The Student Room Group

Scroll to see replies

Original post by KloppOClock
lmao i got 45


Why? And what questions couldn't you complete?
Original post by Marxist
Why? And what questions couldn't you complete?


i didnt get question 3 and asked my teacher who said it was core 4 stuff.

question 4 i was sort of on the right track but ran out of time so scored lowly

and i sort of just failed the multiple choice, im going through it now though and getting a lot of the right solutions this time
Reply 82
Original post by KloppOClock
i didnt get question 3 and asked my teacher who said it was core 4 stuff.


That's untrue.
Reply 83
Original post by KloppOClock
i didnt get question 3 and asked my teacher who said it was core 4 stuff.


I agree with Zacken - look at the solution, there's no C4 knowledge needed.
Original post by KloppOClock
i didnt get question 3 and asked my teacher who said it was core 4 stuff.

question 4 i was sort of on the right track but ran out of time so scored lowly

and i sort of just failed the multiple choice, im going through it now though and getting a lot of the right solutions this time


Your teacher is wrong obviously. The MAT only requires you to have C1-C2 knowledge. Be careful of using C3-C4 knowledge on certain questions - you end up making things more complicated than they seem. You should be able to do all questions with what you learnt from C1-C2.
Hi folks, I'm preparing for the MAT right now, for Maths and Computer Science. I'm doing a paper a week right now, which should give enough time to check things inbetween.

I'm in Scotland, and C1/C2 doesn't match up exactly with Higher. I'm just learning a few things during Advanced Higher (like sums of series and the Binomial Thereom), but is there anything that I'm missing that's not in Higher or Advanced Higher? I'm aware of the trapezium rule, and most likely periodicity of functions (like sin(sqrt(x)) has the waves get further and further apart), but I think I might be missing a few other things.
Reply 86
Original post by lewman99
Hi folks, I'm preparing for the MAT right now, for Maths and Computer Science. I'm doing a paper a week right now, which should give enough time to check things inbetween.

I'm in Scotland, and C1/C2 doesn't match up exactly with Higher. I'm just learning a few things during Advanced Higher (like sums of series and the Binomial Thereom), but is there anything that I'm missing that's not in Higher or Advanced Higher? I'm aware of the trapezium rule, and most likely periodicity of functions (like sin(sqrt(x)) has the waves get further and further apart), but I think I might be missing a few other things.


You can just google the maths specification for a given exam board and compare it with yours.


Posted from TSR Mobile
Original post by lewman99
Hi folks, I'm preparing for the MAT right now, for Maths and Computer Science. I'm doing a paper a week right now, which should give enough time to check things inbetween.

I'm in Scotland, and C1/C2 doesn't match up exactly with Higher. I'm just learning a few things during Advanced Higher (like sums of series and the Binomial Thereom), but is there anything that I'm missing that's not in Higher or Advanced Higher? I'm aware of the trapezium rule, and most likely periodicity of functions (like sin(sqrt(x)) has the waves get further and further apart), but I think I might be missing a few other things.


I've attached the MAT specification (from the Oxford Maths Dept website).
Can I get some help on Q3 on spec paper A? (ii) onwards..
Original post by Hussain919
Can I get some help on Q3 on spec paper A? (ii) onwards..


okay at first glance this question seems to be about how the minimum of a function may not necessarily be at the x value where dydx=0 \displaystyle \frac{dy}{dx} = 0 due to restrictions in the domain. (you may not have learnt what a domain is yet, its basically just all the x values the function can take)

so for example, if I asked you for the minimum point of f(x)=x2+4f(x) = \displaystyle x^2 + 4 you would do dydx=0 \displaystyle \frac{dy}{dx} = 0 so 2x=0 \displaystyle 2x = 0 , therefore x=0 \displaystyle x=0 and f(0)=02+4=4 \displaystyle f(0) = 0^2 + 4 =4 so the minimum point is (0,4) \displaystyle (0,4)

However if I asked you for the minimum point of x2+4 \displaystyle x^2 + 4 but this time restricted the domain to, lets say, 2x6 \displaystyle 2 \leq x \leq 6 you couldnt just use dydx \displaystyle \frac{dy}{dx} since x=0 doesn't lie in the domain.

I dont think there's a better way to do something like this other than to sketch it. So as you see from the picture I attached, with no restrictions on the domain just use dy/dx \displaystyle dy/dx for your max/minimum howeer if theres a restriction on the domain then you may need to just sketch it and go from there.

so just sketch the function with a value of p greater than 1, (do one of those tables with x values and y values if you really need to) and go from there.

