STEP Maths I, II, III 1996 Solutions

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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?

I think you have made a mistake in the composition of the two transformations. You appear to have evaluated T_1(T_2(x)) instead of T_2(T_1(x))

Reply 83

1996 Paper I.nos9,10,12 & 14.pdf53.8KB

To finish off paper I here are numbers 9,10,12 and 14

Reply 84

1996PAPER II.numbers 9,10,11,13 & 14.pdf81.1KB

Also finishoing paper II
Here are numbers 9,10,11,13 and 14

Reply 85

1996PAPER III.number 12.pdf36.7KB

Finally paper III number 12.
Is 1996 now complete?

Reply 86

This is a matrix (Pure mathematics 6 - old FP3) type solution:

The three linear equations in the question can be represented by matrix and vector multiplication:

[br]\begin{pmatrix} 1 & 1 & a \\1 & a & 1 \\2 & 1 & 1 \end{pmatrix}[br]\begin{pmatrix} x \\ y \\ z \end{pmatrix}[br]= \begin{pmatrix} 2 \\ 2 \\ 2b \end{pmatrix}[br]

Let the 3x3 matrix of coefficients be

\mathbf{M}

We need

\mathbf{M}^{-1}

which I shall find by Laplace Cofactor Expansion. Other methods are applicable.

Firstly find

|\mathbf{M}|= 1(a-1)-1(-1)+a(1-2a)=2a(1-a)

This means there are the degenerate cases

a=0

and

a=1

which we must consider separetely.

For now, back to inversion. The matrix of minors is:

\begin{pmatrix} a-1 & -1 & 1-2a \\1-a & 1-2a & -1 \\1-a^2 & 1-a & a-1 \end{pmatrix}

The matrix of cofactors is:

\begin{pmatrix} a-1 & 1 & 1-2a \\a-1 & 1-2a & 1 \\1-a^2 & a-1 & a-1 \end{pmatrix}

Therefore

\mathbf{M}^{-1}=\frac{1}{2a(1-a)}\begin{pmatrix} a-1 & a-1 & 1-a^2 \\1 & 1-2a & a-1 \\1-2a & 1 & a-1 \end{pmatrix}

Back to solvng the equations:

[br]\begin{pmatrix} 1 & 1 & a \\1 & a & 1 \\2 & 1 & 1 \end{pmatrix}[br]\begin{pmatrix} x \\ y \\ z \end{pmatrix}[br]= \begin{pmatrix} 2 \\ 2 \\ 2b \end{pmatrix}[br]

Multiply both sides by

\mathbf{M}^{-1}

[br]\begin{pmatrix} x \\ y \\ z \end{pmatrix} =[br]\frac{1}{2a(1-a)}[br]\begin{pmatrix} a-1 & a-1 & 1-a^2 \\ 1 & 1-2a & a-1 \\ 1-2a & 1 & a-1 \end{pmatrix}[br]\begin{pmatrix} 2 \\ 2 \\ 2b \end{pmatrix}

[br]\begin{pmatrix} x \\ y \\ z \end{pmatrix}=[br]\frac{1}{a(1-a)}[br]\begin{pmatrix} 2a-2+b-a^2b \\ 2-2a+ab-b \\ 2-2a+ab-b \end{pmatrix}

[br]\begin{pmatrix} x \\ y \\ z \end{pmatrix}=[br]\frac{1}{a}[br]\begin{pmatrix} ab+b-2 \\ 2-b \\ 2-b \end{pmatrix}[br]

Therefore:

x=\frac{ab+b-2}{a}

y=z=\frac{2-b}{a}

Remembering that a cannot equal zero or one.

If

a=0

the system of equations is:

x+y=2

(1)

x+z=2

(2)

2x+y+z=2b

(3)

(1)-(2) => (4)

y-z=0

Therefore

y=z

(1)+(2) => (5)

2x+y+z=4

Compare this to (3)

2x+y+z=2b

So if

b=2

there are an infinite number of solutions where:

y=z=2-x

b\not=2

then there are no solutions as the equations are inconsistent.

If

a=1

the system of equations is:

x+y+z=2

(1)

x+y+z=2

(2)

2x+y+z=2b

(3)

Adding

x

to both sides of (1) gives
(4)

2x+y+z=2+x

Therefore

2b=2+x

by comparison with (3)
Therefore

x=2b-2

There are an infinite number of solutions with:

x=2b-2

and

y+z=4-2b

Reply 87

Edavies: It is not true that a cannot equal zero or one. E.g.

When a = 1, we have the equations x+y+z = 2, 2x+y+z = 2b. Then there are an infinite number of solutions, given by x = 2b-2, y + z = 2.

Matrix solutions are usually not a good idea for these questions, because they don't handle the 'degenerate solutions' case very well.

Reply 88

Hmm. Sorry, I thought that the 'degenerate cases' where an infinite number of solutions are valid were considered 'no solutions'. I can see that if a=0 or a=1 the system reduces to simply having 2 equations. I'll add this to the answer and explain.

Reply 89

It's not terribly well worded, but it (used to be) a very standard question from the Cambridge examiners (you'd get a similar Tripos question every year), and you'd be expected to distinguish (and deal appropriately with) the "no solutions", "1 unique solution", "an infinite number of solutions" cases.

Reply 90

I've edited my solution and think I've dealt with the cases effectively....

Reply 91

With the a=0 case you have:

x+y = 2, x+z = 2, 2x+y+z = 2b.
But adding the first 2 equations gives you 2x+y+z = 4. So for the equations to be consistent, you must have b = 2 (and then y=z=2-x for an infinite number of solutions); otherwise there are no solutions.

For the a=1 case, you write:

(1) x+y+z = 2
(2) x+y+z = 2

and then add to get "2x+y+z = 4". Which isn't right.

Reply 92

DFranklin

I'd like to blame these stark errors on my grasp of LaTeX. I Confused the two cases and threw in a bit of complete stupidity to boot. I've had another go...

Reply 93

Still not really right, I'm afraid. "2b=2+x" isn't a valid condition, because x is an unknown variable.

Reply 94

So what would be the conclusion for the a=1 case? It's beyond me, I simply haven't done enough linear algebra. The proper technique for linear equation systems and matrices are not rigorously (if at all) taught in my syllabus.

Reply 95

What I said in post #88.

The way to do these questions doesn't actually involve any special technique. You just solve the 3 simultaneous equations (by eliminating variables), being very careful to examine what happens in cases where you want to divide by zero. (And this is why using the "invert the matrix" technique isn't a great plan; you end up having to solve the simultaneous equations for the degenerate cases anyhow, so you might as well not bother with the matrix).

Reply 96

DFranklin

Still not really right, I'm afraid. "2b=2+x" isn't a valid condition, because x is an unknown variable.

I see what you mean about why matrices are a bad idea, but am still confused about the case a=1. I deduced the statement 2b=2+x but it isn't a valid condition....? It might just be a bit of simple simultaneous equation solving, but to me it seems some care, even technique is needed when deciding what to make of the statements that simple algebraic manipulations yield.

Its making the distinctions between x, y and z, the variables for which I solve the equations, and a and b, the 'constants which vary' and form the solution which is a subtlety I've not been taught and I'am having difficulty working out on my own.

As far as I can see a better course of action would be that since I see "2b=2+x", I decide that x=2b-2 and get an infinite number of solutions for y and z as I don't have enough information to uniquely define them.

Thanks for your time and advice.

Reply 97