SimonM
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(Up-to-date as of post #115)

FINISHED!

Well done everbody!

Introduction/goal
Solutions for earlier STEP papers are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved (red above); if you see any mistakes in solutions posted, please point them out.

(Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.)

Link to the papers:
STEP I - STEP II - STEP III

STEP I:
1 - solved by generalebriety, alternative by Dystopia, also see thread
2 - part (i) solved by eponymous, part (ii) solved by eponymous.
3 - solved by insparato, partial alternative solution by calcium878
4 - solved by ukgea
5 - part (i) solved by calcium878, part (ii) solved by calcium878
6 - solved by coffeym
7 - solved by Dystopia
8 - solved by calcium878
9 - solved by coffeym
10 - part (i) solved by Square, part (ii) solved by DFranklin
11 - solved by DFranklin
12 - solved (rather sketchily) by ukgea
13 - solved by DFranklin
14 - solved by DFranklin

STEP II:
1 - solved by SsEe
2 - solved by Chewwy
3 - solved by insparato
4 - solved by generalebriety
5 - solved by generalebriety
6 - solved by Chewwy
7 - solved by ukgea
8 - solved by Dystopia
9 - solved by DFranklin
10 - solved by Dystopia
11 - solved by Dystopia
12 - solved by DFranklin
13 - solved by SsEe
14 - solved by coffeym

STEP III:
1 - solved by generalebriety, also see thread
2 - solved by Speleo
3 - solved by generalebriety
4 - solved by insparato
5 - solved by DFranklin
6 - solved by ukgea
7 - solved by generalebriety
8 - solved by generalebriety
9 - solved by DFranklin
10 - solved by DFranklin
11 - solved by DFranklin
12 - solved by SsEe
13 - solved by DFranklin
14 - solved by DFranklin



Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
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ukgea
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STEP II Q6, solution by chewwy, shamelessly stolen from the STEP II thread:

http://www.thestudentroom.co.uk/atta...4&d=1183161671
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insparato
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I'll do Question 3 STEP I. (What a gem)
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generalebriety
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An excellent idea.

III/1:

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\tan (\theta_1 + \theta_2) + \tan (\theta_3 + \theta_4)}{1 - \tan (\theta_1 + \theta_2)\tan (\theta_3 + \theta_4)}

\displaystyle = \frac{\frac{t_1+t_2}{1-t_1t_2} + \frac{t_3+t_4}{1-t_3t_4}}{1 - \left(\frac{t_1+t_2}{1-t_1t_2}\right) \left(\frac{t_3+t_4}{1-t_3t_4}\right)}

\displaystyle = \frac{(t_1+t_2)(1-t_3t_4) + (t_3+t_4)(1-t_1t_2)}{ (1-t_1t_2)(1-t_3t_4) - (t_1+t_2)(t_3+t_4)} .

\displaystyle at^4 + bt^3 + ct^2 + dt + e = a(t-t_1)(t-t_2)(t-t_3)(t-t_4)

\displaystyle \Rightarrow t_1+t_2+t_3+t_4 = -b/a,\quad t_1t_2 + t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 + t_3t_4 = c/a,
\displaystyle t_1t_2t_3 + t_1t_2t_4 + t_1t_3t_4 + t_2t_3t_4 = -d/a, \quad t_1t_2t_3t_4 = e/a.

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{(t_1+t_2+t_3+t_4) - (t_1t_2t_3 + t_1t_2t_4 + t_1t_3t_4 + t_2t_3t_4)}{t_1t_2t_3t_4 + 1 - (t_1t_2 + t_1t_3 + t_1t_4 + t_2t_3 + t_2t_4 + t_3t_4)}

\displaystyle = \frac{(-b/a) + (d/a)}{(e/a) + 1 - (c/a)}

\displaystyle = \frac{d-b}{a+e-c} . \quad (*)

--

\displaystyle \tan \theta_i = t_i \Rightarrow \sin \theta_i = \frac{t_i}{\sqrt{1 + t_i^2}} ,\quad \cos \theta_i = \frac{1}{\sqrt{1+t_i^2}} .

