(Updated as far as #100.) SimonM  23.03.2009
For explanation, scroll down:
STEP Mathematics I
1: Solution by DFranklin
2: Solution by Insparato
3: Solution by Insparato and DFranklin
4: Solution by Speleo
5: Solution by Speleo
6: Solution by Speleo
7: Solution by DFranklin
8: Solution by Speleo
9: Solution by SimonM
10: Solution by ad absurdum
11: Solution by waxwing
12: Solution by SimonM
13: Solution by SimonM
14: Solution by SimonM
STEP Mathematics II
1: Solution by Speleo
2: Solution by DFranklin
3: Solution by Rabite
4: Solution by Speleo
5: Solution by DFranklin
6: Solution by Speleo
7: Solution by Speleo
8: Solution by Speleo
9: Solution by waxwing
10: Solution by brianeverit
11: Solution by waxwing
12: Solution by nota bene
13: Solution by nota bene
14: Solution by brianeverit
STEP Mathematics III
1: Solved In Siklos Booklet, uploaded by khaixiang
2: Solution by Insparato
3: Solution by Speleo
4: Solution by Speleo
5: Solution by Speleo
6: Solved By Khaixiang, waxwing
7: Solution by DFranklin
8: Solution by DFranklin
9: Solution by DFranklin
10: Solution by DFranklin
11: Solution by DFranklin
12: Solution by brianeverit
13:
14: Solution by DFranklin
Explanation:
In this thread, we will try to collectively provide solutions for the questions from the STEP Maths I, II and III papers from 1996, seeing as no solutions are available on the net.
If you want to contribute, simply find a question which is marked with "unsolved" above (you might want to check for new solutions after the time the list was last updated as well), solve it, and post your solution in the thread. I will try to update the list as often as possible. If you have a solution to a question that uses a different idea from one already posted, submit it as well, as it's often useful to see different approaches to the same problem.
If you find an error in one of the solutions, simply say so, and hope that whoever wrote the solution in the first place corrects it.
Solutions written by TSR members:
1987  1988  1989  1990  1991  1992  1993  1994  1995  1996  1997  1998  1999  2000  2001  2002  2003  2004  2005  2006  2007
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STEP Maths I, II, III 1995 Solutions watch

SimonM
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 05032007 17:34
Last edited by SimonM; 21072011 at 20:07. 
insparato
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 05032007 17:54
Last edited by insparato; 08052010 at 17:43. 
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 05032007 17:56
STEP I Question 4
Im[cos5t + isin5t] = Im[(cost + isint)^5]
sin5t = Im[cos^5t + 5isintcos^4t  10sin^2tcos^3t  10isin^3tcos^2t + 5sin^4tcost + isin^5t]
sin5t = 5sintcos^4t  10sin^3tcos^2t + sin^5t
sin5t = sint[5cos^4t  10cos^2t + 10cos^4t + 1 + cos^4t  2cos^2t]
sin5t = sint[16cos^4t  12cos^2t + 1]
Let t = pi/5
0 = sint[16cos^4t  12cos^2t + 1]
sint =/= 0
16cos^4t  12cos^2t + 1 = 0
cos^2t = [12 + sqrt[144  64]]/32
= [12 + sqrt80]/32
= [3 + sqrt5]/8
A glance at the graph of cos^2 shows that the larger root is wanted (the other is very small).
cos^2t = [3 + sqrt5]/8
Consider (1/4)(1 + rt5)
= sqrt[(1/16)(6 + 2rt5)] = [(3 + sqrt5)/8]
Therefore cost = [1 + sqrt5]/4 
insparato
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 05032007 18:03
Having a go at STEP I Question 3 now.
3 i)
Sum of Differences method
f(r)  f(r1) = .....
f(1)  f(0)
f(2)  f(1)
f(3)  f(2)
.
.
.
f(n1)  f(n2)
f(n)  f(n1)
All terms cancel except two f(n) and f(0)
Therefore
ii)
Therefore
iii)
Hmm ive gone wrong somewhere on the last part.
