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DFranklin

I did A-levels a long time ago, just before STEP, and did quite a few of the old Cambridge Colleges Exam questions. My experience was that I learned a lot of maths outside the A-level syllabus from doing the CCE; my guess is it would have been the same with STEP circa 1990.

As well of the syllabus, some of these questions really do require an awful lot of manipulation - far more than the typical exam questions you'd get in the actual Tripos. (And I did the tripos in 1988-1990, so slightly before these STEP papers).

As well of the syllabus, some of these questions really do require an awful lot of manipulation - far more than the typical exam questions you'd get in the actual Tripos. (And I did the tripos in 1988-1990, so slightly before these STEP papers).

I wonder how long STEP mathematics has been going for? Infact i know there was STEP science too. I wouldnt mind looking at some CCE papers but something tells me i wont be able to find them anymore.

insparato

Not to nitpick, but I disagree a little with the solution to Q8 on STEP II. Or perhaps I've just forgotten something, anyway.

For the second part, you come out with

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.

You get the same result either way, but I think this way is right, since the one in post...#19 or whatever it was uses the equal roots of aux quadratic thing.

For the second part, you come out with

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.

You get the same result either way, but I think this way is right, since the one in post...#19 or whatever it was uses the equal roots of aux quadratic thing.

Got more of a chance of getting four questions than I do, that's for sure...

Anyway, here's a half-baked solution to STEP III Q3 while I work on my missing '2's.

[edit] Oh, fine. Someone's already solved it apparently.

Anyway, here's a half-baked solution to STEP III Q3 while I work on my missing '2's.

[edit] Oh, fine. Someone's already solved it apparently.

Rabite

Got more of a chance of getting four questions than I do, that's for sure...

Anyway, here's a half-baked solution to STEP III Q3 while I work on my missing '2's.

[edit] Oh, fine. Someone's already solved it apparently.

Anyway, here's a half-baked solution to STEP III Q3 while I work on my missing '2's.

[edit] Oh, fine. Someone's already solved it apparently.

DFranklin

Solved it, yes, written it up, no - nice job. I didn't see: did you manage the last bit? I thought it surprisingly tricky; needing to find the value of tan(3pi/8) was a tad annoying...

Im sure there was a question about finding tan 3pi/8 using an argand diagram and some real and imaginary points. its root 2 + 1 something like that right?

Close, close. sqrt 2 - 1 apparently.

Just muck around with tan pi/4 = 1 and the compound angle formulae.

Here's the second bit.

Omg I did a question. Hell freezes over.

Just muck around with tan pi/4 = 1 and the compound angle formulae.

Here's the second bit.

Omg I did a question. Hell freezes over.

Here's III, Q7:

$\alpha^2 = (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+za) = x^2+y^2+z^2 +2\beta$

$\alpha^3 = (x+y+z)^3 = x^3+y^3+z^3+xy^2+xz^2+yx^2+yz^2+zx^2+zy^2+2\alpha \beta$

So $x^3+y^3+z^3=\alpha^3-3\alpha \beta + 3 \gamma$.

Then $x+y+z = 1 \implies \alpha = 1$, $x^2+y^2+z^2=3 \implies 1 = 3 + 2\beta \implies \beta = -1$, $x^3+y^3+z^3=4 \implies 4=1+3+3\gamma \implies \gamma =0$.

Now x,y,z are the roots of $(t-x)(t-y)(t-z) = t^3 - \alpha t^2 + \beta t - \gamma = t^3-t^2-t = t(t^2-t-1)$.

Now $t^2-t-1$ has roots $(1\pm\sqrt{5})/2$. So $\{x,y,z\} = \{0, (1\pm\sqrt{5})/2\}$.

For the 2nd part, put $\alpha = a+b+c,\quad \beta = ab+bc+ca, \quad\gamma = abc$.

Write $\Delta = s(s-a)(s-b)(s-c).$

If a,b,c are the roots of the equation, then we can read off $\alpha = 16, \beta = 81, \gamma = 128$.

Then $16\Delta = 16(4\cdot 16\cdot 81-16^3-8\cdot 128) = 16(64\cdot 81-64\cdot 64-64\cdot 16)=16.64$.

Then $\Delta = 64$ and so $area = \sqrt{\Delta} = 8$.

