STEP Maths I, II, III 1995 Solutions

If I had to guess, I'd think O&C came first, then the STEP question, then your M4 book, but as you suggested it is possible that the M4 authors used O&C to mean the STEP exam!

Which exam board are you on out of interest?

(Updated as far as #100.) SimonM - 23.03.2009

STEP II Q9
The current link on the first page is the answer for STEP II Q9 1996; STEP II Q9 1995 hasn't yet been posted. So here's mine.

Step I, Q3, last bit:



Alternatively:

.

Having a go at STEP I Question 3 now.

3 i)

Sum of Differences method

f(r) - f(r-1) = .....

f(1) - f(0)
f(2) - f(1)
f(3) - f(2)
.
.
.
f(n-1) - f(n-2)
f(n) - f(n-1)

All terms cancel except two f(n) and f(0)

Therefore

$\displaystyle \sum_{r=1}^n f(r) - f(r-1) = f(n) - f(0)$

ii) $f(r) = r^2(r+1)^2$

$f(r-1) = r^2(r-1)^2$

$f(r) - f(r-1) = r^2(r+1)^2 - r^2(r-1)^2$

$= r^2(r^2+2r+1 -r^2+2r-1) = 4r^3$

Therefore

$\displaystyle \sum_{r=1}^n f(r) - f(r-1) = f(n) - f(0) = n^2(n+1)^2$

$\displaystyle \sum_{r=1}^n f(r)-f(r-1) = \sum_{r=1}^n 4r^3 = n^2(n+1)^2$

$\displaystyle \sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2$

iii)

$\displaystyle \sum_{r=1}^n (2n-1)^3 - \sum_{r=1}^n (2n)^3$

$\frac{1}{4}(2n-1)^2(2n)^2 - \frac{1}{4}(2n)^2(2n+1)^2$

$n^2(4n^2-4n+1 -(4n^2+4n+1))$

$n^2(-8n)$

$-8n^3$

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 (2n-1)^3 - sum(2n)^3 should work ?

STEP III, 1995, Q14.

If $M_T(x)$ is the moment generating function for T, then $M_T(x) = \int_{-\infty}^\infty e^{xt} f(t) dt = \int_0^\infty e^{xt} f(t) dt$ since f(t) = 0 for t < 0.

So $M_T(x) = \int_0^\infty t e^{(x-1)t}dt$. We need x < 1 for convergence, so assume x < 1. Integrating by parts:
$M_T(x) = [\frac{t}{x-1}te^{(x-1)t}]_0^\infty - \frac{1}{x-1}\int_0^\infty e^{(x-1)t} dt = \frac{1}{(x-1)^2}$.

Now if X, Y are any two independently distributed variables, then $M_{X+Y}(x) = M_X(x)M_Y(x).$. So $M_U(x) = \frac{1}{(x-1)^4}$.

Now when we come to find the p.d.f., we're going to find ourselves evaluating $\int_0^\infty t^3 e^{(x-1)t}dt$, and for the p.d.f. of the last bit, evaluating $\int_0^\infty t^N e^{(x-1)t}dt$. So to avoid repeating ourselves, we'll prove a little reduction formula integral:

Define $I_n(y) = \int_0^\infty t^n e^{ty} dt$ (where y < 0). Then $I_n(y) = [\frac{t^n}{y} e^{ty}]_0^\infty - \frac{n}{y} \int_0^\infty t^{n-1} e^{ty} dt = \frac{-n}{y} I_{n-1}(y)$.
Noting that $I_0(y) = \int_0^\infty e^{ty} dt = [\frac{e^ty}{y}]_0^\infty = -\frac{1}{y}$, we have $I_n(y) = \frac{n!}{(-y)^{n+1}}$.

So if $g(t) = \frac{1}{6}t^3e^{-t}$, then $\int_0^\infty e^{xt} g(t)dt = \frac{1}{6} \int_0^\infty t^3 e^{(x-1)t} dt = \frac{1}{6} I_3(x-1)= \frac{1}{6} \frac{3!}{(x-1)^4} = \frac{1}{(x-1)^4} = M_U(x)$. Thus g is the p.d.f. corresponding to the moment generating function $M_U$.

Similarly, in the case of answering n questions, let $M_n(x)$ be the moment generating function of the sum. Then we have $M_n = \frac{1}{(1-x)^{2n}}$. Consider $h_n(t) = t^{2n-1}e^{-t}$. Then $\int_0^\infty e^{xt} h(t)dt = I_{2n-1}(x-1) = \frac{(2n-1)!}{(x-1)^2n}$, and so $g_n(t) = \frac{1}{(2n-1)!}h(t) = \frac{1}{(2n-1)!}t^{2n-1}e^{-t}$ is the p.d.f. for the sum of n questions.

Comment: Are moment generating functions on the current syllabus? I had done them, but I couldn't remember a blind thing about them; I had to look them up in Wiki before doing this question.

Having a go at STEP I Question 3 now.