Substituting into the given equation:

\displaystyle p\cos 2\theta + \cos (\theta -\alpha ) + p = 0

\displaystyle 2p\cos^2\theta - p + \cos\theta\cos\alpha - \sin\theta\sin\alpha + p = 0

\displaystyle \frac{2p}{1+t^2} + \frac{\cos\alpha}{\sqrt{1+t^2}} - \frac{t\sin\alpha}{1 + t^2} = 0

\displaystyle 2p = -(\cos\alpha + t\sin\alpha ) \sqrt{1+t^2}

\displaystyle 4p^2 = (\cos^2\alpha + t\sin 2\alpha + t^2\sin^2\alpha )(1+t^2)

\displaystyle (\sin^2\alpha )t^4 + (\sin 2\alpha )t^3 + t^2 + (\sin 2\alpha )t + (\cos^2\alpha - 4p^2) = 0

Substituting into (*):

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = \frac{\sin 2\alpha - \sin 2\alpha}{a+e-c} = 0

\displaystyle \tan (\theta_1 + \theta_2 + \theta_3 + \theta_4) = 0 \Rightarrow \theta_1 + \theta_2 + \theta_3 + \theta_4 = n\pi

as required.
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ukgea
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STEP III Q2, solution by Speleo, shamelessly stolen from the STEP III thread:

(Original post by Speleo)
2
i
1.3.5.7...(2n-1) = \frac{1.2.3.4...(2n-1)(2n)}{2.4.6.8...(2n)}
= \frac{(2n)!}{(2.1)(2.2)(2.3)(2.4)...(2.n)} = \frac{(2n!)}{2^nn!} as required.
\frac{1}{\sqrt{1-4x}} = (1-4x)^{-\frac{1}{2}}
Binomial expansion valid for |4x| < 1 i.e. |x| < \frac{1}{4} as given.
\frac{1}{\sqrt{1-4x}} = 1 + 2x + ... + \frac{(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2r-1}{2}).(-4x)^r}{r!} + ...
Every successive term is multiplied by two extra negative terms, one fraction from the exponent and one power of -4x, so every term stays positive overall, so all minus signs can be removed.
= 1 + 2x + ... + \frac{(\frac{1}{2})(\frac{3}{2})...(\frac{2r-1}{2}).(4x)^r}{r!} + ...
Using the formula just proved and noting that there is a divisor of 2^r in total from the product of (\frac{3}{2})...(\frac{2r-1}{2}):
= 1 + 2x + ... + \frac{(2r!)4^rx^r}{2^r2^r(r!)^2} + ...
= 1 + 2x + ... + \frac{(2r!)x^r}{(r!)^2} + ...
= 1 + \Sigma_1^{\infty}\frac{(2n!)x^n}{(n!)^2} as required.
Verify that 2x is the first term by setting n = 1 and getting 2x.

ii
Differentiate both sides with respect to x:
\frac{2}{(1-4x)^{\frac{3}{2}}} = \Sigma_1^{\infty}\frac{n(2n!)x^{n-1}}{(n!)^2}
Setting x = \frac{6}{25}:
250 = \Sigma_1^{\infty}\frac{n(2n!)\frac{6}{25}^{n-1}}{(n!)^2}
Multiply both sides by \frac{6}{25}:
\Sigma_1^{\infty}\frac{n(2n!)\frac{6}{25}^n}{(n!)^2} = 60 as required.

iii
This time integrate instead:
-\frac{1}{2}\sqrt{1-4x} = x + \Sigma_1^{\infty}\frac{(2n!)x^{n+1}}{(n+1)(n!)^2} + C
Set x = 0:
 -\frac{1}{2} = 0 + 0 + C \rightarrow C = -\frac{1}{2}
Set x = \frac{2}{9}:
-\frac{1}{6} = \frac{2}{9} + \Sigma_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n}(n+1)!(n!)} - \frac{1}{2}
\Sigma_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n+2}(n+1)!(n!)} = \frac{1}{9}
\Sigma_1^{\infty}\frac{(2n!)2^{n+1}}{3^{2n}(n+1)!(n!)} = 1 as required.
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ukgea
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STEP III Q6:

Firstly,

pp^* = |p|^2 = a^2

and similarly for q.

