The Student Room Group

How to lay out a question properly (STEP)

Some of the more knowledgable people on this forum have complained about the general lack of ability of candidates to lay out their answers properly. I think it would be quite useful for those taking STEP if some of the people who took it in the past could just post some specific things that a lot of people don't do. I could even post a solution I did just now and people could give some constructive criticism because unfortunately I am really bad at laying out answers.
Just a thought, if you don't think this would be beneficial then let this thread flounder.

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Reply 1
Very good idea. I will post some of my solutions to step so people can constructively criticise them.
(edited 13 years ago)
Reply 2
The big one I see is that people fail to connect their lines together. So they'd write something like:

x2+2bx+c=0x^2 +2bx + c = 0

(xb)2+cb2=0(x-b)^2+c-b^2 = 0

xb=±cb2x-b = \pm \sqrt{c-b^2}

x=b±cb2x = b \pm \sqrt{c-b^2}

Now in this case it's not that likely to cause confusion, but in principle an examiner could be looking at this and thinking "Are these 4 separate equations, or are they supposed to be linked?".

You don't need to write much, but something that indicates how the lines are joined together makes it a lot clearer. My personal inclination is to link with words, and lay out horizontally, so you actually need less vertical space overall:

Suppose x2+2bx+c=0x^2 +2bx + c = 0. Then (xb)2+cb2=0(x-b)^2+c-b^2 = 0, so xb=±cb2x-b = \pm \sqrt{c-b^2} and finally x=b±cb2x = b \pm \sqrt{c-b^2}

Diagrams are also helpful (if difficult to impossible on TSR).

Not really to do with "layout", but make sure you're clear about the difference between "A implies B" and "B implies A". I'm amazed how many people doing STEP drop easy marks by getting confused about this.

Other than that, yeah, if you post a solution I'll give feedback.
Reply 3
Original post by DFranklin
The big one I see is that people fail to connect their lines together. So they'd write something like:

x2+2bx+c=0x^2 +2bx + c = 0

(xb)2+cb2=0(x-b)^2+c-b^2 = 0

xb=±cb2x-b = \pm \sqrt{c-b^2}

x=b±cb2x = b \pm \sqrt{c-b^2}

Now in this case it's not that likely to cause confusion, but in principle an examiner could be looking at this and thinking "Are these 4 separate equations, or are they supposed to be linked?".

You don't need to write much, but something that indicates how the lines are joined together makes it a lot clearer. My personal inclination is to link with words, and lay out horizontally, so you actually need less vertical space overall:

Suppose x2+2bx+c=0x^2 +2bx + c = 0. Then (xb)2+cb2=0(x-b)^2+c-b^2 = 0, so xb=±cb2x-b = \pm \sqrt{c-b^2} and finally x=b±cb2x = b \pm \sqrt{c-b^2}

Diagrams are also helpful (if difficult to impossible on TSR).

Not really to do with "layout", but make sure you're clear about the difference between "A implies B" and "B implies A". I'm amazed how many people doing STEP drop easy marks by getting confused about this.

Other than that, yeah, if you post a solution I'll give feedback.


I think using arrows => is more convenient
Reply 4
Original post by Hermione17
I think using arrows => is more convenient
The biggest problem with using     \implies is when your are already manipulating statements with implication signs.

So, if you're doing something with limits, you might want to start from the definition of pointwise convergence:

ϵ>0,Ns.t.n>N    ana<ϵ\forall \epsilon > 0, \exists N {\text s.t. } n > N \implies |a_n - a| < \epsilon.

If I were then to write:

    ϵ>0,Ns.t.n>N    an2a2<ϵan+a\implies \forall \epsilon > 0, \exists N {\text s.t. } n > N \implies |a_n^2 - a^2| < \epsilon |a_n+a|.

Then it quickly gets confusing about what is implying what. (Particularly if you don't use a new line).

