STEP I, II, III 2002 Solutions
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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Re: STEP I, II, III 2002 SolutionsSTEP III, Question 8
this was a bit laborious to type out
Spoiler:Showlet z and w be unit modulus complex numbers with arguments x and y respectively
for
from taking into account quadrants (other quadrants accounted for by periodicity),
from periodicity of tan to assure 
letting
we arrive at the required result
for some integer n (where m is an integer and q,r,s,t are in {0,1})
and similarly:
for some integer n
thus we conclude
and 
are collinear so
if and only if 
we note that
so as required
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Re: STEP I, II, III 2002 SolutionsThat's a nice little proof, looks fine to me. You should definitely check out l'hopitals rule, I use it quite a bit. I'm just worried that they won't let us get away with (lnx)/x tending to zero being obvious so if it comes up i'll probably prove it to stay on the safe side.(Original post by toasted-lion)
How's this:
tends to 0 as x tends to infinity, which we know and seems intuitively obvious (but please post your proof as I've never heard of l'hopital) so, letting 
, 
tends to 0 and also 
tends to 0, so squaring, 
tends to 0. That seems ok to me, now your "l'hopital"?
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Re: STEP I, II, III 2002 SolutionsI think we're safe with (lnx)/x tending to 0 being obvious but I think you're right to be cautious with anything more complicated. I checked out l'hopital's rule but didn't really follow the proof, I think I'll see if my maths teacher can help next week. I think I'm starting to get somewhere with STEP now, I can normally find 3 or 4 questions I'm ok with and scribble something down for another 2 or 3, still not totally confident of getting 2 1s though. It's time as well, I just took over an hour on a STEP II question!(Original post by maltodextrin)
That's a nice little proof, looks fine to me. You should definitely check out l'hopitals rule, I use it quite a bit. I'm just worried that they won't let us get away with (lnx)/x tending to zero being obvious so if it comes up i'll probably prove it to stay on the safe side. -
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Re: STEP I, II, III 2002 SolutionsSorry, but I suppose in Q3, the minimum value of f(x) should be sqrt[(a^2-1)*b]? because in your inequality x>a^2*(x-1), you should solve it like:
(1-a^2)x>-b*a^2...
(Original post by Dadeyemi)
Some more;
Did these quite a while ago I'm afraid some may be partial solutions. -
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Re: STEP I, II, III 2002 SolutionsHow did you decide upon your substitution for part three? It's things like this I would have no chance finding in the actual exam. I managed to get the same result as you using the substitution(Original post by sonofdot)
STEP II 2002 Question 1
![\begin{array}{rl}
\displaystyle\int_{\frac{\pi}{6} }^{\frac{\pi}{4}} \frac{1}{1-\cos 2\theta} \, d\theta
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{1-(1-2\sin^2 \theta)} \, d\theta \\ \br \\
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{2\sin^2 \theta} \, d\theta \\ \br \\
& \displaystyle = \frac12 \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \mathrm{cosec}^2 \theta \, d\theta \\ \br \\
& \displaystyle = \frac12 \left[ -\cot \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \\ \br \\
& \displaystyle = \frac12 \left( \frac{1}{\tan \frac{\pi}{6}} - \frac{1}{\tan \frac{\pi}{4}} \right) \\ \br \\
& \displaystyle = \frac12 \left( \sqrt3 - 1 \right) \\ \br \\
& \displaystyle = \boxed{\frac{\sqrt3}{2} - \frac{1}{2}}
\end{array}](http://www.thestudentroom.co.uk/latexrender/pictures/40/40fa4da3b28a5c6ffcec490aa9f115ff.png "\begin{array}{rl}
\displaystyle\int_{\frac{\pi}{6} }^{\frac{\pi}{4}} \frac{1}{1-\cos 2\theta} \, d\theta
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{1-(1-2\sin^2 \theta)} \, d\theta \\ \br \\
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{2\sin^2 \theta} \, d\theta \\ \br \\
& \displaystyle = \frac12 \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \mathrm{cosec}^2 \theta \, d\theta \\ \br \\
& \displaystyle = \frac12 \left[ -\cot \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \\ \br \\
& \displaystyle = \frac12 \left( \frac{1}{\tan \frac{\pi}{6}} - \frac{1}{\tan \frac{\pi}{4}} \right) \\ \br \\
& \displaystyle = \frac12 \left( \sqrt3 - 1 \right) \\ \br \\
& \displaystyle = \boxed{\frac{\sqrt3}{2} - \frac{1}{2}}
\end{array}")
(which I think followed on from the previous parts of the question better), but I think yours might have reached the result quicker (although I don't really follow your working in the middle).
