could get max(a, b) [from x = 1 and y = 0 or vice versa] and I ruled 2 of the other options out but does anyone have any explanation?

(Original post by Amitbalter)
MCQs:
1) l^105
2) 2-root2
3)didn't do
4)2 for even 3 for odd
5) the one that had a positive value for y when X=1, and also passed through the origin
6) didn't do
7) 3
8) a<(6/5)^4
9) didn't do
10) b
On question 1, does anyone know where I went wrong:

It was a GP with a = 1, r = l, n = 15
Sum = (1-l^15)/(1-l) [which was one of the options]

For (5) I think you're describing (a) - I checked on Wolfram Alpha after I came out the exam and I think that was the right one.

For everything else (that you answered) I got the same, so that's good.

(Original post by DavidBick)
If you calculate them all carefully, they all come to 107. This rhymes with the fact that the function A^n(B^m(x))= (2^n)(3^n)x+(2^n*3^n - 1) which is used in the last question. Note that 107 = 108-1

The latter also shows that it is impossible to reach 214x+96 (or whatever it was) since for each compound function added, the constant is > 1/2 the x coefficient. Hence, since 96 < 1/2 of 214, it is impossible with any sum.
I agree with this. I don't remember explicitly calculating that function though - do you think that me stating without proof "the constant will always be one less than the coefficient of x" would lose me marks?

Also for the last part, I justified it with another reason as well, that 214 = 2 * 107 [which are prime], and 214 =/= 2^m * 3^n for any integer m,n, so there was no way of combining the functions to get 214x + c. Is this true?

(Original post by m0.4444)
What did people get for q6?

It was the dancers one about off-beat dances. if the sum is odd. If i remember for the very last parts i got something about there being 2 rings if n was even and 1 ring if n was odd?
So I ran out of time on part (vi) of this but for the rest:
(i) 0 - you can't have an off-beat dance with 4 people
(ii) 3? or 6? - if A is 1 then that forces the numbers of the other five people; I think the pattern was 100100 so you could have 100100, 001001 or 010010 [starting with A]. Might be getting confused with a different part though.
(iii) Explanation - has to be a multiple of 3 because we need 100 to be repeated an integer number of times.
(iv) n is a multiple of 3? If it was even, you get two smaller groups of n/2 holding hands as in the first parts of the question, so n has to be a multiple of 3. If n was odd, you get a longer chain that starts with A, goes to the end and loops back round to B and carries on. You need n to be a multiple of 3 either way so I think that was the answer.
(v) 8 ways (either 1 or 3 of them have a 1)
8. for that polynomial i got n is even
9. (Original post by 11234)
That graph was a right?
what did it look like
10. (Original post by RuairiMorrissey)
what did it look like
one that had positive value of y for x=1 and through origin and was symmetirc
11. (Original post by louisforrest)
Is that not the answer to B
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You worked out a but u need to find x
12. for the max value of ax + by it was sqrt(a^2+b^2).
=
Proof.
By cauchy schwarz (a2+b2)(x2+y2)>(ax+by)2 and x2+y2<1 so less than sqrt a2+b2
13. (Original post by ShatnersBassoon)
What did people put for the penultimate multiple choice question (about ax+by) and why? I put "max(a, b)" but it was a slight guess - I know you could get max(a, b) [from x = 1 and y = 0 or vice versa] and I ruled 2 of the other options out but does anyone have any explanation?
The inequality was x^2 + y^2 <= 1 which basically says (x, y) lies within the unit circle. To maximise (ax+by), it is simple to note that it is essentially 45 degrees in the unit circle i.e. x = y = root(2)/2.

Therefore the max you can get is root(a^2 + b^2)
14. I would say it was similar to last yhears paper. So maybe similar boundaries??
15. (Original post by ShatnersBassoon)

On question 1, does anyone know where I went wrong:

It was a GP with a = 1, r = l, n = 15
Sum = (1-l^15)/(1-l) [which was one of the options]

...

I agree with this. I don't remember explicitly calculating that function though - do you think that me stating without proof "the constant will always be one less than the coefficient of x" would lose me marks?

Also for the last part, I justified it with another reason as well, that 214 = 2 * 107 [which are prime], and 214 =/= 2^m * 3^n for any integer m,n, so there was no way of combining the functions to get 214x + c. Is this true?
For question 1, sadly it wasn't a GP... they asked for the product of the terms (nearly caught me).

The tricky thing with the last part of that question is it's not a single function, it's an arbitrary sum of them. So your first statement would be right for a solitary compound function of that form, but for a sum the difference between the x coeff. and constant will be equal to the number of functions added. You might still get a mark though!

Equally, the (semi) primality of 214 is irrelevant in light of it being a sum. (2^1)(3^0)+(2^0)(3^1) is a basic example which produces the prime 5.

I'm sure you'll still get marks - sorry don't want to be downer. You seem to have more of an intuition on it than others on this thread, so I imagine you're a high scorer overall.
16. (Original post by collmr)
The inequality was x^2 + y^2 <= 1 which basically says (x, y) lies within the unit circle. To maximise (ax+by), it is simple to note that it is essentially 45 degrees in the unit circle i.e. x = y = root(2)/2.

Therefore the max you can get is root(a^2 + b^2)
yea boi
17. (Original post by 11234)
one that had positive value of y for x=1 and through origin and was symmetirc
Was it not the squigily one?
18. (Original post by louisforrest)
Is that not the answer to B
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You solved for a, not x. x is 1 - a so answer is 2 - sqrt2
19. (Original post by RuairiMorrissey)
Was it not the squigily one?
nvm. got any predictions??
20. For Q4 the geometry one: I got (cot alpa, 1) and then alpha = pi/6. Then I got the radius of the big cirle is 9. Then something like 4sqrt3-11pi/6 cant remember the exacts

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