Oxford MAT 2013/2014

Original post by jadoreétudier

I'll wait patiently for your solution :smile:

I attempted part 2 in question 3 again with no luck.

My answer for part 3 is c=1 giving a minimum value of 32/9.

For part 1 how would you show that f(x) has one maximum and minimum? Could you just say that it's a cubic function and all cubic functions have one maximum and one minimum? I imagine not.. ?

Part 3 is correct. I got confused about the + or - and didn't complete parts 1 and 2. To show that it's got one maximum and one minimum, you can try to get the first and second derivatives. Equating the first derivative to zero will give you the maxima and minima. You'll get two, it'll be a quadratic, but you still don't know which is which. So now you have to plug in those values into the second derivative and demonstrate that one is positive (that'd correspond to the minimum) and the other is negative (that'd correspond to the maximum). I didn't actually do all the work, though. :tongue:

(edited 10 years ago)

Reply 682

jadoreétudier

Original post by souktik

Part 3 is correct. I got confused about the + or - and didn't complete parts 1 and 2. To show that it's got one maximum and one minimum, you can try to get the first and second derivatives. Equating the first derivative to zero would give you the maxima and minima. You'll get two, it'll be a quadratic, but you still don't know which is which. So now you have to plug in those values into the second derivative and demonstrate that one is positive .that'd correspond to the minimum) and the other is negative (that'd correspond to the maximum). I didn't actually do all the work, though. :tongue:

Ok doke! So not all cubics have a maximum and minimum turning point. y=x^3 just has a point of inflection at the origin.

(edited 10 years ago)

Reply 683

Original post by jadoreétudier

Ok doke! So not all cubics have a maximum and minimum turning point. y=x^3 just has a point of inflection at the origin.

Tell me if this approach gives any weird results or just becomes really ugly. (The latter isn't even that unlikely, I must admit.)

It's 2.30 am here, I'm gonna call it a night. :smile:

Reply 684

IceKidd

Original post by souktik

Tell me if this approach gives any weird results or just becomes really ugly. (The latter isn't even that unlikely, I must admit.)

It's 2.30 am here, I'm gonna call it a night. :smile:

Original post by jadoreétudier

Ok doke! So not all cubics have a maximum and minimum turning point. y=x^3 just has a point of inflection at the origin.

it comes out right. thats what i did, just takes a page of algebra and calculus -.-

Reply 685

IceKidd

Original post by yl95

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Posted from TSR Mobile

from bak in th day choones cuz. dat fazer mannn

Reply 686

Original post by IceKidd

it comes out right. thats what i did, just takes a page of algebra and calculus -.-

Haha, then I'm glad I didn't do it. My Boring-work-to-be-done-meter was pinging from the moment I started thinking about the problem. :tongue:

Reply 687

Original post by souktik

5 is perhaps the toughest MAT question I've seen yet. If they'd asked us to prove the answers for parts (ii) and (iii) it could have been an easy olympiad combinatorics problem. :tongue:

I'll post a detailed solution as soon as I can. Till then, you guys might try looking at parity (odd or even) of the number of total moves required to get to winning combinations. Part (i) involves the order in which you move each of the three beads and the number of times they each move.

Posted from TSR Mobile

hi I'm very curious about your solutions
particularly (i)(c)

my answers were
(i)(a)90 (b)0 (c)4656
(ii) yes, no, yes, no
(iii) no, yes, yes, yes

can anybody tell me if I am correct?
(this is 2002 Q5 for anybody else reading)

(edited 10 years ago)

Reply 688

Original post by sun_tzu

hi I'm very curious about your solutions
particularly (i)(c)

my answers were
(i)(a)90 (b)0 (c)5916
(ii) yes, no, yes, no
(iii) no, yes, yes, yes

can anybody tell me if I am correct?
(this is 2002 Q5 for anybody else reading)

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

Reply 689

yup, you used pretty much same logic as me, I got 2520 for ways with 3F and 1B
however I believe you undercounted because you need to consider situations where a counter moves the side

Reply 690

Original post by sun_tzu

yup, you used pretty much same logic as me, I got 2520 for ways with 3F and 1B
however I believe you undercounted because you need to consider situations where a counter moves the side

Err, I don't quite understand what you mean by a counter moving the side. The only moves allowed are forward and backward, right?

Posted from TSR Mobile

Reply 691

oh wow I really overcomplicated that haha
but I'm glad to have worked it out anyway

Reply 692

Original post by sun_tzu

oh wow I really overcomplicated that haha
but I'm glad to have worked it out anyway

Oh, okay, I can understand what you did. I'll work that version out when I'm free, haha. :tongue:

Reply 693

jadoreétudier

Original post by sun_tzu

I got the same as you in part 2.
In part 3 I got yes yes no yes. And I don't know who's right :P
I'll look at it again tonight. I've got to go to work :frown:

Reply 694

uhh now I've read the question properly my revised answers for part 3 are no, yes, no, yes

Reply 695

Original post by sun_tzu

uhh now I've read the question properly my revised answers for part 3 are no, yes, no, yes

Then our answers are in agreement. Can you please remind me what the questions were for parts (iii) and (iv)?

Posted from TSR Mobile

Reply 696

Q5
http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%202002.pdf

Reply 697

Q5
http://www.mathshelper.co.uk/Oxford%20Admissions%20Test%202002.pdf

Original post by sun_tzu

Lol, I do have the link. But pdf's look weird on my phone at times. Never mind, I'll take a look later. Thanks anyway. :tongue:

Reply 698

TheGoldenRatio

can someone explain the largest value parts please??

Reply 699