Oh looking at the paper 1^3  2^3 + 3^3 .......(2n+1)^3 hmmm where does (2n+1)^3
surely sum (2n1)^3  sum(2n)^3 should work ?Last edited by insparato; 05032007 at 18:27. 
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 05032007 18:09
STEP I Question 5
f'(x) = n  nC2x + nC3x^2  nC4x^3 + ... + nCn(1)^(n1).x^n1
But (1x)^n = 1  nx + nC2x^2  nC3x^3 + ... + (1)^(n)x^n
1  (1x)^n = nx  nC2x^2 + nC3x^3 + ... + (1)^(n1)x^n
[1  (1x)^n]/x = n  nC2x + nC3x^2  nC4x^3 + ... + nCn(1)^(n1).x^n1
as required.
f(x) = INT f'(x) dx limits 0 and x
f(x) = INT [1  (1x)^n]/x dx limits 0 and x
Let y = 1  x; dy/dx = 1
> INT [1  y^n]/[1  y] dy limits 1 and 1  x
= INT [1  y^n]/[1  y] dy limits 1x and 1
f(x) = INT [1y^n]/[1y] dy = INT 1 + y + y^2 + ... + y^n1 dy
= [y + y^2/2 +y^3/3 + ... + y^n/n]
x = 1 so limits are 0>1
f(1) => 1 + 1/2 + 1/3 + 1/4 + ... + 1/n 
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 05032007 18:32
STEP I Question 6
i) y^2.dy/dx + y^1 = e^2x
Let u = 1/y
du/dy = y^2
dy/dx = du/dx.dy/du = y^2.du/dx = u^2.du/dx
du/dx + u = e^2x
du/dx  u = e^2x
(e^x)du/dx  (e^x)u = e^(x)
d/dx(ue^x) = e^x
ue^x = A  e^x
u = Ae^x  e^2x
y = 1/[Ae^x  e^2x]
ii) y^3.dy/dx + y^2 = e^2x
Let u = y^2
du/dy = 2y^3
dy/dx = du/dx.dy/du = du/dx.(1/2)y^3
= (1/2)u^(3/2).du/dx
(1/2)u^(3/2)u^(3/2).du/dx + u = e^2x
du/dx  2u = 2e^2x
(e^2x)du/dx  (e^2x)u = 2
d/dx[ue^2x] = 2
ue^2x = B  2x
u = Be^2x  2xe^2x
y = 1/sqrt[Be^2x  2xe^2x] 
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 05032007 19:12
STEP I Question 8
y'' + fy' + gy = h
y=x
f + gx = h
y=1
g = h
f = g(1x)
y=1/x
2/x^3  f/x^2 + g/x = h
2/x^3  g(1x)/x^2 + g/x = g
2/x^3 = g(1  1/x + (1x)/x^2)
2 = g(x^3  x^2  x^2 + x)
g = h = 2/[x^3  2x^2 + x] = 2/x(x1)^2
f = 2/x(x1)
y = ax + b + c/x
y' = a  c/x^2
y'' = 2c/x^3
2c/x^3  [2/x(x1)][a  c/x^2] + [2/x(x1)^2][ax + b + c/x] = [2/x(x1)^2]
2c  [2/x1][ax^2  c] + x[2/(x1)^2][ax^2 + bx + c] = 2x^2/(x1)^2
2c(x1)^2  (x1)(2ax^2  2c) + 2x(ax^2 + bx + c) = 2x^2
2cx^2  4cx + 2c  2ax^3 + 2ax^2 + 2cx  2c + 2ax^3 + 2bx^2 + 2xc = 2x^2
2cx^2 + 2ax^2 + 2bx^2 = 2x^2
a + b + c = 1 
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 05032007 19:30
Question 1 STEP III, 1995 is found in Siklos' Advanced Problems in Mathematics, question 14. (Not the booklet that can be downloaded for free)
EDIT: I've attached it here for those who didn't order the booklet.Last edited by khaixiang; 05032007 at 19:40. 
