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STEP 2016 Solutions

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Original post by StrangeBanana
Isn't the point that the "gravitational part" of their trajectories (so to speak) is the same, so you can just forget about it? It's like having the same term present on two sides of an equation, allowing you to cancel.


Yu took it banana boy. Me sed it first


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Original post by mikelbird
Perhaps I am not understanding what you mean by cartesian trajectories...the actual trajectories are certainly are not straight lines and so I feel further explanation is needed. I suppose you are making the point about the fact that gravity seems to disappear in the second part when establishing the relationships with regard to theta but then you are not really dealing with a cartesian graph anymore (or the rescaling of the height coordinate needs some explanation).

Yes, I had shown that rbullet=rtarget    rA=rB\mathbf{r}^{\text{bullet}} = \mathbf{r}^{\text{target}} \iff \mathbf{r}^A=\mathbf{r}^B i.e. that gravity is irrelevant and that it is necessary (and sufficient, but that's not important here) that A,BA, B collide in order for the bullet to hit the target.

Furthermore, by "cartesian trajectory", I meant the equation of motion that relates xx and yy and I presume this is what you meant by "actual trajectory". The trajectories of AA and BB (that is, of the particles of part (i)) are straight lines - this is immediate from the parametric forms of part (i) by eliminating tt (mentally or otherwise) and indeed what one expects if gravity does not influence the motion. So, for the bullet to hit the target, it is therefore necessary that these lines must intersect in the upper right quadrant. My conclusion follows.*

I must say this is all in my explanation for the last part, and I felt this was the whole point of the question. I'm still unsure where you feel an issue lies.
Original post by Farhan.Hanif93
Yes, I had shown that rbullet=rtarget    rA=rB\mathbf{r}^{\text{bullet}} = \mathbf{r}^{\text{target}} \iff \mathbf{r}^A=\mathbf{r}^B i.e. that gravity is irrelevant and that it is necessary (and sufficient, but that's not important here) that A,BA, B collide in order for the bullet to hit the target.

Furthermore, by "cartesian trajectory", I meant the equation of motion that relates xx and yy and I presume this is what you meant by "actual trajectory". The trajectories of AA and BB (that is, of the particles of part (i)) are straight lines - this is immediate from the parametric forms of part (i) by eliminating tt (mentally or otherwise) and indeed what one expects if gravity does not influence the motion. So, for the bullet to hit the target, it is therefore necessary that these lines must intersect in the upper right quadrant. My conclusion follows.*

I must say this is all in my explanation for the last part, and I felt this was the whole point of the question. I'm still unsure where you feel an issue lies.


I understand what you are saying completely....perhaps I worded what I said too strongly but on first reading I did not feel what you were saying was crystal clear...so probably my response was 'over enthusiastic'!!
I Think this solves it!!

Spoiler

(edited 5 years ago)
Original post by mikelbird
I Think this solves it!!


The other method is fine aswell. By putting z=1 you aren't dividing by zero if that was your issue since all you are doing is plugging z=1 into two equal polynomials. All you were noting was that those polynomials are equal due to the factored forms.


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step is bulsht
Reply 326
Original post by Llewellyn
Well, actually, I think it could maybe be done.

https://docs.google.com/document/d/1bB7weoe8IFmfGUsulb9bLEsrXrBpTife2IicyPG-3xc/edit?usp=sharing

Very tricky to know the exact mark allocation... I'm tempted to reduce marks by line for each part and add an overall "get the gist of the question" or "bonus mark(s) at the end for getting a full solution"

I'll add in the rest of the pure section later. Anyone else free to edit/ improve.


Was the rest of this ever made?

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Reply 327
Original post by fa991
Was the rest of this ever made?

