STEP Maths I, II, III 1994 Solutions

(Updated as far as #120.) SimonM - 29.04.2009

For explanation, scroll down:

Questions:
(Up to date as of post 89, 23:30 Saturday 2007-10-13 GMT)

STEP I:
1: Solution by eponymous
2: Solution by SimonM
3: Solution by khaixiang
4: Solution by zrancis
5: Solution by brooker
6: (Incomplete) Solution by zrancis
7: Solution by nota bene /DFranklin
8: Solution by ad absurdum and Speleo
9: Solution by ad absurdum
10: Solution by ad absurdum
11: Solution by zrancis
12: Solution by nota bene
13: Solution by nota bene
14: Solution by Glutamic Acid

STEP II:
1: Solution by ad absurdum
2: Solution by Rabite and DFranklin
3: Solution by Speleo
4: Solution by Rabite
5: Solution by Rabite
6: Solution by ad absurdum
7: Solution by brianeverit
8: Solution by brianeverit
9: Solution by *bobo* (1) (2)
10: Solution by *bobo*
11: Solution by *bobo* (1) (2) (3)
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit

STEP III:
1: Solution by khaixiang
2: Solution by khaixiang
3: Solution by khaixiang
4: Solution by speedy_s
5: Solution by nota bene and DFranklin
6: Solution by khaixiang and DFranklin
7: Solution by DFranklin
8: Solution by nota bene
9: Solution by *bobo* (1) (2)
10: Solution by SimonM
11: Solution by brianeverit
12: Solution by SimonM
13: Solution by Glutamic Acid
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 1994, 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

Scroll to see replies

Reply 1

nota bene

I thought it had gone some time since we finished all Pure on the other papers, and Stat+Mechanics proceed very slowly, so we might as well start a new thread so that everyone can contribute again. :smile:

I have not scanned through Siklos booklet yet, so if anyone wants to add which ones are solved there - feel free to do so!

I think I have STEP I Q 2 covered, and will attempt Q 7. I'll post it up later.

Reply 2

khaixiang

STEP III, Question 1

$\\ \displaystyle \int_{0}^{x} sech(t) dt\\=\int_{0}^{x}\frac{\cosh t}{\cosh^2 t} dt\\=\int_{0}^{x}\frac{\cosh t}{1+\sinh^2 t}dt$

Using the substitution $u=\sinh t$

$\displaystyle \\=\int_{0}^{\sinh x}\frac{1}{1+u^2}du\\=\arctan(\sinh x)$

To find the reduction formula:

$\displaystyle \\I_{n}=\int_{0}^{x}sech^n(t)dt\\=\int_{0}^{x}(sech^2(t))(sech^{n-2}(t))dt\\=\int_{0}^{x}(1-\tanh^2 t)(sech^{n-2}(t))dt\\=\int_{0}^{x}sech^{n-2}(t)dt-\int_{0}^{x}(\tanh t)(\tanh t \times sech^{n-2}(t))dt$
Using integration by parts,
$\displaystyle \\I_{n}=I_{n-2}+[\frac{tanh(t)\times sech^{n-2}(t)}{n-2}]_{0}^{x}-\int_{0}^{x}\frac{sech^{2}(t)\times sech^{n-2}(t)}{n-2}dt\\ \therefore I_{n}=(\frac{n-2}{n-1})I_{n-2}+(\frac{1}{n-1})tanh(x)\times sech^{n-2}(x)$

So,

$\displaystyle \\I_{5}=\frac{3}{8}I_{1}+\frac{3tanh(x)sech(x)}{8}+\frac{tanh(x)sech^{3}(x)}{4}\\ I_{5}=\frac{3}{8}\arctan(\sinh x) +\frac{3tanh(x)sech(x)}{8}+\frac{tanh(x)sech^{3}(x)}{4}$

and,

$\displaystye \\I_{6}=\frac{8}{15}I_{2}+\frac{4}{15}tanh(x)sech^{2}(x)+\frac{1}{5}tanh(x)sech^{4}(x)\\ I_{6}=\frac{8}{15}\tanh x+\frac{4}{15}tanh(x)sech^{2}(x)+\frac{1}{5}tanh(x)sech^{4}(x)$

