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Oxford Maths Test (2002 paper) - Help?

I had a go at the 2002 paper yesterday as a mock, but didn't complete all of it. There were some questions where I started off but got completely stuck at some point.

Here's the paper: http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%202002.pdf

In particular I struggled with questions 1 (A, D, F and J) and got completely stuck on questions 3 and 5 and so couldn't continue. I've done question 2 before since it was on a specimen paper, and question 4 was nice and straightforward.

If someone is willing to help I'll be very grateful - this test is driving me NUTS! Does anybody else think they're going to fail the admissions test?

Reply 1

omegaSQU4RED

To be a bit more specific, on Question 3 I tried using the fact that if you make the expression simpler to solve (so it becomes a quadratic) by differentiating with respect to x, you can argue that the discriminant of that quadratic indicates unique or repeated (or perhaps none) turning points (e.g. if the discriminant > 0 then there are two distinct turning points, one a maximum and a minimum). However by doing that I ended up with

which can't be solved.

Question 5 I simply ran out of time on, I'll look at it again some other time, but in a question like that, do you need to just draw out every single possibility or is it worth having knowledge about combinations (i.e. in S1)?

Reply 2

refref

Better post this in the oxford admissions test solution thread :smile:

1A I dont actually know, try a few values of a I assume, but this one I didnt get.

D you need to know the chain rule which is in C3

F Write out all the powers in an inequality and take logs base 2.

J You can try different values of alpha I think

Question 3 was a hard one because it was so fiddly, and we dont have a solution for question 5 yet (couldnt do question 5). Look on the other thread. (2002 was one of the harder ones in my opinion)

Reply 3

DFranklin

Q3:

(i) f(x)=(c-1/c - x)(4-3x^2) = 3x^3-3(c-1/c)x^2 -4x + 4(c-1/c)
f'(x) = 9x^2 -6(c-1/c)x-4
This has roots

\frac{6(c-1/c) \pm \sqrt{36(c-1/c)^2+36.4}}{18} = \frac{(c-1/c)\pm\sqrt{(c-1/c)^2+4}}{3}

A simple sketch shows f' is decreasing for the smaller root (which is therefore a maximum) and increasing for the larger one (whcih is therefore a minimum).

= \frac{(c-1/c)\pm \sqrt{c^2 -2 + 1/c^2 + 4}}{3} = \frac{(c-1/c)\pm \sqrt{(c+1/c)^2}}{3} = \frac{(c-1/c)\pm (c+1/c)}{3}

(ii)Write f'(x) = 9(x-a)(x-b) = 9(x-a)(x-a+(a-b)) = 9(x-a)^2 - 9(x-a)(b-a) where a<b are the roots we just found.
Note that

f(b)-f(a) = \int_a^b f'(x) \,dx = [3(x-a)^3 - \frac{9}{2}(x-a)^2(b-a)]_a^b= -\frac{3}{2}(b-a)^3

But (b-a) is 2(c+1/c)/3

So

|f(b)-f(a)| = \left|-\frac{3.8}{2.27}(c+1/c)^3\right| = \frac{4}{9} (c+1/c)^3.

[Or you could just plug the 2 roots into the original equation. I thought this way was more interesting.]

(iii) This is obviously minimized by minimizing (c+1/c), and since c is positive, this is equivalent to minimizing (c+1/c)^2.
But (c+1/c)^2 = (c-1/c)^2 + 2, which is obviously minimized by setting c = 1.

Then c+1/c = 2 and so we have a final answer of 32/9.