$\alpha^2 = (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+za) = x^2+y^2+z^2 +2\beta$

$\alpha^3 = (x+y+z)^3 = x^3+y^3+z^3+xy^2+xz^2+yx^2+yz^2+zx^2+zy^2+2\alpha \beta$

Unparseable latex formula:

.\text{Now }xy^2+xz^2+yx^2+yz^2+zx^2+zy^2 \\[br]= y(xy+yz)+z(zx+yz)+x(xy+zx) \\[br]= y(xy+yz+zx)+z(xy+yz+zx)+x(xy+yz+zx) - 3xyz \\[br]= \alpha \beta - 3\gamma

So $x^3+y^3+z^3=\alpha^3-3\alpha \beta + 3 \gamma$.

Then $x+y+z = 1 \implies \alpha = 1$, $x^2+y^2+z^2=3 \implies 1 = 3 + 2\beta \implies \beta = -1$, $x^3+y^3+z^3=4 \implies 4=1+3+3\gamma \implies \gamma =0$.

Now x,y,z are the roots of $(t-x)(t-y)(t-z) = t^3 - \alpha t^2 + \beta t - \gamma = t^3-t^2-t = t(t^2-t-1)$.

Now $t^2-t-1$ has roots $(1\pm\sqrt{5})/2$. So $\{x,y,z\} = \{0, (1\pm\sqrt{5})/2\}$.

For the 2nd part, put $\alpha = a+b+c,\quad \beta = ab+bc+ca, \quad\gamma = abc$.

Write $\Delta = s(s-a)(s-b)(s-c).$

Unparseable latex formula:

\text{Then }16\Delta = \alpha(\alpha -2a)(\alpha-2b)(\alpha-2c)\\[br]=\alpha(\alpha^3 - 2(a+b+c)\alpha^2+4(ab+bc+ca)\alpha-8abc)[br]=\alpha(4\beta \alpha-\alpha^3-8\gamma).

If a,b,c are the roots of the equation, then we can read off $\alpha = 16, \beta = 81, \gamma = 128$.

Then $16\Delta = 16(4\cdot 16\cdot 81-16^3-8\cdot 128) = 16(64\cdot 81-64\cdot 64-64\cdot 16)=16.64$.

Then $\Delta = 64$ and so $area = \sqrt{\Delta} = 8$.

DFranklin

Here's III, Q7:

$\alpha^2 = (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+za) = x^2+y^2+z^2 +2\beta$

$\alpha^3 = (x+y+z)^3 = x^3+y^3+z^3+xy^2+xz^2+yx^2+yz^2+zx^2+zy^2+2\alpha \beta$

So $x^3+y^3+z^3=\alpha^3-3\alpha \beta + 3 \gamma$.

Then $x+y+z = 1 \implies \alpha = 1$, $x^2+y^2+z^2=3 \implies 1 = 3 + 2\beta \implies \beta = -1$, $x^3+y^3+z^3=4 \implies 4=1+3+3\gamma \implies \gamma =0$.

Now x,y,z are the roots of $(t-x)(t-y)(t-z) = t^3 - \alpha t^2 + \beta t - \gamma = t^3-t^2-t = t(t^2-t-1)$.

Now $t^2-t-1$ has roots $(1\pm\sqrt{5})/2$. So $\{x,y,z\} = \{0, (1\pm\sqrt{5})/2\}$.

For the 2nd part, put $\alpha = a+b+c,\quad \beta = ab+bc+ca, \quad\gamma = abc$.

Write $\Delta = s(s-a)(s-b)(s-c).$

If a,b,c are the roots of the equation, then we can read off $\alpha = 16, \beta = 81, \gamma = 128$.

Then $16\Delta = 16(4\cdot 16\cdot 81-16^3-8\cdot 128) = 16(64\cdot 81-64\cdot 64-64\cdot 16)=16.64$.

Then $\Delta = 64$ and so $area = \sqrt{\Delta} = 8$.

$\alpha^2 = (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+za) = x^2+y^2+z^2 +2\beta$

$\alpha^3 = (x+y+z)^3 = x^3+y^3+z^3+xy^2+xz^2+yx^2+yz^2+zx^2+zy^2+2\alpha \beta$

Unparseable latex formula:

.\text{Now }xy^2+xz^2+yx^2+yz^2+zx^2+zy^2 \\[br]= y(xy+yz)+z(zx+yz)+x(xy+zx) \\[br]= y(xy+yz+zx)+z(xy+yz+zx)+x(xy+yz+zx) - 3xyz \\[br]= \alpha \beta - 3\gamma

So $x^3+y^3+z^3=\alpha^3-3\alpha \beta + 3 \gamma$.

