STEP Maths I, II, III 1996 Solutions

This one looks like it may be wrong :/

STEP I Question 7

i) dy/dt = -ky; k>=1
lny = C - kt
y = e^(C-kt) = Ae^(-kt)
When t = 0, y = 1
1 = A
y = e^(-kt)
y = [e^(-k)]^t
Let b = e^(-k)
y = b^t
Since k>=1, b<=1/e, b<1

ii) dy/dt = a - ky
dy/dt = a + ylnb
dy/dt - ylnb = a
Integrating factor = e^(-tlnb)
e^(-tlnb)dy/dt + e^(-tlnb)y(-lnb) = ae^(-tlnb)
d/dt[ye^(-tlnb)] = ae^(-tlnb)
ye^(-tlnb) = -(a/lnb)e^(-tlnb) + C
When t = 0. y = 1
1 = C - (a/lnb)
C = 1 + (a/lnb)
ye^(-tlnb) = -(a/lnb)e^(-tlnb) + (a/lnb) + 1
y = -(a/lnb) + (a/lnb)e^(tlnb) + e^(tlnb)
y = [1 + (a/lnb)]b^t - (a/lnb)

Reply 8

STEP I Question 8

i) S = 0.38383838...
= 38/100 + 38/(100)^2 + 38/(100)^3 + ...
= [38/100]/[1-(1/100)]
= (38/100)/(99/100)
= 38/99.

ii) x = 0.a1a2...aNb1b2b3...bkb1b2...bkb1..
= 0.a1a2...aN + 0.000...000b1b2...bkb1...
= a1a2...aN/(10^N) + 0.b1b2...bkb1.../(10^N)
But 0.b1b2...bkb1...
= b1b2...bk/(10^k) + b1b2...bk(10^k)^2 + ...
= [b1b2...bk/10^k]/[1 - 1/(10^k)]
= b1b2...bk/(10^k - 1)
So x = a1a2...aN/(10^N) + b1b2...bk/(10^N)(10^k - 1)
Since a1a2...aN and b1b2...bk integers, x is rational.

Reply 9

I haven't checked to see if the last part is consistent with the Siklos book.

STEP II Question 2

Let a = yz, b = zx, c = xy.

I: 2a + b - 5c = 2
II: a - b + 2c = 1
III: a - 2b + 6c = 3

I-2II = 3b - 9c = 0, b = 3c
I-2III = 5b - 17c = -4
-2c = -4
c = 2
b = 6
a = 1 - 4 + 6 = 3
abc = 36
(xyz)^2 = 36
xyz = +- 6

Factors of +- 6 must be +- 1, 2, 3 or 1, 1, 6.
Since yz = 3 and xy = 2; 1, 1, 6 is not possible.
Therefore a, b, c are +- 1, 2, 3.
|y| = 1 clearly, so |z| = 3 and |x| = 2.
Since xy, yz, zx are all > 0,
(x,y,z) = (2,1,3) or (-2,-1,-3).

Reply 10

This is ridiculous, TSR won't post this in one go:

STEP II Question 8

g(0) = f(0) + f(-0) = 2

g'(x) = f'(x) - f'(-x)
g'(0) = -1 + 1 = 0
g''(x) = f''(x) + f''(-x)
f''(x) = x + 3cos2x - f(-x)
f''(-x) = -x + 3cos2x - f(x)
g''(x) = 6cos2x - f(x) - f(-x)
g''(x) = 6cos2x - g(x)
g''(x) + g(x) = 6cos2x

Aux quadratic -> m^2 + 1 = 0
m = +-i
g(x) = Acosx + Bsinx + ...
g(x) = pcos2x
-4p + p = 6
p = -2
g(x) = Acosx + Bsinx - 2cos2x
g(0) = 2.
A - 2 = 2
A = 4
g'(x) = -4sinx + Bcosx + 4sin2x
g'(0) = 0 = B
g(x) = 4cosx - 2cos2x

Reply 11

h(x) = f(x) - f(-x)
h(0) = 0
h'(x) = f'(x) + f'(-x)
h'(0) = -2
h''(x) = f''(x) - f''(-x)
f''(x) = x + 3cos2x - f(-x)
f''(-x) = -x + 3cos2x - f(x)
h''(x) = 2x - f(-x) + f(x)
h''(x) = 2x + h(x)
h''(x) - h(x) = 2x.
Aux quadratic is m²-1 = 0 ie m=±1.
If h(x) = kx, then h(x) = -2x.
So h(x) = Ae^x + Be^-x -2x.
h(0) = 0=A+B -0
h`(0) = -2 =A-B - 2
So A+B = 0 and A-B = 0, so A and B are both 0.
h(x) = -2x

2f(x) = g(x) + h(x)

f(x) = 2cosx - cos2x - x

Correction in italics thanks to Rabite :smile:

Reply 12

After several attempts, I have a solution to Paper I, Q5. I think I must have missed something obvious though - this was very hard work. (Though largely because it is so easy to make a mistake; the written up answer probably won't look too bad).

