Oxford MAT 2013/2014

The former.

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Seriously? I am so. screwed.

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Yeah, I'm pretty sure. I believe they wanted to see if we could apply part (iv) and get the answer directly.

Yeah, I'm pretty sure. I believe they wanted to see if we could apply part (iv) and get the answer directly.

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bro what were your anwers for the 2nd and 4th parts of the q2?

That's annoying. Other people I know got the same answer as me. This paper has told me that I can't read properly.

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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".

Do you have a copy of the paper?

It's online on the MAT page alreadyn

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can someone explain multiple choice part G??

Do you have a copy of the paper?

Ok, here I go trying to count my scores.

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Have to say, I really don't like 1C. Wrong/right answers depending on whether you interpret $f''(2x) = \dfrac{d^2}{dt^2} f(t)\Big|_{t=2x}$ or $f''(2x) = \dfrac{d^2}{dt^2} f(2t)\Big|_{t=x}$. I confess I'm not sure what the correct interpretation is myself!

I know, right! I used the second interpretation and I'm all set to lose 4 marks.

Have to say, I really don't like 1C. Wrong/right answers depending on whether you interpret $f''(2x) = \dfrac{d^2}{dt^2} f(t)\Big|_{t=2x}$ or $f''(2x) = \dfrac{d^2}{dt^2} f(2t)\Big|_{t=x}$. I confess I'm not sure what the correct interpretation is myself!