Then $x+y+z = 1 \implies \alpha = 1$, $x^2+y^2+z^2=3 \implies 1 = 3 + 2\beta \implies \beta = -1$, $x^3+y^3+z^3=4 \implies 4=1+3+3\gamma \implies \gamma =0$.

Now x,y,z are the roots of $(t-x)(t-y)(t-z) = t^3 - \alpha t^2 + \beta t - \gamma = t^3-t^2-t = t(t^2-t-1)$.

Now $t^2-t-1$ has roots $(1\pm\sqrt{5})/2$. So $\{x,y,z\} = \{0, (1\pm\sqrt{5})/2\}$.

For the 2nd part, put $\alpha = a+b+c,\quad \beta = ab+bc+ca, \quad\gamma = abc$.

Write $\Delta = s(s-a)(s-b)(s-c).$

Unparseable latex formula:

\text{Then }16\Delta = \alpha(\alpha -2a)(\alpha-2b)(\alpha-2c)\\[br]=\alpha(\alpha^3 - 2(a+b+c)\alpha^2+4(ab+bc+ca)\alpha-8abc)[br]=\alpha(4\beta \alpha-\alpha^3-8\gamma).

If a,b,c are the roots of the equation, then we can read off $\alpha = 16, \beta = 81, \gamma = 128$.

Then $16\Delta = 16(4\cdot 16\cdot 81-16^3-8\cdot 128) = 16(64\cdot 81-64\cdot 64-64\cdot 16)=16.64$.

Then $\Delta = 64$ and so $area = \sqrt{\Delta} = 8$.

What is t when you're trying to find x,y,z? Id found alpha,beta and gamma but did not no how to proceed to find x,y,z.

insparato

What is t when you're trying to find x,y,z? Id found alpha,beta and gamma but did not no how to proceed to find x,y,z.

All 't' is is the "variable" in a polynomial (I can't use 'x' for the variable because that letter is already taken). We specifically choose the polynomial to be (t-x)(t-y)(t-z) because that polynomial has roots x,y,z (not that we have any idea yet what x,y,z are, but if we can solve the polynomial we will know!).

But if you multiply out the cubic polynomial in t:

(t-x)(t-y)(t-z), you get $t^3-\alpha t^2 + \beta t - \gamma$. So since x,y,z are the roots of the LHS (by construction), they are also the roots of the RHS. But we know alpha, beta, gamma, so we can explictly construct the polynomial on the RHS. Solving that polynomial gives us x, y, z.

Hmm...

For Q5, the first bit is easy.

But I have no clue for the last bit - any ideas? Also I have a quibble with the middle bit.

Here's what I did anyway:

First bit is standard - conjure up Pascal's triangle to the row for ^7 expansions. Write the powers of 'i' in a margin somewhere and don't tell anyone about them. Expand it normally, and equate the real bit to cos7$\theta$ and the imaginary bits to sin7$\theta$.

Divide the two, then divide top and bottom by cos^7.

Next bit: put denominator equal to zero, ignoring the first 't'. Note that 't' occurs in even powers. In a cubic with 1 as the x³ coefficient, the constant bit is -(product of roots). So I think the answer is sqrt(7).

But how do you justify that the roots of this cubic are as given, since tan7$\theta$ has more than three roots obviously.

For the last bit...do they want you to consider tan14$\theta$?

Using Tan2T = 2TanT/(1-Tan²T) only gives the same equation on top.

*sigh* I really don't see myself getting any marks in these papers.

[edit] Does the existence of this thread mean that there are, easily available, solutions to all other papers?

For Q5, the first bit is easy.

But I have no clue for the last bit - any ideas? Also I have a quibble with the middle bit.

Here's what I did anyway:

First bit is standard - conjure up Pascal's triangle to the row for ^7 expansions. Write the powers of 'i' in a margin somewhere and don't tell anyone about them. Expand it normally, and equate the real bit to cos7$\theta$ and the imaginary bits to sin7$\theta$.

Divide the two, then divide top and bottom by cos^7.

Next bit: put denominator equal to zero, ignoring the first 't'. Note that 't' occurs in even powers. In a cubic with 1 as the x³ coefficient, the constant bit is -(product of roots). So I think the answer is sqrt(7).

But how do you justify that the roots of this cubic are as given, since tan7$\theta$ has more than three roots obviously.

For the last bit...do they want you to consider tan14$\theta$?

