Oxford MAT 2013/2014

Wait, guys, is 1J) really B? Still means I got 7/10 then from my lucky guess.

2.ii.a.
$f(t)-f(1-t)=t$ (for all $t$)
implies: $f(1-t)-f(t)=1-t$ (replacing $t$ by $1-t$)
Add the two equations and you get $t+(1-t)=0$, that is $1=0$.


2.ii.c.
I used $f(t)=t.(2t-1)^3$. In general, $f(t)=t.g(t)$ is a valid example as long as the condition in 2.ii.b. is followed. A simpler example would have been $f(t)={\frac{(2t-1)^3}{2}}$, or $f(t)={\frac{g(t)}{2}}$ in general.




Well, it was confusing. In an exam hall, it's easy to misinterpret "one digit at least 5" as "at least one digit 5".

yes it is

My answers for Question 1:

I is (c)

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I think you'll be fine. There are people who got in the 50-60s who are worrying much more than you.

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Does this mean they're likely to allow for 2 answers?

I left the last part of q2 man. Was it harder / longer than the other parts? I'm screwed if the last part was worth 6 marks!

Really? How did you get that?



I'm not worried, just a little annoyed. Meh.



Unlikely. The first interpretation seems to be correct. They'd put those options on purpose, possibly to check if we're familiar with the notation. I find that a little unfair, but there's no point in cribbing.



What do you mean by the last part? 2.ii.c, right? What did you get for 2.i, 2.ii.a and 2.ii.b? I don't think not doing 2.ii.c will cost you 6 marks if you got the other parts correctly. 4+3+4+4 might be the division of marks.

Just compared the multiple choice questions from this year with those of the other years and I still feel like they were way harder this year. Does anyone feel the same?

Just compared the multiple choice questions from this year with those of the other years and I still feel like they were way harder this year. Does anyone feel the same?

No way bro I found 2012 paper really hard for some reason. Especially the multiple choice and q5. In this paper other than the last multiple choice all were fine.

https://www.maths.ox.ac.uk/system/files/attachments/test13.pdf

Paper's up. I got BCDDDBDBBB for multiple choice, might be wrong tho!

2.ii.a.
$f(t)-f(1-t)=t$ (for all $t$)
implies: $f(1-t)-f(t)=1-t$ (replacing $t$ by $1-t$)
Add the two equations and you get $t+(1-t)=0$, that is $1=0$.


2.ii.c.
I used $f(t)=t.(2t-1)^3$. In general, $f(t)=t.g(t)$ is a valid example as long as the condition in 2.ii.b. is followed. A simpler example would have been $f(t)={\frac{(2t-1)^3}{2}}$, or $f(t)={\frac{g(t)}{2}}$ in general.



Well, it was confusing. In an exam hall, it's easy to misinterpret "one digit at least 5" as "at least one digit 5".

https://www.maths.ox.ac.uk/system/files/attachments/test13.pdf

Paper's up. I got BCDDDBDBBB for multiple choice, might be wrong tho!

Does this mean they're likely to allow for 2 answers?

I don't always troll!

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I doubt it, since the fact they gave both "possible" answers implies they wanted you to choose the "correct" one.

I'm just arrogant enough to think that if I'm not sure what the correct answer is, it's not a very fair question. [Although I could just be being stupid or forgetful about a standard definition].

I'm interested to see what RichE has to say about it.