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The 2012 STEP Results Discussion Thread

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Original post by Xero Xenith
I think it really depends on how long it takes before you start getting many questions right, and how fast you can do them. Not sure what the average is, but I'd love to know this as well!

Previous years' STEP-ers: how long did you spend on it per week when you were preparing? :smile:

When I took STEP I, I tried to stick to one paper a week (i.e. all 13-16 questions depending on the age of the paper). Although, in year 12 this was initially quite hard to stick to as I had to self-teach parts of C3 and C4 alongside doing the papers, which was time consuming.

I'm not going to talk about year 13 as I didn't work as hard on it as I perhaps should have done. :p:
Original post by Farhan.Hanif93
When I took STEP I, I tried to stick to one paper a week (i.e. all 13-16 questions depending on the age of the paper). Although, in year 12 this was initially quite hard to stick to as I had to self-teach parts of C3 and C4 alongside doing the papers, which was time consuming.

I'm not going to talk about year 13 as I didn't work as hard on it as I perhaps should have done. :p:


That's about two questions a day for STEP I - mind if I ask how many months you were doing that for? :smile: It's really useful to know how much time previous people spent.

Actually, it would be useful to know how much wasn't "enough"... No worries if you really don't want to say though :tongue:
Original post by Xero Xenith
That's about two questions a day for STEP I - mind if I ask how many months you were doing that for? :smile: It's really useful to know how much time previous people spent.

Like I said, this wasn't really happening at first because I was self teaching and also just starting out with STEP, which is always a bit of a struggle. I'd say that I got into that cycle at about March-April time and from Jan-March, I got through 4 or 5 papers. This is all pretty rough as that was quite a while back now.

Actually, it would be useful to know how much wasn't "enough"... No worries if you really don't want to say though :tongue:

It was literally a case of me picking up a STEP II or III paper whenever I got bored of Geography and doing it. I didn't really structure it in any fixed way and consequently didn't do too many papers completely nor did I work on exam technique until later in the year. There was a month of so where I did actually go all out but that fizzled away because I had no pressure on my performance.

For that reason, I'm probably not the best example. :p:
Quick question on S1 2000 Q4i:
After finding dy/dx I tried to show that dy/dx is increasing between 0 and 1, after doing that I showed that when f(0) = 0 and f(1) = 1/16 and thus 1/16 is the largest point on the curve between 0 and 1.
It is different to the solution posted on TSR that can be found on the first link but I am just wondering whether what I have done satisfies the question ?
And any ideas on how to show that dy/dx is always increasing between 0 and 1 ?
Reply 104
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Thrug
3
Original post by الجبر

And any ideas on how to show that dy/dx is always increasing between 0 and 1 ?


Generally, wouldn't you just need to show it is always positive in that range?
Original post by Thrug
Generally, wouldn't you just need to show it is always positive in that range?


yeah but how do I do that ? lol:confused:
Original post by الجبر
Quick question on S1 2000 Q4i:
After finding dy/dx I tried to show that dy/dx is increasing between 0 and 1, after doing that I showed that when f(0) = 0 and f(1) = 1/16 and thus 1/16 is the largest point on the curve between 0 and 1.
It is different to the solution posted on TSR that can be found on the first link but I am just wondering whether what I have done satisfies the question ?
And any ideas on how to show that dy/dx is always increasing between 0 and 1 ?


EDIT: realised the question is not "show dy/dx is increasing" - my bad, misunderstood the above

If you need to show that dy/dx is increasing, you just need to show that it ends up higher than it begins (you've done this), AND that there are no points of inflection.

So, show that d2ydx2\frac{d^2y}{dx^2} is never 0 in that range and you're done :smile:
(edited 12 years ago)
Reply 107
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Thrug
3
Alternatively, you could solve for the values of x where dy/dx is greater than 0 and show 0<x<1 is in this region.

I.e if x<2 => dy/dx is always positive then 0<x<1 must always be positive.
Original post by Xero Xenith
If you need to show that dy/dx is increasing, you just need to show that it ends up higher than it begins (you've done this), AND that there are no points of inflection.

So, show that d2ydx2\frac{d^2y}{dx^2} is never 0 in that range and you're done :smile:


Wait, just because f(1) is larger than f(0) it doesn't mean that the gradient is always positive between them (there could be a max and a min), I need to show that the gradient is always positive for:
6x5(x2+1)4−8x7(x2+1)3(x2+1)8\dfrac{6x^5(x^2+1)^4 - 8x^7(x^2+1)^3}{(x^2+1)^8} between 0 and 1.
Any ideas ?
Original post by Thrug
Alternatively, you could solve for the values of x where dy/dx is greater than 0 and show 0<x<1 is in this region.

I.e if x<2 => dy/dx is always positive then 0<x<1 must always be positive.


Smart way of looking at it, thnx a lot !:smile:
Reply 110
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Thrug
3
Original post by Xero Xenith
If you need to show that dy/dx is increasing, you just need to show that it ends up higher than it begins (you've done this), AND that there are no points of inflection.