3 i)

Sum of Differences method

f(r) - f(r-1) = .....

f(1) - f(0)
f(2) - f(1)
f(3) - f(2)
.
.
.
f(n-1) - f(n-2)
f(n) - f(n-1)

All terms cancel except two f(n) and f(0)

Therefore

$\displaystyle \sum_{r=1}^n f(r) - f(r-1) = f(n) - f(0)$

ii) $f(r) = r^2(r+1)^2$

$f(r-1) = r^2(r-1)^2$

$f(r) - f(r-1) = r^2(r+1)^2 - r^2(r-1)^2$

$= r^2(r^2+2r+1 -r^2+2r-1) = 4r^3$

Therefore

$\displaystyle \sum_{r=1}^n f(r) - f(r-1) = f(n) - f(0) = n^2(n+1)^2$

$\displaystyle \sum_{r=1}^n f(r)-f(r-1) = \sum_{r=1}^n 4r^3 = n^2(n+1)^2$

$\displaystyle \sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2$

iii)

$\displaystyle \sum_{r=1}^n (2n-1)^3 - \sum_{r=1}^n (2n)^3$

$\frac{1}{4}(2n-1)^2(2n)^2 - \frac{1}{4}(2n)^2(2n+1)^2$

$n^2(4n^2-4n+1 -(4n^2+4n+1))$

$n^2(-8n)$

$-8n^3$

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 (2n-1)^3 - sum(2n)^3 should work ?

STEP II, 1995, Q2:

$r_{n+1} = s_n$ because if we have an acceptable colouring of n+1 posts with the first and last posts coloured red, then the next to last post must be coloured other than red, and so removing the last post gives a sequece of n posts with the last post coloured other than red. And vice versa.

$r_n + s_n$ is the number of ways of colouring n posts so that the first post is red and no two posts have the same colour (with no extra condition about the colour of the last post). Colouring the posts in order, for each post other than the first, there are 2 choices for its colour (either of the colours that differ from the previous post). So $r_n+s_n = r_{n+1}+r_n = 2^{n-1}$.

Then $r_{n+1} = 2^{n-1} -r_{n}$. Then if $r_k = \frac{2^{k-1} + 2(-1)^{k-1}}{3}$ we have:


So since $r_1 = 1 = \frac{2^0 + 2(-1)^0}{3}$, result follows by induction.

Of course, there's nothing special about painting the first post red (as opposed to blue or white), so the number of ways of painting n+1 posts so no two posts are the same colour and the first and last posts are coloured the same is just 3 times what we've just calculated. But then by identifying the 1st and last posts (which are the same colour), we are left with the same situation as n posts in a circle. (e.g. if we consider n=13, then we can think of having posts 1,2,...,12 as the hours on a clock, and we consider post 13 to be the same as post 1).

So the required answer is just $3r_{n+1} = 2^n+2(-1)^n$.

Edit: Mistake spotted by DVS.

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^2-1)t)

dx/dt = -kBe^(-kt)sin(sqrt(k^2-1)t) + sqrt(k^2-1)Be^(-kt)cos(sqrt(k^2-1)t) = 0 at max/min.
ksin(sqrt(k^2-1)t) = sqrt(k^2-1)cos(sqrt(k^2-1)t)
tan(sqrt(k^2-1)t) = sqrt(k^2-1)/k
sqrt(k^2-1)t = arctan[sqrt(k^2-1)/k] + mpi
t = [1/sqrt(k^2-1)]arctan[sqrt(k^2-1)/k] + mpi/sqrt(k^2-1)

When dividing successive x terms, the difference in the e term will just be pi/sqrt(k^2-1).
x_n+1/x_n = e^(-kpi/sqrt(k^2-1)) sin[arctan[sqrt(k^2-1)/k] + (m+1)pi]/sin[arctan[sqrt(k^2-1)/k] + mpi]
sin(t + pi) = -sinpi
x_n+1/x_n = -e^(-kpi/sqrt(k^2-1))
e^(-kpi/sqrt(k^2-1)) = a.
-kpi/sqrt(k^2-1) = lna
k^2/(k^2-1) = (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

STEP III Question 5 - Credit to DFranklin for helping with the final part

(1-x^2)y'' = xy' + 2m^2(1 - 2y)

Let m arcsin x = A
y = sin^2(m arcsin x)
y = sin^2A
y' = sin2A.m/sqrt(1-x^2)
y'' = 2cos2A.m^2/(1-x^2) + xsin2A.m/(1-x^2)^(3/2)
2m^2(1-2y) = 2m^2cos2A
xy' = xsin2A.m/sqrt(1-x^2)
(1-x^2)y'' = 2cos2A.m^2 + xsin2A.m/sqrt(1-x^2)
Clearly LHS = RHS, as required.

