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Oxford MAT 2013/2014 Watch

1. How long do Oxford take to get back to applicants about interviews? I'm applying for Imperial and not Oxford, however I'm assuming the time frame should only be slightly longer - i.e. the time Imperial take to give some offers should be similar to how long Oxford take to give interview invitations.
2. can anyone help me with 2011 q4 using geometry rules. why does the maximum value occur at the tangental point and why is this point x=y?
3. (Original post by sun_tzu)
I got pi/9 - root3 /12
same as you so we are wrong together (or right hopefully)
I got same as you. Pencil? Not ink pen?
4. (Original post by CD315)
How long do Oxford take to get back to applicants about interviews? I'm applying for Imperial and not Oxford, however I'm assuming the time frame should only be slightly longer - i.e. the time Imperial take to give some offers should be similar to how long Oxford take to give interview invitations.
Early December

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5. (Original post by yl95)
Early December

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No worries, thanks.
6. (Original post by IceKidd)
but im stumped on how you could find the little bit!
wow im far too dumb to have thought of that lol.

nicely done !
8. (Original post by daniyalfaiz)
can anyone help me with 2011 q4 using geometry rules. why does the maximum value occur at the tangental point and why is this point x=y?
The line y=x is the radius of the circle, which will give you the maximum because then a line x+y=a will be a tangent at that point. That probably wasn't a good explanation so here is a picture (not identical to the question because they use a quarter disc, but the same principle

From that graph, x+y=c will always be smaller than x+y=a, likewise with x+y=b. So the maximum values will be on the tangent to the circle - the line x+y=a.
Attached Images

9. Anyone know what time the test actually is?
10. (Original post by Republic1)
The line y=x is the radius of the circle, which will give you the maximum because then a line x+y=a will be a tangent at that point. That probably wasn't a good explanation so here is a picture (not identical to the question because they use a quarter disc, but the same principle

From that graph, x+y=c will always be smaller than x+y=a, likewise with x+y=b. So the maximum values will be on the tangent to the circle - the line x+y=a.
Just did this paper. I had the right answer, but I couldn't find the words to argue that it was true. Would just saying "This occurs at x=y -> 2x^2=1 -> x=.5sqrt2=y -> x+y=sqrt2" net me the full points here?
11. (Original post by dutchmaths)
Just did this paper. I had the right answer, but I couldn't find the words to argue that it was true. Would just saying "This occurs at x=y -> 2x^2=1 -> x=.5sqrt2=y -> x+y=sqrt2" net me the full points here?
That sounds fine to me - I followed your working anyway so I guess an examiner would. I'm probably the worst person to ask about how to set your workings out, I'm terrible.

(I guessed 5sqrt2 was a typo, you meant 1/sqrt2 ?)
12. (Original post by Republic1)
That sounds fine to me - I followed your working anyway so I guess an examiner would. I'm probably the worst person to ask about how to set your workings out, I'm terrible.

(I guessed 5sqrt2 was a typo, you meant 1/sqrt2 ?)
.5*sqrt2 = 0.5*sqrt2= (2^-1)*(2*0.5) = 2^-0.5 = 1/sqrt2
13. (Original post by Republic1)
The line y=x is the radius of the circle, which will give you the maximum because then a line x+y=a will be a tangent at that point. That probably wasn't a good explanation so here is a picture (not identical to the question because they use a quarter disc, but the same principle

From that graph, x+y=c will always be smaller than x+y=a, likewise with x+y=b. So the maximum values will be on the tangent to the circle - the line x+y=a.
thanks!
14. Can someone explain to me the reasoning behind this:

The equation has no solutions?

I sketched the graph on Wolfram and it makes sense, there is no x-intercept. But, in my head I can't understand it. Surely |x-1| means either -(x-1) or +(x-1), and there are solutions for when it is +(x+1)? Sorry if I'm not being clear
15. (Original post by ainsley94)
Anyone know what time the test actually is?
I've been told 9.15!

(Original post by Mike_Ross)
Can someone explain to me the reasoning behind this:

The equation has no solutions?

I sketched the graph on Wolfram and it makes sense, there is no x-intercept. But, in my head I can't understand it. Surely |x-1| means either -(x-1) or +(x-1), and there are solutions for when it is +(x+1)? Sorry if I'm not being clear
x^2 > 0, |x-1| > 0 => x^2 + |x-1| cannot be 0.
16. (Original post by CD315)

x^2 > 0, |x-1| > 0 => x^2 + |x-1| cannot be 0.
Ah thank you, classic case of over analysis
17. (Original post by Mike_Ross)
Ah thank you, classic case of over analysis
Welcome to my world!
I'm struggling to make sense of this diagrammatically. Can anyone help?Attachment 251378

Also, have I gone mad, or is this forgetting the little bit at the corner?Attachment 251377
That silly mistake I discovered a while ago and fixed it (as you'll see in the 'MAT Geometry Worksheet'). But I must have forgotten to upload the corrected version of the slides. Oops. Will rectify tomorrow morning.
19. (Original post by dutchmaths)
.5*sqrt2 = 0.5*sqrt2= (2^-1)*(2*0.5) = 2^-0.5 = 1/sqrt2
Ah didn't see the decimal
20. (Original post by CD315)
Welcome to my world!

Also another quick question; is there a quick way of knowing that the minimum value of is -1? It seems there should be some logic to it but differentiating and setting equal to zero gives a painful cubic...

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