But of course, if you want to use     \implies that's fine. Do be aware, however, that there are no prizes for the person with the highest "mathematical symbol to English words" ratio. (Overuse of symbols and underuse of English is, IMHO, another thing people tend to do badly when laying out mathematics).
Reply 5
Original post by DFranklin
The biggest problem with using     \implies is when your are already manipulating statements with implication signs.

So, if you're doing something with limits, you might want to start from the definition of pointwise convergence:

ϵ>0,Ns.t.n>N    ana<ϵ\forall \epsilon > 0, \exists N {\text s.t. } n > N \implies |a_n - a| < \epsilon.

If I were then to write:

    ϵ>0,Ns.t.n>N    an2a2<ϵan+a\implies \forall \epsilon > 0, \exists N {\text s.t. } n > N \implies |a_n^2 - a^2| < \epsilon |a_n+a|.

Then it quickly gets confusing about what is implying what. (Particularly if you don't use a new line).

But of course, if you want to use     \implies that's fine. Do be aware, however, that there are no prizes for the person with the highest "mathematical symbol to English words" ratio. (Overuse of symbols and underuse of English is, IMHO, another thing people tend to do badly when laying out mathematics).


Thanks for that :smile:
I often use a new line. I also use } to group all the points that implies something else to avoid confusion

The reason why I prefer using that symbol is just because I feel it's more convenient than using "so", "then", "and finally", it's also what I am used to using, and what I was taught to use. In a STEP exam, thinking about the solutions is hard already, so I always write them down in an acceptable and more convenient way.

Also, while writing down my thoughts, such as A => B, then later realise than B => A is also true, and it's needed for my next step, then I would just need to add < to make it <=>. It saved valuable time for me many times
Reply 6
Please point out anything I could do better.


Step iii 1988 question 1.

sketch the function h, where h(x)=ln(x)xh(x) =\frac{ln(x)}{x} for x>0
Hence or otherwise find all pairs of distinct integers nm=mn n^m=m^n


Let h(x)=ln(x)x=0h(x) =\frac{ln(x)}{x}=0 then x=ex=e The graph has a zero at x=e

x0x\rightarrow 0 Therefore ln(x)x\frac{ln(x)}{x}\rightarrow -\infty So h approaches -infinity when x approaches 0.

xx\rightarrow \infty then ln(x)x\frac{ln(x)}{x}\rightarrow 0. So h approaches 0 when x approaches infinity.




h(x)=lnxxh(x)=\frac{lnx}{x}then h(x)=(1xxln(x))x2=1lnxx2h'(x)=\frac{(\frac{1\cdot x}{x}-ln(x))}{x^2}=\frac{1-lnx}{x^2}

Therefore.
h(x)=0h'(x)=0 when x=e1x=e^1
h(x)>0h'(x)>0 when x<e x<e
h(x)<0h'(x)<0 when x>e x>e


h(x)=1xx22x(1lnx))x4=3+2ln(x)x2{\Large h''(x)=\frac{\frac{-1}x{x^2}-2x(1-lnx))}{x^4}}=\frac{-3+2ln(x)}{x^2}

Therefore
h(x)>0h''(x)>0 when x>e32 x>e^{\frac{3}{2}}
h(x)<0h''(x)<0 when x<e32 x<e^{\frac{3}{2}}

When x=0, h(x) is at -infinity. h(x) is increasing at a decreasing rate until x=e1x=e^1, then h(x) begins to decrease because h(x)<0h'(x) <0. The decrease in h(x) is at first at a decreasing rate, until x=e3/2x=e^{3/2} at which point h(x) becomes convex to the origin and decreases at a decreasing rate. h converges to 0.