Reply 3

khaixiang

STEP III, Question 2

2(i)
$\displaystyle \\x^4+10x^3+26x^2+10x+1=0\\x^2+10x+26+\frac{10}{x}+\frac{1}{x^2}=0\\ (x+\frac{1}{x})^2-2+10y+26=0\\y^2+10y+24=0\\(y+6)(y+4)=0\\x+\frac{1}{x}=-4\\x=-2\pm\sqrt{3}\\x+\frac{1}{x}=-6\\x=-3\pm\sqrt{10}$

2(ii)
$\displaystyle \\x^4+x^3-10x^2-4x+16=0\\x^2+x-10-\frac{4}{x}+\frac{16}{x^2}=0\\ \text{ set } y=x-\frac{4}{x}\\(x-\frac{4}{x})^2+8+(x-\frac{4}{x})-10=0\\y^2+y-2=0\\(y-1)(y+2)=0\\x-\frac{4}{x}=1\\x=\frac{1\pm\sqrt{17}}{2}\\x-\frac{4}{x}=-2\\x=-1\pm\sqrt{5}$

Reply 4

Speleo

STEP II Question 3

f(0) = 1
f(x-y) = f(x)f(y) - f(a-x)f(a+y)

Let x = a, y = 0
f(a) = f(a)f(0) - f(0)f(a) = 0

i) f(x-y) = f(x)f(y) - f(a-x)f(a+y)
So f(a-x) = f(a)f(x) - f(0)f(a+x)
f(a-x) = -f(a+x)
By symmetry, f(a+y) = -f(a-y)

f(y-x) = f(x)f(y) - f(a-y)f(a+x)
= f(x)f(y) - [-f(a+y)][-f(a-x)]
f(y-x) = f(x-y)
f(t) = f(-t)

ii) f(2a) = f(a - (-a)) f(a)f(-a) - [f(0)]^2
= - (1)^2
= -1

iii) f(2a - t) = f(2a)f(t) - f(-a)f(a+t)
= -f(t) - f(a)f(a+t)
= -f(t)

iv) f(4a) = f(2a - (-2a)) = f(2a)f(-2a) - f(-a)f(a)
= 1 - f(a)f(a)
= 1

f(3a) = f(2a - (-a)) = f(2a)f(-a) - f(-a)f(0)
= 0

f(4a + t) = f(4a - (-t)) = f(4a)f(-t) - f(-3a)f(a-t)
= f(t) - f(3a)
= f(t)

---

f(y-x) = f(x)f(y) - f(a-x)f(a+y)

f = cos(t)

cos(y-x) = cosxcosy + sinxsiny
But sinx = cos(pi/2 - x) and siny = -cos(pi/2 + x)
so cos(y-x) = cosxcosy - sin(pi/2 - x)sin(pi/2 + x)

f = cos(Bt + C)
When t = -2, f = 0
cos(C - 2B) = 0
C - 2B = pi/2 (say)
When t = 0, f = 1
cos(C) = 1
C = 0
B = -pi/4
f = cos(-pi/4.t) = cos(pi/4.t)

f(x-y) = cos(xpi/4 - ypi/4) = cos(xpi/4)cos(ypi/4) + sin(xpi/4)sin(ypi/4)
But sin(ypi/4) = cos(pi/2 - ypi/4) = cos[(pi/4)(2 - y)] = cos[(pi/4)(-2 + y)] = f(a+y)
sin(xpi/4) = -cos(pi/2 + xpi/4) = - cos[(pi/4)(2 + x)] = - cos[(pi/4)(-2 - x)] = -f(a-x)
f(x-y) = f(x)f(y) - f(a-x)f(a+y) as required.