Using Tan2T = 2TanT/(1-Tan²T) only gives the same equation on top.

*sigh* I really don't see myself getting any marks in these papers.

[edit] Does the existence of this thread mean that there are, easily available, solutions to all other papers?

Rabite

But how do you justify that the roots of this cubic are as given, since tan7$\theta$ has more than three roots obviously.

For the last bit...do they want you to consider tan14$\theta$?

*sigh* I really don't see myself getting any marks in these papers.

Rabite

[edit] Does the existence of this thread mean that there are, easily available, solutions to all other papers?

There are solutions on the net for 1999-onwards... 2001- are on miekelriggs

Edit: I just had a look at I Q13 I am not quite sure I understand what it asks, P(N=number of cups) must be 1/n! But then when the question asks for E(N) it says it equals e, what e? 2.7....? If so I have no clue how to get there... The variable must be discrete as we can't have any 1.5 cups or anything... So can someone clear up what this e is?

STEP III

Q5.

(Here, q denotes theta, c denotes cos(q) and s denotes sin(q).)

(cos(q) + isin(q))^7 = cos(7q) + isin(7q)

Expand the LHS using the binomial theorem, then equate real and imaginary parts to get

cos(7q) = c^7 - 21 c^5 s^2 + 35 c^3 s^4 - 7 c s^6

sin(7q) = 7 c^6 s - 35 c^4 s^3 + 21 c^2 s^5 - s^7

Thus

tan(7q) = (7 c^6 s - 35 c^4 s^3 + 21 c^2 s^5 - s^7)/(

c^7 - 21 c^5 s^2 + 35 c^3 s^4 - 7 c s^6)

= (7t - 35t^3 + 21t^5 - t^7)/(1 - 21t^2 + 35t^4 - 7t^6)

= t(t^6 - 21t^4 + 35t^2 - 7)/(7t^6 - 35t^4 + 21t^2 - 1), as required.

(i)

If q=pi/7, 2pi/7, 3pi/7, then tan(7q)=0 while tan(q) != 0. So,

0 = tan^6(q) - 21tan^4(q) + 35tan^2(q) - 7

That is, tan^2(pi/7), tan^2(2pi/7) and tan^2(3pi/7) are roots of the equation

x^3 - 21x^2 + 35x - 7 = 0

The product of the roots of this equation is 7. That is,

tan^2(pi/7) tan^2(2pi/7) tan^2(3pi/7) = 7

(ii)

This time we want to find a cubic whose roots are tan^2(pi/14), tan^2(3pi/14) and tan^2(5pi/14). Notice that if q=pi/14, 3pi/14, 5pi/14, then tan(7q) is undefined while tan(q) is. So,

7tan^6(q) - 35tan^4(q) + 21tan^2(q) - 1 = 0

will do the trick.

Thus, the 3 values of tan are roots of the equation

7x^3 - 35x^2 + 21x^2 - 1 = 0

The sum of the roots of this equation is 35/7 = 5. That is,

tan^2(pi/14) + tan^2(3pi/14) + tan^2(5pi/14) = 5

----------------------------

Q6.

(Group theory is no longer on the A-level syllabus, so this question is more or less irrelevant now. I'll do it anyway.)

(i)

Let A be the 2x2 matrix whose entries are all 1. Note that A^2 = 2A.

So S is simply { xA : x is a nonzero real }. Let's show that it's closed under matrix multiplication. Let xA and yA be in S. Then:

xA * yA = xy A^2 = 2xy A, which is in S.

To show that S forms a group under matrix multiplication, first we note that matrix multiplication is associative. We take the identity element to be (1/2)A. Indeed, if xA is in S then 1/2 A * xA = xA * 1/2 A = x/2 A^2 = x A. And, finally, if yA is in S, then because y is nonzero, we have that 1/(2y) A is its inverse. Thus all the axioms are satisfied.

(ii)

Suppose A in G is singular. Since G is a group, A has a group inverse B. That is, AB = E; whence,

det(E) = det(AB) = det(A)det(B) = 0

So det(E) = 0. Thus, if C is any matrix in G, then det(C) = det(CE) = det(C)det(E) = 0, and therefore C is singular.

For an example, look back at S in part (i). This time take the set of 3x3 matrices of that form - call it T. Here, if we let B be the 3x3 matrix whose entries are all 1, then B^2 = 3B. So showing that T is a group is analagous to show that S is. The identity in this case is B/3, which is a singular matrix.

