STEP Mathematics Problem-Solving Society!

Reply 20

*Subscribes*

Problem 4 (April 30th)

Paper: STEP Mathematics II (Further Mathematics A)
Year: 1990
Question: 3

I think that a good question which gets some discussion going on curve sketching can only be a good thing. Comfort with curve sketching is an absolute must as it comes up frequently both in its own right and as part of other questions - in the latter case, discomfort with the idea can often put you off of an otherwise straightforward question.

Reply 22

hrickards

2

It turns out I actually the wrong question --- 2007 Q3 STEP III, but I don't have time tonight to do another.

Reply 23

Scientific Eye

Hello there!

I've tried Problem 2 and I've come to a solution for it. I don't know if it is entirely correct, since I'm still a newbie at STEP II, and unfortunately I don't know any LaTex at all, but I hope to start learning tomorrow.

Reply 24

Jkn

OP

11

Original post by hrickards

It turns out I actually the wrong question --- 2007 Q3 STEP III, but I don't have time tonight to do another.

Ah I've just done that myself so I don't mind posting it if no-one else offers. I found it a bit nasty though, I normally avoid curve-sketching like the plague!

You could always try it tomorrow midday?

Original post by Scientific Eye

Hello there!

I've tried Problem 2 and I've come to a solution for it. I don't know if it is entirely correct, since I'm still a newbie at STEP II, and unfortunately I don't know any LaTex at all, but I hope to start learning tomorrow.

Okay awesome, just make sure you don't polish it by looking at the solution bank or anything first haha :smile:

Great! I only started using it a few months ago too. It's easy if you copy and past chunks from the TSR-tutorial and then edit them once their in the reply box (after a few days you'll have the basics memorised).

Useful in that problem:

Wrap all code in the "tex" brackets.

\pi =

\pi

\frac{t+t^{2013}}{j^2+k^2+n^2} =

\frac{t+t^{2013}}{j^2+k^2+n^2}

\sqrt{2013} =

\sqrt{2013}

\displaystyle\int_0^{2013} x^{hello} =

\displaystyle\int_0^{2013} x^{hello}

Basically put "\displaystyle" at the start of each "tex" bracket and everything will be slighter bigger and more awesome!
Oh and btw....

Spoiler

(edited 10 years ago)

Joining!

OP

Original post by CD315

Joining!

Are ye now... :colone:

You have to do this first...

Prove that

\displaystyle(\frac{27(x^2+y^2+z ^2)}{4})^{\frac{1}{4}} \geq \sum_{cyc}\frac{x}{\sqrt{x+y}} \geq (\frac{27(xy+yz+zx)}{4})^{\frac{ 1}{4}}

for positive real numbers x, y and z.

Edit: Sarcasm guys :rolleyes:

(edited 10 years ago)

Reply 27

Jkn

OP

11

Are people ready to post solutions then? How about you guys do them today:

Original post by MathsNerd1

...

Problem 1

Original post by Scientific Eye

...

Problem2

Original post by hrickards

...

Problem3

Original post by Felix Felicis

...

Problem 4

Reply 28

Felix Felicis

13

Original post by Jkn

Problem 4

Curses! I was hoping for the product one :colone:

Will post the solution up for this later tonight but do you mind if I 'un-subscribe' temporarily until Junish so I can get my ASs out of the way as I'm setting STEP aside for a while until then? :redface:

Reply 29

Jkn

OP

11

Original post by Felix Felicis

Curses! I was hoping for the product one :colone:

Will post the solution up for this later tonight but do you mind if I 'un-subscribe' temporarily until Junish so I can get my ASs out of the way as I'm setting STEP aside for a while until then? :redface:

Mwuhaha

Well you said you were gonna do the STEP II ones and no-one had shown interest in that one yet :colone:

Okay awesome :smile:

Not sure how you're gonna do the sketches :tongue:

Computer image, take a photo of your paper, just a description? Nooo idea :lol:

Wow, you're preparing for STEP early! Do what you will good sir!

Reply 30

31541

2

Original post by Jkn

Are ye now... :colone:

You have to do this first...

Prove that

\displaystyle(\frac{27(x^2+y^2+z ^2)}{4})^{\frac{1}{4}} \geq \sum_{cyc}\frac{x}{\sqrt{x+y}} \geq (\frac{27(xy+yz+zx)}{4})^{\frac{ 1}{4}}

for positive real numbers x, y and z.

Edit: Sarcasm guys :rolleyes:

Don't even know what

\sum_{cyc}

means

Reply 31

username937760

16

Original post by CD315

Don't even know what

\sum_{cyc}

means

You don't need to at present either.

Reply 32

Scientific Eye

My solution to Problem 2:

Part 1

Spoiler

\displaystyle i)\ a)

Spoiler

\displaystyle b)

Spoiler

\displaystyle ii)

Spoiler

I don't think I've ever written so much code in my life, haha.

I skipped a few bits of algebraic manipulation in order to save having to type all of it out. I can only hope it doesn't cause too much confusion.

(edited 10 years ago)

Reply 33

Nick_

Original post by Scientific Eye

My solution to Problem 2:

I don't think I've ever written so much code in my life, haha.

I skipped a few bits of algebraic manipulation in order to save having to type all of it out. I can only hope it doesn't cause too much confusion.

Yep looks good to me, except you should be able to work out arctan(root(3)) without a calculator. And for the final integral you could have factorized out 3 from the bottom and then used the initial result with a=1/root(3), but your way seems just a quick.

(edited 10 years ago)

Reply 34

Scientific Eye

Original post by Nick_

Yep looks good to me, except you should be able to work out arctan(root(3)) without a calculator. And for the final integral you could have factorized out 3 from the bottom and then used the initial result with a=1/root(3), but your way seems just a quick.

Ah, good point. It probably would have been faster to just factor it out, though. Then you wouldn't have to change the limits or even set

\displaystyle y = \sqrt3 sin \theta

. Thanks.

Reply 35

Jkn

OP

11

Original post by Scientific Eye

Spoiler

Excellent solution my friend :biggrin:

(and great job with the latex code!)

I did it exactly the same as you (included the last integral which was awkward either way) except for the final answer.

Noting that

tanA=\sqrt{3} \Rightarrow \frac{sinA}{cosA}=\frac{(\frac{ \sqrt{3}}{2})}{(\frac{1}{2})}

give you a possible solution of

A=\frac{\pi}{3}

which yields an answer of

I=\frac{\pi\sqrt{3}}{18}

I believe you are likely to have lost 2 (perhaps 1) mark for this.

My best guess would be 18 (does anyone agree?) so long as you went into more detail for the algebraic manipulation in part b, which I have no doubt that you did on paper. So well done :biggrin:

How did you find the other questions? (3 was a little nasty I thought!)

Original post by CD315

Don't even know what

\sum_{cyc}

means

Basically symmetrical across the variables in a systematic way (i.e. in this case it's shorthand for 3 fractions).

But I was joking bro don't worry haha :lol: