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# STEP maths I, II, III 1991 solutions Watch

1. Here's another solution to the first bit of the vector / matrix question. It splits into two parts. The first part is actually pretty useful in general situations, and worth knowing IMHO. The 2nd bit is just pretty.

Suppose OA intersects with BC. Then we can find with . Dot everything with :

Nearly everything vanishes, and we're left with

, so (since ). This is basically the end of the useful part; it's a nice approach if you need to solve the intersection of 2 lines in vector form.

For the second part, we want to show can't happen. Expand the LHS and we get:

. Take out the factors b,a,c from rows 1,2,3 to get
. Subtract row1 from rows 2 and 3:
. Take out factors (a-b), (c-b) from rows 2,3:
. Finally expand what's left to get

But we know a,b,c,0 are all distinct, so . QED.

Edit: cursed LaTeX...
2. Anyone wanna, uh, do any 1991 questions?
3. (Original post by DFranklin)
ax(c-b)
Nice
4. (Original post by General)
Anyone wanna, uh, do any 1991 questions?
That was so like sixteen years ago.
Not like 1990 though, that's like so modern and contemporary, it's in the future.
5. Go open a thread now I know you want it

(I'm finished with II/1)
6. (Original post by nota bene)
I know you want it
Nota, my innocent ears!
7. STEP 1, Q1.
Alternate solution to ii).

We need to show is a solution to

8. Paper I number 2
Attached Images
9. 1991STEP I no 2.pdf (38.4 KB, 179 views)
10. Paper I number 7
Not too sure about this one. Would welcome confirmation or correction.
Attached Images
11. 1991STEP I no 7.pdf (38.9 KB, 186 views)
12. paper I numbner 9
Attached Images
13. 1991STEP I no 9.pdf (33.5 KB, 254 views)
14. Paper I nos.10,11,12 and 13
Attached Images
15. 1991STEP I no 10.pdf (31.3 KB, 207 views)
16. 1991STEP I no 11.pdf (29.2 KB, 185 views)
17. 1991STEP I no 12.pdf (45.2 KB, 192 views)
18. 1991STEP I no 13.pdf (40.9 KB, 156 views)
19. Paper I nos.14,15 and 16
Attached Images
20. 1991STEP I no 14.pdf (32.8 KB, 157 views)
21. 1991STEP I no 15.pdf (37.8 KB, 131 views)
22. 1991STEP I no 16.pdf (21.3 KB, 164 views)
23. The bulk of II/11. First time I've ever done one of these, and it could be totally wrong, but I'll try:

It is projected with an initial velocity u. The value of u doesn't really matter, since it's the angle we're after and it's not specified, so for ease of calculation I'll use u = 1.
The horizontal component is cos θ, and the vertical component is sin θ.
Using suvat, we want SV which will occur when v = 0, and I'll take gravity to be 10ms^-2.
So
To find the time this happens in using t = (v-u)/a.

Multiply this by the horizontal component to get the horizontal distance.

Using Pythagoras for the distance.
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
s = \sqrt{(\dfrac{\sin^2 \theta}{20})^2 + (\dfrac{\sin \theta \cos \theta}{10})^2.

Now basically we want to maximize .

Differentiate it to get
Setting it to zero and simplifying.
*
Therefore if

.
// The negative square root can be disregarded, since a negative angle has no meaning in this scenario.

Therefore
Also, looking at *, we have solutions of 0 and pi/2. Of course, if the angle is 0 then the shell won't go anywhere. But to show pi/2 isn't a maximum:
Spoiler:
Show

Differentiate dy/dθ to get
Substituting pi/2 into it we get 0 - 0 + 4 = 4. Therefore this isn't a maximum.
24. (Original post by Glutamic Acid)
Differentiate it to get
Setting it to zero and simplifying.

Therefore if
Therefore
I haven't checked most of your working but you're definitely losing solutions here and you haven't shown that you've got a maximum.

Also, are you sure that II/11? My copies of the paper don't seem to relate to what you've written there. That could just be that my papers are mislabelled.
25. (Original post by Swayam)
I haven't checked most of your working but you're definitely losing solutions here and you haven't shown that you've got a maximum.

Also, are you sure that II/11? My copies of the paper don't seem to relate to what you've written there. That could just be that my papers are mislabelled.
If cos θ sin θ = 0, then either cos or sin can be 0, so the angle can be 0 or pi/2. An angle of 0 can be disregarded, and I suppose I should show that pi/2 isn't (or is?) a solution. Also - root 2 has no application in this scenario. I'll edit these in in the morning. As far showing it's a maximum, that's probably a good idea although I don't think it's necessary.

I think this is II/11, my filename is STEP I 1991 though, but the front page says it is 'Further Mathematics Paper A'.
26. (Original post by Glutamic Acid)
If cos θ sin θ = 0, then either cos or sin can be 0, so the angle can be 0 or pi/2. An angle of 0 can be disregarded, and I suppose I should show that pi/2 isn't (or is?) a solution. Also - root 2 has no application in this scenario
If sin x = 0, x = 0 or pi (for 0 <= x < 2pi). So what about the angle pi? I don't know if 0 can be disregarded since I haven't read the question but if you say so.

If cosx = 0, x = pi/2 or 3pi/2 and those definitely are solutions to that equation.

Why can you disregard the tanx = -root2 solution? Maybe I'm missing something here but 2.186 radians certainly works.

You'll have to show which of 0, pi/2, pi, 3pi/2, pi and the other angles give the maximum height (or show that some of those angles are invalid as far as the question is concerned).
27. (Original post by Swayam)
If sin x = 0, x = 0 or pi (for 0 <= x < 2pi). So what about the angle pi? I don't know if 0 can be disregarded since I haven't read the question but if you say so.

If cosx = 0, x = pi/2 or 3pi/2 and those definitely are solutions to that equation.

Why can you disregard the tanx = -root2 solution? Maybe I'm missing something here but 2.186 radians certainly works.

You'll have to show which of 0, pi/2, pi, 3pi/2, pi and the other angles give the maximum height (or show that some of those angles are invalid as far as the question is concerned).
The question is about a projection into the air, so the smallest angle to the ground will be acute. sin pi will give a smallest angle of 0, so the projectile won't travel anywhere, the same reason why 0 can be disregarded. So we're looking at a solution in the range 0 < x <= pi/2.

Since it's an angle of projection, the negative angle will be projected into the ground, which can't be done in this situation.

I think that only leaves x = pi/2, which I'll do in the morning.
28. (Original post by Rabite)
I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.
I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.
I do not agree with all of your solution. In the range -1 to 0 the function should be 3x^2+1 (Not 11)
There is no minimumvalue at (1,-4) the minimum point is at (2,-5)
and the inverse function for x>1 should be the same as for x<-1 since it is part of the same parabola. My complete solution is attached
Attached Images
29. 1991PAPER fmb.3.pdf (36.5 KB, 99 views)
30. (Original post by Rabite)
III/6

God, this one took ages to type out.
No one's even gonna read it, but hey...

Could someone else at least try the question to see what the real answer is?
I agree with the answer but certainly the first two parts are easier if you take the parametric equations to be x=cosht, y=sinht ratrher than the trig ones.
31. Paper III number 13. (first part)
Would someone like to finish it please?
Attached Images
32. 1991PAPERIII.13(part).pdf (30.1 KB, 110 views)

Updated: May 19, 2016
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