Reply 5

khaixiang

Speleo

f = cos(Bt + C)
When t = -2, f = 0
cos(C - 2B) = 0
C - 2B = pi/2 (say)
When t = 0, f = 1
cos(B + C) = 1
B + C = 0 (say)
B = -C
3C = pi/2
C = pi/6
B = -pi/6

f = cos(pi/6 - t.pi/6)

f(x)f(y) - f(a-x)f(a+y)
= cos(pi/6 - x.pi/6)cos(pi/6 - y.pi/6) - cos(pi/6 + x.pi/6 + pi/3)cos(pi/6 - y.pi/6 + pi/3)
= cos(pi/6 - x.pi/6)cos(pi/6 - y.pi/6) - cos(pi/2 + x.pi/6)cos(pi/2 - y.pi/6)
= cos(pi/6 - x.pi/6)cos(pi/6 - y.pi/6) - sin(-x.pi/6)sin(y.pi/6)
= cos(pi/6 - x.pi/6)cos(pi/6 - y.pi/6) + sin(x.pi/6)sin(y.pi/6)

f(x-y) = cos(pi/6 - x.pi/6 + y.pi/6)
= cos(pi/6 - x.pi/6)cos(y.pi/6) - sin(pi/6 - x.pi/6)sin(y.pi/6)
= cos(pi/6 - x.pi/6)cos(y.pi/6) + sin(x.pi/6 - pi/6)sin(y.pi/6)

^ Last bit is incomplete for now. I will return to it, but anyone else is welcome to finish it.

I dont think the function is f(t)=cos(pi/6-t.pi/6), note that f(0) is not 1... f(t-8), so I think it suffices to find a cosine function with a period of 8 :smile:

Reply 6

Speleo

I fail at substituting t = 0 into Bt + C >_<

edit: Think it's complete now.

Reply 7

DFranklin

STEP III, Q14. I thought this was a really interesting problem, but I also found it extremely difficult. I went down at least two false tracks, and even now, there's a certain amount of "I know what the answer has to be, so let's work out why I need to multiply my result by 6 to get the right answer".

The first observation I made was that if you have a circle center O, and two points on the circumference P, Q with $\angle POQ = \alpha$ , then $\angle OPQ = \angle OQP = (\pi - \alpha / 2)$ .

Now let P,Q,R be the vertices of the triangle. We consider the case $\frac{1}{2} \le k \le 1$ .

Let $\angle POQ = \alpha, \angle QOR = \beta$ . Then $\alpha, \beta$ are uniformly distributed on [0,2pi].

The 2nd observation I made (well, actually about 17th, but the 2nd useful observation!) is that in this case, only one angle can exceed $k\pi$ , (since the sum of all 3 angles is pi). Suppose that angle to be Q, and suppose also that going clockwise the order of the points is PQR. Then we have a scenario as shown in the "pt 1" diagram attached. Note that O is not inside the triangle PQR; this can be seen looking at the boundary case where PR is a diameter, since then $\angle PQR = \pi/2$ , which is the minimal value for $k\pi$ .

So we want $\angle PQR$ in terms of $\alpha, \beta$ , and from observation 1, we find $\angle PQR = ((\pi - \alpha)+(\pi-\beta))/2 = \pi - (\alpha+\beta)/2$ .

Write $x=\alpha / 2\pi, y = \beta /2\pi$ (so x,y and uniformly distributed on [0,1]). Then $\angle PQR > k\pi \implies 2\pi - (\alpha+\beta) > 2k\pi \implies 1-(x+y) > k \implies x+y < 1-k$ .

So fix x, we must have $0\le y < 1-k-x$ . This is possible only for $x < 1-k$ , when it occurs with probability $1-k-x$ .

So $\mathbb{P}(x+y < 1-k) = \int_0^{1-k}(1-k-x)dx = \frac{1}{2}(1-k)^2$ .

Now this is the probability for the points in the order PQR with Q the largest angle. To get the full probability we must multiply by 2 for the possible order RQP, and by 3 for the possibilities that P or R are the largest angle. Thus our final answer is $3(1-k)^2$ as required.

