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The 2012 STEP Results Discussion Thread

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Reply 1280
Avatar for beelz
beelz
If you have two vectors a and b, am I right in thinking that

( lblb - lala )^2 = lbl^4 - 2lallbla.b + lal^2

If so, why does (lblb)^2 become lbl^4?

(Confused about this after doing STEP II 2008 Q8)
Original post by beelz
If you have two vectors a and b, am I right in thinking that

( lblb - lala )^2 = lbl^4 - 2lallbla.b + lal^2

If so, why does (lblb)^2 become lbl^4?

(Confused about this after doing STEP II 2008 Q8)

I haven't checked the question but your expression is not quite correct (check the last term).
As for your other question:

(∣b∣b)2=(∣b∣b)⋅(∣b∣b)=∣b∣2(b⋅b)=∣b∣2×∣b∣2=∣b∣4(|\mathbf{b}|\mathbf{b})^2 = (|\mathbf{b}|\mathbf{b})\cdot(| \mathbf{b}| \mathbf{b}) = |\mathbf{b}|^2 (\mathbf{b}\cdot \mathbf{b}) = |\mathbf{b}|^2 \times |\mathbf{b}|^2 = |\mathbf{b}|^4

EDIT: I should also mention that you should be wary of using "^2" notation on a vector due to the possible (although it would be nitpicking) ambiguity as there is more than one way to multiply vectors together. I'm not sure whether STEP examiners will punish this or not as it is pretty obvious what you're trying to say in this context.
(edited 12 years ago)
Reply 1282
Avatar for Roy064
Roy064
1
Time to join this thread I guess, starting STEP preparation today for STEP 1 and potentially 2 as well depending on how things progress.
Original post by Aurum
Your factorization is incorrect, the other factor should be
[br]ab2+b(a−1)+a−1[br][br]ab^2+b(a-1)+a-1[br]


Thanks. But still don't know what to do with discriminant.
D=(a−1)2−4a(a−1)=−3a2+2a+1 D=(a-1)^2-4a(a-1)=-3a^2+2a+1
Original post by Dog4444
Thanks. But still don't know what to do with discriminant.
D=(a−1)2−4a(a−1)=−3a2+2a+1 D=(a-1)^2-4a(a-1)=-3a^2+2a+1


D=(1−a)(1+3a)D=(1-a)(1+3a)

You're not looking for an easy factorisation here...just bang it in to the quadratic formula and divide through by 1−a1-a to find k. The final answer is not 'nice'
Original post by Roy064
Time to join this thread I guess, starting STEP preparation today for STEP 1 and potentially 2 as well depending on how things progress.


Cambridge offer?
Reply 1286
Avatar for Roy064
Roy064
1
Original post by wcp100
Cambridge offer?


I wish mate, Warwick.

You Cambridge?
Original post by TheMagicMan
D=(1−a)(1+3a)D=(1-a)(1+3a)

You're not looking for an easy factorisation here...just bang it in to the quadratic formula and divide through by 1−a1-a to find k. The final answer is not 'nice'


Not nice at all. I usually expect a nice answer, it usually means that you're on a right way.
:confused:
Is it just me or are the old STEP papers (before 2004) much easier? 2011 paper 2 seems way harder than some of the old ones.
(edited 12 years ago)
Original post by oh_1993
Is it just me or are the old STEP papers (before 2004) much easier? 2011 paper 2 seems way harder than some of the old ones.


Question 7 is easy, if you're not put off by the nested sum, and...

Spoiler



Question 6 is fine if you can do the very first part properly. Questions 12 and 13 are also godsends, but then I typically find the stats very easy on a STEP paper (which is probably reassuring given I am an actuary!)
Reply 1290
Avatar for beelz
beelz
Original post by Farhan.Hanif93
I haven't checked the question but your expression is not quite correct (check the last term).
As for your other question:

(∣b∣b)2=(∣b∣b)⋅(∣b∣b)=∣b∣2(b⋅b)=∣b∣2×∣b∣2=∣b∣4(|\mathbf{b}|\mathbf{b})^2 = (|\mathbf{b}|\mathbf{b})\cdot(| \mathbf{b}| \mathbf{b}) = |\mathbf{b}|^2 (\mathbf{b}\cdot \mathbf{b}) = |\mathbf{b}|^2 \times |\mathbf{b}|^2 = |\mathbf{b}|^4

EDIT: I should also mention that you should be wary of using "^2" notation on a vector due to the possible (although it would be nitpicking) ambiguity as there is more than one way to multiply vectors together. I'm not sure whether STEP examiners will punish this or not as it is pretty obvious what you're trying to say in this context.


