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STEP Maths I,II,III 1987 Solutions

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Square
Okies, the pennies work, but its still not working on paper. Have you actually drawn a diagram, or are you just trying to imagine the sides of the triangle? I think you've made a couple of errors that should be reasonably obvious if you draw a proper diagram. (I suggest you don't make the 'small' circle too small, it should make things clearer).
im that poor that i dont have 7 pennies
Square
Okies, the pennies work, but its still not working on paper. If it helps, I just noticed the "7 penny case" has R=2r (because according to the question's definition of R, the inner radius of the 'center circle' is R-r, not R). (No, I haven't bothered drawing a diagram either!)
Reply 43
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coffeym
10
I did Q3 from STEP I on the first page, you might like to update the first post so others don't waste time doing it too :biggrin:
STEP I, Q2: (spoilered as people seem to be still working on it...)

Spoiler

that makes alot more sense haha

i wouldnt of realised that the small angle was pi/n though on my own :frown:

You know on the second part where you're subtracting the intercetion of the outer circle with the polygon

i keep getting the angle for the sector to be pi2−pi/n \frac{pi}{2} - {pi}/{n}


no wait ive realised it was because i was only using one triangle for the sector, and therefore only subtracting half of the sector
Reply 46
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kabbers
4
ok typing step1/8 :smile:
Reply 47
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kabbers
4
Q I/8:

don't blame me for any mistakes, i've learnt c3&4 myself in the past month and a bit, still a little slow in some areas! :wink: please check for any!

part one : [thanks nota bene!]: x = 1/t will not work when x = 0, which is sure to happen since the integration is between bounds of -1 & 1.


Part two part one:

Integrate ∫−111(1+x2)2 dx\displaystyle\int^1_{-1} \frac{1}{(1+x^2)^2} \, \mathrm{d}x

Let's try a substitution of xx with tanθtan\theta

Thus

x=tanθx = tan\theta
dxdθ=sec2θ\frac{\mathrm{d}x}{\mathrm{d}\theta} = sec^2 \theta
dx=sec2θ⋅dθ\mathrm{d}x = sec^2 \theta \cdot \mathrm{d}\theta

Find the bounds of the new integral:
1=tanθ1 = tan\theta, so the top one will be π/4\pi/4
−1=tanθ-1 = tan\theta, so the bottom one will be −π/4-\pi/4

Perform the substitution:

=∫−π/4π/4sec2θ(1+tan2θ)2 dθ=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{sec^2 \theta}{(1+tan^2 \theta)^2} \, \mathrm{d}\theta

tan2θ+1=sec2θtan^2 \theta + 1 = sec^2 \theta , so:

=∫−π/4π/4sec2θsec4θ dθ=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{sec^2 \theta}{sec^4 \theta} \, \mathrm{d}\theta

=∫−π/4π/41sec2θ dθ=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{1}{sec^2 \theta} \, \mathrm{d}\theta

=∫−π/4π/4cos2θ dθ=\displaystyle\int^{\pi/4}_{-\pi/4} cos^2 \theta \, \mathrm{d}\theta

Use the identity cos2θ=cos2θ−sin2θcos 2\theta = cos^2 \theta - sin^2 \theta, cos2θ=0.5(cos(2θ)+1)=0.5cos(2θ)+0.5cos^2 \theta = 0.5(cos (2\theta) + 1) = 0.5cos (2\theta) + 0.5

=∫−π/4π/40.5cos(2θ)+0.5 dθ=\displaystyle\int^{\pi/4}_{-\pi/4} 0.5cos (2\theta) + 0.5 \, \mathrm{d}\theta

=[0.25sin(2θ)+0.5θ]−π/4π/4=[0.25sin (2\theta) + 0.5\theta]^{\pi/4}_{-\pi/4}

=0.5+π/4= 0.5 + \pi/4



Part two part two:


Integrate ∫−11−t2(1+t2)2 dt\displaystyle\int^1_{-1} \frac{-t^2}{(1+t^2)^2} \, \mathrm{d}t

Let's try (again) a substitution of tt with tanθtan\theta

Thus

t=tanθt = tan\theta
dtdθ=sec2θ\frac{\mathrm{d}t}{\mathrm{d}\theta} = sec^2 \theta
dt=sec2θ⋅dθ\mathrm{d}t = sec^2 \theta \cdot \mathrm{d}\theta