For $\frac{1}{3}\le k \le \frac{1}{2}$ we have the scenario shown in the second diagram. For convenience we introduce $\gamma = \angle POR$ , but note that $\alpha + \beta + \gamma = 2\pi$ , so although $\alpha, \beta$ are still uniformly distributed, $\gamma$ is in fact fixed.

This time all 3 angles are candidates for exceeding $k\pi$ . We find:

$\angle PQR = (\pi - \alpha)/2 + (\pi - \beta) / 2 = \pi - (\alpha+\beta) / 2.$

$\angle QRP = (\pi - \beta) / 2 + (\pi - \gamma) / 2= (\pi-\beta-\pi+\alpha+\beta)/2 = \alpha / 2.$

$\angle RPQ = (\pi - \alpha) / 2 + (\pi - \gamma) / 2= (\pi-\alpha-\pi+\alpha+\beta)/2 = \beta / 2.$

Because any/all of the angles may exceed $k\pi$ , we will find the probability that all the angles are $\le k\pi$ . Now

$\angle PQR \le k\pi \Leftrightarrow \pi - (\alpha+\beta)/2 \le k\pi \Leftrightarrow \alpha+\beta \ge 2(1-k)\pi$
$\angle QRP \le k\pi \Leftrightarrow \alpha \le 2k\pi$
$\angle RPQ \le k\pi \Leftrightarrow \beta \le 2k\pi$

Again write $x=\alpha / 2\pi, y = \beta /2\pi$ . Then the conditions above become

$x \le k, y \le k, x+y \ge 1-k$ . Again, we fix x and find that the range of possible values for y is $1-k-x \le y \le k$ .

So we must have $1-k-x \le k$ i.e. $1-2k\le x$ . For these values of x we have, the chance of y falling in the appropriate range is just $k - (1-k - x) = 2k + x - 1$ .

So we find the global probability
$\mathbb{P}(1-k-x \le y \le k) = \int_{1-2k}^k (2k-1)+x dx = [ (2k-1)x + x^2/2]_{1-2k}^k \\[br]= (3k-1)(2k-1)+\frac{k^2-(1-2k)^2}{2}$
$= \frac{1}{2}[2(6k^2-5k+1)-3k^2+4k-1] \\= \frac{1}{2}[9k^2-6k+1] \\= \frac{1}{2}(3k-1)^2$ .

Again, this is the probability for the configuration where the order of the points is PQR. We need to double to allow for the configuration RQP. (In this case we don't multiply by 3 because our calculations have always allowed for any of PQR to exceed $k\pi$ ). So the probability that no angle exceeds $k\pi$ is $(3k-1)^2$ .

So the probability than an angle exceeds $k\pi$ is $1- (3k-1)^2$

Comment: Outside of exam conditions, this is a great question, but I really wouldn't want to try it in an actual exam. I've definitely found this hardest out of all the questions I've done, partly because I went down completely the wrong track. In particular, I spent a long time working on this before realising that in the first case, O wasn't actually inside the triangle PQR. And the multiplying the final answers by 6 and 2 feels a bit of a fudge. I think it's justifiable, but if I'm honest, if I'd needed to multiply by 3 instead I'm sure I'd have found a justification for that as well.

I'm pretty sure the actual answer to the last part is correct; it agrees with the first part at k=1/2 and I have also verified by computer simulation (something else not terribly practical under exam conditions).

(edited 6 years ago)

S3_1994_Q14_pt1.png32.7KB

S3_1994_Q14_pt2.png32.8KB

Reply 8

zrancis

Step I - Question 2

Okay so i'm just about getting the hang of STEP I, please pick apart this solution if there are any problems with it, I do think there may be some simplification needed to neaten a few of the solutions up but I personally can't see how to sort it out. Here goes:

Part i) $y = x^a$
$\frac{dy}{dx} = ax^{(a-1)}$

Part ii)
$\ y = a^x}$
$lny = ln(a^x)$
$lny = xlna$
$\frac{dy}{dx} \frac{1}{y} = lna$
$\frac{dy}{dx} = ylna$
$\frac{dy}{dx} = a^xlna$

