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1. Re: 1987 Specimen STEP solutions thread
Lol, all 4 mechanics questions ftw.
Does anyone else want to do the honours of finishing off III?
I might be able to bu right now I'm feeling sleepy and don't trust my answers. A second ago I thought the area of a triangle was base*height ...
2. Re: 1987 Specimen STEP solutions thread
STEP III Q9
...
Let and and . Clearly

Closure
Note that

Hence is closed under matrix multiplication.

Identity

Note that s.t. , which is easily demonstrated to be unique. Hence has a unique identity element, as required.

Inverse

Note that the cofactor matrix of is given by .

Multiplying the corresponding first row elements of and yields that . This means that .

Note that all nxn matrices are invertible iff their determinant is nonzero, which is the case here. Hence all matrices in are invertible.

Take an invertible nxn matrix such that and

implies that by the associativity of matrix multiplication and thus, by , . If follows that any invertible nxn matrix has a unique multiplicative inverse.

Hence it is shown that s.t. , as required.

It follows that is a group under matrix multiplication, as required. .

Let be the subset of containing the 3x3 matrices of the above form with .

Note that . This means that and therefore is closed.

Taking implies that there exists a unique identity in H.

Also , hence , which can be shown to be unique for each element of .

It follows that

Considering the group , note that . Taking ensures that there exists a matrix of the form in s.t , as required(?).

[To be honest, I'm not entirely sure what they want me to say for this last part.]

In fact, I've done very little on groups so it would be nice if anyone more knowledgeable could check through this to make sure it's right.

3. Re: 1987 Specimen STEP solutions thread
(Original post by Farhan.Hanif93)
STEP III Q9
...
Let and and . Clearly

Closure
Note that

Hence is closed under matrix multiplication.

Identity

Note that s.t. , which is easily demonstrated to be unique. Hence has a unique identity element, as required.

Inverse

Note that the cofactor matrix of is given by .

Multiplying the corresponding first row elements of and yields that . This means that .

Note that all nxn matrices are invertible iff their determinant is nonzero, which is the case here. Hence all matrices in are invertible.

Take an invertible nxn matrix such that and

implies that by the associativity of matrix multiplication and thus, by , . If follows that any invertible nxn matrix has a unique multiplicative inverse.

Hence it is shown that s.t. , as required.

It follows that is a group under matrix multiplication, as required. .

Let be the subset of containing the 3x3 matrices of the above form with .

Note that . This means that and therefore is closed.

Taking implies that there exists a unique identity in H.

Also , hence , which can be shown to be unique for each element of .

It follows that

Considering the group , note that . Taking ensures that there exists a matrix of the form in s.t , as required(?).

[To be honest, I'm not entirely sure what they want me to say for this last part.]

In fact, I've done very little on groups so it would be nice if anyone more knowledgeable could check through this to make sure it's right.

Unfortunately I'm not one of those knowledgeable people but I don't understand your notation in the last part. What does mean?

What I did for the last part was a bit simple but I simply multiplied two matrices in this form and then inverted the order and equated entries. This showed that one of the entries didn't matter so if you let the other two be 0 but this other one non zero then commutativity would hold for all B. I don't know whether it's valid...
4. Re: 1987 Specimen STEP solutions thread
What either of you have done is fine for "justfiy your answers". Justify here means "you're not going to get marks for just writing 'yes'".

The only other thing I'll say is that there are "minimal" axioms for a group: The closure + associativity axioms are the same, but the other two become:

(i.e. there's a left identity)

(i.e. there's a left inverse)

It can be helpful to state that G is a group if these conditions apply and then you have less to verify.
5. Re: 1987 Specimen STEP solutions thread
(Original post by ben-smith)
Unfortunately I'm not one of those knowledgeable people but I don't understand your notation in the last part. What does mean?
In an nxn matrix M, is the ij-th entry i.e. the element in the i-th row, j-th column.

What I did for the last part was a bit simple but I simply multiplied two matrices in this form and then inverted the order and equated entries. This showed that one of the entries didn't matter so if you let the other two be 0 but this other one non zero then commutativity would hold for all B. I don't know whether it's valid...
This sounds similar to what I did.

Question did feel a bit pointless, though. I was expecting a catch at the end i.e. either H wasn't a subgroup or G wasn't commutative under matrix multiplication but that wasn't the case.

