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I hate maths
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#21
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#21
(Original post by Sonechka)
Do you think it would get 20? The solutions should be pretty much flawless (I'm not sure how signficant an error/omission needs to be for a mark to be dropped).
I'd think he would get 18 or 19. Commenting something to the effect of "without loss of generality, let points X_0,X_1,X_2,...,X_{n-1} be represented by complex numbers 1,\omega,\omega^2,...,\omega^{n-1} as rotating all points on the circle with respect to the origin does not change the distances of each point from each other" or something like that would be safer. The idea that the rotation is distance preserving is critical for the question to work.
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I hate maths
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#22
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#22
(Original post by Zacken)
I did this in my exam because of how easy it was, despite having done no mechanics preparation.



All good, last part is confusingly written, in my opinion (It starts off by assuming the result). All that's needed is:

\displaystyle

\begin{align*} \lambda(X_1 - X_2) &= 2 \log \frac{9}{5} - 1 - \frac{4}{3}\ln \frac{9}{8} \\ & = \frac{4}{3} \log 3 - 2 \log 5 - 1 + 4 \log 2 \\ & > \frac{1}{3} (4 \times 3.17  - 6 \times 1.61 - 3)  > 0 \end{align*}

and \lambda \geqslant 0 so X_1 > X_2.
I agree this question was a gift. However question 12 was even more of a gift, it might just be the shortest STEP question I've ever answered. Also about the assumption, as steps are reversible I used iff statements so it's cool, if I assumed X_1<X_2 instead it would just become a proof by contradiction since X_1<X_2 would imply something absurd. Thank you for improving on it though, I suppose we want people to see the neatest version.
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Zacken
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#23
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#23
(Original post by I hate maths)
I agree this question was a gift. However question 12 was even more of a gift, it might just be the shortest STEP question I've ever answered. Also about the assumption, as steps are reversible I used iff statements so it's cool, if I assumed X_1<X_2 instead it would just become a proof by contradiction since X_1<X_2 would imply something absurd. Thank you for improving on it though, I suppose we want people to see the neatest version.
I agree that the second paragraph is technically correct due to the iffs, but the first sentence “Suppose that X1 > X2” isn’t.
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I hate maths
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#24
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#24
(Original post by Zacken)
I agree that the second paragraph is technically correct due to the iffs, but the first sentence “Suppose that X1 > X2” isn’t.
Ah yes you're right about that, I didn't use that assumption in the end, it was something I wrote for myself when I didn't know which direction the proof was going in or what kind of proof it would be but forgot to remove it.
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etothepiiplusone
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#25
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#25
I'm not sure if this thread is still active, but the coverage for STEP II 2006 Q2 is poor, the penultimate part of the question is a graph sketch, which TSR, the official hints and answers thing and integral maths fail to attempt. Here's my $0.02:

Spoiler:
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We have that \mathrm{\frac{dy}{dx}} = 6e^{2x} - \frac{42}{4 - 3x}. At x=\frac12, \mathrm{\frac{dy}{dx}} = 6e - \frac{84}{5} < 6 \left( \frac{67}{24} \right) - \frac{84}{5} = - \frac{1}{10} < 0, using the upper bound for e we derived in the previous part. Likewise, at x=1, \dydx = 6e^2 - 42 > 6 \left( \frac83 \right)^2 - 42 = \frac23 > 0, using the lower bound for e. Thus \mathrm{\frac{dy}{dx}} changes sign between x=\frac12 and x=1; it must be somewhere between these two values since it's clearly continuous in this interval.

Desmos, in checking your answer, is your friend from here onwards.

Let's first consider the behaviour of y at the extreme values. It's undefined at x=\frac43, but as x approaches \frac43, \ln approaches negative infinity so y approaches negative infinity. As x instead approaches negative infinity, again the 3e^{2x} term does nothing significant, but the \ln term here approaches positive infinity, albeit as \mathrm{\frac{dy}{dx}} decreases; the graph is concave as opposed to convex.

Let's now consider further the stationary points of y. We know there exists one between x=\frac12 and x=1, but there in fact exists another: consider \mathrm{\frac{dy}{dx}} = 6e^{2x} - \frac{42}{4 - 3x} = 0 or 6e^{2x} = \frac{42}{4-3x}; the graphs y=6e^{2x} and y=\frac{42}{4-3x} intersect somewhere between a half and one of course, but also somewhere else at a smaller value of x since \frac{42}{4-3x} does not grow smaller faster than 6e^{2x} (seems overkill for STEP II, but is necessary for determining the shape of the graph). Considering these graphs visually moreover, we see that these are the only two stationary points of y.