Reply 13

Step QI, P5.

(i)

(r+s\sqrt{3})=4-2\sqrt{3} \implies r^2+3s^2+2rs\sqrt{3} = 4-2\sqrt{3}

. Suppose

rs \neq -1

. Then

r^2+3s^2-4 = -2(1+rs)\sqrt{3}

would give us a rational expression for

\sqrt{3}

, contradiction.

So

Unparseable latex formula:

rs=1, \text{ so } s = -1/r \text { and so } r^2+3/r^2 = 4 \implies r^4-4r^2+3 = 0\\[br]\implies r^2 = 1, 3

.

As r is rational,

r^2 = 1, \text{ so } r = \pm 1, s = -r

. (i.e. our solutions are

\pm(1-\sqrt{3})

.

(ii)

(p+qi)^2 = p^2-q^2+2pqi = 3-2\sqrt{3} + 2(1-\sqrt{3})i

. So

p^2-q^2 = 3-2\sqrt{3}, pq = 1-\sqrt{3}

. Again, we set up the quadratic and get:

p^4-(3-2\sqrt{3})p^2-(1-\sqrt{3})^2 = 0

. Use the quadratic formula to get:

2p^2 = (3-2\sqrt{3}) \pm \sqrt{(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2}

Now

(3-2\sqrt{3})^2 = 3^2+4\cdot 3 - 12\sqrt{3} = 21 - 12\sqrt{3}, \quad (1-\sqrt{3})^2 = 1 + 3 - 2 \sqrt{3} = 4-2\sqrt{3}

.
So

(3-2\sqrt{3})^2+4(1-\sqrt{3}))^2 = 37 - 20 \sqrt{3}

.

As in (i), we need to find the square root of this, so look for rational

\alpha, \beta

with

(\alpha+\beta\sqrt{3})^2 = 37 - 20\sqrt{3}

.

We find

\beta = -10/\alpha

and end up with a quadratic

\alpha^4 - 37\alpha^2+300=0

. Solve to find

\alpha^2 = 12, 25

and since we want

\alpha

rational deduce

\alpha = \pm 5

and so the square root is

\pm(5-2\sqrt{3})

.

Plugging this back in, we have

2p^2 = (3-2\sqrt{3}) \pm (5-2\sqrt{3}) = -2, 8-4\sqrt{3}

.

Since p is real, we take the latter root, dividing by 2 and using (i) we find

p = \pm (1-\sqrt{3})

. Use

pq = (1-\sqrt{3})

to get final solutions

\pm (1-\sqrt{3}+i)

.

(iii) Again, use the quadratic formula:

Unparseable latex formula:

2z=2\pm\sqrt{4 - 4(1+i)(2\sqrt{3}-2)} \\[br]\implies z = 1 \pm \sqrt{1 - (1+i)(2\sqrt{3}-2)} = 1 \pm \sqrt{3-2\sqrt{3}+2(1-\sqrt{3})i)

.

Using (ii),

z = 1 \pm (1-\sqrt{3}+i) = \sqrt{3} - i, 2-\sqrt{3}+i

.

Reply 14

ukgea

6

On the first page of the tests:

"To be brought by the candidate: electronic calculators, [...]"

Wtf? Were they allowed to use calculators back then? On all the later papers it says "Candidates may not use electronic calculators" instead. But anyone know why were they allowed previously and not now?

Reply 15

Step II, Q3:

F0 = 0, F1 =1, F2 = F0+F1 = 1, F3 = F1 + F2 = 2, F4 = F3 + F2 = 3, F5 = 5, F6 = 8, F7 = 13.

F0.F2-F1^2 = -1, F1.F3-F2^2 = 1, F2.F4-F3^2 = -1.

Claim

F_{n+1}F_{n-1}-F_n^2 = (-1)^n

. Proof by induction on n.
By above, true for n<=3. Assume true for n = k. Then

Unparseable latex formula:

F_{k+2}F_k - F_{k+1}^2 \\[br]= (F_{k+1}+F_k)F_k - F_{k+1}^2 \\ [br]= F_k^2 +F_{k+1}F_k - F_{k+1}^2 = F_k^2 - F_{k+1}(F_{k+1} - F_k) \\[br]= [F_{k+1}F_{k-1}+(-1)^{k+1}] - F_{k+1}(F_{k+1} - F_k) \\[br]= (-1)^{k+1} + F_{k+1}F_{k-1} - F_{k+1}(F_{k-1})\\[br]= (-1)^{k+1}F_{k+1}^2

So true for n=k+1 and so true for all n by induction.

Want to show

F_{n+k} = F_kF_{n+1}+F_{k-1}F_n

. Assume true for 1 < k <=m. Then

Unparseable latex formula:

F_{n+k+1} = F_{n+k}+F_{n+k-1}\\[br]= F_kF_{n+1}+F_{k-1}F_n + F_k-1F_{n+1}+F_{k-2}F_n\\[br]=(F_k+F_{k-1})F_{n+1}+(F_{k-1}+F_{k-2}F_n \\[br]=F_{k+1}F_{n+1}+F_k F_n

.

