STEP 2017 Solutions

If anybody is willing to LaTeX this up, please do!

Agree with above solution. Here's an 'or otherwise' solution to get the second part directly:

$r$th 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.

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.

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.

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.

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.

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.

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.

STEP 2, Year 2012

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,
$0 = \sec^2\alpha - (-2)\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(\frac{du}{d\alpha}\cos\alpha - u\sin\alpha).$
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,
$0 = \sec^2\alpha + 2\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(-u\sin\alpha)$
$\implies 0 = \frac{1}{\cos^2\alpha} - \frac{gdu\sin\alpha}{u^3\cos^3\alpha}$
$\implies \frac{gd\sin\alpha}{u^2\cos\alpha} = 1$
$\implies u^2 = gd\tan\alpha$
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 $u^2=gd\tan\alpha$,
$\tan\gamma = \tan\alpha - \frac{2gd}{2gd\tan\alpha\cos^2\alpha}$
$\implies \tan\gamma = \frac{\sin\alpha}{\cos\alpha} - \frac{1}{\sin\alpha\cos\alpha} = \frac{\sin^2\alpha - 1}{\sin\alpha\cos\alpha}.$
And since $\cos^2 x + \sin^2 x \equiv 1$,
$\tan\gamma = \frac{-\cos^2\alpha}{\sin\alpha\cos\alpha}$
$\implies \tan\gamma = -\frac{\cos\alpha}{\sin\alpha} = -\cot\alpha$
and so by the same reasoning as before,
$\tan(-\gamma) = \tan(\frac{\pi}{2} - \alpha)$
$\implies \gamma = \alpha - \frac{\pi}{2}$.