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 05032007 19:31
STEP II Question 4
By parts:
u_n = [sin^(n1)tcost] + (n1) INT sin^(n2)tcos^2t dt
Putting cos^2t = 1  sin^2t and using limits on the [...] gives:
u_n = 0 + (n1)(u_n2  u_n)
u_n = (n1)(u_n2)  (n1)u_n
nu_n = (n1)(u_n2)
nu_n.u_n1 = (n1)u_n1.u_n2
^ This proves that mu_m.u_m1 is the same for all m, so it is only necessary to prove it for one case, n = 2, and the rest follow by induction.
2 INT sin^2t dt . INT sint dt
= INT 1  cos2t dt . [cost]
= [t  1/2 sin2t].[1]
= [pi/2  0]
= pi/2
*graph*. sin^(n1) is a flattened sin curve and sin^n is one flattened slightly more, so just below it but having the same end points.
Since sin^nt is between 0 and 1 in the range of the integrals, higher powers mean lower areas, but they will all still be greater than 0, so 0 < u_n < u_n1.
n.u_n.u_n1 = pi/2
But u_n < u_n1, so:
n.(u_n)^2 < pi/2
and
n.{u_n1)^2 > pi/2
Second half of the next inequality has already been shown.
Changing n to n+1 we have pi/2 < (n+1)(u_n)^2 by the second half of the previous inequality.
npi/2 < n(n+1)(u_n)^2
(n/n+1)pi/2 < n(u_n)^2
So the entire inequality has been shown.
As n> infinity, n/n+1 = 1  1/n goes to 1.
Therefore, pi/2 <= n(u_n)^2 <= pi/2
n(u_n)^2 = pi/2 
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 10
 05032007 19:41
STEP II Question 7
i) AIB = (1/2)rc; BIC = (1/2)ra; CIA = (1/2)rb
Area of triangle = Z = r(1/2)(a+b+c) = rs
ii) Z = (1/2)bcsinA
Z^2 = (1/4)b^2c^2sin^2A
Z^2 = (1/4)(b^2c^2  (bccosA)^2)
Z^2 = (1/16)(4b^2c^2  (2bccosA)^2)
But 2bccosA = b^2 + c^2  a^2
Z^2 = (1/16)(4b^2c^2  (b^2 + c^2  a^2)^2)
Difference of two squares:
Z^2 = (1/16)(2bc + b^2 + c^2  a^2)(2bc  b^2  c^2 + a^2)
Z^2 = (1/16)((b+c)^2 a^2)(a^2  (bc)^2)
Difference of two squares again:
Z^2 = (1/16)(b+ca)(b+c+a)(ab+c)(a+bc)
Z^2 = (1/16)(a+b+c2a)(a+b+c)(a+b+c2b)(a+b+c2c)
Z^2 = (sa)s(sb)(sc)
Z = sqrt[s(sa)(sb)(sc)]
iii) Draw a diagram, x^2 = R^2  r^2
But r = Z/s = sqrt[(1/s)(sa)(sb)(sc)]
x^2 = R^2  (1/s)(sa)(sb)(sc)
x = sqrt[R^2  (1/s)(sa)(sb)(sc)] 
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 05032007 20:02
STEP II Question 8
dx/dt = (2Kx)x
INT 1/(2Kx)x dx = INT 1 dt
Partial fractions, 1/(2Kx)x = A/(2Kx) + B/x
1 = Ax + B(2Kx)
2Kb = 1
b = 1/(2K)
0 = A  B
A = 1/(2K)
INT 1/2K(2Kx) + 1/2Kx dx = INT 1 dt
(1/2K)ln[2K(2Kx)] + (1/2K)ln[2Kx] = t + C
ln[Ax] = lnx as far as we are concerned (an arbitrary constant is involved).
