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STEP I, II, III 2002 Solutions Tweet

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1. Re: STEP I, II, III 2002 Solutions
2002 STEP III question 11

Attached Thumbnails

2. Re: STEP I, II, III 2002 Solutions
2002 STEP III question 12

Last edited by brianeverit; 03-06-2012 at 10:31.
3. Re: STEP I, II, III 2002 Solutions
2002 STEP III question 13

4. Re: STEP I, II, III 2002 Solutions
2002 STEPIII question 14

I think this will complete the 2002 solutions

5. Re: STEP I, II, III 2002 Solutions
(Original post by toasted-lion)
Question 5, I'll pick up where Dadeymi left off:

Now,

So

I think it should be

6. Re: STEP I, II, III 2002 Solutions
(Original post by klausw)
I think it should be

Nice one Klaus from the hill near a mill .
7. Re: STEP I, II, III 2002 Solutions
Thx!!
8. Re: STEP I, II, III 2002 Solutions
(Original post by Glutamic Acid)
II/7: (Scary scary vectors.)
??????????

Let the lines have direction vector which I'll denote by r, and without loss of generality let |r| = 1 so a^2 + b^2 + c^2 = 1;

.

b = 1 - a; c = 1 - a so
The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.

m3 has direction ; m4 direction .

Considering gives .

(i) A has position vector ; B .
Let P = , and Q =

AQ . BP = 0 so

. Showing that the latter is > 0; 1 - sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.

(ii)

So

The second equation gives . Substituting into the third gives . Substituting these into the first:
; multiplying out gives , so no non-zero solutions.

However I think there is a mistake in the final part:

Firstly, after the line
The second equation gives .
When you substitute into the third it gives not but
and if you substitute these values of alpha and beta into the first equation you will find the terms don't cancel out. Finally you will get a root

But this is not correct as makes meaningless and also if you substitute it into the second equation you get 0=2/3.

The correct "zero value" of should arise when you substitute into the third equation. Here you must have cancelled a from both side and this is exactly the "zero value" of required by last part of the question.

I don't know if I have made any stupid mistakes here and I hope someone can check my work. And if I am right, I need to say that I still don't know why the root arises here.
Last edited by klausw; 11-04-2012 at 21:38.
9. Re: STEP I, II, III 2002 Solutions
Step 2 question 2

the one where you sub

followed most of the soln on here but for the second part of the question, after dividing the whole thing by i ended up with the same equation

and then i just guessed factors which worked to factorise it into

and put each of the values of w into

but i got 8 solutions (each value of w giving a +/- value from quadratic formula.. where have i gone wrong?
10. Re: STEP I, II, III 2002 Solutions
(Original post by 8inchestall)
Step 2 question 2

the one where you sub

followed most of the soln on here but for the second part of the question, after dividing the whole thing by i ended up with the same equation

and then i just guessed factors which worked to factorise it into

and put each of the values of w into

but i got 8 solutions (each value of w giving a +/- value from quadratic formula.. where have i gone wrong?
Just did this question. Instead of guessing factors I used

And got

and then

and then got to the same solutions of w as you.

And yes i got 8 solutions, should you not get 8 if the polynomial is of order 8? I need someone to confirm this.

Spoiler:
Show

(four solutions)

(two solutions)

(two solutions)

Last edited by desijut; 12-05-2012 at 18:29.
11. Re: STEP I, II, III 2002 Solutions
(Original post by desijut)
Just did this question. Instead of guessing factors I used

And got

and then

and then got to the same solutions of w as you.

And yes i got 8 solutions, should you not get 8 if the polynomial is of order 8? I need someone to confirm this.

Spoiler:
Show

(four solutions)

(two solutions)

(two solutions)

i got the same solutions as you except for this one:

Spoiler:
Show

where i got

12. Re: STEP I, II, III 2002 Solutions
(Original post by 8inchestall)
i got the same solutions as you except for this one:

Spoiler:
Show

where i got

Yes, my bad, copied it down wrong, i got that too
13. Re: STEP I, II, III 2002 Solutions
(Original post by SimonM)
STEP III, Question 1

Spoiler:
Show
The area is

As , so the area tends to infinity as well

Volume of the solid of revolution is

As , the volume tends to 2\pi
Just a typo

As , the volume tends to
Last edited by klausw; 19-05-2012 at 21:51.
14. Re: STEP I, II, III 2002 Solutions
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.
Agree with most of the III Q4 solution here but for the final bit the only solutions I get is (0, 0), (-1, 0), and(0, 1), any one any comments?
15. Re: STEP I, II, III 2002 Solutions
(Original post by klausw)
Agree with most of the III Q4 solution here but for the final bit the only solutions I get is (0, 0), (-1, 0), and(0, 1), any one any comments?
They are correct (assuming you've written it as y^3 - x^3 = y^2 + x^2 and your solutions are of the format (x,y)). But surely you can see that (1, -1) (and vice versa) is also a solution. Even if you don't see where it comes from you can see that it works.

You don't really need much confirmation from other people. Sub it in, if it works then it's a solution.
16. Re: STEP I, II, III 2002 Solutions
(Original post by hassi94)
They are correct (assuming you've written it as y^3 - x^3 = y^2 + x^2 and your solutions are of the format (x,y)). But surely you can see that (1, -1) (and vice versa) is also a solution. Even if you don't see where it comes from you can see that it works.

You don't really need much confirmation from other people. Sub it in, if it works then it's a solution.
You are right, I forget the (1, -1)solution. Thx!
17. Re: STEP I, II, III 2002 Solutions
(Original post by tommm)
STEP II 2002 Q10

Spoiler:
Show
after t hours, the competitor has km left to run

the remainder is run at , in a time (T-t) hours

using speed = distance/time:

which rearranges to

This is a quadratic in t. Clearly t is real, therefore it has real roots, so

which rearranges to give for all values of t.

If T = 3, then

which has the (repeated) solution .

Therefore competitor one runs at 13km/h for 3/4 hours
then (29/2)km/h for the remains 9/4 hours

Therefore distance run by competitor one after time x

(The 9/8 is obtained by observing that the two expressions must agree when x = 3/4.)

Now the speed of competitor 2 after time x

we can integrate to find the distance:
distance
from considering the distance travelled initially (0) and after 3 hours (42+3/8), we find

Therefore distance =

therefore the distance between the two

Via differentiation, we find that the first expression has no maxima in the required range, but the second expression has a maximum at .

Substituting this in, we find that the maximum distance is km.

Just wondering, isn't the answer 81/40?
18. Re: STEP I, II, III 2002 Solutions
(Original post by Extricated)
Just wondering, isn't the answer 81/40?
I can't see at all where I've got my result from, must've posted that years ago!
19. Re: STEP I, II, III 2002 Solutions
(Original post by tommm)
I can't see at all where I've got my result from, must've posted that years ago!

(Original post by Extricated)
Just wondering, isn't the answer 81/40?
20. Re: STEP I, II, III 2002 Solutions
(Original post by ben-smith)
ah lol

well I told extricated there was a slight error and he posted it up on the thread

I thought i would do a few questions for m2 rev

you did that ques when you were in y12 :O