Part iii)
$y = x^x$
$lny = ln(x^x)$
$lny = xlnx$
$\frac{dy}{dx} \frac{1}{y} = lnx + 1$
$\frac{dy}{dx} = x^x(lnx + 1)$

Part iv)
$y = x^x^x$
$lny = ln(x^x^x)$
$lny = x^xlnx$
$\frac{dy}{dx} \frac{1}{y} = x^x(lnx + 1)lnx + \frac{x^x}{x}$ (From part iii)
$\frac{dy}{dx} \frac{1}{y} = x^x((lnx)^2 + lnx) + x^{(x-1)}$
$\frac{dy}{dx} = yx^x((lnx)^2 + lnx + \frac{1}{x})$
$\frac{dy}{dx} = x^{(x^x + x)}((lnx)^2 + lnx + \frac{1}{x})$

Part v)
$y = (x^x)^x$
$y = x^x^2$
$lny = ln(x^x^2)$
$lny = x^2lnx$
$\frac{dy}{dx} \frac{1}{y} = 2xlnx + x$
$\frac{dy}{dx} = x^x^2(2xlnx + x)$
$\frac{dy}{dx} = x^{(x^2+1)}(2lnx + 1)$

Not too bad I thought, maybe a fatal error or too thrown in there though :biggrin:

Moderator edit: converted [tex] tags to LaTeX ones.

Reply 9

nota bene

zrancis

Not too bad I thought, maybe a fatal error or too thrown in there though :biggrin:

I think you're right - just what I was going to write up after finishing my essay, I think we've reached the same conclusions=) Although I wrote the simplification of the last one a bit differently... $\frac{dy}{dx}=x(x^x)^x(2ln(x)+1)$

Reply 10

zrancis

nota bene

Good stuff!
I'm going to have a look at some of the other questions on STEP I also, they seem quite doable so i'll post back when I get through some more.

Reply 11

nota bene

I'll type this up and someone else can fill in the first part of the question, which I am not seeing the general formula for...
STEP I Q 7

For the first part I note the pattern, which means the next line would be
17+18+19+20+21+22+23+24+25=64+125
i.e. we add 2 more numbers to the LHS per new line. and The sum of line n is $(n-1)^3 + n^3$ .

Part two of the question is very standard, a simple proof by induction:
$1^3+2^3+3^3+\dots+n^3=\frac{n^2(n+1)^2}{4}$
for n=1
$1^3=\frac{1^2(1+1)^2}{4}$ which clearly is true.
Assume true for n=k
$1^3+2^3+3^3+\dots+k^3=\frac{k^2(k+1)^2}{4}$
Adding the next term (k+1)^3 and starting working on the RHS gives:
$\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{k^2(k+1)^2}{4}+\frac{4(k+1)^3}{4}=\frac{(k+1)^2(k^2+4(k+1))}{4}= \frac{(k+1)^2(k+2)^2}{4}$
This is exactly what substituting k+1 into the original assumption gives, hence by mathematical induction $\mathbb{P}(k)\mathbb{\rightarrow P}(k+1) \mathrm{Q.E.D.}$

Reply 12

DFranklin

nota bene

I'll type this up and someone else can fill in the first part of the question, which I am not seeing the general formula for...
STEP I Q 7

For the first part I note the pattern, which means the next line would be
17+18+19+20+21+22+23+24+25=64+125The pattern is $(n^2+1)+(n^2+2)+...+(n+1)^2 = n^3+(n+1)^3$ .

Proof: Evaluate the LHS:
$\displaystyle\sum_{k=1}^{(n+1)^2} k - \sum_{k=1}^{n^2} k = \frac{1}{2}[(n+1)^2((n+1)^2+1) - n^2(n^2+1)^2] \\=[br]((n+1)^4+(n+1)^2 - n^4 - n^2) / 2 \\=[br](n^4+4n^3+6n^2+4n+1 + n^2 +2n+1 - n^4-n^2) / 2\\=[br](4n^3+6n^2+6n+2) / 2 = 2n^3 + 3n^2+3n+1 = n^3 + (n+1)^3$

Reply 13

khaixiang

Question 3, STEP III, 1994

i)A plane can intersect a sphere at a point when the plane is tangential to the sphere.
ii)A plane can intersect a sphere where the intersection forms a circle.

