STEP 2013 Solutions

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I was thinking the exact same thing really, thought I screwed it up completely but hopefully I haven't :redface:

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!

Reply 81

MathsNerd1

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 :redface:

Reply 82

Alexn159

Original post by FJacob

I think there might be a mistake on ii b

Spoiler

I got x= 2+- root 10 as well.

I took y = root ( (X+2)^2 -11/2) and worked through to that.

Reply 83

Mastermind2

Original post by FJacob

I think there might be a mistake on ii b

Spoiler

I agree

Reply 84

Slowbro93

Original post by FJacob

I think there might be a mistake on ii b

Spoiler

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 :tongue:

Reply 85

Mastermind2

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?

Reply 86

Slowbro93

Fixed my solution to Q1 (made working out so much nicer :tongue:

)

Going to update OP :yep:

Reply 87

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:

Original post by cpdavis

Fixed my solution to Q1 (made working out so much nicer :tongue:

)

Going to update OP :yep:

DJMayes stacked up solutions to Q9, 10, 11 in that one post. :tongue:

Reply 88

Slowbro93

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. :tongue:

Just finished a solution to 4, want to race? :colone:

Will also put other solutions from DJ :ahee:

Reply 89

FJacob

3.
i)

Spoiler

ii)

Spoiler

iii)

Spoiler

iv)

Spoiler

Here's my solution to this neglected question, which happened to be my favorite. Please let me know of any mistakes.

(edited 10 years ago)

Reply 90

mikelbird

When you have all finished putting up your solutions could someone put the paper up please....

Reply 91

Pyoro

Original post by cpdavis

Just finished a solution to 4, want to race? :colone:

Will also put other solutions from DJ :ahee:

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. :tongue:

Reply 92

Pyoro

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 n^th 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}{16}\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)[br]

(edited 10 years ago)

Reply 93

Slowbro93

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

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. :tongue:

I've just finished the first two integrals :ahee:

If anything, I could just put both solutions up (I've it in the past :wink:

)

Reply 94

MathsNerd1

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}[br] [br]\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}[br][br]

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 -.-

Reply 95

GeorgeXV

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).

Reply 96

msmith2512

Original post by cpdavis

I've linked the paper in the OP :yep:

I've just finished the first two integrals :ahee:

If anything, I could just put both solutions up (I've it in the past :wink:

)

You might have linked the paper but it is difficult to read and impossible to print ....

.... A scanned copy would still be great.

Reply 97

mikelbird

Original post by cpdavis

I've linked the paper in the OP :yep:

I've just finished the first two integrals :ahee:

If anything, I could just put both solutions up (I've it in the past :wink:

)

Thank you very much!!

Reply 98

Appeal to reason

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 :redface:

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.