Announcements
10 years ago
#81
(Original post by DeanK22)
III 5 is basically I 3 2006 with a tincy bit at the end;
I thought it seemed familiar! How strange that they repeated a III question on I.
0
#82
STEP II, Question 12

Spoiler:
Show

Therefore

since each of the remaining will spit out particles.

Therefore

Therefore if and if
0
#83
STEP II, Question 13

Spoiler:
Show

The sum over all probabilities will be 1, therefore we have:

Consider the probability generating function

Therefore

And

Since

And the formula for
0
10 years ago
#84
(Original post by SimonM)
STEP III, Question 6

Spoiler:
Show
A trivial consequence of the product to sum formulae

Let S be the sum

So

Therefore or
I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as
x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi
0
10 years ago
#85
STEP II Question 11

First Part:

Spoiler:
Show

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

Let w = the time after projection at which the direction of motion makes an angle of 45 degrees with the horizontal.

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

Second Part:

Spoiler:
Show

Okay so the situation at is this:

P_2's speed along y is m.
P_2 is h above the ground.

and then that doubles to .

so

so

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

*

cos2a < 0 because 90 < 2a < 180

so and time can't be negative so yeah, the second root can be disregarded.

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

4
10 years ago
#86
STEP II, Question 9

Solution One:

Spoiler:
Show

This is the algebra cesspool...

Let the reaction forces on the rod at A and C be and respectively, so that the friction forces at these points are and .

Resolving horizontally gives:

Resolving vertically:

From , rearranging gives us in terms of :

Substituting this for in , multiplying by gives:

Using compound-angle formulae reduces this to:

Rearranging for :

Finally, substituting this back into and rearranging for gives:

We want the magnitude of the friction force, , which is therefore:

QED

Solution Two:

Spoiler:
Show

Significantly cleaner.

As before, let the reaction forces on the rod at A and C be and respectively, so that the friction forces at these points are and .

and are the so-called "angles of friction", and describe the angle between the reaction forces and the friction forces.

Since our two friction forces and the weight of the rod result in an equilibrium, we can draw a force triangle.

Then:

and

Simply applying the sine rule leads to the result.
1
9 years ago
#87
STEP II 2003 Q5
Spoiler:
Show
E = a/2, F = (p+b)/2, G = b/2, H = (p+a)/2 =>
EF = a/2+x[p+b-a]/2
GH= b/2+y[p+a-b]/2
for which there is a solution EF = GH = (p+a+b)/4 when x = y = 1/2
so we have s = (d+t)/2 = (p+a+b)/4 so the position vector of T is t = (p+a+b)/4-d/2
the plane OAB is the x-y plane i.e. z = 0 so the component of T in the k direction is 0, so considering the vectors in the k direction we have: 0 = n/4-d/4 <=> d = 2n
I'm fairly sure the answer is
Spoiler:
Show
d=n/2

Consider that the z co-ordinate of H is the average of the z co-ordinates of A (0) and P (n). So H has a a z co-ordinate of n/2. Similarly, S, the midpoint of AH, has a z co-ordinate of n/4. If S is the midpoint between D and T (which we are told has a z co-ordinate of 0), then D must have a z co-ordinate of n/2. Therefore d=n/2
.
0
9 years ago
#88
(Original post by Mark13)
I'm fairly sure the answer is
Spoiler:
Show
d=n/2

Consider that the z co-ordinate of H is the average of the z co-ordinates of A (0) and P (n). So H has a a z co-ordinate of n/2. Similarly, S, the midpoint of AH, has a z co-ordinate of n/4. If S is the midpoint between D and T (which we are told has a z co-ordinate of 0), then D must have a z co-ordinate of n/2. Therefore d=n/2
.
Yea I just made some rediculously silly mistakes (e.g. forgetting to multiply by two) I corrected it and get the same answer.
0
9 years ago
#89
(Original post by maltodextrin)
I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as
x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi
I did using 'otherwise' and I got that answer as well.
0
9 years ago
#90
(Original post by tommm)
STEP III 2003, Q4

Spoiler:
Show

, therefore

Using , we obtain the equation of the tangent:

This meets the curve again when . Substituting this in and rearranging to find a quadratic in , we obtain:

By the factor theorem, we find that is a root. We can factorise and divide by (because ) to obtain

.

Using the quadratic formula, we obtain

One of these gives , which we can disregard.

Therefore, as required.

The double angle formula for tangent is:

Therefore,

So, if , then

Therefore, if , then

We can also apply the formula similarly to the cotangent function, so

Using the identities and , we get

as required.

Another value of which satisfies the required properties is .

I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?

If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...
0
9 years ago
#91
(Original post by maltodextrin)
I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?