Find the bounds of the new integral:

t=tanθt = tan\theta
1=tanθ1 = tan\theta, so the top one will be π/4\pi/4
−1=tanθ-1 = tan\theta, so the bottom one will be −π/4-\pi/4

Perform the substitution:

=∫−11−tan2θ(1+tan2θ)2 dt=\displaystyle\int^{1}_{-1} \frac{-tan^2 \theta}{(1+tan^2 \theta)^2} \, \mathrm{d}t

=∫−π/4π/4−tan2θsec4θ⋅sec2θ dθ=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{-tan^2 \theta}{sec^4 \theta} \cdot sec^2\theta \, \mathrm{d}\theta

=∫−π/4π/4−tan2θsec2θ dθ=\displaystyle\int^{\pi/4}_{-\pi/4} \frac{-tan^2 \theta}{sec^2 \theta} \, \mathrm{d}\theta

=∫−π/4π/4−tan2θ⋅cos2θ dθ=\displaystyle\int^{\pi/4}_{-\pi/4} -tan^2\theta \cdot cos^2 \theta \, \mathrm{d}\theta

=∫−π/4π/4−sin2θ dθ=\displaystyle\int^{\pi/4}_{-\pi/4} -sin^2 \theta \, \mathrm{d}\theta

Use the identity cos2θ=cos2θ−sin2θcos 2\theta = cos^2 \theta - sin^2 \theta, sin2θ=0.5(1−cos(2θ))=−0.5cos(2θ)+0.5sin^2 \theta = 0.5(1 - cos (2\theta)) = - 0.5cos (2\theta) + 0.5

=∫−π/4π/4−(−0.5cos(2θ)+0.5) dθ=\displaystyle\int^{\pi/4}_{-\pi/4} -(-0.5cos (2\theta) + 0.5) \, \mathrm{d}\theta

=∫−π/4π/40.5cos(2θ)−0.5 dθ=\displaystyle\int^{\pi/4}_{-\pi/4} 0.5cos (2\theta) - 0.5 \, \mathrm{d}\theta

=[0.25sin(2θ)−0.5θ]−π/4π/4=[0.25sin (2\theta) - 0.5\theta]^{\pi/4}_{-\pi/4}

=0.5−π/4= 0.5 - \pi/4
kabbers
Q8: still working on part one of this q, thinking of reciprocal difficulties..?

hint (at least this was the reason I thought of when I did this question last night):

Spoiler

Reply 49
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kabbers
4
nota bene
hint (at least this was the reason I thought of when I did this question last night):

Spoiler



that makes sense, slightly more tangible than 'reciprocal difficulties'! it being an 'explain' question, would this be enough?

thanks! :smile:
Reply 50
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Swayum
18
Square
Ooops, sorry.


Still not right. It's linking to question 3's solution by coffeym.
Reply 51
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ukgea
6
kabbers
that makes sense, slightly more tangible than 'reciprocal difficulties'! it being an 'explain' question, would this be enough?

thanks! :smile:


I think I would have demanded a more elaborate explanation, although it has to do with using the reciprocal, I think there are subtler processes at work here, making the substitution not working.

In fact, the substitution changes the first integral into an integral not entirely different from the second. If you can find it, were I the examiner I would probably be satisfied.

Also, kabbers: there's an error in your next-to-last line.

Otherwise, well done!
Reply 52
Avatar for kabbers
kabbers
4
ukgea
I think I would have demanded a more elaborate explanation, although it has to do with using the reciprocal, I think there are subtler processes at work here, making the substitution not working.

In fact, the substitution changes the first integral into an integral not entirely different from the second. If you can find it, were I the examiner I would probably be satisfied.

Also, kabbers: there's an error in your next-to-last line.

Otherwise, well done!


i see, i worked through the original integral substituting x = 1/t, this however did turn out the other integral they gave.. same thing happened in reverse =/ i may have missed something , will try it again!

thanks, [i think i've] fixed that mistake :smile:


edit: another thought, this kind of ignores their requirement of referring to x = 1/t, but would it be possible to say if the two integrals were indeed equal, -t^2 would be equal to 1, which is clearly not the case [without complex numbers anyway? even then probably still not the case?]
Reply 53
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Square
13
I'm possibly getting somewhere with question 5 of STEP I.