The intersection of plane P1 and plane P2 produces line L with equation

$\left[ \begin {array}{c} x\\\noalign{\medskip}y\\\noalign{\medskip}z\end {array} \right] =\left[ \begin {array}{c} 1\\\noalign{\medskip}2\\\noalign{\medskip}t\end {array} \right]$

where t is a parameter.

Given $\\S_{1}: x^2+y^2+z^2=7\\S_{2}: x^2+y^2+z^2-6y-4z+10=0\\S_{2}: x^2+(y-3)^2+(z-2)^2=3$

To find intersection of line L with both spheres, we substitute x=1 and y=2 respectively into both S1 and S2 and solve for z. Hence,

L intersect sphere S1 at
$(1, 2, \sqrt{2}) \text{ and } (1, 2, -\sqrt{2})$

L intersect sphere S2 at
$(1, 2, 1) \text{ and } (1, 2, 3)$

Since line L also lies on plane P1, the circle C1 then contains $(1, 2, \sqrt{2}) \text{ and } (1, 2, -\sqrt{2})$ and the line joining these 2 points is a chord of C1.

Also, line L lies on plane P2, the circle C2 then contains $(1, 2, 1) \text{ and } (1, 2, 3)$ and the line joining these 2 points is a chord of C2.

Since there exists a chord of circle C1 which intersects (from $(1, 2, 1) \text{ to } (1, 2, \sqrt{2})$ ) with a chord of C2, the 2 circles are linked.

Comments by DFranklin:

If I was going to nitpick, my main criticism would be how little working you have provided. If you get the right answer, it probably wouldn't affect your marks, but if you go wrong, it's going to be hard for an examiner to work out where you made the mistake; he might easily think you used a completely incorrect method as opposed to just making a simple algebraic error. (I appreciate that it is difficult writing up questions on the forum, so if it's just that you couldn't be bothered typing everything out, that's fine). In an exam, you want to give an examiner every reason to believe you know what you're doing. "Hiding" your working out to make it look like you can do 20 lines of algebra in your head isn't going to impress them, and it means you're unlike to salvage many marks from any mistakes. (The whole "answer full questions" thing tends to be taken too seriously: if you actually look at the exam results, it's often how many marks you scrape on the questions you didn't complete that actually determines which grade you get).

My other comments are really about the way you worded things.

I found it a little hard to interpret your descriptions of the intersections: e.g.

[indent]"A plane can intersect a sphere at a point when the plane is tangential to the sphere."[/indent]

Does this mean: "A plane can intersect a sphere at a point. In this case the plane is tangential to the sphere."? Or does it mean "A plane can intersect a sphere at a point which is tangential to the sphere"?

To be fair, I think these descriptions can be difficult to make without coming out with tautologies like "a sphere intersects with a plane at the points where they intersect", so I certainly don't think you did a bad job. But it might be something you want to think about.

For what it's worth, the best I can come up with for this question is:

[INDENT]"The typical intersection of a plane with a sphere is a circle. However, in the degenerate case where the plane is tangent to the sphere, the intersection is a single point."[/INDENT]

The other thing is that at the end you have two overlapping chords, rather than intersecting chords. Not that it makes a huge difference, and I don't think you'd lose any marks for it, but it did confuse me for a moment.

Anyhow, this is being really nitpicky, which in a sense is unfair: anyone can always find something to complain about in someone else's proof. I think you've done a very good job.

Reply 14

DFranklin

STEP III, Q7 (again, sketch answer for completeness; I understand groups are no longer on syllabus).