Graph looks like this
http://www.wolframalpha.com/input/?i=ln%28x%29%2Fx





Hence find all distinct pairs of positive integers, nm=mnn^m=m^n

nm=mnnln(m)=mln(n)n^m=m^n \Rightarrow n\cdot ln(m)=m\cdot ln(n) \Rightarrow
ln(m)m=ln(n)n\frac{ln(m)}{m}=\frac{ln(n)}{n}


Here is where the graph comes in handy. We are looking for distinct pairs sonm n \neq m
Thus we can imagine drawing a horizontal line through our graph. The intersections are
real solutions to the equation. ln(m)m=ln(n)n\frac{ln(m)}{m}=\frac{ln(n)}{n}

Therefore we only need to look as far as x=e^1, since this is the maximum of h. Therefore the n, m which are solutions satisfy min(n,m)2min(n,m)\leq 2.

so m^2 =2^m or m=1. But no solutions to m=1, because n=m.

m^2=2^m. We find that m=4 satisfies the equation. This is a unique solution (excluding m=2). Hence

(4,2) is the only pair n, m that satisfy.
(edited 13 years ago)
Original post by Hermione17

Also, while writing down my thoughts, such as A => B, then later realise than B => A is also true, and it's needed for my next step, then I would just need to add < to make it <=>. It saved valuable time for me many times


It saves time, but it's worth being careful. There are occasions, usually when the question consists of mere algebraic manipulation, that our arguments can be reversed in order to obtain and 'if and only if' relationship. However, there are many more situations where one cannot, and we require separate consideration of our hypotheses and our desired conclusion. I say this because it's something that becomes very important at university level, whereby one has many theorems floating around in ether, requiring various hypotheses and giving several results that may be stronger, weaker, equivalent, etc., yet such consideration is almost never required at A level. And naturally the examiner is keen to test awareness of this. In my opinion, if you want to maximize your marks, unless it is obvious your argument can be reversed, I would break A iff B down into A -> B and B -> A and give equal care to both. I'm not saying you're guilty of this, your justification being notational, but it's salient to STEP exams.
Original post by methusaleh

Hence there are no solutions.


I need to go somewhere so can't look at your whole solution, but this isn't true!
Reply 9
Original post by methusaleh
x


There is a silly mistake here

h(x) = 0 => lnx = 0 => x=1 not x=e
made another silly mistake but rectified it before any of you could notice, hopefully.
there may be other mistakes up for grabs who knows.
(edited 13 years ago)
Original post by methusaleh
Please point out anything I could do better.


Step iii 1988 question 1.

sketch the function h, where h(x)=ln(x)xh(x) =\frac{ln(x)}{x} for x>0
Hence or otherwise find all pairs of distinct integers nm=mn n^m=m^n




Hence there are no solutions.


I remember answering a similar question for my entrance exam - looked it up as Maths II 23rd Nov 1973 Q1. On that occasion it asked "how many positive values of x correspond to a given value of h?" And it asked for a sketch of x^y = y^x.

I recall 2^4 = 4^2.
Original post by ian.slater
I remember answering a similar question for my entrance exam - looked it up as Maths II 23rd Nov 1973 Q1. On that occasion it asked "how many positive values of x correspond to a given value of h?" And it asked for a sketch of x^y = y^x.

I recall 2^4 = 4^2.



Oh bother, some1 caught me out. I did make the correction before your post mr slater, I'd just like to add :P
Original post by methusaleh
x0x\rightarrow 0 Therefore ln(x)x\frac{ln(x)}{x}\rightarrow -\infty So h approaches -infinity when x approaches 0.


I think we should justify why h approaches - infinity instead of + infinity

maybe something like this:
when 0<x<1, lnx <0 => lnx/x <0
=> it approaches - infinity when x approaches 0?
Original post by Hermione17
I think we should justify why h approaches - infinity instead of + infinity

maybe something like this:
when 0<x<1, lnx <0 => lnx/x <0
=> it approaches - infinity when x approaches 0?