(1-x^2)y'' = xy' + 2m^2(1 - 2y)
Differentiate:
(1-x^2)y''' - 2xy'' = xy'' + y' - 4m^2y'
(1-x^2)y(3') = (2 + 1)xy(2') + (1 - 4m^2)y(1')

Now assume:
(1-x^2)y(n+2') = (2n + 1)xy(n+1') + (n^2 - 4m^2)y(n')
Differentiate:
(1-x^2)y(n+3') - 2xy(n+2') = (2n+1)y(n+1') + (2n + 1)xy(n+2') + (n^2 - 4m^2)y(n+1')
(1-x^2)y((n+1)+2') = (2(n+1) + 1)xy((n+1)+1') + ((n+1)^2 - 4m^2)y((n+1)')
Proof by induction complete.

y(x) = y(0) + y'(0)x + y''(0)x^2/2! + ... + y(n')(0)x^n/n! + ...

y(n+2')(0) = (n^2 - 4m^2)y(n')(0)
and
y''(0) = 2m^2(1 - 2y(0))
y''(0) = 2m^2 - 4m^2y(0)
From the first few lines of working we have expressions for y and y', and both y(0) and y'(0) = 0.
So the odd derivatives of y all = 0 when x = 0.
y''(0) = 2m^2
y(4')(0) = 2m^2.(2^2 - 4m^2)
y(6')(0) = 2m^2.(2^2 - 4m^2).(4^2 - 4m^2)
y(8')(0) = 2m^2.(2^2 - 4m^2).(4^2 - 4m^2).(6^2 - 4m^2)
y(2n')(0) = 2m^2. [PROD (2k)^2 - 4m^2] from k=1 to n.

$y = m^2x^2 + \frac{2m^2.(2^2 - 4m^2)x^4}{4!} + ... + \frac{2m^2\Pi_{k=1}^n[(2k)^2 - 4m^2]x^{2n}}{(2n)!} + ...$

Question 6, STEP III, 1995.
(Thanks to DFranklin for his guidance on this question)

$\\|z-i|=1, \text{ let } z=x+iy\\|x+i(y-1)|=1\\x^2+(y-1)^2=1$
$\\ \text{ let } |z|=r, \text{ and } \arg(z)=\theta\\ \text{ then } r\cos\theta=x, r\sin\theta=y\\r^2\cos^2\theta+(r\sin\theta-1)^2=1\\r(r-2\sin\theta)=0\\ \implies r=2\sin\theta \text{ and } r=0, r=0 \text{ when } \theta=0,\pi$

$\\ \displaystyle |\frac{1}{z}|={|1|}/{|z|}\\=\frac{1}{2\sin\theta}$

$\\ \arg(1/z)=\arg(1)-\arg(z)\\=-\theta$

$\\ \therefore \displaystyle |\frac{1}{z}|=-\frac{1}{2\sin(\arg(\frac{1}{z}))}\\ \text{ But } |\frac{1}{z}|\sin(\arg(\frac{1}{z}))=y, \therefore y=-\frac{1}{2} \text{ is the cartesian equation of the locus of } \frac{1}{z}$

So loci of points in an argand diagram which represents z and 1/z are a circle of radius 1 centre at (0,1) and a horizontal line y=-1/2.




Locus of C in argand diagram is a circle with centre (-1/2,0) and radius 1/2.

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)]/(1-x)

ii) 1 + 2x + 3x^2 + ... + nx^(n-1) = [-(1-x)(n+1)(x^n) - (x^(n+1) - 1)]/(1-x)^2
Let x = -1
1 - 2 + 3 - ... + (-1)^(n-1)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 + (n-1)/2 + n^2
= (n-1)/2 - (n-1)((n-1)/2 + 1) + n^2
= n/2 - 1/2 - (1/2)(n-1)(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

You have a condition called STEPitis. Its quite contagious no cure im afraid.

STEP I Question 12

Part (i)

Part (ii)

Part (iii)

Question 3 is more like a Sudoku than a STEP question...

Anyway.
Call the number of points each guy got A, B, C, D and E.
A+B+C+D+E = 15*5 = 75, just adding up how many points are available in total.

Also. A>B>C>D>E.
A=24, so B+C+D+E = 75-24 = 51.
If E>11, then the least value of B+C+D+E = 12+13+14+15 > 51. So that's not possible.
Also we know that E = (???+??+?)+3+5. The least score he could possible have earned is 1+1+1+3+5 = 11.
So E cannot be greater than 11 or less than 11. So E = 11.

E = 11 gives
B+C+D = 40.
Remembering B>C>D, B = 15, C = 13 and D = 12 (no other combination works).
A must have achieved 5+5+5+5+4 =24 in that order, as E has 5 in the last stage.
Here's a chart:

Stage
1 2 3 4 5
---------
5 5 5 5 4 ... A = 24
? ? ? ? ? ... B = 15
? ? ? ? ? ... C = 13
? ? ? ? ? ... D = 12
1 1 1 3 5 ... E = 11

We are told that C = 4a+b.
C is odd, and all the '5's are used up, so C must get 3+3+3+1+3 in that order.
B is odd, so it must include a 1, 3 or a 5. 3 and 5 are all used up, so it includes a 1 - and the only space for a 1 is at the end.
Thus C scored 1 in the last stage.