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# STEP 2017 Solutions watch

1. STEP I (paper here)
1: Solution by Zacken
2: Solution by Zacken
3: Solution by Rishab_01
4: Solution by Zacken
5: Solution by Rishab_01
6: Solution by Zacken
7: Solution by harlem_basket
8: Solution by Zacken
9: Solution by FractalSteinway
10: Solution by FractalSteinway
11: Solution by JohnCraig
12: Solution by Farhan.Hanif93
13: Solution by 13 1 20 8 42

STEP II (paper here)
1: Solution by Zacken
2: Solution by Farhan.Hanif13
3: Solution by Slice of Pi
4: Solution by Slice of Pi and SolC
5: Solution by Rishab_01
6: Solution by IrrationalRoot
7: Solution by Zacken
8: Solution by Farhan.Hanif13
9: Solution by atsruser
10: Solution by Forecast
11: Solution by DFranklin
12: Solution by DFranklin
13: Solution by 13 1 20 8 42

STEP III (paper here):
1: Solution by Zacken
2: Solution by SliceofPi
3: Solution by Jamvicious
4: Solution by DFranklin
5: Solution by ThatPerson
6: Solution by DFranklin
7: Solution by DFranklin
8: Solution by IrrationalRoot
9: Solution by Stelios98
10: Solution by DFranklin
11: Solution by DFranklin
12: Solution by Zacken
13: Solution by Zacken
2. STEP I, Question 1:

Let so that . Hence the integral is

Let so that . Hence the integral is

Let so that . Hence the integral is

Let so that . Hence the integral is

This was embarrassingly easy for a STEP question, so it's hard to predict the spread of marks but something like: 3/4/5/8 or 3/4/6/7 is what I'd guess.
1. STEP IV (paper here):
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
3. (Original post by Zacken)
Question 1:

Let so that . Hence the integral is

Let so that . Hence the integral is

Let so that . Hence the integral is

Let so that . Hence the integral is

This was embarrassingly easy for a STEP question, so it's hard to predict the spread of marks but something like: 3/4/5/8 or 3/4/6/7 is what I'd guess.
zain set me link for good copy for the paper
4. Question 3
Attached Images

5. Do you have the paper?
6. (Original post by A Slice of Pi)
Do you have the paper?
No but i remember this question, it was pretty standard. If you happened to study parabola then its a piece of cake else I would rate it moderate.
7. Question 2:

Note that, since integration preserves rank, we have for all . This gives

for all . If then replace the above with to get

by multiplying both sides by . This completes (i) and (ii). Again. Integration preserves rank so for all we have

Hence

so that from above and from below by some trivial rearranging. Hence we get

for all . For simply invert and you get the same thing after some trivial rearranging, completing (iii).

I'd predict a split of about 5/7/8.
8. (Original post by solC)
Question 11: (Credit to JohnCraig)
The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?

(Original post by A Slice of Pi)
Do you have the paper?
Yep, but just a bunch of badly taken pictures - I'll rotate/crop and pdf them at some point today.

(Original post by Rishabh_01)
...
If anybody is willing to LaTeX this up, please do!
9. (Original post by Zacken)
The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?

Yep, but just a bunch of badly taken pictures - I'll rotate/crop and pdf them at some point today.

If anybody is willing to LaTeX this up, please do!
dished why u blanking me
10. (Original post by Zacken)
The attachment doesn't seem to be working for me - could you try uploading it so something like imgur and then inject the image here using the "insert image via URL"?
Oh right, sorry. I'll try to fix it now.

What are your opinions on the paper thus far, from the questions you've seen?
11. Question 5
Attached Images

12. What do you thing you be the marks boundries for grade 1 and grade S
13. Question 11: (credit to JohnCraig)

If this doesn't work I give up
14. STEP I, Question 8:

Base case is trivial. Assume then

(i) Assume that and . Then and

Now let so that , so that as we have unbounded above so and assuming that gives , i.e since .

(ii) Assume that , then since we have so .

Hence since we have giving as required.

Now note that from the recurrence for we can compute so .
15. If anyone has a copy of the paper can you send it to me? I could write up a few solutions.
17. STEP I, Question 6:

(i) Assume that takes only (wlog) non-negative values in the interval . Let be such that . Then by continuity we have an interval where . But then . Contradiction. (really contraposition)

(ii) We have . Hence must take both positive and negative values on (or identically and we are done anyway). But since squares are non-negative this means that must take positive and negative values on hence must hit at some point.

We get and so two possible solutions are . Both of which hit 0 at some point in (0,1) which you can check using obvious inequalities. e.g .

(iii) Set then . Also . And finally .

So apply (ii) with and get somewhere in .
18. Could you add my solution of question 5 to pinned thread
19. (Original post by Rishabh_01)
Could you add my solution of question 5 to pinned thread

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