(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
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Maths
>< Maths Exams

STEP Maths I, II, III 1995 Solutions watch

SimonM
 Follow
 13 followers
 18 badges
 Send a private message to SimonM
 Thread Starter
Offline18ReputationRep: Follow
 1
 05032007 17:34
Last edited by SimonM; 21072011 at 20:07. 
insparato
 Follow
 6 followers
 13 badges
 Send a private message to insparato
Offline13ReputationRep: Follow
 2
 05032007 17:54
Last edited by insparato; 08052010 at 17:43. 
 Follow
 3
 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
 Follow
 6 followers
 13 badges
 Send a private message to insparato
Offline13ReputationRep: Follow
 4
 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. 
 Follow
 5
 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 
 Follow
 6
 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] 
 Follow
 7
 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 
 Follow
 8
 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. 
 Follow
 9
 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 
 Follow
 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)] 
 Follow
 11
 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 
 Follow
 12
 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
 Follow
 6 followers
 13 badges
 Send a private message to insparato
Offline13ReputationRep: Follow
 13
 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
 Follow
 64 followers
 18 badges
 Send a private message to DFranklin
Offline18ReputationRep: Follow
 14
 05032007 20:28

insparato
 Follow
 6 followers
 13 badges
 Send a private message to insparato
Offline13ReputationRep: Follow
 15
 05032007 20:30
Grrr i should revise finite series . That question was definitely not beyond me.

 Follow
 16
 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
 Follow
 64 followers
 18 badges
 Send a private message to DFranklin
Offline18ReputationRep: Follow
 17
 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. 
 Follow
 18
 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 
 Follow
 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
 Follow
 6 followers
 13 badges
 Send a private message to insparato
Offline13ReputationRep: Follow
 20
 05032007 22:06
(Original post by Speleo)
STEP II Question 6
What the hell is wrong with me?
Related discussions
 STEP Maths I, II, III 1997 Solutions
 STEP Maths I, II, III 1993 Solutions
 2013 STEP thread mark II
 STEP Prep Thread 2016 (Mark. II)
 STEP Prep Thread 2015
 Oxford MAT 19921995 solutions thread.
 STEP Prep Thread 2014
 STEP Prep Thread 2016
 STEP Prep Thread 2017
 STEP Mathematics ProblemSolving Society!
Related university courses

Mathematics with Finance
University of Leeds

Mathematics, Statistics and Financial Economics
Queen Mary University of London

Mathematics
HeriotWatt University

Mathematics
University of Edinburgh

Mathematics with Finance
University of Brighton

Mathematics
Liverpool Hope University

Mathematics for Finance (with a year abroad)
Swansea University

Mathematics with a Placement Year
University of Reading

Mathematics with Education
University of Plymouth

Mathematics with Sandwich Placement
University of Wolverhampton
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 Lemur14
 brainzistheword
 Quirky Object
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 Labrador99
 EmilySarah00
 thekidwhogames
 entertainmyfaith
 Eimmanuel
 Toastiekid
 CinnamonSmol
 Qer
 The Empire Odyssey
 RedGiant
 Sinnoh