STEP 2017 Solutions Watch

FractalSteinway
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#81
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#81
(Original post by Zacken)
If anybody is willing to LaTeX this up, please do!
For lack of anything better to do this morning I've written up the question paper in LaTeX. Do tell me if you catch any typos.
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math42
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#82
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(Original post by amd87)
Agree with above solution. Here's an 'or otherwise' solution to get the second part directly:

rth slice makes a sandwich iff (r-1)th slice starts a sandwich. This happens if, at the (r-1)th slice, you choose to make a sandwich (with probability q) AND you have the option to choose (which you have so as long as your are not forced to finish a sandwich you started with the previous slice).

Hence s_r = q(1-s_{r-1})
So you agreed that the sr formula holds for r = n too? Logically I feel it should, after all, it all depends on your decision or lack thereof at the n-1th step, which has no special characteristics relative to other steps. It just seems a bit odd the way the question is constructed.
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A Slice of Pi
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#83
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(Original post by 13 1 20 8 42)
So you agreed that the sr formula holds for r = n too? Logically I feel it should, after all, it all depends on your decision or lack thereof at the n-1th step, which has no special characteristics relative to other steps. It just seems a bit odd the way the question is constructed.
I looked at this question yesterday, and I thought the wording and structure was generally poor, unusually so for STEP. As for the sr formula, if I remember rightly, it was formed from the addition of two simultaneous equations, which were defined for different inequalities in r. That may well be why the endpoint is n-1 and not n.
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math42
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#84
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(Original post by A Slice of Pi)
I looked at this question yesterday, and I thought the wording and structure was generally poor, unusually so for STEP. As for the sr formula, if I remember rightly, it was formed from the addition of two simultaneous equations, which were defined for different inequalities in r. That may well be why the endpoint is n-1 and not n.
Yeah but, at least the way I did it, because you're subbing in t_r-1 rather than t_r, your permissible range for r becomes 3 </= r </= n (just shifting the original range 2 </= r </= n -1) and you have to show it for r = 2 separately since this involves t_1 which is not given by the formula. So then you've shown it for the whole range 2 </= r </= n.
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Rishabh_01
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#85
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#85
(Original post by FractalSteinway)
For lack of anything better to do this morning I've written up the question paper in LaTeX. Do tell me if you catch any typos.
Wow that's a pitch perfect replication of the original.
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Rishabh_01
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#86
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Q-1 completed
Q-2 completed the first two parts and in the third part integrated correctly but could not reach to the desiered inequality.
Q-3 completed
Q-5 completed but made a error in finding one part of the question.
Q-6 completed the first two parts and was unable to do the third part plus was i found the values if a and b in terms of alpha and not a numerical value
Q-4 make the graph and did the first part completly

Q-9 only wrote the equation of trajectory and simplified but was unable to reach to the aswer of the first part
Q-11 wrote only the equations of force of balance

Can someone tell me which grade i would get.
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A Slice of Pi
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#87
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#87
(Original post by Rishabh_01)
Q-1 completed
Q-2 completed the first two parts and in the third part integrated correctly but could not reach to the desiered inequality.
Q-3 completed
Q-5 completed but made a error in finding one part of the question.
Q-6 completed the first two parts and was unable to do the third part plus was i found the values if a and b in terms of alpha and not a numerical value
Q-4 make the graph and did the first part completly

Q-9 only wrote the equation of trajectory and simplified but was unable to reach to the aswer of the first part
Q-11 wrote only the equations of force of balance

Can someone tell me which grade i would get.
Should be OK for an S.
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Rishabh_01
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#88
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#88
I just 1 for my imperial offer, s would be great though .
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IrrationalRoot
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#89
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#89
(Original post by Rishabh_01)
I just 1 for my imperial offer, s would be great though .
Is this grade 1 in STEP in Imperial offers a new thing? Because when I applied last year, all I had to do was sit the MAT (and apparently not even do very well) and then I just got the standard A*A*A offer.
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Zacken
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#90
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(Original post by IrrationalRoot)
Is this grade 1 in STEP in Imperial offers a new thing? Because when I applied last year, all I had to do was sit the MAT (and apparently not even do very well) and then I just got the standard A*A*A offer.
FWIW, I got a 1,2 in II, III offer from Imperial last year. But I specifically asked for a STEP offer and didn't sit the MAT.
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Rishabh_01
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#91
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they asked for grade 1 in step 1 and grade 2 in step 2, competition is becoming tougher and tougher I guess these days so I guess that's why they are asking for step
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Rishabh_01
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#92
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#92
(Original post by Zacken)
FWIW, I got a 1,2 in II, III offer from Imperial last year. But I specifically asked for a STEP offer and didn't sit the MAT.
Do you study at Imperial Zacken, if yes then which branch and I would glad if you could help me out with some queries about Imperial College.
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alfmeister
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#93
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1. Completed all but stupidly miss differentiated the substitution for the last 2 integrals so didn't get the half required for both of the last ones. Apart from that its was fine.