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Re: STEP I, II, III 2002 SolutionsIf you look at the limits it'll give you an idea about why that substitution was considered. It's certainly not something I spotted the first time I did this question(Original post by mikru)
How did you decide upon your substitution for part three? It's things like this I would have no chance finding in the actual exam. I managed to get the same result as you using the substitution (which I think followed on from the previous parts of the question better), but I think yours might have reached the result quicker (although I don't really follow your working in the middle).
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Re: STEP I, II, III 2002 SolutionsThe clue, I suppose, is in the limits of the equation. You have, in part iii), the limits in part ii) - but in reciprocal form.(Original post by mikru)
How did you decide upon your substitution for part three?
The easiest way to turn part iii)'s limits into part ii)'s is to put y = 1/x. This immediately leads to limits of √3/2 and 1, but in the wrong order. The swapping of 'dy's for 'dx's fixes this by introducing a negative sign, thereby flipping the limits. Then a bit of algebra transforms the integrand into exactly the same thing as part ii)'s.
Your cosec 2θ substitution is exactly the same thing, though, if you think about it. If we're saying y = 1/x, and part ii) works by putting x = sin 2θ, then you are, essentially, saying that y = cosec 2θ. Doing this just leads to a bit of extra work if you don't immediately think "oh, hang on a sec*, I've done this bit of working before...
".
*no pun intended. -
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Re: STEP I, II, III 2002 SolutionsI see now, that does indeed make alot of sense - and does cut down the work by alot. I'm clearly not in maths-mode today as I even thought to myself "Those limits are reciprocals.. hmm.. let's try cosec!", even though by now I should be used to the final part of integration questions leading to the same result as the previous part.(Original post by Adje)
The clue, I suppose, is in the limits of the equation. You have, in part iii), the limits in part ii) - but in reciprocal form.
The easiest way to turn part iii)'s limits into part ii)'s is to put y = 1/x. This immediately leads to limits of √3/2 and 1, but in the wrong order. The swapping of 'dy's for 'dx's fixes this by introducing a negative sign, thereby flipping the limits. Then a bit of algebra transforms the integrand into exactly the same thing as part ii)'s.
Your cosec 2θ substitution is exactly the same thing, though, if you think about it. If we're saying y = 1/x, and part ii) works by putting x = sin 2θ, then you are, essentially, saying that y = cosec 2θ. Doing this just leads to a bit of extra work if you don't immediately think "oh, hang on a sec*, I've done this bit of working before...".
*no pun intended.
Oh well. Thank you
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Re: STEP I, II, III 2002 Solutionstommm is there any chance you could help me out with how you got from(Original post by tommm)
STEP II 2002 Q9
Spoiler:Show
There's loads of ugly algebraic manipulation involved in this question, so apologies if I don't type out full working in getting from one stage to the next.
Considering horizontal motion:
= horizontal displacement after time t
Using
we obtain 
Considering vertical motion:
which gives
(*)
when
or 
, because the particle just clears the walls
Therefore
rearranging, we obtain
which rearranges to give
substituting this into (*), and using the identity
, we obtain
which rearranges to give the required result.
From our expression for tan(alpha), we get
Using the formula for tan(A+B) and rearranging:
Because
is small, we can disregard its term in the denominator, giving the required result.
Now we similarly obtain:
replacing the part of the last fraction that equals 2V^2/g and rearranging:
(A)
Now the LHS =
, of which we can disregard the 
term because it is very small. Therefore LHS 
.
