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MAT Prep Thread - 2nd November 2016

Also for the multiple choice with integrals and also the one with A and B being max or something can someone show how they got they're answers?

Also for the last multiple choice what was option A and option B and was it asking which one is false?

here's wot i put earlier about the integral question...

my thoughts for question 1H:
1) y = a - x^2
2) y = x^4 - a

set y=0 and find x-intercepts:
1) root(a), -root(a)
2) 4throot(a), -4throot(a)

integrate both between limits (x-intercepts):
1) [ax - x^3/3] = (a^1.5 - (a^1.5)/3) - (-a^1.5 - (-a^1.5)/3) = 2a^1.5 - 2a^1.5/3 = 4/3 * a^1.5 = 4a.root(a)/3
2) [x^5/5 - ax] = ((a^1.25)/5 - a^1.25) - ((-a^1.25)/5 - (-a^1.25)) = 2a^1.25/5 - 2a^1.25 = -8/5 * a^1.25 = -8a.fourthroot(a)/5

set the two areas equal to each other:
4/3 * a^1.5 = 8/5 * a^1.25
a^0.25 = 6/5 a = (6/5)^4
only one answer to do with (6/5)^4

not sure about question 1I about max(ax+by)...

Reply 1581

riemmanmath

9

Original post by 0x3bfc9a1

I am not so sure what you mean by (214-96)*6>214. For Q4 you didn't get alpha's simplest form, I don't know how you are supposed to get the area correct since you will need the angle to calculate the sector area.

Well i just used alpha=arctan sqrt(3)/3 instead of pi/6. In question 2, A^nB^m(x)=2^n3^mx+2^n3^m-1 so there are 214-96=118 sums (k=118) and each x coefficient is greater or equal than 2^13^1=6 because m_i and n_i were positive. But 6*118>214 so is not possible.

(edited 7 years ago)

Reply 1582

boombox111

1

Original post by riemmanmath

Well i just used alpha=arctan sqrt(3)/3 instead of pi/6. In question 2, A^nB^m(x)=2^n3^mx+2^n3^m-1 so there are 214-96=118 sums (k=118) and each x coefficient is greater or equal than 2^13^1=6 because m_i and n_i were positive. But 6*118>214 so is not possible.

214x - 96 was the sum of many (A^nB^m(x)), not just one.

Reply 1583

riemmanmath

9

Original post by boombox111

214x - 96 was the sum of many (A^nB^m(x)), not just one.

Precisely, what is your point? You have that all the A^nB^m(x) are lineal polynomial of the form ax+a-1 so there have to be 214-96 sums of A^nB^m(x) to get the polynomial 214x+96.

Reply 1584

boombox111

1

Original post by riemmanmath

Precisely, what is your point? You have that all the A^nB^m(x) are lineal polynomial of the form ax+a-1 so there have to be 214-96 sums of A^nB^m(x) to get the polynomial 214x+96.

so it's impossible.
yeah?!
i think we're arguing the same point.

Reply 1585

riemmanmath

9

Original post by boombox111

so it's impossible.
yeah?!
i think we're arguing the same point.

Yes it is impossible, but I just was saying that by a wide margin (i.e 214x+120 wouldnt be possible either)

Reply 1586

SM-

6

How many marks would be the last two parts of each of 2 3 and 5 as well as the part about explaining integral in question 3?

Reply 1587

Mystery.

21

How did people do Q4? I got the angle as pi/3??

Reply 1588

LaserRanger

2

Here I will like to explain some questions I found many ppl got so confused:
1A come on you are supposed to find the product not the sum
The MC with ax+by: Plug in a and b both equal to 1, then only option C will remain
The so complicated graph of (x-1)^2-cos(pix): Plug in (0,0) then D and E eliminated
Differentiate to find out that the graph has one local max only hence B and C eliminated
The one with X^2+1: Since there are only even powers we can let y=x^2 then it should be crystal clear
x(n) and pi(n): Note if pi(n)=1, there is only one prime factor hence this must be prime number hence A is wrong
Q2: (only about the 214x+92) that one was trivial using brute force since coefficient of x must be multiple of 6. Note that m and n cannot be 0
because the question says m and n are positive integers
Q3v) Note whole graph is symmetrical along x=alpha by the definition of bilateral functions.

Feel free to ask xp

Reply 1589

RuairiMorrissey

5

Original post by Mystery.

How did people do Q4? I got the angle as pi/3??

I don't really remember the question tbh, any idea when the paper will be put up?

Reply 1590

riemmanmath

9

Original post by LaserRanger

Here I will like to explain some questions I found many ppl got so confused:
1A come on you are supposed to find the product not the sum
The MC with ax+by: Plug in a and b both equal to 1, then only option C will remain
The so complicated graph of (x-1)^2-cos(pix): Plug in (0,0) then D and E eliminated
Differentiate to find out that the graph has one local max only hence B and C eliminated
The one with X^2+1: Since there are only even powers we can let y=x^2 then it should be crystal clear
x(n) and pi(n): Note if pi(n)=1, there is only one prime factor hence this must be prime number hence A is wrong
Q2: (only about the 214x+92) that one was trivial using brute force since coefficient of x must be multiple of 6. Note that m and n cannot be 0
because the question says m and n are positive integers
Q3v) Note whole graph is symmetrical along x=alpha by the definition of bilateral functions.

Feel free to ask xp

The question 2 I got very mad afterwards (although I solved it) because I inmediatly supposed 214 would be divisible by 6 but isn't...xD. I also feel that the problem should say m,n nonnegative to make it more interesting.

Reply 1591

LaserRanger

2

Original post by riemmanmath

The question 2 I got very mad afterwards (although I solved it) because I inmediatly supposed 214 would be divisible by 6 but isn't...xD. I also feel that the problem should say m,n nonnegative to make it more interesting.