----------------------------

Q8.

(I don't think this question is relevant either.)

If T(x) = x, then

ax - b = x(cx - d)

=> ax - b = cx^2 - dx

=> cx^2 - (a+d)x + b = 0

=> x = [(a+d) +/- sqrt((a+d)^2 - 4bc)]/2c

If T(x) = y, then

ax - b = y(cx - d)

=> ax - b = cxy - dy

=> ax - cxy = b - dy

=> x(a - cy) = b - dy

=> x = (b - dy)/(a - cy)

So T^{-1}(x) = (dx - b)/(cx - a). Thus, a'=d, b'=b, c'=c and d'=a.

If c!=0, then T(x) always gives us a real number unless cx-d=0 => x=d/c. So we take r=d/c. To find the image of S_r, let x be in S_r and let y = T(x). Then,

x = (dy - b)/(cy - a)

So y != a/c. Thus, T(S_r) = { x in R : x != a/c }.

On the other hand, if c=0, then T(x) = (-a/d)x + b/d. So if d=0 then we have a problem; otherwise, T is defined for all real numbers x. So T(S_r) is all of R minus the point (-a/d)r + b/d.

Moving on.

T_1(T_2(x)) = [(a_1 a_2 - b_1 c_2)x - (a_1 b_2 - b_1 d_2)]/[(c_1 a_2 - d_1 c_2)x - (c_1 b_2 - d_1 d_2)], so it's of the same form.

Thus, T^2(x) is the identity iff the following conditions are satisfied:

a^2 - bc = 1

ab - bd = 0

ac - dc = 0

bc - d^2 = -1

Finally, for the last bit, if we have for instance (a_3)^2 - b_3 c_3 = 0, then T_3 is not the identity. Let's re-express this:

(a_3)^2 = b_3 c_3

=> (a_1 a_2 - b_1 c_2)^2 = (a_1 b_2 - d_1 d_2) (c_1 a_2 - d_1 c_2)

Now let's take T_1(x) = 1-x. That is,

a_1 = b_1 = -1, c_1 = d_1 = 0

Then T_2 must satisfy the extra constraint

(c_2 - a_2)^2 = b_2

Let's take a_2 = 0. Then (c_2)^2 = b_2, and b_2 c_2 = -1. So b_2 = 1 and c_2 = -1; also, b_2 c_2 - (d_2)^2 = -1 implies d_2 = 0.

Thus T_2(x) = -1/x, and T_3(x) = 1 + 1/x.

Let's verify:

T_1(x)^2 = 1 - (1 - x) = x

T_2(x)^2 = -1/(-1/x) = x

T_3(x)^2 = 1 + 1/(1 + 1/x) = 1 + x/(1 + x) != x

----------------------------

Is it just me or is III/1 unbearably long?

Q5.

(Here, q denotes theta, c denotes cos(q) and s denotes sin(q).)

(cos(q) + isin(q))^7 = cos(7q) + isin(7q)

Expand the LHS using the binomial theorem, then equate real and imaginary parts to get

cos(7q) = c^7 - 21 c^5 s^2 + 35 c^3 s^4 - 7 c s^6

sin(7q) = 7 c^6 s - 35 c^4 s^3 + 21 c^2 s^5 - s^7

Thus

tan(7q) = (7 c^6 s - 35 c^4 s^3 + 21 c^2 s^5 - s^7)/(

c^7 - 21 c^5 s^2 + 35 c^3 s^4 - 7 c s^6)

= (7t - 35t^3 + 21t^5 - t^7)/(1 - 21t^2 + 35t^4 - 7t^6)

= t(t^6 - 21t^4 + 35t^2 - 7)/(7t^6 - 35t^4 + 21t^2 - 1), as required.

(i)

If q=pi/7, 2pi/7, 3pi/7, then tan(7q)=0 while tan(q) != 0. So,

0 = tan^6(q) - 21tan^4(q) + 35tan^2(q) - 7

That is, tan^2(pi/7), tan^2(2pi/7) and tan^2(3pi/7) are roots of the equation

x^3 - 21x^2 + 35x - 7 = 0

The product of the roots of this equation is 7. That is,

tan^2(pi/7) tan^2(2pi/7) tan^2(3pi/7) = 7

(ii)

This time we want to find a cubic whose roots are tan^2(pi/14), tan^2(3pi/14) and tan^2(5pi/14). Notice that if q=pi/14, 3pi/14, 5pi/14, then tan(7q) is undefined while tan(q) is. So,

7tan^6(q) - 35tan^4(q) + 21tan^2(q) - 1 = 0

will do the trick.

