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STEP Prep Thread 2015

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Is there any hyperbolic stuff in STEP?
Original post by ubisoft
Is there any hyperbolic stuff in STEP?


Yh in III


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STEP I 2001 Q3
The last part to shown it is not necessary can't I just say if g^2=4h then the result still holds. Where the cubic has one double root and another root x=k.
????

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(edited 9 years ago)
Original post by ThatPerson
I managed to prove the first limit by the same method you posted - I should've noticed there was a reason they asked me to show 12ln⁡x<x \dfrac{1}{2} \ln x < \sqrt{x} first.

I'm guessing the second limit they want me to show (xlnx --> 0 as x --> 0) follows from lnx/x ---> 0. Don't see it at the moment though. I thought I could change both into 1/x , but that didn't help. Have a feeling that this is obvious...


You are absolutely correct that it follows, and in the way that you do it. Remember that log⁡(1x)=−log⁡(x) \log(\frac{1}{x}) = -\log(x) and try to develop your idea of changing into 1x \frac{1}{x} . You may find it useful to very explicitly substitute y=1x y = \frac{1}{x} and re-arrange the limit accordingly.

As for whether it's obvious; I think it's probably moreso in retrospect. I remember doing that limit several times in STEP and never thinking to use the other limit you have; I always did it more directly using substitutions in a way such as this:

Spoiler



Again, continue what you're doing and derive it using the limits you've already shown as that is the best method. The spoilered stuff is just an alternative for completeness.

Original post by lllllllllll
I haven't done this question but you could also show it by rearranging the equation to ln(x1/x)

so as x --> ∞ ; 1/x --> 0. Therefore x1/x--> 1 so ln(x1/x) --> ln(1) = 0


No, you cannot do this. The idea of taking a limit as a variable approaches some value intuitively means that that variable is changing as it approaches the limit, so you are not allowed to assume some function dependent on it is not also changing.

A counter-example is as follows:

1=lim⁡x→∞xx 1 = \lim_{x \to \infty} \frac{x}{x}

Hopefully this is obvious. However if you ignored the x x on the top and considered 1x \frac{1}{x} separately, you'd end up getting a limit of 0. Alternatively, ignoring the denominator and taking the limit of the numerator would get you ∞ \infty !
(edited 9 years ago)
Original post by DJMayes
You are absolutely correct that it follows, and in the way that you do it. Remember that log⁡(1x)=−log⁡(x) \log(\frac{1}{x}) = -\log(x) and try to develop your idea of changing into 1x \frac{1}{x} . You may find it useful to very explicitly substitute y=1x y = \frac{1}{x} and re-arrange the limit accordingly.

As for whether it's obvious; I think it's probably moreso in retrospect. I remember doing that limit several times in STEP and never thinking to use the other limit you have; I always did it more directly using substitutions in a way such as this:

Spoiler



Again, continue what you're doing and derive it using the limits you've already shown as that is the best method. The spoilered stuff is just an alternative for completeness.


Thanks! Just did it - had a massive facepalm moment; before I wrote−lnx1x\dfrac{-ln x}{\frac{1}{x}} instead of −ln⁡xx \dfrac{-\ln x}{x} .
Original post by physicsmaths
STEP I 2001 Q3
The last part to shown it is not necessary can't I just say if g^2=4h then the result still holds. Where the cubic has one double root and another root x=k.
????

Posted from TSR Mobile


This is what I did, specifically noting when g=2hg=2\sqrt{h} the inequality you get from the differentiated cubic is the same as the one quoted but with g=2hg=2\sqrt{h}.
Original post by SamKeene
This is what I did, specifically noting when g=2hg=2\sqrt{h} the inequality you get from the differentiated cubic is the same as the one quoted but with g=2hg=2\sqrt{h}.


Good to know. I used a counterexample aswell.


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Original post by physicsmaths
Good to know. I used a counterexample aswell.


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It's been a while, but I think I just did a sketch to show that a cubic can have two stationary points without having three roots, which is a counterexample with barely two lines of algebra.

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Original post by Krollo
It's been a while, but I think I just did a sketch to show that a cubic can have two stationary points without having three roots, which is a counterexample with barely two lines of algebra.