2. Completed except considering y > 1

3. Everything except area of opq

4. Didn't show either of the quadratics

6. Most unsure about my proof here but got g(x) and showed g(x) takes a value of 0.

8. Just the induction.

How many do marks do you think I got, I know it's difficult to tell but thanks anyway.
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Zacken
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#94
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(Original post by Rishabh_01)
Do you study at Imperial Zacken, if yes then which branch and I would glad if you could help me out with some queries about Imperial College.
No, sorry. I'm a Cambridge mathmo.
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jkim1998
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#95
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#95
Is there anyone who can estimate my score or grade? I just need to get grade 2. Thanks

1. Completed except last integral
2. Completed the first two parts but took the wrong approach in the 3rd part
3. Completed Q-3 completed
5. Showed the the equation of area, didn't obtain an expression for s, did little bit of differentiating A but didn't get the required equation, but acquired s = x(1+sec B)
8. Completed except proving c(n) > root(2) - 1
9. Only showed that u^2 = gd tan a
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Rishabh_01
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#96
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#96
All the best for tomorrow guys whosoever is taking step -2
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MC11V33N
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#97
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#97
(Original post by Zacken)
I seemed to think it was familiar as well - but the only thing I could find was
You don't happen to know the year and paper this was from do you? Had a pop and would be nice to check, cheers.
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domgreen69
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#98
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(Original post by MC11V33N)
You don't happen to know the year and paper this was from do you? Had a pop and would be nice to check, cheers.
STEP 2, Year 2012
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MC11V33N
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#99
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#99
(Original post by domgreen69)
STEP 2, Year 2012
Ta
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Major-fury
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#100
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(Original post by FractalSteinway)
Question 9

Part (i)

Applying the constant acceleration formulae horizontally and vertically, we have y = u \sin \alpha t - \frac{1}{2}gt^2 and x = u\cos\alpha t where x and y are the horizontal and vertical displacements respectively of the particle from point O at time t. The latter gives t = \frac{x}{u\cos\alpha} which when substituted into the first equation yields
y = x \tan \alpha - \frac{gx^2}{2u^2\cos^2\alpha}.
We know the particle passes through point P, so (d,d\tan\beta) must be a solution. Hence,
d\tan\beta = d\tan\alpha - \frac{gd^2}{2u^2\cos^2\alpha}
and so as d \ne 0 for the situation to be physical,
tan\beta = \tan\alpha - \frac{gd}{2u^2\cos^2\alpha}.
Differentiating implicitly with respect to \alpha with d,\beta held constant,
0 = \sec^2\alpha - (-2)\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(\frac{du}{d\alpha}\cos\a  lpha - u\sin\alpha).
Now as u is minimised over \alpha, we set \frac{du}{d\alpha} = 0 (we will not be required to show this is a minimum) and so,
0 = \sec^2\alpha + 2\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(-u\sin\alpha)
\implies 0 = \frac{1}{\cos^2\alpha} - \frac{gdu\sin\alpha}{u^3\cos^3\a  lpha}
\implies \frac{gd\sin\alpha}{u^2\cos\alph  a} = 1
\implies u^2 = gd\tan\alpha
as was to be shown.

Now substituting this back into our initial path equation,
tan\beta = \tan\alpha - \frac{gd}{2(gd\tan\alpha)\cos^2\  alpha}
\implies \tan\beta = \tan\alpha - \frac{1}{2\sin\alpha\cos\alpha}
\implies \tan\beta = \frac{\sin\alpha}{\cos\alpha} - \frac{1}{2\sin\alpha\cos\alpha}
\implies \tan\beta = \frac{2\sin^2\alpha - 1}{2\sin\alpha\cos\alpha}.
and since \sin 2x \equiv 2\sin x \cos x, and \cos 2x \equiv \cos^2 x - \sin^2 x \equiv 1 - 2 \sin^2 x,
\tan\beta = \frac{- \cos 2\alpha}{\sin 2\alpha} = -\cot 2\alpha
and since the tangent function is odd,
\tan(-\beta) = \cot 2\alpha.
But by the definition of the cotangent function, \cot x \equiv \tan (\frac{\pi}{2} - x) and so
\tan(-\beta) = \tan(\frac{\pi}{2} - 2 \alpha
and since the angles involved are acute,
-\beta = \frac{\pi}{2} - 2\alpha
\implies 2\alpha = \beta + \frac{\pi}{2}
as desired.

Part (ii)

Let \gamma be the angle asked for. So, the gradient of the path's curve at point P must be \tan \gamma. Differentiating our initial path equation with respect to x to find this gradient (holding all other values constant),
\frac{dy}{dx} = \tan\alpha - \frac{2gx}{2u^2\cos^2\alpha}
and so at point P where x=d,
\tan\gamma = \tan\alpha - \frac{2gd}{2u^2\cos^2\alpha}
and therefore substituting in u^2=gd\tan\alpha,
\tan\gamma = \tan\alpha - \frac{2gd}{2gd\tan\alpha\cos^2\a  lpha}
\implies \tan\gamma = \frac{\sin\alpha}{\cos\alpha} - \frac{1}{\sin\alpha\cos\alpha} = \frac{\sin^2\alpha - 1}{\sin\alpha\cos\alpha}.
And since \cos^2 x + \sin^2 x \equiv 1,
\tan\gamma = \frac{-\cos^2\alpha}{\sin\alpha\cos\alp  ha}
\implies \tan\gamma = -\frac{\cos\alpha}{\sin\alpha} = -\cot\alpha
and so by the same reasoning as before,
\tan(-\gamma) = \tan(\frac{\pi}{2} - \alpha)
\implies \gamma = \alpha - \frac{\pi}{2}.
Hey there, could you tell me why you differentiated, i've never been too sure on those :/ thanks
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