Thus, the 3 values of tan are roots of the equation

7x^3 - 35x^2 + 21x^2 - 1 = 0

The sum of the roots of this equation is 35/7 = 5. That is,

tan^2(pi/14) + tan^2(3pi/14) + tan^2(5pi/14) = 5

----------------------------

Q6.

(Group theory is no longer on the A-level syllabus, so this question is more or less irrelevant now. I'll do it anyway.)

(i)

Let A be the 2x2 matrix whose entries are all 1. Note that A^2 = 2A.

So S is simply { xA : x is a nonzero real }. Let's show that it's closed under matrix multiplication. Let xA and yA be in S. Then:

xA * yA = xy A^2 = 2xy A, which is in S.

To show that S forms a group under matrix multiplication, first we note that matrix multiplication is associative. We take the identity element to be (1/2)A. Indeed, if xA is in S then 1/2 A * xA = xA * 1/2 A = x/2 A^2 = x A. And, finally, if yA is in S, then because y is nonzero, we have that 1/(2y) A is its inverse. Thus all the axioms are satisfied.

(ii)

Suppose A in G is singular. Since G is a group, A has a group inverse B. That is, AB = E; whence,

det(E) = det(AB) = det(A)det(B) = 0

So det(E) = 0. Thus, if C is any matrix in G, then det(C) = det(CE) = det(C)det(E) = 0, and therefore C is singular.

For an example, look back at S in part (i). This time take the set of 3x3 matrices of that form - call it T. Here, if we let B be the 3x3 matrix whose entries are all 1, then B^2 = 3B. So showing that T is a group is analagous to show that S is. The identity in this case is B/3, which is a singular matrix.

----------------------------

Q8.

(I don't think this question is relevant either.)

If T(x) = x, then

ax - b = x(cx - d)

=> ax - b = cx^2 - dx

=> cx^2 - (a+d)x + b = 0

=> x = [(a+d) +/- sqrt((a+d)^2 - 4bc)]/2c

If T(x) = y, then

ax - b = y(cx - d)

=> ax - b = cxy - dy

=> ax - cxy = b - dy

=> x(a - cy) = b - dy

=> x = (b - dy)/(a - cy)

So T

If c!=0, then T(x) always gives us a real number unless cx-d=0 => x=d/c. So we take r=d/c. To find the image of S_r, let x be in S_r and let y = T(x). Then,

x = (dy - b)/(cy - a)

So y != a/c. Thus, T(S_r) = { x in R : x != a/c }.

On the other hand, if c=0, then T(x) = (-a/d)x + b/d. So if d=0 then we have a problem; otherwise, T is defined for all real numbers x. So T(S_r) is all of R minus the point (-a/d)r + b/d.

Moving on.

T_1(T_2(x)) = [(a_1 a_2 - b_1 c_2)x - (a_1 b_2 - b_1 d_2)]/[(c_1 a_2 - d_1 c_2)x - (c_1 b_2 - d_1 d_2)], so it's of the same form.

Thus, T^2(x) is the identity iff the following conditions are satisfied:

a^2 - bc = 1

ab - bd = 0

ac - dc = 0

bc - d^2 = -1

Finally, for the last bit, if we have for instance (a_3)^2 - b_3 c_3 = 0, then T_3 is not the identity. Let's re-express this:

(a_3)^2 = b_3 c_3

=> (a_1 a_2 - b_1 c_2)^2 = (a_1 b_2 - d_1 d_2) (c_1 a_2 - d_1 c_2)

Now let's take T_1(x) = 1-x. That is,

a_1 = b_1 = -1, c_1 = d_1 = 0

Then T_2 must satisfy the extra constraint

(c_2 - a_2)^2 = b_2

Let's take a_2 = 0. Then (c_2)^2 = b_2, and b_2 c_2 = -1. So b_2 = 1 and c_2 = -1; also, b_2 c_2 - (d_2)^2 = -1 implies d_2 = 0.

Thus T_2(x) = -1/x, and T_3(x) = 1 + 1/x.

Let's verify:

T_1(x)^2 = 1 - (1 - x) = x

T_2(x)^2 = -1/(-1/x) = x

T_3(x)^2 = 1 + 1/(1 + 1/x) = 1 + x/(1 + x) != x

----------------------------

Is it just me or is III/1 unbearably long?

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