STEP 2013 Solutions Watch

tinto99
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#81
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#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!
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MathsNerd1
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#82
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#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
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Alexn159
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#83
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(Original post by FJacob)
I think there might be a mistake on ii b
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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.
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Mastermind2
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#84
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(Original post by FJacob)
I think there might be a mistake on ii b
Spoiler:
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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
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Slowbro93
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#85
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#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 :facepalm:

Fixing atm
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Mastermind2
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(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?
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Slowbro93
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#87
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#87
Fixed my solution to Q1 (made working out so much nicer )

Going to update OP :yep:
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Pyoro
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#88
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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 :rolleyes:.

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

Going to update OP :yep:
DJMayes stacked up solutions to Q9, 10, 11 in that one post.
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Slowbro93
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(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 :rolleyes:.



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 :ahee:
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FJacob
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3.
i)
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X*Y=Y*X

\Rightarrow \lambda\mathbf{x} + (1-\lambda)\mathbf{y} = \lambda\mathbf{y} + (1-\lambda)\mathbf{x}

1-\lambda = \lambda

\

Which is only true if:

\lambda = 1/2


ii)
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(X*Y)*Z = X*(Y*Z)

\Rightarrow (\lambda\mathbf{x} + (1-\lambda)\mathbf{y})*Z = X*(\lambda\mathbf{y} + (1-\lambda)\mathbf{z})

\lambda^2\mathbf{x} + (\lambda - \lambda^2)\mathbf{y} + (1-\lambda)\mathbf{z} = \lambda\mathbf{x} + (\lambda - \lambda^2)\mathbf{y} + (1-\lambda)^2\mathbf{z}

\lambda^2 = \lambda

\

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


\lambda = 0 \ or \ \lambda = 1


iii)
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(X*Z)*(Y*Z) = (\lambda\mathbf{x} + (1-\lambda)\mathbf{z})*(\lambda y + (1-\lambda)\mathbf{z})

=\lambda^2\mathbf{x} + (\lambda - \lambda^2)\mathbf{y} + (1-\lambda)\mathbf{z}

=(X*Y)*Z



The corresponding identity is X*(Y*Z) = (X*Y)*(X*Z), and is proven in the same way.


iv)
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For \lambda < 1, X*Y represents the point which divides XY in (1-\lambda):\lambda.
This way we find that:


P_2 \ divides \ P_1Y \ in \ (\lambda-\lambda^2):\lambda^2 \ and \ in \  general\  P_{n+1} \ divides \ P_nY \  in\  (\lambda^n-\lambda^{n+1}):\lambda^{n+1}



Thus, P_n \  will \ divide \ the \ whole \ of \ XY \ in \ (1-\lambda^n):\lambda^n.




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

Will also put other solutions from DJ :ahee:
:eek: 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.
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Pyoro
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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.

 \displaystyle \int_0^\frac{\pi}{4} \tan^n(x)\sec^2(x) \mathrm{d}x=\frac{1}{n+1} \left[\tan^{n+1}(x)\right]_0^\frac{\pi}{4} = \frac{1}{n+1}


 \displaystyle\int_0^\frac{\pi}{4  } \sec^n(x)\tan(x) \mathrm{d}x= \displaystyle\int_0^\frac{\pi}{4  } \sec^{n-1}(x)(\sec(x)\tan(x)) \mathrm{d}x = \frac{1}{n} \left[\sec^n(x)\right]_0^\frac{\pi}{4} \\ = \frac{1}{n} \left( \sec^n\left(\frac{\pi}{4}\right) - \sec^n(0)\right) = \frac{\sqrt{2} ^n - 1}{n}


(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.

\displaystyle\int_0^\frac{\pi}{4  } x\sec^4(x)\tan(x) \mathrm{d}x = \frac{1}{4}\left[x\sec^4(x)\right]^\frac{\pi}{4}_0 - \frac{1}{4} \displaystyle\int_0^\frac{\pi}{4  } \sec^4(x) \mathrm{d}x \\ = \frac{1}{4} \frac{\pi\sqrt{2}^4}{4} - \frac{1}{4}\left( \displaystyle\int_0^\frac{\pi}{4  } \sec^2(x) \mathrm{d}x + \displaystyle\int_0^\frac{\pi}{4  } \sec^2(x)\tan^2(x) \mathrm{d}x \right) \\ = \frac{\pi}{4} - \frac{1}{4}\left( \left[\tan(x)\right]^\frac{\pi}{4}_0 + \frac{1}{3} \left[\tan^3(x)\right]^\frac{\pi}{4}_0 \right)= \frac{\pi}{4} - \frac{1}{4}\left(1 + \frac{1}{3}\right) = \frac{\pi}{4} - \frac{1}{3}


\displaystyle\int_0^\frac{\pi}{4  } x^2\sec^2(x)\tan(x) \mathrm{d}x = \frac{1}{2}\left[x^2\sec^2(x)\right]^\frac{\pi}{4}_0 - \displaystyle\int_0^\frac{\pi}{4  } x\sec^2(x) \mathrm{d}x \\ = \frac{1}{2}\left(\frac{2\pi^2}{1  6}\right) - \left(\left[x\tan(x)\right]_0^\frac{\pi}{4} - \displaystyle\int_0^\frac{\pi}{4  } \tan(x) \mathrm{d}x\right) \\ = \frac{\pi^2}{16} - \frac{\pi}{4} - \left[\ln|\cos(x)|\right]_0^\frac{\pi}{4} = \frac{\pi^2}{16} - \frac{\pi}{4} + \frac{1}{2}\ln(2)
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Slowbro93
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#94
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(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 :yep:

(Original post by Pyoro)
:eek: 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.
:ahee: I've just finished the first two integrals :ahee: If anything, I could just put both solutions up (I've it in the past )
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MathsNerd1
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(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.

\displaystyle\int_0^\frac{\pi}{4  } \tan^n(x)\sec^2(x) \mathrm{d}x=\frac{1}{n+1} \left[\tan^{n+1}(x)\right]_0^\frac{\pi}{4} = \frac{1}{n+1}

 

\displaystyle\int_0^\frac{\pi}{4  } \sec^n(x)\tan(x) \mathrm{d}x= \displaystyle\int_0^\frac{\pi}{4  } \sec^{n-1}(x)(\sec(x)\tan(x)) \mathrm{d}x = \frac{1}{n} \left[\sec^n(x)\right]_0^\frac{\pi}{4} = \frac{1}{n} (\sec^n(\frac{\pi}{4}) - \sec^n(0)) = \frac{\sqrt{2} ^n - 1}{n}

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 -.-
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GeorgeXV
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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).
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msmith2512
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(Original post by cpdavis)
I've linked the paper in the OP :yep:



:ahee: I've just finished the first two integrals :ahee: 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.
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mikelbird
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(Original post by cpdavis)
I've linked the paper in the OP :yep:



:ahee: I've just finished the first two integrals :ahee: If anything, I could just put both solutions up (I've it in the past )
Thank you very much!!
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Appeal to reason
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(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.
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Slowbro93
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#100
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#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).
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