# STEP 2018 Solutions

Watch**STEP I (**paper here)

1: Solution by JeremyC

2: Solution by _gcx

3: Solution by Aditya33

4: Solution by Zacken

5: Solution by gnsbrg

6: Solution by 123Master321

7: Solution by JeremyC

8: Solution by Notnek

9: Solution by Notnek

10: Solution by mikelbird

11: Solution by mikelbird

12: Solution by etothepiplusone

13: Solution by Lordkpaj

**STEP II**(paper here)

1: Solution by Aditya33

2: Solution by I hate maths

3: Solution by JeremyC

4: Solution by roboliffe

5: Solution by I hate maths

6: Solution by IrrationalRoot

7: Solution by mikelbird

8: Solution by Zacken

9: Solution by mikelbird

10: Solution by FractalSteinway

11: Solution by ciberyad

12: Solution by Integer123

13: Solution by Gregorius

**STEP III**(paper here):

1: Solution by mikelbird

2: Solution by Zacken

3: Solution by jtSketchy

4: Solution by Garjun

5: Solution by Zacken

6: Solution by mikelbird

7: Solution by DFranklin

8: Solution by Zacken

9: Solution by jtSketchy

10:Solution by jtSketchy

11: Solution by jtSketchy

12: Solution by jtSketchy

13: Solution by Zacken

Last edited by Sir Cumference; 1 year ago

1

reply

Report

#4

(Original post by

Zacken the link to the STEP III paper in your post is going to make a lot of people very excited

**Notnek**)Zacken the link to the STEP III paper in your post is going to make a lot of people very excited

2

reply

Report

#5

4

reply

(Original post by

You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are.

**Student1256**)You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are.

1

reply

Report

#8

(Original post by

It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.

**Zacken**)It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.

0

reply

Report

#9

**Zacken**)

It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.

In the question,

vs

for the normal sine and cosine functions.

I agree with Student1256 - it's an awesome question.

0

reply

Report

#10

(Original post by

You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are. Also I’m not sure if it was *necessary* to give the second last integral in a form not involving inverse functions because the other part made you give the answer in s^-1 form

**Student1256**)You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are. Also I’m not sure if it was *necessary* to give the second last integral in a form not involving inverse functions because the other part made you give the answer in s^-1 form

I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?

0

reply

Report

#11

I can do Q12, it'll be 30 minutes, watch this space!

EDIT: Q12 parts i) and ii) : (. I need to comne back to this tomorrow to add parts iii) and iv), because bizarrely I can't do the algebra without the time pressure, when I was able to do it with pressure in the exam ...

EDIT EDIT: Done Zacken ! Although I will need to proofread this.

i)

ii)

iii)

iv)

EDIT: Q12 parts i) and ii) : (. I need to comne back to this tomorrow to add parts iii) and iv), because bizarrely I can't do the algebra without the time pressure, when I was able to do it with pressure in the exam ...

EDIT EDIT: Done Zacken ! Although I will need to proofread this.

i)

ii)

iii)

iv)

1

reply

(Original post by

That's what I thought when I first saw the question, but it's slightly different.

In the question,

vs

for the normal sign and cosine functions.

**JeremyC**)That's what I thought when I first saw the question, but it's slightly different.

In the question,

vs

for the normal sign and cosine functions.

1

reply

Report

#13

(Original post by

**Zacken**)**Note that we have so that we can see that when .****Let , then for we have****For any with , we have****i)**So**ii)**Check on Desmos.
0

reply

0

reply

Report

#15

STEP I Q5

I)

where is a non-zero real constant.

II)

From I) we know

again non-zero

If ,then ,which is impossible.

Given , then ,that is

Therefore,

Using this, it is plain that verifies

III)

Since the coefficient of is 1,

This is impossible if are distinct and integers as 17 is prime and three factors should equal 1.

b)

implies

Taking into account , we find that there are 5 sets of integers

Thanks to @DFranklin for some corrections in this last bit.

I)

where is a non-zero real constant.

II)

From I) we know

again non-zero

If ,then ,which is impossible.

Given , then ,that is

Therefore,

Using this, it is plain that verifies

III)

Since the coefficient of is 1,

This is impossible if are distinct and integers as 17 is prime and three factors should equal 1.

b)

implies

Taking into account , we find that there are 5 sets of integers

Thanks to @DFranklin for some corrections in this last bit.

2

reply

Report

#16

(Original post by

I agree, it is a well put together question.

So you're saying you could leave the penultimate integral as s/c? I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?

**Notnek**)I agree, it is a well put together question.

So you're saying you could leave the penultimate integral as s/c? I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?

0

reply

Report

#17

(Original post by

Previously it said show that the integral is equal to s^-1(u) +c and that’s it. So I’m pretty sure if you left the other integrals in this form it’d be okay. I’m not saying to leave it in x. You still have to substitute back u

**Student1256**)Previously it said show that the integral is equal to s^-1(u) +c and that’s it. So I’m pretty sure if you left the other integrals in this form it’d be okay. I’m not saying to leave it in x. You still have to substitute back u

0

reply

Report

#18

(Original post by

Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?

**Notnek**)Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?

This is what I answered (plus the constant) but I think your answer is probably nicer as it avoids the inverse function - although I think both should hopefully get the marks.

A similar approach can be used for the second part of (v).

0

reply

Report

#19

**Notnek**)

Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?

So I= u/c(s^-1(u)) + constant.

0

reply

Report

#20

(Original post by

I agree, it is a well put together question.

I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?

**Notnek**)I agree, it is a well put together question.

I suppose it's technically what they asked for but I'm unsure if it would get all the marks. DFranklin what do you think?

But it wouldn't coat more than that, for sure.

0

reply

X

### Quick Reply

Back

to top

to top