Oxford MAT 2013/2014

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Does anybody know if tasks like in question 2011/5 (v) require an explanation? I am always a bit unsure of how much writing is actually required (especially since 6 credits are awarded for this mentioned question) :/

It's normally a good idea to give some explanation. That way, if you're approaching the question in a sensible way, but have made some slip, then the marker will still give you some credit. But I suspect that just writing down the sum and its solution would get you the marks.

Reply 741

TheFuture001

Original post by TheFuture001

I know that, I do have the solutions. But I did it a different way and arrived at the same conclusion, so was wondering whether mathematically what I did makes sense.

So no mathmos around that know? I guess I'll ask my teacher then..

Reply 742

BankOfPigs

I can't confirm this but I'm fairly sure you can't just call the variable x^10 a constant when you have a "quadratic " in x.

Reply 743

jadoreétudier

my answers for 2006:

Spoiler

Reply 744

Republic1

Not sure if I'm having a thick moment here but I'm stuck on 2010 Question 4 part ii).

I went for

\sqrt {1+4h^{2}}<4

to give

h<\sqrt{15}/2

But they do

1+4h^{2}<4

to give the correct answer of

h<\sqrt{3}/2

Why, when the distance from

(0,0)

(1,2h)

\sqrt {1+4h^{2}}<4

?

Maybe I'm just being thick, this 2010 paper has stressed me out - it seems a lot harder :tongue:

(edited 10 years ago)

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Reply 745

MK_

Original post by Republic1

Not sure if I'm having a thick moment here but I'm stuck on 2010 Question 4 part ii).

I went for

\sqrt {1+4h^{2}}<4

to give

h<\sqrt{15}/2

But they do

1+4h^{2}<4

to give the correct answer of

h<\sqrt{3}/2

Why, when the distance from

(0,0)

(1,2h)

\sqrt {1+4h^{2}}<4

?

Maybe I'm just being thick, this 2010 paper has stressed me out - it seems a lot harder :tongue:

You want the distance between the points to be less than the radius of the circle, so the distance squared must be less than the radius squared.

Reply 746

Republic1

Original post by MK_

You want the distance between the points to be less than the radius of the circle, so the distance squared must be less than the radius squared.

Thanks, I knew I was being stupid, I didn't look back at the question and was using 4 as the radius instead of 2 :tongue:

That'll teach me

Reply 747

jadoreétudier

Original post by Yezi_L

For the 1999 paper:

Q1 I got the same as you did except

iii for c), ii for e) and for k) I did: 3*((4*48*47)/(52*51*50)) which is approx= 0.2 Not sure about this though

Q2 Same as you

Q3 The same for part a)

For b) I got the equation "y=x+1" since the slope of the line would be m=1 and by using the coordinates (1,2)

c) I didn't quite get the question :frown:

I started using the change in x over change in y to represent the slope and got a messy equation, so it's probably wrong...

Q4

a) it is 2/3 for both
c) The total area is 2/4. What I did was to split the area in two part and measure each positively (from -1 to 0 and from 1 to 0), otherwise the areas cancel out to give 0. The top curve minus the bottom curve is right.
d) I'm not sure for this part either :s-smilie:

Maybe we had to spot that both integrals over the interval was the same (2/3) and that the area bounded between them was also the same above and below the x-axis(that's why they cancelled out each other) and arrive to some conclusion from there

I didn't really get into question 5 yet, but I suppose that probability is not part of the syllabus for our test on Wednesday? :wink:

I agree with all your MC answers. K you have the right answer, 0.2. But you would do 3*1/13*12/13*12/13.. because you don't know which card will be the King.

For 3.b), the slope (gradient) of the line in the question is a, so the slope of the perpendicular line will be -1/a. Not sure where you got m=1 from. for c), the algebra is quite messy. You're trying to find the two tangents to the circle x^2 + y^2 =1 (in order for the perpendicular distance from the origin to equal one) which go through the point (1,2). for d) try looking at the difference in areas you have worked out

Reply 748

BankOfPigs

Original post by jadoreétudier

my answers for 2006:

Spoiler

I have a few different answers.

Spoiler

(edited 10 years ago)

Reply 749

IceKidd

you know in the old paper they had like a Multiple choice probability question. since probavility at this stage only technically requires common sense, could a prob MCP q come up in our paper, even though probability isnt mentioned on the spec?

Reply 750

jadoreétudier

Original post by BankOfPigs

I have a few different answers.

Spoiler

Thank you.

I have some issues with 2, i agree it's not a wrapped inequality, so it's y is smaller than -2/root3 or larger than 2/root3. But, why then, in part 1, when y =1, which isn't in our range for y, give two real answers for x? in iii why would the largest value for y be 2/root3?

in 4i i'm assuming you got y=-0.5tx + 0.25t^2
sorry i typed my answers up wrong

Reply 751

1298977

Original post by Republic1

Not sure if I'm having a thick moment here but I'm stuck on 2010 Question 4 part ii).

I went for

\sqrt {1+4h^{2}}<4

to give

h<\sqrt{15}/2

But they do

1+4h^{2}<4

to give the correct answer of

h<\sqrt{3}/2

Why, when the distance from

(0,0)

(1,2h)

\sqrt {1+4h^{2}}<4

?

Maybe I'm just being thick, this 2010 paper has stressed me out - it seems a lot harder :tongue:

Well, you're not getting the correct result as you're comparing the distance from

(0,0)

(1,2h)

, which is

\sqrt {1+4h^{2}}

, with the square of the radius. Your inequality should actually be either

\sqrt {1+4h^{2}}<2

1+4h^{2}<4

, not

\sqrt {1+4h^{2}}<4

Reply 752

jadoreétudier

Original post by IceKidd

Probably not, because then it might be too easy for those who have studied probability. :P I'm just guessing though. I can't think of anything which has come up which isn't on the syllabus

Reply 753

BankOfPigs

Original post by jadoreétudier

Actually I'm terribly sorry.

Relooking at the equation did you get 4/3 > 3y^2?

Because in that case it indeed is a wrapped inequality. I actually got the correct answer in my working but then read it wrong :L.

In such a case yeah the original equation with y=1 definitely works.

in part 3 with the wrapped inequality definitely y naturally become 2/root 3 because it is the largest value in the inequality.

Again, sorry for the mistake :tongue:

(yeah I got that for q4.)

(edited 10 years ago)

Reply 754

jadoreétudier

Original post by BankOfPigs

(yeah I got that for q4.)

That's fine!!!! It make sense now :smile:

Reply 755

Doratch

Would someone explain what is meant by "Convergence condition for infinite geometric progressions" in MAT Syllabus?I am unsure of the term, and a few North-American English-speaker friends were confused about it as well.

(edited 10 years ago)

Reply 756

RichE

Original post by Doratch

That

1 + x + x^2 + x^3 + ...

converges when |x| < 1 and converges to 1/(1-x)

Reply 757

sun_tzu

geometric progressions are like
2 6 18 54 ...
or 2 1 0.5 0.25 .... etc
or even 5, -10, 20, -40
ie after the first term(a) each term is multiplied by a constant amount(r)

if you take these sequences and continue to an infinite amount of terms, then sum up every term,
what does it add up to?
mostly it will be infinity but if the common ratio(r) is less than 1
2 1 0.5 0.25
this sequence converges to 4

sum to n terms is given by S=a(1-r^n)/(1-r)
sum to infinity is given by S=a/1-r

you can see where the second equation comes from by imagining n to be extremely large in the first