Ahh, thank you very much monsieur Hanif, I think I was being a bit stupid on that question as I'd used the fact that b.b = lbl^2 in an earlier part but somehow forgot this. I am curious about why when you expand out the brackets you are supposed to use the dot product instead of the cross product.
Original post by oh_1993
Is it just me or are the old STEP papers (before 2004) much easier? 2011 paper 2 seems way harder than some of the old ones.


wouldn't say much easier.....I would just say quicker to do once you notice how to solve the problem. your referring to the pure section I presume mainly?
Original post by beelz
Ahh, thank you very much monsieur Hanif, I think I was being a bit stupid on that question as I'd used the fact that b.b = lbl^2 in an earlier part but somehow forgot this. I am curious about why when you expand out the brackets you are supposed to use the dot product instead of the cross product.

Well I assumed that by (bb−aa)2(b\mathbf{b} - a\mathbf{a})^2, you meant (bb−aa)⋅(bb−aa)(b\mathbf{b}-a\mathbf{a})\cdot (b\mathbf{b} - a\mathbf{a}) (based on the form of the expansion you put forward). Given that I haven't worked the question myself and the fact the vector questions tend to have multiple possible approaches, I can't tell you why you're using the dot product here unless you tell me more about where the expression that you're "squaring" came from.

One weak reason for using the dot product is that cross product isn't on the syllabus so it's likely that the question was designed to avoid it. Again, that depends on the route you took.
Original post by Roy064
I wish mate, Warwick.

You Cambridge?


Year 12 lurking really :tongue: I want to do STEP next year for the fun of it. Chem. Eng. prospective applicant.
Reply 1294
Avatar for Aurum
Aurum
Original post by Dog4444
Not nice at all. I usually expect a nice answer, it usually means that you're on a right way.
:confused:


It's correct, but if you wanted it to look a bit more 'nice', you could divide throughout by 1-a, and then the only not 'nice' thing you would be left with would be under the square root sign. Either way, your answer is correct.
Original post by Rahul.S
wouldn't say much easier.....I would just say quicker to do once you notice how to solve the problem. your referring to the pure section I presume mainly?


yeah take 1996 STEP 2 for example, every pure question seems to have less to it than recent papers.
Original post by oh_1993
yeah take 1996 STEP 2 for example, every pure question seems to have less to it than recent papers.


I have done that paper....umm I actually think it was a decent paper. But yh once you get into a question in that paper.....there is a high chance of a complete solution. I think the recent papers just take abit longer....on the other hand STEP II 2010 was a very nice paper :tongue:

did you apply to Trinity by the way? I think I remember your name somewhere on some list.....Im guessing you got an offer :cool:
Quoted this in a STEP question. overkill? :tongue:
Original post by Rahul.S
I have done that paper....umm I actually think it was a decent paper. But yh once you get into a question in that paper.....there is a high chance of a complete solution. I think the recent papers just take abit longer....on the other hand STEP II 2010 was a very nice paper :tongue:

did you apply to Trinity by the way? I think I remember your name somewhere on some list.....Im guessing you got an offer :cool:


I applied to Queens' but was unsuccessful after interview although I will likely be reapplying next year because I have decided to take a year out. I am taking all 3 STEP papers for Warwick offer and possible reapplication. Waaaay more fun than Maths and Further Maths was. :smile:
Original post by ben-smith
Quoted this in a STEP question. overkill? :tongue:


I think so haha :tongue: I think you could just state that it must intersect without quoting a theorem.. They wouldn't want you to prove it, anyway! :tongue:

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