# STEP 2017 Solutions

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**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

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**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.

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#4

(Original post by

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.

**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.

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#7

(Original post by

Do you have the paper?

**A Slice of Pi**)Do you have the paper?

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**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.

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(Original post by

Do you have the paper?

**A Slice of Pi**)Do you have the paper?

(Original post by

...

**Rishabh_01**)...

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#10

(Original post by

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!

**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!

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#11

(Original post by

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"?

**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"?

What are your opinions on the paper thus far, from the questions you've seen?

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**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 .

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#16

If anyone has a copy of the paper can you send it to me? I could write up a few solutions.

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**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 .

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(Original post by

Could you add my solution of question 5 to pinned thread

**Rishabh_01**)Could you add my solution of question 5 to pinned thread

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