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STEP Prep Thread 2015

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Original post by raff97
Doing the same things as in the earlier parts, what can you deduce about p(-1) and the factorisation of p'(x)?


I figured the whole question out straight after posting. I deduced the p(-1) but then used the fact the later derivitives are also zero at -1 etc


Posted from TSR Mobile
For II 2006 Q3 in the first part. I'm not really sure how to approach the question. I could turn the number into one fraction then expand all the terms to show it's an integer but surely there must be a more efficient method?


Posted from TSR Mobile
Original post by lllllllllll
For II 2006 Q3 in the first part. I'm not really sure how to approach the question. I could turn the number into one fraction then expand all the terms to show it's an integer but surely there must be a more efficient method?


Posted from TSR Mobile


(5+root24)^-4 = (... - ...)^4
Original post by lllllllllll
For II 2006 Q3 in the first part. I'm not really sure how to approach the question. I could turn the number into one fraction then expand all the terms to show it's an integer but surely there must be a more efficient method?

Spoiler

Original post by Brubeckian
(5+root24)^-4 = (... - ...)^4


Original post by DFranklin

Spoiler



Thanks, got the answer now. Very simple :smile:


Posted from TSR Mobile
I need some help with STEP III 1993 Q9 (iv).

My generalisation is

(n1)i=1nai22aiaj \displaystyle (n-1)\sum^{n}_{i=1} a^2_i \geq 2 \sum a_i a_j

I'm not quite sure how to prove it. My first thought was induction, but I think the inductive step will be quite complicated here.

Also, I wasn't sure on the notation for the sum on the RHS. The part after this uses the following to define the summation so I think it works here as well "where the latter sum is taken over all pairs (i,j) (i,j) with 1i<jn 1 \leq i < j \leq n ".
Original post by ThatPerson
I need some help with STEP III 1993 Q9 (iv).

My generalisation is

(n1)i=1nai22aiaj \displaystyle (n-1)\sum^{n}_{i=1} a^2_i \geq 2 \sum a_i a_j

I'm not quite sure how to prove it. My first thought was induction, but I think the inductive step will be quite complicated here.
I think direct is easier here. I would consider an individual term on the RHS and show it's <= (a suitable sum of squares) / 2.

Also, I wasn't sure on the notation for the sum on the RHS. The part after this uses the following to define the summation so I think it works here as well "where the latter sum is taken over all pairs (i,j) (i,j) with 1i<jn 1 \leq i < j \leq n ".
If you want to be careful, you'd do something like 1i<jnaiaj\displaystyle \sum_{1\leq i < j \leq n} a_i a_j but i<jaiaj\displaystyle \sum_{i<j} a_i a_j is sufficient here, really.
Step 3 2012 (i) is giving me a head ache, when it says 'touch' does that include intersections? Or literally just when they touch (no intersection)?
Original post by Brubeckian
Step 3 2012 (i) is giving me a head ache, when it says 'touch' does that include intersections? Or literally just when they touch (no intersection)?


They mean touch as in the same way you would say a tangent "touches" the the curve it is tangent to, but importantly in this case it never meets with the other line again apart from that one point where they touch.
Would anyone know if this is the last year for AEA? The information seems quite mixed.
Original post by Krollo
Would anyone know if this is the last year for AEA? The information seems quite mixed.


Yes it is.
Original post by Jai Sandhu
Yes it is.


Well ****. Is it worth doing it this year then?
Hello clever STEP people

I am stuck with a problem and need to show something which I do not think is too bad
(me being too tired and having a bad day...)

In order to process the problem I need to show the result in the attachment.
It has to be true because then the problem works.
I tried the obvious substitution/translation to the left but I am stuck
anyone please?
Original post by Jai Sandhu
Yes it is.


It isnt. People have found out from edexcel that it isnt.


Posted from TSR Mobile
Q8 II 2012 stuck on the last part.
Using the first resul I have got to
Un+1-Un>Un-Un-1
Not sure what to do from here.


Posted from TSR Mobile
Original post by Krollo
Well ****. Is it worth doing it this year then?


No point if your doing I.


Posted from TSR Mobile
Original post by TeeEm
Hello clever STEP people

I am stuck with a problem and need to show something which I do not think is too bad
(me being too tired and having a bad day...)

In order to process the problem I need to show the result in the attachment.
It has to be true because then the problem works.
I tried the obvious substitution/translation to the left but I am stuck
anyone please?
Did you try the sub u=2Lxu = 2L - x?
Original post by DFranklin
Did you try the sub u=2Lxu = 2L - x?


I tried u = x - L
Original post by TeeEm
I tried u = x - L
I think u = 2L-x works.
Original post by DFranklin
I think u = 2L-x works.


thanks
I will try it

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