We now have all the information to get the shape of the graph. My sketch can (hopefully) be viewed (without axises, since the location of the intersections with the axises was not determined, nor the location of the stationary points) at https://ibb.co/if3GFK
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Quirky Object
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#26
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#26
(Original post by etothepiiplusone)
I'm not sure if this thread is still active, but the coverage for STEP II 2006 Q2 is poor, the penultimate part of the question is a graph sketch, which TSR, the official hints and answers thing and integral maths fail to attempt. Here's my $0.02:

Spoiler:
Show



We have that \mathrm{\frac{dy}{dx}} = 6e^{2x} - \frac{42}{4 - 3x}. At x=\frac12, \mathrm{\frac{dy}{dx}} = 6e - \frac{84}{5} < 6 \left( \frac{67}{24} \right) - \frac{84}{5} = - \frac{1}{10} < 0, using the upper bound for e we derived in the previous part. Likewise, at x=1, \dydx = 6e^2 - 42 > 6 \left( \frac83 \right)^2 - 42 = \frac23 > 0, using the lower bound for e. Thus \mathrm{\frac{dy}{dx}} changes sign between x=\frac12 and x=1; it must be somewhere between these two values since it's clearly continuous in this interval.

Desmos, in checking your answer, is your friend from here onwards.

Let's first consider the behaviour of y at the extreme values. It's undefined at x=\frac43, but as x approaches \frac43, \ln approaches negative infinity so y approaches negative infinity. As x instead approaches negative infinity, again the 3e^{2x} term does nothing significant, but the \ln term here approaches positive infinity, albeit as \mathrm{\frac{dy}{dx}} decreases; the graph is concave as opposed to convex.

Let's now consider further the stationary points of y. We know there exists one between x=\frac12 and x=1, but there in fact exists another: consider \mathrm{\frac{dy}{dx}} = 6e^{2x} - \frac{42}{4 - 3x} = 0 or 6e^{2x} = \frac{42}{4-3x}; the graphs y=6e^{2x} and y=\frac{42}{4-3x} intersect somewhere between a half and one of course, but also somewhere else at a smaller value of x since \frac{42}{4-3x} does not grow smaller faster than 6e^{2x} (seems overkill for STEP II, but is necessary for determining the shape of the graph). Considering these graphs visually moreover, we see that these are the only two stationary points of y.

We now have all the information to get the shape of the graph. My sketch can (hopefully) be viewed (without axises, since the location of the intersections with the axises was not determined, nor the location of the stationary points) at https://ibb.co/if3GFK


Thanks! This is just a place for people to flag up any incorrect solutions rather than a 'project' as such so it'll be active for as long as people keep submitting their corrections.
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DFranklin
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#27
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#27
(Original post by Quirky Object)
..
(Original post by Sonechka)
..
Just to inform you that the TSR redesign has broken the behaviour of the [expand] tag (when nested), which renders many of the STEP solutions unreadable.

And for the record, I put it to you that it really falls on TSR to fix the broken behaviour, rather than expect everyone to go around redoing the solutions.
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Quirky Object
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#28
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#28
(Original post by DFranklin)
Just to inform you that the TSR redesign has broken the behaviour of the [expand] tag (when nested), which renders many of the STEP solutions unreadable.

And for the record, I put it to you that it really falls on TSR to fix the broken behaviour, rather than expect everyone to go around redoing the solutions.
I've never expected everyone to go around redoing solutions...the point of this thread was for people to easily flag up any mistakes they happen to come across while using the TSR solutions as part of their preparation. There's nothing I can do about the [expand] tags until TSR fixes it but thanks for letting me know.
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DFranklin
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#29
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#29
(Original post by Quirky Object)
I've never expected everyone to go around redoing solutions...the point of this thread was for people to easily flag up any mistakes they happen to come across while using the TSR solutions as part of their preparation. There's nothing I can do about the [expand] tags until TSR fixes it but thanks for letting me know.
The thing is, (as I posted earlier on in the thread), there are more "invalid" solutions because of TSR changing the formatting behaviour than there are solutions with major mathematical issues. And pretty much none of them have been fixed. So I just wanted to emphasize that the only way the new issues will get fixed is if TSR fixes what they've broken. (Currently the post editing seems so broken I wouldn't even dare to try to fix any of them for fear of making it worse!).
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Quirky Object
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#30
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#30
(Original post by DFranklin)
The thing is, (as I posted earlier on in the thread), there are more "invalid" solutions because of TSR changing the formatting behaviour than there are solutions with major mathematical issues. And pretty much none of them have been fixed. So I just wanted to emphasize that the only way the new issues will get fixed is if TSR fixes what they've broken. (Currently the post editing seems so broken I wouldn't even dare to try to fix any of them for fear of making it worse!).
Ahh I see what you mean, the TSR website is in a bit of a state. I guess we'll just have to wait...
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I hate maths
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#31
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#31
Hi, just wondering if the TSR solution to Step 2 2007 question 7(ii)(b) is correct.

We are tasked with minimising x^5+y^5+z^5-5xyz.