Elements of $S_3$ : Use standard cycle notation: (123) is the permutation that sends 1 to 2, 2 to 3 and 3 to 1; (23) is the permutation that sends 2 to 3 and 3 to 2. Then we have
(1) (the identity permutation, order = 1).
Three elements of order 2: (12), (23), (13)
Two elements of order 3: (123), (132)

Elements of $Z_6$ : These are just the numbers 0,1,2,3,4 and 5.
0 is the identity and has order 1.
1 and 5 have order 6
2 and 4 have order 3
3 has order 2.

S3 is not isomorphic to Z6 as S3 has no element of order 6, unlike Z6.

Z6 is isomorphic to C6 by the explicit isomorphism $\theta(n) = e^{in\pi/3}$ . Proof: $\theta(n)\theta(m) = e^{in\pi/3}e^{im\pi/3}=e^{i(m+n)\pi/3} = e^{iq\pi/3}$ where q=(m+n) mod 6, since $e^{i6\pi/3} = 1$ . So $\theta(n)\theta(m)=\theta(nm)$ and so \theta is a group homomorphism. Then $\theta(m) = \theta(n) \implies e^{im\pi/3} = e^{in\pi/3} \implies m = n$ . Thus $\theta$ is injective, and so since Z6, C6 both have 6 elements, $\theta$ is a bijection and therefore an isomorphism.

There is no subgroup of $\mathbb{C}$ (under either multiplication or addition) that is isomorphic to S3, since addition and multiplication in $\mathbb{C}$ are commutative, while S3 is not. E.g. (12)(23) = (1 3 2), (23)(12) = (1 2 3).

Comment: the group questions are ridiculously easy compared with the rest of the paper. Pity they are obsolete...

Reply 15

Rabite

I did Q4 in STEP II, but probably not the way they wanted it to be done...
I'll post it in the morning for everyone's verdict.
Okay, I uploaded it - but I did it by integration, whereas I think they wanted a more geometric approach...

And Q2, but I have no idea if it's actually right.

Q4.pdf54.9KB

Reply 16

nota bene

I think I have STEP I Q4 covered (part (i) definitely is...will have to look at (ii) again)
Will edit this post with soluition when I wake up...

Reply 17

khaixiang

Rabite

Question 2 is perfectly fine (and very presentable :biggrin:

) Except for the first part (where I used calculus), our workings are more or less the same. I think it's fine as long as you draw them some diagrams showing where the region is, as defined by the question, then proceed to use integration to find the area (which is what I'd done).

The hard part of question 2 is the various proving, really easy to lose marks here... Prove too vigorous will result in using up a lot of time, and giving too short a proof might end up losing a few marks, it's hard to know how much and what they expected of you. I am curious as to what are the "some other ways of doing this question" mentioned in the question. Dislike this question in all(long and lots of traps for algebraic mistakes), but the result at the end is rather nice. Avoid this kind of question at all cause during STEP exam I'd say :p:

Reply 18

Rabite

STEP II:
Here's a half-baked Q2 (the stupid one with cos^{2n+1} and what not) - I would probably bet money that it's wrong though.
[edit] Oops, first PDF was crapped up since it went over a page. I should really learn how to use Latex tags so I dont' have to use this crappy program.

[edit again] Also, I think Q5 is straightforward (attached), but I may have overlooked something...

Q2thing.pdf50.9KB

Qsomething.pdf49.4KB

Reply 19

khaixiang

Rabite

For question 2, the final result is correct, well done :biggrin:

.. But you missed out the part where they asked you to show that P_n is a polynomial of degree n. Other than that, there's an incorrect assertion: "The first part of the question shows that (x^2-1) will be a factor of all derivatives", the very first result proven didnt show this, particularly it can be shown that the nth derivative and above of $(x^2-1)^n$ is not zero when $x=\pm1$ . I have to prove all over again that $f^{(r)}(\pm1)=0 \text{ for } n-1\geq r \geq 1 \text{ where } f(x)=(x^2-1)^n$ so that this result can be used to assert that the square bits are in fact zero.

I think there're other more efficient methods to this question, as the question mentioned "however some other ways of doing this question do not use this [reduction] formula". Hopefully someone can spot it.

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