Along with the fact that a lot of people have done nothing on groups by the time they get on to STEP, I can understand why they got rid of them.
6. Re: 1987 Specimen STEP solutions thread
(Original post by DFranklin)
What either of you have done is fine for "justfiy your answers". Justify here means "you're not going to get marks for just writing 'yes'".

The only other thing I'll say is that there are "minimal" axioms for a group: The closure + associativity axioms are the same, but the other two become:

(i.e. there's a left identity)

(i.e. there's a left inverse)

It can be helpful to state that G is a group if these conditions apply and then you have less to verify.
7. Re: 1987 Specimen STEP solutions thread
(Original post by DFranklin)
...
Can you have a look at the last bit of question 13 please. I haven't really convinced myself there...
8. Re: 1987 Specimen STEP solutions thread
(Original post by ben-smith)
Can you have a look at the last bit of question 13 please. I haven't really convinced myself there...
I think I would have to do the question myself to be give a "proper answer", but I'd be surprised if the correct answer isn't something along the lines of "at this point, the y' equals the acceleration due to gravity, and it follows that the reaction (which is acting against gravity) must be 0. After this point, the hoop will leave the table".
9. Re: 1987 Specimen STEP solutions thread
STEP I Q2

Cosine rule:

Note that
so we require an x such that . (obviously will be a max because min occurs a x=infinity).
Last edited by ben-smith; 13-01-2012 at 23:47.
10. Re: 1987 Specimen STEP solutions thread
(Original post by ben-smith)
...
Which paper are you working from?
11. Re: 1987 Specimen STEP solutions thread
STEP I Q3

Using the initial conditions z(0)=1 so C_1=1/2. Now, to solve for y:
12. Re: 1987 Specimen STEP solutions thread
(Original post by Farhan.Hanif93)
Which paper are you working from?
I managed to acquire the papers missing. I'll upload in the OP.
13. Re: 1987 Specimen STEP solutions thread
STEP I Q4

Let:
Last edited by ben-smith; 3 Weeks Ago at 11:53.
14. Re: 1987 Specimen STEP solutions thread
STEP I - Q7

.

Hence .

Note that

.
15. Re: 1987 Specimen STEP solutions thread
STEP I Q7

16. Re: 1987 Specimen STEP solutions thread
STEP I Q6
Consider the sector formed by the points A,D and E and let Q be the centre of the whole coin. We want to find the area of the segment subtended on A and the area of the triangle QDE. Let QD=QE=x and 2y=DE:

Adding the two areas we mentioned earlier we get:

Multiplying by 7 gives the required result.
Last edited by ben-smith; 14-01-2012 at 01:04.
17. Re: 1987 Specimen STEP solutions thread
STEP I - Q11
Let the cross section of the cylinder have centre O and radius . Let the contact force exerted by the ramp on the cylinder have magnitude R. Note that there is no reaction force exerted by the horizontal surface as the cylinder is on the point of losing contact with it. Friction must be limiting on the vertical surface. On the point of rolling:

(1)

(2)

Subbing (3) into (1): , as required.

Note that , as required.

Furthermore, in order for the cylinder to be rolling up the slope, we require the clockwise moment to be greater than or equal to the anti-clockwise moment.

By (2):

We require

By the earlier result, so it's safe to divide through by it to obtain the required result .
18. Re: 1987 Specimen STEP solutions thread
STEP I - Q12
Let A have initial speed and immediately before the collision, speed . Immediately after the collision, let A have speed and let B have speed .

By the conservation of energy:

, as required.

Whilst sliding from O to disc B, disc A is subject to a constant deceleration of . Hence

Also by conservation of momentum: (1)

By Newton's Law of Restitution: . (2)

(1)+(2):

, as required.

19. Re: 1987 Specimen STEP solutions thread
STEP I - Q13
Let the string have modulus . Whilst in equilibrium, note that .

Whilst the particle moves in the described horizontal circle, let the string have length ; be inclined at an angle to the downwards vertical and exert a tension T on the particle. It follows that the circle must have radius .

(1)
(2)

Note that from (2), . Furthermore, i.e. In order for the particle to be moving in such a circle, the string must stretched to a longer length than what it was when hanging in equilibrium, as required.

From (1):

.

Observe that and that .

Hence the required result follows.
20. Re: 1987 Specimen STEP solutions thread
STEP I Q1

or

Coefficient of x^n =

Expansion valid for: |x|<1
Last edited by oh_1993; 14-04-2012 at 23:14.
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