Thus we have

pq(p^* - q^*) = pp^*q - qq^*p = a^2(q-p)

and the first result follows.

Then, note that if PQ and RS are perpendicular,

\displaystyle \frac{p-q}{s-r} = \frac{|p-q|}{|s-r|}i

or

\displaystyle \frac{p-q}{s-r} = -\frac{|p-q|}{|s-r|}i

Either way, we have

\displaystyle \frac{p^* - q^*}{s^* - r^*} = \left(\frac{p-q}{s-r}\right)^*

\displaystyle \frac{p^* - q^*}{s^* - r^*} = -\frac{p-q}{s-r}

and thus

\displaystyle \frac{s-r}{s^* - r^*} = - \frac{p-q}{p^*-q^*}

\displaystyle \frac{s-r}{s^* - r^*} + \frac{p-q}{p^*-q^*} = 0

from which it follows that

pq + rs

\displaystyle = -a^2 \left(\frac{p-q}{p*-q*} + \frac{s-r}{s^* - r^*}\right)

 = 0

as required.

For the next part, we have in the case n=3: (let a_i, b_i represent A_i, B_i in the Argand plane, with the circle centred in the origin and with radius a)

\displaystyle \left\{\begin{array} a_1a_2 = -b_1b_2  \\ a_2a_3 = - b_2b_3 \\ a_3a_1 = -b_3b_1 \end{array}\right.

Multiplying the first and the third equations, and then dividing by the second, we get

\displaystyle a_1^2 = -b_1^2

there are obviously two distinct b_1 that satisfy this, as required.

For n = 4, it's instead

\displaystyle \left\{\begin{array} a_1a_2 = -b_1b_2  \\ a_2a_3 = - b_2b_3 \\ a_3a_4 = -b_3b_4 \\ a_4a_1 = -b_4b_1 \end{array}\right.

In this case, any b_1 can satisfy this, becayse if you choose a b_1, you can always choose a b_2 so that the first equation is satisfied, and then a b_3 so that the second one is satisfied, and then a b_4 so that the third is satisfied, and then, and here's the catch, equation 4 follows from the other three: (multiply the first and third equations and divide by the fourth) and so the fourth equation will automatically satified.

For larger n, we have exactly the same sitution as in n = 3 when n is odd (you can use the same reasoning, just alternatingly multiply and divide all the equations together) and as in n = 4 when n is even (again the same reasoning, the last equation can always be obtained by alternatingly multiplying and dividing the first ones together.
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Square
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Don't suppose the papers are up on the 'net yet are they?
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ukgea
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(Original post by Square)
Don't suppose the papers are up on the 'net yet are they?
Not really, but they are stilled linked in the first post. :tsr2:
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Square
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Gonna try some of the mechanics questions
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generalebriety
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Ooooooh. III/7. Mine.

(i)
\displaystyle t(x) = \int_0^x \frac{1}{1+u^2} \text{d} u.

u = v^{-1} \Rightarrow \text{d} u = -v^{-2} \text{d} v

\displaystyle \therefore t(x) = \int_{u=0}^{u=x} \frac{1}{1+v^{-2}} \cdot -v^{-2} \text{d} v = -\int_\infty^{1/x} \frac{v^{-2}}{1 + v^{-2}} \text{d} v

\displaystyle = \int_{1/x}^\infty \frac{1}{1+v^2} \text{d} v

\displaystyle \therefore t(1/x) + t(x) = \int_{x}^\infty \frac{1}{1+v^2} \text{d} v + \int_0^x \frac{1}{1+u^2} \text{d} u = \int_0^\infty \frac{1}{1+u^2} \text{d} u = p/2.

Choose x = 1 and result follows.