I might not be though :p:.
Reply 54
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Swayum
18
You appear to have updated stuff a bit but my link is still going to question 3.
Reply 55
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coffeym
10
After having read the STEP I paper properly now, I can't believe the level of difficulty of Q4! It takes about 4 lines of working :p: There simply is no way anything like that could appear nowadays.
ukgea
I think I would have demanded a more elaborate explanation, although it has to do with using the reciprocal, I think there are subtler processes at work here, making the substitution not working.I agree you probably need a bit more detail for full marks, but I think you'd only lose a mark or 2 provided you spotted the singularity at t=0.

As far as the integrals go, the following trick is worth knowing:

Consider I=∫−1111+t2 dtI = \int_{-1}^1 \frac{1}{1+t^2} \,dt. It is a standard result that I=π2I = \frac{\pi}{2} (the integral is arctan t).

But we can also integrate by parts (with du = 1) to get:

I=[t1+t2]−11+∫−112t2(1+t2)2 dtI = \left[ \frac{t}{1+t^2}\right]_{-1}^1 +\int_{-1}^1 \frac{2t^2}{(1+t^2)^2}\,dt

I=1+2∫−11t2(1+t2)2 dtI = 1 + 2\int_{-1}^1 \frac{t^2}{(1+t^2)^2}\,dt

So ∫−11t2(1+t2)2 dt=I−12=π−24\int_{-1}^1 \frac{t^2}{(1+t^2)^2}\,dt = \frac{I-1}{2} = \frac{\pi-2}{4}

Also, I=∫−11t2(1+t2)2 dt+∫−111(1+t2)2 dtI = \int_{-1}^1 \frac{t^2}{(1+t^2)^2}\,dt + \int_{-1}^1 \frac{1}{(1+t^2)^2}\,dt, from which

∫−111(1+t2)2 dt=I−π−24=π+24\int_{-1}^1 \frac{1}{(1+t^2)^2}\,dt = I - \frac{\pi-2}{4} = \frac{\pi+2}{4}.

(These manipulations should be reasonably familiar to those who have done reduction formulae).
Reply 57
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Dystopia
10
STEP II, Q5.

i) f(x)=αxf(x) = \alpha x

[y−f(y)]n=yn(1−α)n[y - f(y)]^{n} = y^{n}(1-\alpha)^{n}

⇒dn−1dyn−1[y−f(y)]n=n!y(1−α)n\Rightarrow \frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}} [y-f(y)]^{n} = n! y (1-\alpha)^{n}

∑n=1∞1n!n!y(1−α)n\displaystyle \sum_{n=1}^{\infty}\frac{1}{n!}n!y(1 - \alpha)^{n}

This is an infinite geometric series with a=y(1−α),  r=(1−α)a=y(1-\alpha), \; r=(1-\alpha). Since 0<α<20 < \alpha < 2, it converges to yα−y\frac{y}{\alpha} - y.

⇒f−1(y)=y+yα−y=yα\Rightarrow f^{-1}(y) = y + \frac{y}{\alpha} - y = \frac{y}{\alpha}

As expected.

ii) Let y=f(x)=x−x34y = f(x) = x - \frac{x^{3}}{4}

[y−f(y)]n=y3n22n[y - f(y)]^{n} = \frac{y^{3n}}{2^{2n}}

dn−1dyn−1y3n=(3n)!(2n+1)!y2n+1\frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}} y^{3n} = \frac{(3n)!}{(2n+1)!}y^{2n+1}

f−1(y)=y+∑n=1∞(3n)!y2n+1n!(2n+1)!22n\displaystyle f^{-1}(y) = y + \sum_{n=1}^{\infty} \frac{(3n)!y^{2n+1}}{n!(2n+1)!2^{2n}}

But y=12y=\frac{1}{2}, so

x=12+∑n=1∞(3n)!n!(2n+1)!24n+1\displaystyle x = \frac{1}{2} + \sum_{n=1}^{\infty}\frac{(3n)!}{n!(2n+1)!2^{4n+1}}

Note that (3n)!n!(2n+1)!24n+1=12\frac{(3n)!}{n!(2n+1)!2^{4n+1}}=\frac{1}{2} when n=0.

