STEP Maths I, II, III 1995 Solutions

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

Unless i'm missing something here or am just too tired to function properly, doesn't the second to last line....

k^2[((lna)^2 - (pi)^2)/(pi)^2] = (lna)^2/(pi)^2

imply that the final solution is....

k^2 = (lna)^2/[(pi)^2 - (lna)^2]

Which would be the wrong answer, but i agree with everything else you've done up until that point so i don't know what's up with this final bit.

OK, there's a mistake quite early on in what Speleo's done that follows through everywhere and I think is causing the -ve sign.

We're in the 0 < k < 1 case, and so we want $e^{-kx} \sin x\sqrt{1-k^2}$, not $e^{-kx} \sin x\sqrt{k^2-1}$.

STEP II, 1995, Q2:

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.

Writing out STEP III Question 2 now.

$\displaystyle I_n = \int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}$

$\displaystyle I_0 = \int_0^a x^{\frac{1}{2}}(a-x)^{\frac{1}{2}}$

$x = asin^2u$

$\frac{dx}{du} = 2asinucosu$

$dx = 2asinucosu. du$

$a = asin^2u$

$u = \frac{\pi}{2}$

$0 = asin^2u$

$u = 0$

$\displaystyle I_0 = \int_0^{\frac{\pi}{2}} a^{\frac{1}{2}}sinu(a^{\frac{1}{2}}cosu).2asinucosu. du$

$\displaystyle = 2a^2\int_0^{\frac{\pi}{2}} sin^2ucos^2u$

$\displaystyle = 2a^2\int_0^{\frac{\pi}{2}} \frac{1}{4}(1-cos2u)(1+cos2u)$

$\displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} 1-cos^22u$

$\displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} 1-\frac{1}{2}(1-cos4u)$

$\displaystyle = \frac{a^2}{2} \int_0^{\frac{\pi}{2}} \frac{1}{2}-\frac{1}{2}cos4u$

$= \frac{a^2}{2} [\frac{1}{2}u - \frac{1}{8}sin4u]_0^{\frac{\pi}{2}}$

$= \frac{a^2}{2} ([\frac{\pi}{4} - 0]-[0-0])$

$= \frac{\pi a^2}{8}$

ii)
$\displaystyle I_n = \int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}$

$u = x^{n+\frac{1}{2}}$

$\frac{du}{dx} = (n+\frac{1}{2})x^{n-\frac{1}{2}}$

$\frac{dv}{dx} = (a-x)^{\frac{1}{2}}$

$v = -\frac{2}{3}(a-x)^{\frac{3}{2}}$



$\displaystyle I_n = \frac{2}{3}(n+\frac{1}{2})\int (a-x)^{\frac{3}{2}}x^{n-\frac{1}{2}}$





$I_n = \frac{2}{3}(n+\frac{1}{2})[aI_{n-1} - I_n]$

$3I_n = (2n+1)(aI_{n-1}-I_n)$

$3I_n = 2anI_{n-1} - 2nI_n -I_n + aI_{n-1}$

$(2n+4)I_n = a(2n+1)I_{n-1}$

$I_n = \frac{a(2n+1)}{2n+4}I_{n-1}$

STEP III Question 4[spolier]

Cn(t)sin(1/2)t = cos(0)tsin(1/2)t + cos(1)tsin(1/2)t + cos(2)tsin(1/2)t + ... + cos(n)tsin(1/2)t
= (1/2)[2sin(1/2)t + sin(3/2)t - sin(1/2)t + sin(5/2)t - sin(3/2)t + ... + sin(n + 1/2)t]
= (1/2)[sin(1/2)t + sin(n+ 1/2)t]
= sin[(n+1)/2]t.cos[n/2]t
Cn(t) = sin[(n+1)/2]t.cos[n/2]t/sin(1/2)t
as required.

Sn(t)sin(1/2)t = sin(0)t.sin(1/2)t + sin(1)t.sin(1/2)t + sin(2)t.sin(1/2)t + ... + sin(n)tsin(1/2)t
= (1/2)[cos(3/2)t - cos(1/2)t + cos(5/2)t - cos(3/2)t + ... + cos[n+ 1/2]t]
= (1/2)[cos[(n+ 1/2]t - cos(1/2)t]
= sin[(n+1)/2]t.sin[n/2]t
Sn(t) = sin[(n+1)/2]t.sin[n/2]t/sin(1/2)t

Cn(t) - (1/2) = sin[(n+1)/2]t.cos[n/2]t/sin(1/2)t - 1/2
= {2sin[(n+1)/2]t.cos[n/2]t - sin(1/2)t}/2sin(1/2)t
|Cn(t) - (1/2)| = |{2sin[(n+1)/2]t.cos[n/2]t - sin(1/2)t}/2sin(1/2)t|
But 0 < t < 2pi, so sin(1/2)t > 0
So:
|Cn(t) - (1/2)| = |2sin[(n+1)/2]t.cos[n/2]t - sin(1/2)t|/2sin(1/2)t
Consider the expression in the modulus signs on the RHS of the equation:
2sin[(n+1)/2]t.cos[n/2]t - sin(1/2)t
= sin(n + 1/2)t + sin(1/2)t - sin(1/2)t
= sin(n + 1/2)t
|sin(n+ 1/2)t| <= 1

So:

|Cn(t) - (1/2)| <= 1/[2sin(1/2)t]

STEP I, Question 14



$\sigma^2 = \frac{pq}{n}$



So we want $2.5758 \times \frac{\sqrt{pq}}{n} < 0.01$

Take $p = \frac{1}{2}$, worst case scenario and reasonable for the proportion of women

So $n \ge 129$

Since $\sqrt{pq}$ will decrease, we can take a smaller sample for millionaires over lefties

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 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]

Am I being completely retarded, or is there a slight mistake in question 7 in STEP II? There's a 1/16 missing on the equation under 'Now use the formula a^2 = b^2 + c^2 - 2bcCosA).

Yeah, in my copy of it there's a "letter to supervisors" saying about the mistake

It looks like nobody has fixed the front page links in awhile, but according to my computer, whenever I click the answer for STEP I Q 1, it links me to Insperato's answer for Q3. Probably just a hyperlink thing? If not, then has that question been done?