[ukgea]

II/7 methinks.

[insparato]

All pure on STEP II done.

[ukgea]

STEP I question 1

r = radius of cylinder, h = height













(is increasing for positive r, so gives us the minimum)







At the minimum,







as required.

At the minimum,









which means that the largest sphere to fit inside the tin has radius r, and will be tangent to both the ends of the sphere and the side. It's volume is







as required.

The smallest sphere into which the tin fits will touch the edges of the two circular disks that form the tin's ends, and thus the radius R of this sphere will be the distance from these edges to the centre of the tin. By the Pythagorean theorem:





The volume of this sphere is

[DFranklin]

I can do III, 4,5,6,7, but anyone else who wants to feel free...

[insparato]

(Original post by DFranklin)
I can do III, 4,5,6,7, but anyone else who wants to feel free...

Ill have a bash at number 5 on De Moivre's Theorem, ive just done this in class so... Dont know how far ill get though.

[Speleo]

STEP I Question 2

I = INT (1 + (a-1)x)^n dx
Consider (1 + (a-1)x)^(n+1)
d/dx -> (n+1)(a-1)(1+ (a-1)x)^n
Therefore I = [(1 + (a-1)x)^(n+1)]/[(n+1)(a-1)]
Limits 1 and 0.
-> a^(n+1)/(n+1)(a-1) - 1/(n+1)(a-1)
= [a^(n+1) - 1]/[(n+1)(a-1)] as required.

I = INT (ax + (1-x))^n dx
= INT nC0(ax)^0(1-x)^n + ... + nCk(ax)^k(1-x)^(n-k) + ... + nCn(ax)^n(1-x)^0 dx
= nC0a^0 INT x^0(1-x)^n dx + ... + nCka^k INT x^k(1-x)^(n-k) dx + nCna^n INT x^n(1-x)^0 dx
Therefore the coefficient of a^k is nCk INT x^k(1-x)^(n-k) dx as required.

(1 + (a-1)x)^n clearly is the same as (ax + (1-x))^n
Let J_k = nCk INT x^k(1-x)^(n-k) dx
[a^(n+1) - 1]/[(n+1)(a-1)] = J_0 + a J_1 + ... + a^k J_k + ... + a^n J_n
But (a^(n+1) - 1)/(a-1) = 1 + a + a^2 + ... + a^k + ... + a^n
[1 + a + a^2 + ... + a^k + ... + a^n] = (n+1)J_0 + (n+1)aJ_1 + ... + (n+1)a^kJ_k + ... + (n+1)a^nJ_n
For all a.
Therefore coefficients of powers of a are equal.
1 = (n+1)J_k
1/n+1 = nCk INT x^k(1-x)^(n-k) dx
INT x^k(1-x)^(n-k) dx = k!(n-k)!/n!(n+1)

INT x^k(1-x)^(n-k) dx = k!(n-k)!/(n+1)!

[insparato]

(Original post by insparato)
Ill have a bash at number 5 on De Moivre's Theorem, ive just done this in class so... Dont know how far ill get though.

Got through the first part okay. Dont think i can get the other parts out, ill have a rethink later ive done too much STEP maths this weekend and not enough biology .

[DFranklin]

(Original post by insparato)
Got through the first part okay. Dont think i can get the other parts out, ill have a rethink later ive done too much STEP maths this weekend and not enough biology .

Hint: (in white)

Use the relation between the coefficients of a polynomial and the sum/product of it's roots.

[datr]

STEP I Q4



If |a| > 1 then (a^2 - 1) > 0 and we can square root it to get a real answer.











If |a| < 1 then (1 - a^2) > 0 and we can sqaure root this to get a real answer.





Now consider the compound angle formula for tan





Let A = tana and B = tanb so a = arctan A and b = arctanB









Damn that's a lot of algebra for a STEP I question.

[Speleo]

STEP I Question 6

i) f(x) sin(x/2) = sin(n + 1/2)x
But SUM 2sin(x/2)coskx = sin(n + 1/2)x - sin(n - 1/2)x + sin(n - 1/2)x - ... - sin(x/2)
2sin(x/2) SUM coskx = sin(n + 1/2)x - sin(x/2)
f(x) sin(x/2) = sin(x/2) + 2sin(x/2) SUM coskx
f(x) = 1 + 2 SUM coskx

ii) INT f(x) dx
= INT 1 + 2 SUM coskx dx
= [x] + 2 SUM INT coskx dx
= pi + 2 [sinx + (1/2)sin2x + (1/3)sin3x + ... + (1/n)sinnx]
= pi
(sinkx = 0 at 0, mpi, m E Z)

INT f(x) cosx dx
= INT sin(n + 1/2)x.cosx/sin(x/2) dx
= INT sin(n + 1/2)x.(1 - sin^2(x/2))/sin(x/2) dx
= INT f(x) dx - INT sin(n + 1/2)x.sin(x/2) dx
= pi + (1/2)INT cos(n+1)x - cos(n)x dx
= pi + (1/2)[(1/n+1)sin(n+1)x - (1/n)sin(n)x]
= pi

[Speleo]

That's the pure from STEP I finished now (when datr posts), as DFranklin did 5 on the last page.

EDIT: you've got your listings a little confused, I just did 6 not 4

[insparato]

Typing up STEP III question 2 Now

- eqn 1
- eqn 2
- eqn 3

eqn (2 - 1)



-eqn4

sub eqn 4 into 3


-eqn 5

sub eqn 4 into 2
-eqn6

subtracting 6 from 5





Sub into eqn 5









Sub these back into eqn 1 just to see they are indeed satisfy the system.



Therefore







Not complete, the cases A = 0, A = 1 have to be investigated as the standard solution blows up when A = 0 and A = 1. Ive done alot of maths today so i wont be doing this tonight but i might have a go tomorrow or something... Anyone who feels up to finishing it can do so freely .