If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...
I think it's because you need T=t to be a repeated root, since its a tangent. Not sure though.
0
9 years ago
#92
(Original post by SimonM)
STEP III, Question 6

Spoiler:
Show
A trivial consequence of the product to sum formulae

Let S be the sum

So

Therefore or
Don't think this is right; there shouldn't be any multiples of in the solution set.
0
9 years ago
#93
(Original post by Mark13)
Don't think this is right; there shouldn't be any multiples of in the solution set.
Yeah I think SimonM just used the wrong trig identity. I think I posted the correct version somewhere in the thread.
0
9 years ago
#94
I'd like to ask, should i figure out whether f(1) or f(2) is greater in each case?
(Original post by tommm)
STEP II 2003 Q8

Spoiler:
Show

Stationary values occur when . We will use this in the original differential equation, so we obtain

Now, if , then from our expression for y, , which would lead to a fraction in our equation for becoming infinite. Therefore, we disregard y = 0 as a stationary point.
Therefore, stationary points occur at

So

Therefore as required.

When as

When as

I've used a graphing program to draw the two graphs. The first is in the case ; the second is when . Note the location of the turning points.

0
9 years ago
#95
I got same solutions as yours
(Original post by toasted-lion)
Hmm... yuk! Consider the area of the sector OGH and the triangle OGH, subtracting to get the segment. Nice perseverance, but it's actually pretty easy.

Are you sure we want the distance to the edge and not to the centre? We had d as the distance to centre before. I got:

for (i) (note the x-coord of the centre is irrelevant)
for (ii) (perpendicular distance)
for (iii) (perpendicular distance again)

and I just plugged these straight into the equation from the first part. As you noted (i) and (ii) follow from (iii), but I'm sure they put them in that order for a reason (ie to give you bite-sized chunks of the overall thought-process). Please tell me if I made a mistake

P.S. You got a stray /latex tag :P
0
9 years ago
#96
(Original post by SimonM)
III/2
Spoiler:
Show
What lead you to think this identity was useful?

When I attempted this I saw that (1-p)^(-0.5) wasn't expanding to quite 2rCr, but didn't think it was important to fix. Then went on to using p=1/5 which lead me to as the general term, and gave up after trying other possibilities for a while.
0
9 years ago
#97
(Original post by nota bene)
...
If you expand (1-p)^(-0.5), you should find you get 2rCr but with some extra powers of lying around. Which you can fix by considering x = 4p.

(I'm not totally sure this is what you're asking, but still).
0
9 years ago
#98
(Original post by DFranklin)
If you expand (1-p)^(-0.5), you should find you get 2rCr but with some extra powers of lying around. Which you can fix by considering x = 4p.

(I'm not totally sure this is what you're asking, but still).
No, I know how to fix it. My question was more supposed to be; how do you realise it is necessary to fix?

To me it isn't exactly natural to make that fix and then differentiate (aha, works!). Maybe I'm just bad at seeing things... Making that fix would be simple if I knew I was working towards something seemigly requiring it, but the already differentiated version where p=1/5 has been put in doesn't scream that similar to the series just found in i) although there is obviously some connection.
0
9 years ago
#99
I think the first part is supposed to actually lead you in the right direction. That is, the first part is trying to get you to realise .

From there, it's fairly easy to fiddle with that identity to get something you can differentiate to get (ii).

[As a general comment, in a STEP-type exam, if you've just found a sum (for particular a_n), and you're asked for a sum , where P(n) is a small polynomial in n, it is very likely that you will need to differentiate the first series (or some simple variant of the first series) to get something looking like the 2nd. Similarly, if you're asked for you should be thinking about integration. And in general, the last part of a STEP question usually has some connection to the earlier parts. So I guess I'm saying (ii) should scream out as similar to (i), at least to the degree of thinking "there must be a way of manipulating one of these series to end up with the other one"].

I'm still not sure if that is what you're actually asking. I'd be surprised if any of this was particularly new to you.
0
9 years ago
#100
(Original post by DFranklin)
I think the first part is supposed to actually lead you in the right direction. That is, the first part is trying to get you to realise .
I didn't need to use that in the first part, I just substituted so didn't really make it 'nice' first, because it was easy to just multiply the thing by 1/(2^(2n)). I guess that's the reason I failed to see it was trying to hint me at anything.

I'm still not sure if that is what you're actually asking. I'd be surprised if any of this was particularly new to you.
I'm quite satisfied! No, nothing of this latter part is particularily new to me. It's just that in this case I didn't have the starting point clear and then it's diffícult to do something with it. It's the type of question I'll abandon after staring 5-10 minutes at it in a real exam.
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Arts University Bournemouth
Art and Design Foundation Diploma Further education
Sat, 25 May '19
• SOAS University of London
Wed, 29 May '19
• University of Exeter
Thu, 30 May '19

### Poll

Join the discussion

#### How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (350)
30.54%
The paper was reasonable (460)
40.14%
Not feeling great about that exam... (197)
17.19%
It was TERRIBLE (139)
12.13%