Thankyou, yes I agree. I was also worried about justifying ln(x)/x approaches 0 when x=> infinity. Do you know of any fancy formula I can quote.
How do I justify ln(x)/x is => 0 as x=>infinity? :confused:
Reply 15
Original post by methusaleh
Thankyou, yes I agree. I was also worried about justifying ln(x)/x approaches 0 when x=> infinity. Do you know of any fancy formula I can quote.
How do I justify ln(x)/x is => 0 as x=>infinity? :confused:

Don't quote me on this, but I think you're allowed to assume L'Hopital's rule in STEP. Google it to find out what it is.
Original post by Glutamic Acid
It saves time, but it's worth being careful. There are occasions, usually when the question consists of mere algebraic manipulation, that our arguments can be reversed in order to obtain and 'if and only if' relationship. However, there are many more situations where one cannot, and we require separate consideration of our hypotheses and our desired conclusion. I say this because it's something that becomes very important at university level, whereby one has many theorems floating around in ether, requiring various hypotheses and giving several results that may be stronger, weaker, equivalent, etc., yet such consideration is almost never required at A level. And naturally the examiner is keen to test awareness of this. In my opinion, if you want to maximize your marks, unless it is obvious your argument can be reversed, I would break A iff B down into A -> B and B -> A and give equal care to both. I'm not saying you're guilty of this, your justification being notational, but it's salient to STEP exams.


Thanks :smile:
I was talking about the obvious ones

As I said earlier, it's also how I was taught. My teacher usually uses that symbol, and I am used to using it. Symbols are invented to make things simpler and easier to present the ideas
Reply 17
Original post by methusaleh
Thankyou, yes I agree. I was also worried about justifying ln(x)/x approaches 0 when x=> infinity. Do you know of any fancy formula I can quote.
How do I justify ln(x)/x is => 0 as x=>infinity? :confused:


Tbh, unless it's explicitly in the question, I'd say it's fairly obvious... You could always do something along the lines of l'hopital's rule, and differentiate top and bottom, and consider what will be happening to each, but I wouldn't bother justifying it, unless they ask, when you can generally fudge something, or it's genuinely difficult and it took you doing some algebra/calculus to work out the limit. I'm not really a fan of STEP limits questions because of this.. I'm never really sure how thorough they expect you to be, and so I generally try to justify, but be fairly hand-wavey!
Original post by methusaleh
Thankyou, yes I agree. I was also worried about justifying ln(x)/x approaches 0 when x=> infinity. Do you know of any fancy formula I can quote.
How do I justify ln(x)/x is => 0 as x=>infinity? :confused:

I'm a little concerned by the use of implication signs (=>) rather than straight forward arrows (->) to represent the 'tending to' symbol.

L'Hopital's rule states that, if limxf(x)g(x)\displaystyle\lim_{x\to \infty} \dfrac{f(x)}{g(x)} (where f(x) and g(x) are continuously differentiable to the nth degree) leads to an indeterminate answer (i.e. \dfrac{\infty}{\infty}) then:

limxf(x)g(x)=limxf(x)g(x)\displaystyle\lim_{x\to \infty} \dfrac{f(x)}{g(x)} = \displaystyle\lim_{x\to \infty} \dfrac{f'(x)}{g'(x)}

I may have missed some details and I'm not sure if you're allowed to assume this without proof in STEP but that's seemingly the obvious way to evaluate this limit.
Original post by Mc^3
Tbh, unless it's explicitly in the question, I'd say it's fairly obvious... You could always do something along the lines of l'hopital's rule, and differentiate top and bottom, and consider what will be happening to each, but I wouldn't bother justifying it, unless they ask, when you can generally fudge something, or it's genuinely difficult and it took you doing some algebra/calculus to work out the limit. I'm not really a fan of STEP limits questions because of this.. I'm never really sure how thorough they expect you to be, and so I generally try to justify, but be fairly hand-wavey!

Agreed I always have my reservations about dealing with limits in STEP. I have no idea how I could use algebraic manipulation to rewrite that limit as something I can deal with without L'Hopital's, I could only do it using a geometrical argument which, as you say, is pretty hand-wavey.

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