(1/2K).[lnx  ln(2Kx)] = t + C'
When t = 0, x = K
C' = (1/2K).[ln(K/K)]
C' = 0
ln[x/(2Kx)] = 2Kt
x/(2Kx) = e^2Kt
2Kx = xe^2Kt
2K = x(1 + e^2Kt)
x = 2K/(1 + e^2Kt)
x = (K + Ke^2Kt + K  Ke^2Kt)/(1 + e^2Kt)
x = K + K(1e^2Kt)/(1+e^2Kt)
*graph* starts at x=K, goes up and becomes asymptotic to x=2K.
t> infinity, x> 2K
dx/dt = (2Kx)x  L
dy/dt = dx/dt
dy/dt = (2K  y  K + a)(y + K  a)  L
dy/dt = (K  y + a)(K + y  a)  L
dy/dt = K^2 + Ky  Ka  Ky  y^2 + ay + Ka + ay  a^2  L
dy/dt = K^2  y^2 + 2ay  a^2  L
dy/dt = a^2  y^2 + 2ay  a^2
dy/dt = (2ay)y
When t = 0, y = K  K + a = a
Equations are symmetrical in (x,K) and (y,a)
y = a + a(1e^2at)/(1+e^2at)
x  K + a = a + a(1e^2at)/(1+e^2at)
x = K + a(1e^2at)/(1+e^2at)
t> infinity
x > K + a 
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 05032007 20:10
7 questions is enough for now , but I'll come back and do STEP II 1 and 6 in a bit if noone else has (please do, I don't particularly want to type them out).

insparato
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 05032007 20:12
You've done alot tonight Speleo go watch some tv or something before STEP rots your brains .
Ill have alook at STEP II 1 and 6 however i dont expect to do any of them . 
DFranklin
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 05032007 20:28

insparato
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 05032007 20:30
Grrr i should revise finite series . That question was definitely not beyond me.

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 05032007 20:47
STEP III Question 3
(sorry, it was calling to me )
No marks for clarity unfortunately.
Aux. quadratic:
m^2 + 2km + 1 = 0
m = [2k + sqrt(4k^2  4)]/2
m = k + sqrt(k^2  1)
i) k > 1
x = e^(kt).[Ae^(sqrt(k^2  1)t) + Be^(sqrt(k^2  1)t)]
ii) k = 1
x = (At + B)e^(kt)
iii) 0 < k < 1
x = e^(kt).[Acos(sqrt(k^2  1)t) + Bsin(sqrt(k^2  1)t)]
x(0) = 0
0 = A
x = Be^(kt)sin(sqrt(k^21)t)
dx/dt = kBe^(kt)sin(sqrt(k^21)t) + sqrt(k^21)Be^(kt)cos(sqrt(k^21)t) = 0 at max/min.
ksin(sqrt(k^21)t) = sqrt(k^21)cos(sqrt(k^21)t)
tan(sqrt(k^21)t) = sqrt(k^21)/k
sqrt(k^21)t = arctan[sqrt(k^21)/k] + mpi
t = [1/sqrt(k^21)]arctan[sqrt(k^21)/k] + mpi/sqrt(k^21)
When dividing successive x terms, the difference in the e term will just be pi/sqrt(k^21).
x_n+1/x_n = e^(kpi/sqrt(k^21)) sin[arctan[sqrt(k^21)/k] + (m+1)pi]/sin[arctan[sqrt(k^21)/k] + mpi]
sin(t + pi) = sinpi
x_n+1/x_n = e^(kpi/sqrt(k^21))
e^(kpi/sqrt(k^21)) = a.
kpi/sqrt(k^21) = lna
k^2/(k^21) = (lna)^2/(pi)^2
k^2 = k^2(lna)^2/(pi)^2  (lna)^2/(pi)^2
k^2((lna)^2/(pi)^2  1) = (lna)^2/(pi)^2
k^2[((lna)^2  (pi)^2)/(pi)^2] = (lna)^2/(pi)^2
k^2 = (lna)^2/[(pi)^2 + (lna)^2 
DFranklin
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 05032007 21:45
Paper I, Q1 (no sketch):
(i) f(x) = x^34x^2x4. It's easy to spot f(1),f(1),f(4) = 0. So f(x) = (x+1)(x1)(x4). So f(x)>0 for 1<=x<=1, x>=4.