(ii)

\displaystyle y = \frac{u}{\sqrt{1+u^2}} \Rightarrow y^2(1+u^2) = u^2 \Rightarrow u = \frac{y}{\sqrt{1-y^2}}

\displaystyle \frac{\text{d} u}{\text{d} y} = (1-y^2)^{-1/2} + y^2(1-y^2)^{-3/2} = (1-y^2+y^2)(1-y^2)^{-3/2} = \frac{1}{\sqrt{(1-y^2)^3}}

Performing this substitution:

\displaystyle t(x) = \int_{u=0}^{u=x} \frac{1}{1 + \frac{y^2}{1-y^2}} \cdot \frac{1}{\sqrt{(1-y^2)^3}} \text{d} y

\displaystyle = \int_{u=0}^{u=x} \frac{1-y^2}{\sqrt{(1-y^2)^3}} \text{d} y

\displaystyle = \int_0^{x/\sqrt{1+x^2}} \frac{1}{\sqrt{1-y^2}} \text{d} y = s\left( \frac{x}{\sqrt{1+x^2}} \right) .

Again, choose x = 1 and result follows.

(iii)

\displaystyle z = \frac{u + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} u}

\displaystyle \Rightarrow u = \frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}} z} (detail omitted!)

\displaystyle \Rightarrow \frac{\text{d} u}{\text{d} z} = \frac{(1 + \frac{1}{\sqrt{3}} z) - (z - \frac{1}{\sqrt{3}})(\frac{1}{ \sqrt{3}})}{(1 + \frac{1}{\sqrt{3}} z)^2} = \frac{4/3}{{(1 + \frac{1}{\sqrt{3}} z)^2}}

\displaystyle t(x) = \int_{u=0}^{u=x} \frac{1}{1 + \left( \frac{z - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}} z} \right)^2} \cdot \frac{4/3}{{(1 + \frac{1}{\sqrt{3}} z)^2}} \text{d} z

\displaystyle = \int_{u=0}^{u=x} \frac{4/3}{(1 + \frac{1}{\sqrt{3}} z)^2 + (z - \frac{1}{\sqrt{3}} )^2} \text{d} z

\displaystyle = \int_{u=0}^{u=x} \frac{4/3}{\frac{4}{3} (1 + z^2)} \text{d} z

\displaystyle = \int_{1/\sqrt{3}}^{\alpha (x)} \frac{1}{1+z^2} \text{d} z

where

\displaystyle \alpha (x) = \frac{x + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}} x} .

Then

\displaystyle \alpha \left( \frac{1}{\sqrt{3}} \right) = \sqrt{3}

and so

\displaystyle t\left( \frac{1}{\sqrt{3}} \right) = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \text{d} z.

\displaystyle 3t\left( \frac{1}{\sqrt{3}} \right) = \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \text{d} u + \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \text{d} z + \int_{\sqrt{3}}^\infty \frac{1}{1+v^2} \text{d} v

\displaystyle = \int_0^\infty \frac{1}{1+u^2} \text{d} u = \frac{1}{2} p.
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insparato
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STEP I Question 3
 cos^4\theta - sin^4\theta = (cos^2\theta + sin^2\theta)(cos^2\theta - sin^2\theta) = cos2\theta

 cos^4\theta + sin^4\theta = \frac{1}{4}(cos^22\theta + 2cos2\theta + 1) + \frac{1}{4}(cos^22\theta - 2cos2\theta +1)

 = \frac{1}{2}cos^22\theta + \frac{1}{2} = \frac{1}{2}(1-sin^22\theta) + \frac{1}{2}

 = 1 - \frac{1}{2}sin^22\theta

Right this stinks of I + J adding and substracting these integrals using the identities. But scrap that theres sin^6x so im going for reduction formula.

 I_n = \displaystyle \int_0^{\frac{\pi}{2}} sin^n\theta \hspace5 d\theta

 I_n = \displaystyle \int sin^{n-1}\theta sin \theta  \hspace5 d\theta

By parts

\displaystyle u = sin^{n-1}\theta

\displaystyle \frac{du}{dx} = (n-1)sin^{n-2}\theta cos\theta

\displaystyle \frac{dv}{dx} = sin\theta

\displaystyle v = -cos\theta

\displaystyle I_n = [-sin^{n-1}\theta cos\theta]_0^{\frac{\pi}{2}} - \int_0^{\frac{pi}{2}} -(n-1)sin^{n-2}\theta cos^2\theta  \hspace5 d\theta

