# STEP 2013 SolutionsWatch

Announcements
5 years ago
#81
(Original post by MathsNerd1)
I was thinking the exact same thing really, thought I screwed it up completely but hopefully I haven't
I've plugged it into wolfram alpha and it seems that there is no 'e' or 'ln' term anywhere in the final solution, so hopefully we haven't made a mistake!
0
5 years ago
#82
(Original post by tinto99)
I've plugged it into wolfram alpha and it seems that there is no 'e' or 'ln' term anywhere in the final solution, so hopefully we haven't made a mistake!
That's what I'm hoping and I knew the substitution of y=X^2*U worked when I tried it yesterday so I didn't take a second thought of it really, everything could go hopelessly wrong for me though
0
5 years ago
#83
(Original post by FJacob)
I think there might be a mistake on ii b
Spoiler:
Show

I let:
y = x^2-4x, which gives
y + sqrt(2y-3) - 9 = 0

Then let z = sqrt(2y-3)
y = (z^2 + 3)/2

(z^2 + 3)/2 + z - 9 = 0
z^2 + 2z - 15 = 0
Delta = 64
z = (-2 +- 8)/2 >= 0
z = 3
y = 12/2 = 6

x^2-4x = 6
Delta = 16 + 24 = 40
x = 2 +- sqrt(10)

Which is verified by wolfram.
I got x= 2+- root 10 as well.

I took y = root ( (X+2)^2 -11/2) and worked through to that.
0
5 years ago
#84
(Original post by FJacob)
I think there might be a mistake on ii b
Spoiler:
Show

I let:
y = x^2-4x, which gives
y + sqrt(2y-3) - 9 = 0

Then let z = sqrt(2y-3)
y = (z^2 + 3)/2

(z^2 + 3)/2 + z - 9 = 0
z^2 + 2z - 15 = 0
Delta = 64
z = (-2 +- 8)/2 >= 0
z = 3
y = 12/2 = 6

x^2-4x = 6
Delta = 16 + 24 = 40
x = 2 +- sqrt(10)

Which is verified by wolfram.
I agree
0
#85
(Original post by FJacob)
I think there might be a mistake on ii b
Spoiler:
Show

I let:
y = x^2-4x, which gives
y + sqrt(2y-3) - 9 = 0

Then let z = sqrt(2y-3)
y = (z^2 + 3)/2

(z^2 + 3)/2 + z - 9 = 0
z^2 + 2z - 15 = 0
Delta = 64
z = (-2 +- 8)/2 >= 0
z = 3
y = 12/2 = 6

x^2-4x = 6
Delta = 16 + 24 = 40
x = 2 +- sqrt(10)

Which is verified by wolfram.

(Original post by Vaz_Tê)
I might be wrong but I think your part ii) b of question one is wrong. I got 2 +/- sqrt(10) and have checked this on my calculator. Your solutions give me 3.

(Original post by GeneralOJB)
Erm, for part iib I got y^2 +2y - 15
Just seen my error, I said that-18+3=-1

Fixing atm
0
5 years ago
#86
(Original post by FJacob)
I have a different answer for iii), and Wolfram seems to agree.

I used the same substitution y=ux, and the equation reduces to
x*(du/dx) = x/u + u
Then you do the same thing again and let u=vx and the final solution is something like x*sqrt(6x^2-2x).

EDIT: I just realized that amounts to the same thing as doing y=vx^2 as someone else already suggested.
Hmm, that answer rings a bell. I think I got that in the exam. They'll accept both answers though?
0
#87
Fixed my solution to Q1 (made working out so much nicer )

Going to update OP
0
5 years ago
#88
STEP I Q4 looks like a gift so I'll type up a solution to practise my LaTeX skills. I've literally used it 3 times so far so good luck to me .

(Original post by cpdavis)
Fixed my solution to Q1 (made working out so much nicer )

Going to update OP
DJMayes stacked up solutions to Q9, 10, 11 in that one post.
0
#89
(Original post by Pyoro)
STEP I Q4 looks like a gift so I'll type up a solution to practise my LaTeX skills. I've literally used it 3 times so far so good luck to me .

DJMayes stacked up solutions to Q9, 10, 11 in that one post.
Just finished a solution to 4, want to race?

Will also put other solutions from DJ
0
5 years ago
#90
3.
i)
Spoiler:
Show

ii)
Spoiler:
Show

Thus, they are not distinct unless x = z or:

iii)
Spoiler:
Show

iv)
Spoiler:
Show

For , represents the point which divides in .
This way we find that:

Here's my solution to this neglected question, which happened to be my favorite. Please let me know of any mistakes.
0
5 years ago
#91
When you have all finished putting up your solutions could someone put the paper up please....
0
5 years ago
#92
(Original post by cpdavis)
Just finished a solution to 4, want to race?

Will also put other solutions from DJ
Oh no, please! I'm so slow, I've only just managed the first half! Though to be fair, being slow at STEP is the story of my life... but I did manage the question on paper in 7 minutes.
0
5 years ago
#93
STEP I 2013 – Q4
If you find an error, please do mention it.

(i)
Both of these integrals can be evaluated by substituting for the nth power term, since the integrand includes its derivative. This is done most easily by inspection.

(ii) By parts, differentiating the xn term down to 1. The Pythagorean sec2 = 1 + tan2 identity is the one to use generally when working with integrals featuring sec and tan.

1
#94
(Original post by mikelbird)
When you have all finished putting up your solutions could someone put the paper up please....
I've linked the paper in the OP

(Original post by Pyoro)
Oh no, please! I'm so slow, I've only just managed the first half! Though to be fair, being slow at STEP is the story of my life... but I did manage the question on paper in 7 minutes.
I've just finished the first two integrals If anything, I could just put both solutions up (I've it in the past )
0
5 years ago
#95
(Original post by Pyoro)
STEP I 2013 – Q4
First draft.

i: Both of these integrals can be evaluated by substituting for the n-th power term, since it is being multiplied by its derivative. This is done most easily by inspection.

Time for a refreshing beverage. More soon...
I can't believe I actually struggled with the second part of it now that I've seen how to do it -.-
0
5 years ago
#96
I also got the solution for 7iii as x root(6x^2 - 2x), that is the same form as the other two solutions which seems likely to me (I used y=ux^2).
0
5 years ago
#97
(Original post by cpdavis)
I've linked the paper in the OP

I've just finished the first two integrals If anything, I could just put both solutions up (I've it in the past )
You might have linked the paper but it is difficult to read and impossible to print ....

.... A scanned copy would still be great.
0
5 years ago
#98
(Original post by cpdavis)
I've linked the paper in the OP

I've just finished the first two integrals If anything, I could just put both solutions up (I've it in the past )
Thank you very much!!
0
5 years ago
#99
(Original post by MathsNerd1)
I got my value of c being 6 so I don't think it was the same at all :-/ I'm really doubting my performance in this paper now
I got c=6 as well. If I can be bothered later I'll have a look at writing up a solution. Firstly though, it is time to enjoy some halo now that exams are done.
0
#100
(Original post by msmith2512)
You might have linked the paper but it is difficult to read and impossible to print ....

.... A scanned copy would still be great.
Oh, that was what a user gave. If I get a better copy I will link it (and will also link STEP II this evening).
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Arts University Bournemouth
Art and Design Foundation Diploma Further education
Sat, 25 May '19
• SOAS University of London
Wed, 29 May '19
• University of Exeter
Thu, 30 May '19

### Poll

Join the discussion

#### How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (351)
30.6%
The paper was reasonable (460)
40.1%
Not feeling great about that exam... (197)
17.18%
It was TERRIBLE (139)
12.12%