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Nah
Step III Question 7


mathsquestion.jpg
(edited 5 years ago)
Original post by Zacken
I left my answer was ((7^7 + 1)^3 - sum of 7's)((7^7 + 1)^3 + sum of 7's). :redface:


In your solution how did you go from the power of 6 to 2 as you would surely use the 4th row of the binomial expansion which is 14641
Reply 330
Original post by Salimshady007
In your solution how did you go from the power of 6 to 2 as you would surely use the 4th row of the binomial expansion which is 14641


Not sure what you're asking, but 2*3 = 6. So you can view it as a difference of cubes.
Original post by Zacken
Not sure what you're asking, but 2*3 = 6. So you can view it as a difference of cubes.


sorry just realised how you got that (sigh kmn)
STEP III Question 11

i)

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ii)

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iii)

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Edits: minor corrections with capital letters.
(edited 5 years ago)
STEP III Question 12

i)

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ii)

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STEP III Question 13

Spoiler



i

Spoiler



ii.

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iii

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Inevitably made some mistake so let me know of anything.
(edited 5 years ago)
Original post by _gcx
STEP III Question 13

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i

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ii.

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iii

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Inevitably made some mistake so let me know of anything.


Small things but you said κ=E(X4)\kappa = \mathbb{E}(X^4) in part i? And misspelled kurtosis at beginning of iii. Otherwise I think it looks good. Definitely one of the nastier questions on the paper.
Original post by I hate maths
Small things but you said κ=E(X4)\kappa = \mathbb{E}(X^4) in part i? And misspelled kurtosis at beginning of iii. Otherwise I think it looks good. Definitely one of the nastier questions on the paper.


I think it was fine the way it was (ie before gcx edited it) given the question was about a normal variate with first parameter 0. Knowing the first parameter gives the mean has to be one of the most well known facts about the normal distribution, even at A-level.
Original post by I hate maths
Small things but you said κ=E(X4)\kappa = \mathbb{E}(X^4) in part i? And misspelled kurtosis at beginning of iii. Otherwise I think it looks good. Definitely one of the nastier questions on the paper.


Didn't notice that, thanks for pointing it out. I meant to just write E[(Xμ)4]=E[X4]\mathbb E \left[{(X - \mu)^4}\right] = \mathbb E \left[{X^4}\right], not sure how κ\kappa made it in there. I probably wouldn't have chosen it, I would've probably left part ii until the end at least. Question 12 is very generous, so I would've chosen that over 13 any day.

Original post by shamika
I think it was fine the way it was (ie before gcx edited it) given the question was about a normal variate with first parameter 0. Knowing the first parameter gives the mean has to be one of the most well known facts about the normal distribution, even at A-level.


The question does say the mean is 0.
Original post by shamika
I think it was fine the way it was (ie before gcx edited it) given the question was about a normal variate with first parameter 0. Knowing the first parameter gives the mean has to be one of the most well known facts about the normal distribution, even at A-level.


I was trying to draw attention to the fact that κ\kappa was already used to mean something else in the question (kurtosis). Nothing to do with the mean.

Original post by _gcx
Didn't notice that, thanks for pointing it out. I meant to just write E[(Xμ)4]=E[X4]\mathbb E \left[{(X - \mu)^4}\right] = \mathbb E \left[{X^4}\right], not sure how κ\kappa made it in there. I probably wouldn't have chosen it, I would've probably left part ii until the end at least. Question 12 is very generous, so I would've chosen that over 13 any day.



The question does say the mean is 0.


By the way as an aside to this, it's great that you can do step 3 questions in year 12, but in my opinion you really ought to be saving the newest papers for timed mocks closer to your step exams. Say, 2012 or 2013 and beyond.
Original post by I hate maths
I was trying to draw attention to the fact that κ\kappa was already used to mean something else in the question (kurtosis). Nothing to do with the mean.



By the way as an aside to this, it's great that you can do step 3 questions in year 12, but in my opinion you really ought to be saving the newest papers for timed mocks closer to your step exams. Say, 2012 or 2013 and beyond.


Yeah, you make a fair point, I only did this question because I probably wouldn't have picked it. Q12 I had done previously when I started to get into stats. At least I've only spoiled 2 questions :tongue:

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