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Oxford MAT 2013/2014

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Ah, okay. Thanks for the response. :smile:

Even if that's not completely accurate, a relatively weak college at Oxford is still a relatively weak college at Oxford! :tongue:

Very true

And if I remember well, St John's is a pretty academic college anyway, so I really wouldn't worry about that if I were you :tongue:

Reply 621

1298977

Original post by jadoreétudier

Does anyone want to do the 1999 and or 2002 papers?

I posted yesterday but fear my post has been lost in the conversation.

:smile:

Okay, I'll take a look at the 2002 paper now. :smile:

Reply 622

IceKidd

Original post by souktik

Hey, how is St John's for math? I didn't research much before applying, to be honest.

like what bluebell said i heard st johns and merton are regarded as pretty top academically (and st johns is richerst!)

Original post by jadoreétudier

Does anyone want to do the 1999 and or 2002 papers?

I posted yesterday but fear my post has been lost in the conversation.

:smile:

i did the 1999 paper a few days ago but the paper ive done it on has now been binned! im going to do 2002 paper today though, so ill be sure to compare answers with you later in the evening!

Reply 623

1298977

Original post by IceKidd

like what bluebell said i heard st johns and merton are regarded as pretty top academically (and st johns is richerst!)

i did the 1999 paper a few days ago but the paper ive done it on has now been binned! im going to do 2002 paper today though, so ill be sure to compare answers with you later in the evening!

Okay, thanks, and please tell me if your Part 1 answers are the same as mine. I'm not absolutely confident about a couple, but I hardly feel like checking carefully. :tongue:

Spoiler

Reply 624

Noble.

Original post by yl95

Merton would be one of them!

Also, Balliol, Magdalen and New but I'm not sure how much that holds true. Noble would know better than me!

Posted from TSR Mobile

There's no such thing as an academic powerhouse really. Merton may have had that reputation, but from speaking to two people I've known studying maths at Merton they're more laid back than my college (New) for maths. I think I remember also the user "anyone_can_fly" on here, who studies maths at Merton, saying that last year no-one got more than 40% in collections and it wasn't much of an issue. If you get less than 40% in collections at my college the main tutor will put you on an official warning :lol:

EDIT: I picked out Merton because that's really the only college people think is the most demanding academically, and it isn't really true at all.

Reply 625

1298977

Does anyone else feel that the "-" between c and 1/c in the first line of problem 3 actually be a "+"?

Anyway, here are some of the answers - I'm not writing any solutions or proofs here - that I got:

Spoiler

Reply 626

Mike_Ross

Can someone help me with 2010 Q3 part i? I don't understand the bit about multiplying through the inequality

sin x < x < tan x

cos(x)

Shouldn't that give you

sin x cos x < x cos x < sin x

?

Mark scheme attached

Reply 627

1298977

Original post by shayiqbal

Can someone help me with 2010 Q3 part i? I don't understand the bit about multiplying through the inequality

sin x < x < tan x

cos(x)

Shouldn't that give you

sin x cos x < x cos x < sin x

You only need to multiply the second part of the inequality,

x < tan x

cos(x)

.
As you have noted yourself, that gives you the result

x cos x < sin x

.
Combine this with the first part of the original inequality,

sin x < x

to get the result

x cos x < sin x < x

.
Hope this helps. I'll be happy to clarify any ambiguities in my explanation.

(edited 10 years ago)

Reply 628

Mike_Ross

Original post by souktik

You only need to multiply the second part of the inequality,

x < tan x

cos(x)

.
As you have noted yourself, that gives you the result

x cos x < sin x

.
Combine this with the first part of the original inequality,

sin x < x

to get the result

x cos x < sin x < x

.
Hope this helps. I'll be happy to clarify any ambiguities in my explanation.

Thank you for the reply.

I see now you only multiply the second part of the inequality, which makes sense, but that would give you (if I understand this right)

sinx < xcosx < sinx

which makes no sense? How did you rearrange the terms to get the required result?

Again, thanks for your time!

Reply 629

1298977

Original post by shayiqbal

Thank you for the reply.

I see now you only multiply the second part of the inequality, which makes sense, but that would give you (if I understand this right)

sinx < xcosx < sinx

which makes no sense? How did you rearrange the terms to get the required result?

Again, thanks for your time!

No problem, happy to help. See,

sinx < x < tanx

actually stands for two separate inequalities which have been combined:

sinx < x

(1) and

x < tanx

(2). You take just the second one, multiply it by

cosx

. That gives you the inequality

xcosx < sinx

(3). Now combine (1) and (3)
to get the final result:

xcosx < sinx < x

.
Is this okay?

Reply 630

Mike_Ross

Original post by souktik

No problem, happy to help. See,

sinx < x < tanx

actually stands for two separate inequalities which have been combined:

sinx < x

(1) and

x < tanx

(2). You take just the second one, multiply it by

cosx

. That gives you the inequality

xcosx < sinx

(3). Now combine (1) and (3)
to get the final result:

xcosx < sinx < x

.
Is this okay?

Ah I see! Thank you ever so much :smile:

Reply 631

1298977

Original post by shayiqbal

Ah I see! Thank you ever so much :smile:

You're welcome. :biggrin:

Reply 632

Arcane1729

guys when it says label a turning point. e.g 2006 q3

do you have to write coordinates- its a messy surd which youd have to cube to get the y value which makes even more messy.

Reply 633

seohyun

2012 q5

Spoiler

Could someone explain the underlined part?

Sent from my GT-N7100 using Tapatalk 4

(edited 10 years ago)

Reply 634

1298977

Original post by seohyun

2012 q5

Spoiler

Could someone explain the underlined part?

Sent from my GT-N7100 using Tapatalk 4

Yeah, sure, I'll try.

m0 is 3, right?

That means:
m_1=2.m_0=3.2; m_2=2.m_1=3.2^2; m_3=2.m_2=3.2^3... m_n=2.m_(n-1)=3.2^n

[You]

Is that okay?

Posted from TSR Mobile

Reply 635

seohyun

Original post by souktik

Yeah, sure, I'll try.

m0 is 3, right?

That means:
m_1=2.m_0=3.2; m_2=2.m_1=3.2^2; m_3=2.m_2=3.2^3... m_n=2.m_(n-1)=3.2^n

[You]

Is that okay?

Posted from TSR Mobile

Thanks, it somewhat clearer now. :smile:

Sent from my GT-N7100 using Tapatalk 4

Reply 636

IceKidd

Original post by souktik

Okay, thanks, and please tell me if your Part 1 answers are the same as mine. I'm not absolutely confident about a couple, but I hardly feel like checking carefully. :tongue:

Spoiler

i got exact same for errything :biggrin:

except for I. i put d) but i wernt too sure on that. i didnt get how to use the ii and iv bit.

if ur pretty sure ur right, can u explain it to me please

Reply 637

IceKidd

Original post by jadoreétudier

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:

Spoiler

for no.2 obviously thats in another paper. for number 3 you havent done yet,

for number 4 i agree with ur parts i,ii,iii. on the last part i agree wit ur coord wer the max value occurs and that its 1. but u also need to find the minimum value which i got to be (1/sqrt2) at ({+or-}0.5,{+or-}0.5)

numberr 5 i dont get -.-

Reply 638

1298977

Original post by IceKidd

i got exact same for errything :biggrin:

except for I. i put d) but i wernt too sure on that. i didnt get how to use the ii and iv bit.

if ur pretty sure ur right, can u explain it to me please