[Speleo]

STEP II Question 5

u = z + 1/z
z^4 + z^3 + z^2 + z + 1 = 0
z^2 + z + 1 + 1/z + 1/z^2 = 0
1 + u + u^2 - 2 = 0
u^2 + u - 1 = 0
u = [-1 +- rt5]/2
z^2 + z[1 -+ rt5]/2 + 1 = 0
z = {[-1 +- rt5]/2 +- sqrt[[-1 +- rt5]^2/4 - 4]}/2
z = -(1/4) +- rt(5)/4 +- (1/2)sqrt[-(5/2) -+ (1/2)rt5]
z =
-(1/4) + rt(5)/4 + (1/2)sqrt[-(5/2) - (1/2)rt5]
-(1/4) + rt(5)/4 - (1/2)sqrt[-(5/2) - (1/2)rt5]
-(1/4) - rt(5)/4 + (1/2)sqrt[-(5/2) + (1/2)rt5]
-(1/4) - rt(5)/4 - (1/2)sqrt[-(5/2) + (1/2)rt5]

z^4 + z^3 + z^2 + z + 1 = 0
z^4 + z^3 + z^2 + z = -1

z^5 + z^4 + z^3 + z^2 + z = 0
z^5 - 1 = 0

z = cis(2pi/5), cis(4pi/5), cis(6pi/5), cis(8pi/5)
cos(2pi/5) = Re(z) = [-1 + rt5]/4 (look at cos graph for sign)
sin(2pi/5) = Im(z) =(1/2)sqrt.../i = (1/2)sqrt[(5/2) + (1/2)rt5] = (1/4)sqrt(10 + 2rt5) (see graph to see which is which)

[DFranklin]

Done III, Q3, which was actually quite nice, but felt very "old style Cambridge Entrance Exams".

In fact, I feel looking at these papers in general that 1996 and earlier marks a bit of a "watershed" in terms of content. The questions require more and more algebra grinding, it feels you're supposed to know a lot of stuff outside the syllabus, and so on.

Going back previous to 1996 it seems STEP I, STEP II also take a bit of a quantum leap - I looked at the 1990 STEP I paper and wondered if it was a mislabelled Paper III (questions on moments of inertia?!?).

So my feeling is we're hitting a bit of diminshing returns if you're looking at practice for the current style of STEP exams. It's probably more useful to complete all the later papers (including mechanics / stats).

Any thoughts? (As I'm just doing this for fun, not exam practice, it doesn't hugely matter to me, though the "fun level" of the 1990 questions doesn't seem very high)!

[insparato]

Im having a go at STEP III Question 7 although the first part im doing is very very tedious.

It does seem that the older papers are harder. But i can only imagine this as being because as you go back in time A level was harder 15 years ago, you had to know more for the normal A level so thats why i think the actual content of STEP I in the older papers is more than the newer STEP I papers.

Edit: I give up on Question 7 urgh i cant seem to get anywhere on the second part and there are others that can do this so...

[Speleo]

Many of the older papers are mislabelled (usually III confused with I) on a certain site.

[insparato]

(Original post by Speleo)
Many of the older papers are mislabelled (usually III confused with I) on a certain site.

Oh I didnt know that.

[DFranklin]

(Original post by Speleo)
Many of the older papers are mislabelled (usually III confused with I) on a certain site.

Oh, yes, so I see...!

[DFranklin]

(Original post by insparato)
It does seem that the older papers are harder. But i can only imagine this as being because as you go back in time A level was harder 15 years ago, you had to know more for the normal A level so thats why i think the actual content of STEP I in the older papers is more than the newer STEP I papers.

I did A-levels a long time ago, just before STEP, and did quite a few of the old Cambridge Colleges Exam questions. My experience was that I learned a lot of maths outside the A-level syllabus from doing the CCE; my guess is it would have been the same with STEP circa 1990.

As well of the syllabus, some of these questions really do require an awful lot of manipulation - far more than the typical exam questions you'd get in the actual Tripos. (And I did the tripos in 1988-1990, so slightly before these STEP papers).

[ukgea]

Question 4 Paper III





.

Let be positive integers such that

.

We want to maximise

.

Clearly, there is no point in having one , because then we can replace it with and . The sum of all the numbers stays the same, but the product increases (or at least stays the same) since . Thus, we need only consider cases where all are either 1, 2 or 3.

But let , and suppose we have for some . Then there will also be some such that or (if all are 1, we can replace them with one number , which increases the product). In the former case, we can replace and with a 3, and since , this doesn't change the sum but increases the product. In the latter case, we can replace the 3 and 1 with 2 and 2, and since , this again does not change the sum but increases the product. Thus, for N > 1 (which always holds for the cases we are going to evaluate) we can safely assume that at the maximum, or for all .

Assume now there are three or more 2s among the . Then, we can remove those three 2s and replace them with two 3s. We will then have increased the product (by a factor 9/8) while the sum remained constant. Thus, at the maximum, there will be at most two 2s.

Now, for any integer N, there is a unique way to write them as a sum of 2s and 3s with at most two 2s. Because, consider the rest of the sum when divided by 3. If there is no 2s, the rest is 0, if there is one 2, the rest is 2, if there are two 2s, the rest is 1. Since the rest of N when divided by 3 is exactly one of 0, 1 and 2, it follows that the number of 2s in the sum is uniquely defined by N, and hence the number of 3s is uniquely defined as well. We will then have the formula



Noting that the rests of 5, 6, 7, 8, 9, 1000 when divided by 3 are 2, 0, 1, 2, 0, 1 respectively, plugging these number into the formula give











as required, and finally

.