I have some additions to the solutions to some of the STEP III qu's posted by Dadeyemi that would complete them. Would they be wanted?
Also, SimonM (or anybody else), what is the LaTeX for surrounding particular results in boxes, as you sometimes do for answers to parts of questions? That might make my solutions a little more readable...
Meanwhile,
Question 6, STEP III, 2002
Standard seperating of variables yields:
At this point, we would like to take roots, but notice that we must consider seperately the cases , and , or we will be taking square roots of a negative number.
(a) For :
Resulting in:
Notice that if , it is no longer necessary to consider , and we have:
(b) For :
Resulting in:
Or simply:
Now we consider parts (i) and (ii) of the question:
(i) , therefore we use the second formula to determine :
Substituting into yields,
(ii) , therefore we use the first formula to determine :
Substituting into yields (as if by magic),
.
Therefore, all the solution curves that pass through one, pass through the other point as well, and we have the following two functions:
and
To prove that satisfies the differential equation, we simply substitute them straight in, yielding,
, as desired.
It should be fairly clear that in the case , we have the formula for a circle:
Specifically, for the two functions, circles with centers , and radius 1.
To sketch the functions for , we simply consider what happens as , and notice that they are symmetric in the axis.
I will attach a sketch as soon as I can.
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STEP I, II, III 2002 Solutions
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Why bother with a post grad? Are they even worth it? Have your say!  26102016 

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 21
 11042009 15:30
Last edited by Elongar; 11042009 at 16:55.Post rating:6 
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 11042009 16:28
(Original post by Elongar)
Also, SimonM (or anybody else), what is the LaTeX for surrounding particular results in boxes, as you sometimes do for answers to parts of questions? That might make my solutions a little more readable..
Code: \boxed{Whatever you want in here.}
Next time you're unsure of a code for a symbol, hover your cursor over someone else's LateX which has coded such a symbol, and the LateX coding will appear. 
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 11042009 16:50
Many thanks . I'll bear that in mind.

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 12042009 06:54
Question 8, STEP I, 2002
First PartSecond PartLast edited by Unbounded; 12042009 at 07:31. 
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 13042009 04:17
Question 13, STEP I, 2002
(i)For the roots to be real, the discriminant must be greater than or equal to zero, ie:
Case 1: U = 1
For the inequality to hold, we require V = 1
And the probability for this case is
Case 2: U = 0
For the inequality to hold, we require V = 1
The probability for this case is
Case 3: U = 1
For the inequality to hold, again we require V = 1
The probability for this case is
Adding up the probabilities, we get:
(ii)Going back to cases (and using the fact that we know V must be 1)
Case 1: U = 1
The larger root is and has a probability of
Case 2: U = 0
The larger root is 1, and this has a probability of
Case 3: U = 1
The larger root is and this has a probability of .
The expected value of the large root E(R) is now given by:

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 26
 23042009 23:39
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.Post rating:3 
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 24042009 00:49
STEP II 2002 Q10
Spoiler:Showafter t hours, the competitor has km left to run
the remainder is run at , in a time (Tt) hours
using speed = distance/time:
which rearranges to
This is a quadratic in t. Clearly t is real, therefore it has real roots, so
which rearranges to give for all values of t.
If T = 3, then
which has the (repeated) solution .
Therefore competitor one runs at 13km/h for 3/4 hours
then (29/2)km/h for the remains 9/4 hours
Therefore distance run by competitor one after time x
(The 9/8 is obtained by observing that the two expressions must agree when x = 3/4.)
Now the speed of competitor 2 after time x
we can integrate to find the distance:
distance
from considering the distance travelled initially (0) and after 3 hours (42+3/8), we find
Therefore distance =
therefore the distance between the two
Via differentiation, we find that the first expression has no maxima in the required range, but the second expression has a maximum at .
Substituting this in, we find that the maximum distance is km.
Post rating:2 
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 24042009 14:05
STEP II 2002 Q11
Spoiler:Show
Consider the beam to be made up of tiny sections of length and distance from B.
Weight of one of these sections
now distance of CoM from B
Taking the limit as , it becomes an integral:
We now draw a diagram of the rod against the wall. There are normal reactions at A and B respectively, with corresponding frictional forces (acting towards the base of the wall and vertically upwards respectively.) There is also the weight of the rod, W, at the centre of mass.
Resolving horizontally:
Resolving vertically:
Therefore , which we will use later.
Taking moments about B:
which rearranges eventually to give
Substituting in gives the required result.
If , then
If the beam slides for any , then the greatest possible value of is given when gives limiting equilibrium
Therefore
(ignoring the negative root)
Yay, all three mechanics.Last edited by qgujxj39; 24042009 at 14:08.Post rating:2 
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 26042009 16:30
I/5
Spoiler:Show
Looking at f(0) gives that as desired
Now, consider f(1) and as required.
If you consider f(1) you on one hand get and on the other hand
So, for even n
and for odd n
Polynomial: Guess the root x=2 We can now factorise it to study . Thus Solving the resulting quadratic gives
Thus the roots to the polynomial are x=2, 6 (double) and 8.