So true for k=m+1.

Explictly when k=1 we have

F_{n+1} = F_1F_{n+1}+F_0F_n = F_{n+1}

. So true for k=1, so true for all k.

Reply 16

ukgea

On the first page of the tests:

"To be brought by the candidate: electronic calculators, [...]"

Wtf? Were they allowed to use calculators back then? On all the later papers it says "Candidates may not use electronic calculators" instead. But anyone know why were they allowed previously and not now?

I would guess it's a response to the increasing capabilities of calculators: graphing, algebra, programmability. Rather than try to rule out the "super calculators", simpler to just ban them all, and make sure the questions don't need them.

I think Siklos also says there was a tendancy for the weaker candidates to just blindly throw everything into the calculator without thinking too much.

Reply 17

ukgea

6

Paper II question 6

We can easily see that the proper factors of 12 are 2, 3, 4, 6 and those of 16 are 2, 4, 8.

Let M be a positive factor of N (not necessarily proper), Then f can be written as

M = p_1^{n_1}p_2^{n_2}p_3^{n_3}\cdot P_r^{n_r}

where

0 \leq n_i \leq m_i

for all

1 \leq i \leq r

.

For each of the i, there are

m_i + 1

ways of choosing the integers

n_i

, for a total number of F possible factors of N:

F = (m_1+1)(m_2+1)(m_3+1)\cdot(m_r+1)

Each of these must be unique since prime factorisation is unique. But now we've counted the two "improper" factors 1 and N as well, so we need to remove these, and the total number of proper factors is

(m_1+1)(m_2+1)(m_3+1)\cdot(m_r+1) - 2

.

(i) For the total number of proper factors to be 12, we must have

(m_1+1)(m_2+1)(m_3+1)\cdot(m_r+1) = 14

which means that, r must be 2 and

m_1 + 1 = 2, m_2 + 1 = 7

, or vice versa, or

m_1 + 1 = 14

In the latter case, the smallest integer that has only one prime factor repeated 13 times is

N = 2^13

. In the former case, clearly, to minimise N, we need to take as small prime factors as possible, and make sure that the smaller of the two factors is raised to power 6, i.e. the minimum is

N = 2^6\cdot3

. This is less that

2^13

, and so

N = 2^6\cdot3

is our answer.

(ii) Okay. The smallest integer is

120 = 2^3\cdot3\cdot5

. But it seems that every attempt to prove this either boils down to a huge awful grind, or a complete orgy of hand-waving. The only slightly rigorous proof I've found is the one I'm going to make below, but it uses the inequality between the arithmetic and geometric means, which says (in the case of 3 numbers, which is the case I'm going to use):

For any positive reals a, b, c, we have

\displaystyle \frac{a + b + c}{3} \geq \sqrt[3]{abc}

but then, I'm not sure I'm allowed to use this.

Anyway, basically, we first note that the number of factors depend only on the number of times the prime factors are occuring in N, but not on what the prime factors actually are. To minimise the number N, we thus need to make sure that we use the smallest prime factors available, and make sure that we have the highest exponents in our smallest prime factors, i.e. we want

p_1 = 2, p_2 = 3, p_3 = 5, p_4 = 7, \ldots

and

m_1 \geq m_2 \geq m_3 \geq m_4 \geq \cdots

Now, we note that

2^3\cdot3\cdot5

satisfies the requirements (it has 14 proper factors). This shows that our minimum can't have prime factors higher than or equal to 7, because we have already established that the minimum satisfies

m_1 \geq m_2 \geq m_3 \geq m_4 \geq \cdots

, which means that the smallest possible N with a prime factor of 7 is

2\cdot3\cdot5\cdot7

. But this is larger than

2^3\cdot3\cdot5

and hence is not the minimum. Thus we can assume that

N = 2^{m_1}\cdot3^{m_2}\cdot5^{m_3}

.

Now, because we want 12 or more proper factors, we have to have

(m_1 + 1)(m_2 + 1)(m_3 + 1) \geq 14

Together with the arithmetic-geometric means inequality, this means that

m_1 + 1 + m_2 + 1 + m_3 + 1 \geq 3\sqrt[3]{14}

which means that

m_1 + m_2 + m_3 \geq 5

(calculators were allowed, remember? :p:

)

Now we basically list the numbers with

m_1 + m_2 + m_3 \geq 5

in rising order, and note that none of the ones below

2^3\cdot3\cdot5

can have 12 proper factors or more:

32 = 2^5 has 4 proper factors
48 = 2^4*3 has 8 proper factors
64 = 2^6 has 5 proper factors
96 = 2^5*3 has 10 proper factors
120 = 2^3 * 3 * 5 has 14 proper factors

and thus 120 is our desired minimum.

So, there ought to be a better way of showing this last part. Ideas, anyone?

Reply 18