Another hint if you need it below

Spoiler

https://www.maths.ox.ac.uk/system/files/attachments/test09.pdf
Q1, (D)

I thought this was a relatively straightforward question but I turned out to be wrong and I don't understand why the answer is C (I chose B)

I first summed the series and noticed it had this pattern:
+1,-1,+2,-2,+3,-3..

Seeing this, i realised that the n must be odd (so that it will make the -1^n positive) and then that leaves 1*101 which is > 100.
Could someone tell me what i'm doing wrong please
Original post by DylanJ42
okay at first glance this question seems to be about how the minimum of a function may not necessarily be at the x value where dydx=0 \displaystyle \frac{dy}{dx} = 0 due to restrictions in the domain. (you may not have learnt what a domain is yet, its basically just all the x values the function can take)

so for example, if I asked you for the minimum point of f(x)=x2+4f(x) = \displaystyle x^2 + 4 you would do dydx=0 \displaystyle \frac{dy}{dx} = 0 so 2x=0 \displaystyle 2x = 0 , therefore x=0 \displaystyle x=0 and f(0)=02+4=4 \displaystyle f(0) = 0^2 + 4 =4 so the minimum point is (0,4) \displaystyle (0,4)

However if I asked you for the minimum point of x2+4 \displaystyle x^2 + 4 but this time restricted the domain to, lets say, 2x6 \displaystyle 2 \leq x \leq 6 you couldnt just use dydx \displaystyle \frac{dy}{dx} since x=0 doesn't lie in the domain.

I dont think there's a better way to do something like this other than to sketch it. So as you see from the picture I attached, with no restrictions on the domain just use dy/dx \displaystyle dy/dx for your max/minimum howeer if theres a restriction on the domain then you may need to just sketch it and go from there.

so just sketch the function with a value of p greater than 1, (do one of those tables with x values and y values if you really need to) and go from there.

Another hint if you need it below

Spoiler



i just completed the square
Reply 92
Original post by blankboi
https://www.maths.ox.ac.uk/system/files/attachments/test09.pdf
Q1, (D)

I thought this was a relatively straightforward question but I turned out to be wrong and I don't understand why the answer is C (I chose B)

I first summed the series and noticed it had this pattern:
+1,-1,+2,-2,+3,-3..

Seeing this, i realised that the n must be odd (so that it will make the -1^n positive) and then that leaves 1*101 which is > 100.
Could someone tell me what i'm doing wrong please


I think it would be more productive to view the series like this:

1 + (-2 + 3) + (-4 + 5) + (-6 + 7)
or if you prefer, 1 - (2 - 3) - (4 - 5) - ...
either way you notice you're getting 1 + 1 + 1 + 1 ...
so the sum to the first term is 1, to the third term is 2, the sum to the fifth term is 3, the sum to the seventh term is 4, etc. etc., right?
indeed it would seem apparent that for the odd terms the sum is (n + 1)/2. Clearly it's not going to be an even term you want, as the previous odd term sum is going to be larger. So as you said the n must be odd. Then you simply solve for (n + 1)/2 = 100 (you know this will happen as the series of odd terms goes up in 1s so takes every positive integer)
(edited 7 years ago)
Original post by blankboi
https://www.maths.ox.ac.uk/system/files/attachments/test09.pdf
Q1, (D)

I thought this was a relatively straightforward question but I turned out to be wrong and I don't understand why the answer is C (I chose B)

I first summed the series and noticed it had this pattern:
+1,-1,+2,-2,+3,-3..

Seeing this, i realised that the n must be odd (so that it will make the -1^n positive) and then that leaves 1*101 which is > 100.
Could someone tell me what i'm doing wrong please


1-2+3-4+5....
1-2 = -1
-1+3 = 2
2-4=-2

so adding them up goes 1,-1,2,-2,3,-3 etc so by the time it gets to 99,-99,+100
there would have been (2*99)+1 terms, which is 199.
Original post by KloppOClock
i just completed the square


ah yea completing the square is useful
is the question 4 that was on the first page of this convo actually a part of the MAT?
the one with the stochastic differential equations
Original post by BManMagic
is the question 4 that was on the first page of this convo actually a part of the MAT?
the one with the stochastic differential equations


no
im running out of things to do, how are all you revising for this? I have too much time and not enough work
Reply 98
Original post by KloppOClock
im running out of things to do, how are all you revising for this? I have too much time and not enough work


probably means you should chill out a little...
Original post by Zacken
probably means you should chill out a little...


i need to do well in MAT to compensate for my **** grades lol.

Ive already done 6 papers but dont want to do any more as I need to save some for later

So far I am just doing C1,C2 papers and some aea,smc and nrich questions.

Latest

Trending

Trending