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I stated that but all we need is to show g^2>4h is not necessary and that if g^2=4h (with no triple root) then it is fine. I found the proper solution in Siklos booklet and I did it the same way as him. If you say that g^2>=4h is necassary then that is wrong due to the problem you pointed out of one(distinct) root and two stationary points.


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Is it normal to spend more than a couple of days on 1 question?
Original post by Brubeckian
Is it normal to spend more than a couple of days on 1 question?
Not really. It might be the case when you first start STEP prep, and after that there might be the odd question you can't see how to do but eventually find a way after a day or 2. But it certainly shouldn't be the norm.
Original post by DFranklin
Not really. It might be the case when you first start STEP prep, and after that there might be the odd question you can't see how to do but eventually find a way after a day or 2. But it certainly shouldn't be the norm.


I'm on a gap year, started prep yesterday... I probably just need to get back into the swing of things like identities, general methods etc
Original post by DJMayes


No, you cannot do this. The idea of taking a limit as a variable approaches some value intuitively means that that variable is changing as it approaches the limit, so you are not allowed to assume some function dependent on it is not also changing.

A counter-example is as follows:

1=lim⁡x→∞xx 1 = \lim_{x \to \infty} \frac{x}{x}

Hopefully this is obvious. However if you ignored the x x on the top and considered 1x \frac{1}{x} separately, you'd end up getting a limit of 0. Alternatively, ignoring the denominator and taking the limit of the numerator would get you ∞ \infty !


Ahh yeah you're right. I differentiated ln(x) to show that ln(x) behaves like a constant as x tends to infinity and therefore ln(x)/x will tend to 0 as x--> ∞ \infty ...Not sure if that's okay?

I haven't really seen the question yet so I'm not really sure if I'm missing any previous results

Original post by Brubeckian
I'm on a gap year, started prep yesterday... I probably just need to get back into the swing of things like identities, general methods etc


Nice, I'm on a gap year as well. So not happy to have received a STEP paper as part of an offer!
(edited 9 years ago)
Original post by lllllllllll
Ahh yeah you're right. I differentiated ln(x) to show that ln(x) behaves like a constant as x tends to infinity and therefore ln(x)/x will tend to 0 as x--> ∞ \infty ...Not sure if that's okay?

I haven't really seen the question yet so I'm not really sure if I'm missing any previous results



Nice, I'm on a gap year as well. So not happy to have received a STEP paper as part of an offer!


We're in pretty similar positions then :P What did you apply for? I got an offer for straight Comp Sci at Imperial.
Original post by Brubeckian
We're in pretty similar positions then :P What did you apply for? I got an offer for straight Comp Sci at Imperial.


Omg same, my offer is STEP II grade 1 and A* in FM for straight Comp Sci at Imperial :frown:
Original post by lllllllllll
Omg same, my offer is STEP II grade 1 and A* in FM for straight Comp Sci at Imperial :frown:


Ouch, they're making you resit A-levels during your gap year?
http://www.thestudentroom.co.uk/showthread.php?t=828655&page=4&p=32446708#post32446708

Will someone tell me why he takes moments about the point of contact with the floor rather than at O?

I mean, I don't know what I'm ignoring here, but when doing it his way, letting A be the pivot point, I can't get the distances that he does? E.g. I don't see how Mgcos(α) Mgcos(\alpha) is q q away from A, HOWEVER, I do see it being that far away from O (talking in terms of perpendicular distances).

Any help would be appreciated.
(edited 9 years ago)
Original post by Brubeckian
Ouch, they're making you resit A-levels during your gap year?


It's cool. I did my AS in further maths last year so it's basically halved my work load :smile:
Original post by physicsmaths
STEP I 2001 Q3
The last part to shown it is not necessary can't I just say if g^2=4h then the result still holds. Where the cubic has one double root and another root x=k.
????

Posted from TSR Mobile


For that part you can just say it isn't necessary as the cubic could not intersect the x-axis at all or only intersect once, draw some illustrations and you're done
Original post by C-king
For that part you can just say it isn't necessary as the cubic could not intersect the x-axis at all or only intersect once, draw some illustrations and you're done


Yeh I wrote this aswell. By my condition that I gave is the one siklos said aswell, which I found out after. Nice easy question though did it very quickly.


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