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 26042009 18:17
I/3
Spoiler:ShowSo as desired.
So
=
For stationary points dy/dx=0, thus one solution is
All other solutions will be when:
Thus a=b or a=b
Now we have four solutions to dy/dx=0, check these to verify which point is max/min.
a=b and a=b give the same value for the function y, namely
Thus, if you recall the result proved in the beginning (, produces a minimum and produces a maximum.
As global max is then . Moreover, global minimum is a+b thusLast edited by nota bene; 26042009 at 19:29. 
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 31
 26042009 19:14
(Original post by nota bene)
I/5
Spoiler:Note that this is actually a lot longer than your solution, but I suspect you had some failed "root guesses" that you didn't bother writing up.ShowNote 576 = 600  24 = 25 x 24  24 = 24 x 24 = .
Suppose the roots are ABCD. Then we know ABCD = 3^2 . 2^6
Also, calculate 1 + 22 + 172 + 552 + 576 = 599 + 172 + 552 = 771 + 552 = 1323 = 3 x 441 = 3 x 3 x 3 x 7 x 7.
So we know (A1)(B1)(C1)(D1) = 3 x 3 x 3 x 7 x 7. (*)
And finally calculate 1  22 + 172  552 + 576 = 175 = 7 x 5 x 5.
So we know (A+1)(B+1)(C+1)(D+1) = 7 x 5 x 5. (**)
Looking at (*), to get the '7' factors, we must have two roots that are either 8 (since 81 = 7) or 6 (since 61 = 7). Or we could have an unexpectedly large root, but that isn't likely.
But then looking at (**), we see the '8' case can't work.
So we must have 2 roots that = 6, WLOG, C = D = 6.
Then from (*) we get (A1)(B1) = 27
and from (**) we get (A+1)(B+1) = 7
So either {A+1, B+1} = {1,7} or {1, 7}.
Since we know no root = 0 (since the product of ABCD isn't 0), we must have the {1, 7} case and so 2 additional roots 2, 8.
Note: since we didn't rule out the "unexpectedly large root test", we should still check these actually are the 4 roots.
Spoiler:ShowIn practical terms, I suspect using a hybrid technique to make educated guesses at the roots would be better than either pure guessing, or trying to deduce the roots solely by using the first part of the question. 
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 26042009 19:23

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 26042009 19:35
(Original post by DFranklin)
I think this is your problem:
Thanks! 
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 03052009 11:20
(Original post by Dadeyemi)
Some more;
Did these quite a while ago I'm afraid some may be partial solutions.
For anyone looking for a solution to this question I got:
gives 0 solutions
gives 1 solution
gives 2 solutions
also gives 1 solution
For part (i) note that we have the case where , and require . This is the only solution.
For part (ii) it isn't quite as neat, we have the case where so we're looking for two solutions. One is x=4 which you see quickly on inspection, for the other we have to trawl through some algebra to get x=49/16. Even if the first part of the question isn't to taste I think part two is worth doing to check you know what to do to transform the equation into something managable, namely a standard quadratic. 
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 03052009 14:36
(Original post by Dadeyemi)
Some more;
Did these quite a while ago I'm afraid some may be partial solutions.
By looking at the graph it is clear that:
Conditions for exactly one solution are c = b(a^2)/(a^2  1) or c > a(b^1/2)
For exactly two solutions b(a^2  1) < c^2 < (a^2)b
i) Squaring once gives 8x  14 = 8(x  2)^1/2, squaring again gives 16(x^2)  72x + 81 = 0, by the quadratic formula, x = 9/4
ii) Squaring once 8x  22 = 10(x  3)^1/2 squaring again 16(x^2)  113x + 196 = 0 by the quadratic formula x = 4 or x = 49/16
Seems too easy so i've probably gone wrong. I can't see how the first bit helps with the second, it tells you how many solutions the equations have but you find that out for yourself when you solve them. Any feedback welcome. 
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 03052009 14:38
(Original post by toastedlion)
RE Q3: You lost an a squared very early on and so came to get a minimum on the xaxis. In fact the minimum occurs when , and at this point f(x)>0. Also, your axes are very inappropriately placed seeing as b>0. Thankfully you then gave up, because if you had carried on you would have got into quite a mess!
For anyone looking for a solution to this question I got:
gives 0 solutions
gives 1 solution
gives 2 solutions
also gives 1 solution
For part (i) note that we have the case where , and require . This is the only solution.
For part (ii) it isn't quite as neat, we have the case where so we're looking for two solutions. One is x=4 which you see quickly on inspection, for the other we have to trawl through some algebra to get x=49/16. Even if the first part of the question isn't to taste I think part two is worth doing to check you know what to do to transform the equation into something managable, namely a standard quadratic. 
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 03052009 14:59
I might be missing something obvious but how did you determine that (lna)^2/a and (lna)/a tend to zero as a tends to infinity? I knew that (lna)/a tended to zero but proved it using l'hopital because I thought they'd want to see proof, i didn't know that (lna)^2/a did though so also ended up showing it using l'hopital.

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 03052009 20:54
(Original post by maltodextrin)
I left this page open overnight and didn't refresh, well i guess 2 solutions is better than none lol! 
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 03052009 21:18
(Original post by maltodextrin)
I might be missing something obvious but how did you determine that (lna)^2/a and (lna)/a tend to zero as a tends to infinity? I knew that (lna)/a tended to zero but proved it using l'hopital because I thought they'd want to see proof, i didn't know that (lna)^2/a did though so also ended up showing it using l'hopital. 
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 40
 03052009 22:06
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Updated: November 23, 2015
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