From this, we can conclude that
has the same sign as the RHS of equation (A). Therefore, 
iff
This rearranges to
we can disregard the
term because it is much smaller than all the others, and then we can square root to give the required result. The other result can easily be shown by reversing the > sign.
tan(a + da) = (h + dh)(a + b)/ab to
da(seca)^2/(1 - da(tana)) = dh(a + b)/ab
I'm pretty useless with approximations so i'm probably missing something obvious. -
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Re: STEP I, II, III 2002 SolutionsHmmm... There's too much working missing to say exactly what tommm did (which means, incidentally, that there's too much working missing to get full marks, IMHO).
I would go (without having actually read the question, you understand!):
So
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Re: STEP I, II, III 2002 SolutionsI don't think that's what tommm did because he mentions using the formula for tan(A + B) but your way looks fine to me. Thanks!(Original post by DFranklin)
Hmmm... There's too much working missing to say exactly what tommm did (which means, incidentally, that there's too much working missing to get full marks, IMHO).
I would go (without having actually read the question, you understand!):
So
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Re: STEP I, II, III 2002 SolutionsNo, I don't think that it's what tommm did either. But I don't think tommm's approach was the best one anyhow, so...
Incidentally: the "missing bit" that I think you'd lose marks for in tommm's approach.. At some point, he's obviously done some approximation like "for small
". Without explicitly explaining that, I think you'd be guaranteed to lose marks, even if everything written was actually correct.
Last edited by DFranklin; 26-05-2009 at 12:13. -
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Re: STEP I, II, III 2002 SolutionsDadeyemi's solution to STEP III question 5:
isn't the linear co-efficient (bt-d) not (b-d)?
EDIT: not that it makes a difference to the end answer, as t=1.Last edited by around; 26-05-2009 at 22:57. -
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Re: STEP I, II, III 2002 SolutionsFor STEP III Question 4: Don't we also have x = 0 y = - 1 and x = 1 y = 0 as solutions? The way I did the last part was by saying x = y + n where n is a non negative integer, then you get(Original post by Dadeyemi)
Some more;
Did these quite a while ago I'm afraid some may be partial solutions.
(3n - 2)y^2 + (3n^2 - 2n)y + n^3 - n^2 = 0
It turns out that this only has roots when n < 3. Thus the only solutions are given by x = y + 2 x = y + 1 and x = y. Plugging these in gives all the solutions.
I'm not sure if this is the method they were looking for, or even if its correct. It would be great if someone could have a look for me. -
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Re: STEP I, II, III 2002 SolutionsSTEP I, Question 4
Spoiler:Show
Therefore the equation of the tangent is
Since
is a solution
This gives us the equation,
We have
so the only intersections are 0 and 1.
We have
(given)
The area under the tangent is a trapezium with area
Since the graph is always above it, we get
which is what we want to show.
Considering the half closest to the y axis, and revolving it around that axis we get:
![\displaystyle \pi \int_{1/2}^1 x^2 \, dy = \pi \left [ \ln y - y\right ]_{1/2}^1 = \pi \left ( \ln 2 - \frac{1}{2} \right )](http://www.thestudentroom.co.uk/latexrender/pictures/54/541dd960a14ebcbdbf4025df17928a86.png "\displaystyle \pi \int_{1/2}^1 x^2 \, dy = \pi \left [ \ln y - y\right ]_{1/2}^1 = \pi \left ( \ln 2 - \frac{1}{2} \right )")
and
![\displaystyle \pi \int_{1/2}^1 (2(y-1))^2\, dy = 4\pi \left [ \frac{y^3}{3} \right ]_{-1/2}^0 = \pi \frac{1}{6}](http://www.thestudentroom.co.uk/latexrender/pictures/13/130afc912c435cf8f887f3bef81457a5.png "\displaystyle \pi \int_{1/2}^1 (2(y-1))^2\, dy = 4\pi \left [ \frac{y^3}{3} \right ]_{-1/2}^0 = \pi \frac{1}{6}")
Therefore
Last edited by SimonM; 27-05-2009 at 22:16.