(ii) g(x,y) = x^3  4x^2yxy^2+4y^3 = (x+y)(xy)(x4y) (noting the similarities with (i)). So g(x,y) = 0 on the lines y=x,y=x,y=x/4.
(iii) Draw the 3 lines we found in (ii). Note that g(1,0) > 0, and g is going to change sign every time you cross a line. (Or if not feeling brave in trusting that, find a test point in each region).Last edited by DFranklin; 05032007 at 22:28. 
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 05032007 21:46
STEP II Question 1
I just can't stop myself D:
i) (1  x)(1 + x + x^2 + x^3 + ... + x^n)
= (1 + x + x^2 + x^3 + ... + x^n)  (x + x^2 + x^3 + x^4 + ... + x^(n+1))
= 1  x^(n+1)
1 + x + x^2 + x^3 + ... + x^n = [1  x^(n+1)]/(1x)
ii) 1 + 2x + 3x^2 + ... + nx^(n1) = [(1x)(n+1)(x^n)  (x^(n+1)  1)]/(1x)^2
Let x = 1
1  2 + 3  ... + (1)^(n1)n = [(2)(n+1)(1)^n  (1)^(n+1) + 1]/4
If n is odd,
S = 2(n+1)/4 = (n+1)/2
If n is even,
S = [2(n+1)  2]/4
= [2n  2 + 2]/4 = n/2
iii) Group pairs.
If n is even:
= SUM 1  4r from 1 to n/2
= n/2  n(n/2 + 1)
= n/2  n^2/2  n
= n/2  n^2/2
If n is odd:
= SUM 1  4r from 1 + (n1)/2 + n^2
= (n1)/2  (n1)((n1)/2 + 1) + n^2
= n/2  1/2  (1/2)(n1)(n+1) + n^2
= n/2  1/2  (1/2)(n^2  1) + n^2
= n/2 + n^2/2
i.e. A = B = 1/2 
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 19
 05032007 22:01
STEP II Question 6
What the hell is wrong with me?
z^2 + az + b = 0
and (zu)(zv) = 0
uv = a
uv = b
From this point on a means alpha.
a^7  1 = cis(2pi/7)^7  1 = cis(2pi)  1 = 1  1 = 0
Other roots: a^0, a^2, a^3, a^4, a^5, a^6
(a + a^2 + a^4)^2 + A(a + a^2 + a^4) + B = 0
a^2 + a^4 + a + 2a^3 + 2a^5 + 2a^6 + Aa + Aa^2 + Aa^4 + B = 0
B + (A+1)a + (A+1)a^2 + 2a^3 + (A+1)a^4 + 2a^5 + 2a^6 = 0
Guess other root as a^3 + a^5 + a^6
(a^3 + a^5 + a^6)^2 + A(a^3 + a^5 + a^6) + B = 0
a^6 + a^3 + a^5 + 2a + 2a^2 + 2a^4 + Aa^3 + Aa^5 + Aa^6 + B = 0
B + 2a + 2a^2 + (A+1)a^3 + 2a^4 + (A+1)a^5 + (A+1)a^6 = 0
Clearly B=2 and A=1, as the sum of the powers of a is 0.
Take one from the other if you want confirmation (I think).
z^2 + z + 2 = 0 has root a + a^2 + a^4
z = [1 + sqrt(7)]/2
Looking at the diagram of the 7th roots of unity, a + a^2 + a^4 has Im(z) > 0, so a = [1 + isqrt7]/2
Re(a) = sum of cosines = (1/2)
Im(a) = sum of sines = (sqrt7)/2 
insparato
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 05032007 22:06
(Original post by Speleo)
STEP II Question 6
What the hell is wrong with me?
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