So, show that d2ydx2\frac{d^2y}{dx^2} is never 0 in that range and you're done :smile:


Surely just dy/dx?
Original post by Thrug
Alternatively, you could solve for the values of x where dy/dx is greater than 0 and show 0&lt;x&lt;1 is in this region.

I.e if x&lt;2 =&gt; dy/dx is always positive then 0&lt;x&lt;1 must always be positive.


He said dy/dx is increasing, not just positive :confused:

Original post by الجبر
Wait, just because f(1) is larger than f(0) it doesn't mean that the gradient is always positive between them (there could be a max and a min), I need to show that the gradient is always positive for:
6x5(x2+1)4−8x7(x2+1)3(x2+1)8\dfrac{6x^5(x^2+1)^4 - 8x^7(x^2+1)^3}{(x^2+1)^8} between 0 and 1.
Any ideas ?



EDIT: SORRY - MISREAD THE QUESTION - THE BELOW DOESN'T APPLY

(x²+1)^8 is always positive, clearly.

At x=0, it's equal to zero. I'm guessing it's not including x=0?

We now have to solve
6x5(x2+1)4−8x7(x2+1)3>06x^5(x^2+1)^4-8x^7(x^2+1)^3>0
divide by x5x^5 (as x is not 0)

6(x2+1)4−8x2(x2+1)3>06(x^2+1)^4-8x^2(x^2+1)^3>0

divide by (x2+1)3(x^2+1)^3 which is always positive

6(x2+1)−8x2>06(x^2+1)-8x^2>0

6−2x2>06-2x^2>0

6>2x26>2x^2

3>x23>x^2

which is true in the region.

(This seems correct, anyway :colondollar:)
(edited 12 years ago)
Reply 112
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gff
5
Original post by Xero Xenith
AND that there are no points of inflection.


Why would you think that inflection points matter? You don't need second derivative here.
You only need to show is that the function is increasing.
Original post by gff
Why would you think that inflection points matter? You don't need second derivative here.
You only need to show is that the function is increasing.


He said dy/dx is increasing - not just that the function is increasing. Not sure if he meant that but I'm going by what he said:

Original post by الجبر
Quick question on S1 2000 Q4i:
After finding dy/dx I tried to show that dy/dx is increasing between 0 and 1, after doing that I showed that when f(0) = 0 and f(1) = 1/16 and thus 1/16 is the largest point on the curve between 0 and 1.
It is different to the solution posted on TSR that can be found on the first link but I am just wondering whether what I have done satisfies the question ?
And any ideas on how to show that dy/dx is always increasing between 0 and 1 ?


EDIT: Scrap that, just looked at the question. Sorry :colondollar:
(edited 12 years ago)
I dont think much of the solution in the solutions thread to part (i).

The solution in the accompanying file is clearer
Document1.pdf39.6KB
Original post by mikelbird
I dont think much of the solution in the solutions thread to part (i).

The solution in the accompanying file is clearer


Thnx !!
But where did you get it from, all I have is the megapack v3 from the first page and it only has solutions from 2004 onwards :frown:
Could you please post the original file and any other pdf's like it that aren't in the megapack on the first page.
Reply 116
Avatar for Thrug
Thrug
3
Original post by الجبر
Thnx !!
But where did you get it from, all I have is the megapack v3 from the first page and it only has solutions from 2004 onwards :frown:
Could you please post the original file and any other pdf's like it that aren't in the megapack on the first page.


He most likely just made it now for you :smile:
Original post by Thrug
He most likely just made it now for you :smile:


WOW - I can't thank him enough then lol.
Can everyone else rep him for me and I'll rep you for doing that :biggrin:
Original post by Xero Xenith
He said dy/dx is increasing, not just positive :confused:




EDIT: SORRY - MISREAD THE QUESTION - THE BELOW DOESN'T APPLY

(x²+1)^8 is always positive, clearly.

At x=0, it's equal to zero. I'm guessing it's not including x=0?

We now have to solve
6x5(x2+1)4−8x7(x2+1)3>06x^5(x^2+1)^4-8x^7(x^2+1)^3>0
divide by x5x^5 (as x is not 0)

6(x2+1)4−8x2(x2+1)3>06(x^2+1)^4-8x^2(x^2+1)^3>0

divide by (x2+1)3(x^2+1)^3 which is always positive

6(x2+1)−8x2>06(x^2+1)-8x^2>0

6−2x2>06-2x^2>0

6>2x26>2x^2

3>x23>x^2

which is true in the region.

(This seems correct, anyway :colondollar:)


May I make a point about this way of laying out reasoning? You start with an inequality you want to solve, and deduce a true statement from it. You probably mean to use <=> between each line, but should make this clear.
Original post by ian.slater
May I make a point about this way of laying out reasoning? You start with an inequality you want to solve, and deduce a true statement from it. You probably mean to use &lt;=&gt; between each line, but should make this clear.


In truth I'm not sure how to do the "implies" and "implies and is implied by" symbols on LaTeX... I should look this up :colondollar:

OK, it's just \implies - now I feel a bit silly. Thanks for the pointer :smile:

∃[Me]:Me  ⟹  silly\exists [Me] : Me \implies silly

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