By AM-GM:

\displaystyle 



\begin{align*} \frac{x^5+y^5+z^5+n+n^{-1}}{5} &\geqslant (x^5y^5z^5nn^{-1})^{\frac{1}{5}} \\ x^5+y^5+z^5-5xyz &\geqslant -(n+n^{-1}), \forall n \in \{ \mathbb{R} |n>0\} (*)\end{align*}

Then notice that again by AM-GM:

\displaystyle



\begin{align*}\frac{n+n^{-1}}{2} &\geqslant \sqrt{nn^{-1}} \\ n+n^{-1} &\geqslant 2 \end{align*}

with equality achieved when n=1. So the maximum of -(n+n^{-1}) is -2, and since (*) must also be satisfied by n=1 we have:

x^5+y^5+z^5-5xyz \geqslant -2 with equality achieved when x=y=z=1, so the minimum required is -2.

The answer that's already there does not consider the general n but jumps straight to n=1, is there an easy-to-see reason I'm not seeing for not considering other n? To my amateur eyes, the solution is incomplete as it stands.
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etothepiiplusone
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#32
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#32
(Original post by I hate maths)
the solution is incomplete as it stands
It's hardly incomplete, just lacking a bit of motivation.

You're given earlier in the question that in Jensen's inequality, there's equality iff all the variables are equal, so in your notation, n=n^{-1} at the point of equality. Because we need to use positive numbers in the inequality, n=1 instead of -1.

EDIT: Oops, I used $'s instead of [tex]'s. Changed.
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I hate maths
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#33
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#33
(Original post by etothepiiplusone)
It's hardly incomplete, just lacking a bit of motivation.

You're given earlier in the question that in Jensen's inequality, there's equality iff all the variables are equal, so in your notation, n=n^{-1} at the point of equality. Because we need to use positive numbers in the inequality, n=1 instead of -1.
In hindsight I actually now realise I really overcomplicated it, to prove a minimum it's enough to show f(x,y,z) \geqslant a and that f(x,y,z)=a for some values x,y,z. For some reason I confused myself into thinking considering general n was important to show.

I did not use n=-1 anywhere in my writing.
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etothepiiplusone
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#34
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#34
STEP II 1999 Q3 doesn't really have a solution, just some progress and then some extended and useful discussion, which I can't see actually culminating in a full proof. So here's my go at the parts from the first mention of 'induction' onwards:

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'Prove by induction on n that \mathrm{S}_n(x) is a polynomial.'

Base case: It's clear to see that \mathrm{S}_0(x) = e^{x^3}e^{x^{-3}} = 1, which is a polynomial.

Inductive step: Suppose \mathrm{S}_k(x) is a polynomial for some k. Then we have that \frac{\mathrm{d}^k}{\mathrm{d}x^  k}\left(e^{- x^3}\right) = e^{-x^3}\mathrm{S}_k(x) since \mathrm{S}_k(x) is a polynomial, so that the (non-polynomial) e^{x^3} in front of the derivative is cancelled. Thus (using the product rule in the third equality) we have

\mathrm{S}_{k+1}(x) = e^{x^3}\frac{\mathrm{d}^{k+1}}{d x^{k+1}} \left(e^{- x^3}\right) = e^{x^3} \frac{\mathrm{d}}{\mathrm{d}x} \left(e^{-x^3} \mathrm{S}_k(x) \right) = e^{x^3} \left(e^{-x^3} \mathrm{S}_k'(x) - 3x^2 e^{-x^3} \mathrm{S}_k(x) \right) = \mathrm{S}_k'(x) - 3x^2 \mathrm{S}_k(x) (*).

(TSR LaTeX is terrible (apart from anything else). The relevant part here is the relation \mathrm{S}_{k+1}(x) = \mathrm{S}_k'(x) - 3x^2 \mathrm{S}_k(x) (*).)

And since the derivative of a polynomial is a polynomial, we indeed have that \mathrm{S}_{k+1}(x) is a polynomial too, and our induction is complete.

The order of \mathrm{S}_n(x) is 2n. This is because the -3x^2 in front of the \mathrm{S}_k(x) term in (*) is what provides the largest power of x as the other term has lesser degree. The fact that \mathrm{S}_0(x) = 1 has order 0 gives us a base case that completes our argument.

The coefficient of the largest power of x, which is 1 in \mathrm{S}_0(x), is multplied by -3 each successive iteration due to (*). Thus the coefficient of \mathrm{S}_n(x) is (-3)^n.

By (*),

\mathrm{S}_n(a)\mathrm{S}_{n+1}(  a) = \mathrm{S}_n(a)\mathrm{S}_n'(a) - 3a^2 \mathrm{S}_n(a)^2 = - 3a^2 \mathrm{S}_n(a)^2 \leq 0

where the second equality follows from the fact that \mathrm{S}_n'(a)=0 and the inequality from the fact that squares are non-negative.

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etothepiiplusone
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#35
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#35
https://www.thestudentroom.co.uk/sho...5#post18366432 is another case of a TSR user posting a wrong solution that no one seems to have checked. In the OP or the post I just linked to, someone should link the integral maths solution - https://www.thestudentroom.co.uk/sho...5#post18366432 - since it's actually correct
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