x=∑n=0∞(3n)!n!(2n+1)!24n+1\displaystyle x = \sum_{n=0}^{\infty}\frac{(3n)!}{n!(2n+1)!2^{4n+1}}

iii) Let y=f(x)=x−eλxy = f(x) = x - e^{\lambda x}

[y−f(y)]n=eλny[y - f(y)]^{n} = e^{\lambda n y}

dn−1dyn−1[y−f(y)]n=(λn)n−1eλny\frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}} [y-f(y)]^{n} = (\lambda n)^{n-1} e^{\lambda n y}

f−1(y)=y+∑n=1∞(λn)n−1n!eλny\displaystyle f^{-1}(y) = y + \sum_{n=1}^{\infty} \frac{(\lambda n)^{n-1}}{n!} e^{\lambda n y}

But y = 0, so

x=∑n=1∞(λn)n−1n!\displaystyle x = \sum_{n=1}^{\infty} \frac{(\lambda n)^{n-1}}{n!}
kabbers
edit: another thought, this kind of ignores their requirement of referring to x = 1/t, but would it be possible to say if the two integrals were indeed equal, -t^2 would be equal to 1, which is clearly not the case [without complex numbers anyway? even then probably still not the case?]
No.

∫02t dt=∫021 dt\int_0^2 t \, dt = \int_0^2 1\, dt but this doesn't mean t = 1.

On the other hand, you can correctly argue that the -t^2 integral is negative and the other integral is positive, so they can't be equal. But even then, all you are really saying is "these two integrals aren't the same, so something must have gone wrong". To get the marks, I think you need to identify that something.
Reply 59
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coffeym
10
STEP I Q5

Let
Unparseable latex formula:

\displaystyle \psi=\int^{\beta}_{\alpha}{\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,dx



Now if we let x=αcos⁡2θ+βsin⁡2θ\displaystyle x=\alpha\cos^2{\theta}+\beta\sin^2{\theta} then we can say:

(x−α)=(β−α)sin⁡2θ\displaystyle (x-\alpha)=(\beta-\alpha)\sin^2{\theta} and

(β−x)=(β−α)cos2θ\displaystyle (\beta-x)=(\beta-\alpha)cos^2{\theta}

Also, dx=2(β−α)cos⁡θsin⁡θ\displaystyle dx=2(\beta-\alpha)\cos{\theta}\sin{\theta} (best not to write this in a simpler form)

Substituting all of these results into the integral, and changing limits as required, we reach the result:

ψ=∫0π22(β−α)sin⁡θcos⁡θ(β−α)2sin⁡2θcos⁡2θ dθ\displaystyle \psi=\int^{\frac{\pi}{2}}_0{\frac{2(\beta-\alpha)\sin{\theta}\cos{\theta}}{\sqrt{(\beta-\alpha)^2\sin^2{\theta}\cos^2{\theta}}}}\,d\theta

  ⟹  ψ=2∫0π2 dθ=π\displaystyle \implies \psi=2\int^{\frac{\pi}{2}}_0{}\,d\theta=\pi as required by the first part.

If α>β\displaystyle \alpha>\beta, then simply ψ=−π\psi=-\pi

For the next part, let

Unparseable latex formula:

\displaystyle \omega=\int^{\beta}_{\alpha}{\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,dx



and make the substitution x=1y  ⟹  dx=−1y2dy\displaystyle x=\frac{1}{y}\implies dx=-\frac{1}{y^2}dy

Hence, after making the substitution, and swapping limits with help from the minus sign, we reach:

Unparseable latex formula:

\displaystyle \omega=\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{y\sqrt{(\frac{1}{y}-\alpha)(\beta-\frac{1}{y})}}\,dx



Multiplying the lone yy inside the square root, we find that

ω=∫1β1α1(1−αy)(βy−1) dy\displaystyle \omega=\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{\sqrt{(1-\alpha y)(\beta y-1)}}}}\,dy

Unparseable latex formula:

\displaystyle =\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{\sqrt{\alpha\beta (\frac{1}{\alpha}-y)(y-\frac{1}{\beta})}}}\,dy



Now note that, taking out the factor of 1αβ\displaystyle \frac{1}{\sqrt{\alpha\beta}}, that the integral is simply the same as the one we evaluated in the first part, with certain symbols being inverted. This does not change the value of the integral, as everything is consistent. Hence

ω=1αβψ\displaystyle \omega=\frac{1}{\sqrt{\alpha\beta}}\psi

  ⟹  ω=παβ\displaystyle \implies \omega=\frac{\pi}{\sqrt{\alpha\beta}} as required.

This completes the question.

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