Oxford MAT 2013/2014

I've done 1999 and 2002 from this website:
http://www.mathshelper.co.uk/oxb.htm
Although there aren't solutions. Would anyone else like to do them too and then we can compare our answers?
My answers for both papers:

Hello! I'm new to the forum.

I've just recently started to prepare for this test. I was ill for the past few months and have only recently got better... I'm extremely worried now!! The past papers seem impossible!!!

Okay, for part (i), this is how I counted-
To get to the given position, each bead must move forward 2 steps net. So the number of forward moves is 6+k and the number of backward moves is k. Total is 6+2k, an odd number. This immediately shows us that the answer to b is 0. As for a, one can notice that each way to reach that configuration corresponds uniquely to an arrangement of 2 A's, 2 B's and 2 C's. (If we move the left bead we write A, middle bead - B, right bead - C.) So the number of ways is 6C2 times 4C2 times 2C2, or 90. For part c, two of the beads are moved 2 times forward, the other bead is moved forward thrice and backward once. There are 3C1 ways to determine the bead that moves 3F, 1B. It can move FFBF or FBFF, but not BFFF or FFFB, as it can't go backward from its initial position or move forward 3 steps. (Here, F=forward, B=backward.) Now for the arrangement. 8C4 ways to pick the times when we move the bead that moves 3F,1B. The 4 remaining moves can be assigned to the two other beads in 4C2 times 2C2 ways. Total number of ways = 3.2.8C4.4C2 = 2520. I'm not sure if I got the same answer last time, might have messed up the calculations. Please tell me if I'm undercounting or overcounting, somehow, or if I've made mistakes in my calculation.

Posted from TSR Mobile

Could you please explain Q1 A and G ? Is there a particular formula to get the answer that I've missed out

Q1 A (on 2002)
differentiate to find has turning points at: -a+root(a^2+3) all over 3
since a<root(a^2+3) always
then it will have one turning point negative x and one turning point positive x
consider the graph of positive cubic with y-intercept (0,-2)
will have 1 sol. only when x>0

Q1 G (2002)
consider a equilateral triangle slice of your hexagon. By setting side length 2, the line down the centre(the radius of the inner circle) is of length root3
so we have radius1=2 and radius2=root3
area1=4*PI area2=3*PI
therefore ratio 4:3

I could easily have made a mistake but I hope this helps

I'd really appreciate it if someone could explain that part to me.
2012 1.H


Sent from my GT-N7100 using Tapatalk 4

You're looking at the integral from ZERO to x, not from pi to x. For x greater than pi, that integral would be equal to the integral from 0 to pi, plus the integral from pi to x. The integral from 0 to pi is, say, X. X is clearly positive. The integral from pi to x is Y, suppose. Y is clearly negative. Now look at the graph. As long as x hasn't reached 2pi, the lower negative area is less than the full positive area between 0 and pi. So the sum of the areas of the positive and negative sections remains positive. At x=2pi, the positive and negative areas become equal and cancel out, giving a total of 0.

This is a lousy explanation, I understand. Tell me if I need to clarify anything.

Posted from TSR Mobile

Ahhhh, thank you, I get it now!

Sent from my GT-N7100 using Tapatalk 4

Yeah, I did the very same things, if I remember correctly. Hey, sun_tzu, what's your score range like? 90ish?

Posted from TSR Mobile

once you realise sin(sin(x)) looks pretty much the same as sin(x) with the same x-intercepts its pretty clear


yeah usually, I found it pretty daunting at first actually but came to realise that the questions always have the same themes pop up

I don't know whether to wish bad luck or good luck because surely if everybody else does well then relatively my performance will be worse xD

Q1 A (on 2002)
differentiate to find has turning points at: -a+root(a^2+3) all over 3
since a<root(a^2+3) always
then it will have one turning point negative x and one turning point positive x
consider the graph of positive cubic with y-intercept (0,-2)
will have 1 sol. only when x>0

Q1 G (2002)
consider a equilateral triangle slice of your hexagon. By setting side length 2, the line down the centre(the radius of the inner circle) is of length root3
so we have radius1=2 and radius2=root3
area1=4*PI area2=3*PI
therefore ratio 4:3

I could easily have made a mistake but I hope this helps

yeah usually, I found it pretty daunting at first actually but came to realise that the questions always have the same themes pop up

I've found this too. Stuff like functions and a bit of complex trig always always seems come up.

If you're from the UK and score 90+, I can't think of any plausible scenario in which you DON'T get into Oxford.

Posted from TSR Mobile

Did you arrive at this conclusion because:
-a+root(a^2+3) would give a positive turning point and
-a-root(a^2+3) would give a negative one?

...and this would allow you to draw the graph of a positive cubic(inc=>dec.>inc) intersecting the y-axis at (0,-2) as it decreases?

Thanks a lot for your help! Now everything's much clearer (I feel stupid for drawing that triangle but not spotting its relation with the radius of the smaller circle)

haha thanks but how about the scenario where I'm applying to cambridge and imperial not oxford xD

Then there's the scenario where you'll be getting offers from both Have you been practicing with STEP questions instead of the MAT ones? Btw, which college in Cambridge did you apply to if it's alright to ask?

yes you've got it exactly, glad to have helped. By considering the best way to explain it, it actually makes it clearer in my mind so I understand it better too

haha thanks but how about the scenario where I'm applying to cambridge and imperial not oxford xD