\displaystyle = (n-1)\int_0^{\frac{\pi}{2}}sin^{n-2}\theta (1-sin^2\theta)  \hspace5 d\theta

\displaystyle = (n-1)\int_0^{\frac{\pi}{2}}sin^{n-2}\theta - sin^n\theta \hspace5 d\theta

\displaystyle I_n = (n-1)(I_{n-2} - I_n)

\displaystyle I_n = nI_{n-2} - nI_n - I_{n-2} + I_n

\displaystyle nI_n = (n-1)I_{n-2}

\displaystyle I_n = \frac{n-1}{n} I_{n-2}

Let

\displaystyle I_n = \int_0^{\frac{\pi}{2}} cos^n\theta \hspace5 d\theta = \int_0^{\frac{\pi}{2}} cos^{n-1}\theta cos\theta  \hspace5 d\theta

\displaystyle u = cos^{n-1}\theta

\displaystyle \frac{du}{dx} = -(n-1)cos^{n-2}sin\theta

\displaystyle \frac{dv}{dx} = cos \theta

\displaystyle v = sin\theta

\displaystyle I_n = [cos^{n-1}\theta sin\theta]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} -(n-1)cos^{n-2}sin^2\theta  \hspace5 d\theta

\displaystyle I_n = (n-1)\int_0^{\frac{\pi}{2}} cos^{n-2}\theta(1-cos^2\theta) \hspace5 d\theta

\displaystyle I_n = (n-1)\int_0^{\frac{\pi}{2}} cos^{n-2}\theta - cos^n\theta \hspace5 d\theta

\displaystyle I_n = (n-1)(I_{n-2} - I_n)

\displaystyle I_n = nI_{n-2} - nI_n - I_{n-2} + I_n

\displaystyle nI_n = (n-1)I_{n-2}

\displaystyle I_n = \frac{(n-1)}{n} I_{n-2}

\displaystyle \int_0^{\frac{\pi}{2}} sin^4\theta \hspace5 d\theta

\displaystyle I_4 = \frac{3}{4}I_2

\displaystyle I_2 = \frac{1}{2}I_0

\displaystyle I_0 = \frac{\pi}{2}

\displaystyle I_4 = \frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

\displaystyle I_4 = \frac{3\pi}{16}

\displaystyle \int_0^{\frac{\pi}{2}} cos^4\theta \hspace5 d\theta

\displaystyle I_4 = \frac{3}{4}I_2

\displaystyle I_2 = \frac{1}{2}I_0

\displaystyle I_0 = \frac{\pi}{2}

\displaystyle I_4 = \frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

\displaystyle I_4 = \frac{3\pi}{16}

\displaystyle \int_0^{\frac{\pi}{2}} sin^6\theta \hspace5 d\theta

\displaystyle I_6 = \frac{5}{6}I_4

\displaystyle I_4 = \frac{3}{4}I_2

 I_2 = \frac{1}{2}I_0

 I_0 = \frac{\pi}{2}

 I_6 = \frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

 I_6 = \frac{5\pi}{32}


\displaystyle \int_0^{\frac{\pi}{2}} cos^6\theta \hspace5 d\theta

\displaystyle I_6 = \frac{5}{6}I_4

\displaystyle I_4 = \frac{3}{4}I_2

 I_2 = \frac{1}{2}I_0

 I_0 = \frac{\pi}{2}

 I_6 = \frac{5}{6}.\frac{3}{4}.\frac{1}{2}.\frac{\pi}{2}

 I_6 = \frac{5\pi}{32}
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generalebriety
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I'm also claiming III/8. Might be my last contribution of the night then. Urgh, this will be messy to latex.

(i) Putting u = x:

0 + \text{a} (x) + x\text{b} (x) = 0

Putting u = exp(-x):

1 - \text{a} (x) + \text{b} (x) = 0

Adding:

1 + (x+1)\text{b} (x) = 0 \Rightarrow \text{b} (x) = \frac{-1}{x+1}

\therefore \text{a} (x) = \frac{x}{x+1}

General solution: u = Ax + Be^{-x} .

\displaystyle \frac{\text{d} y}{\text{d} x} + 3y^2 + \frac{x}{1+x} y = \frac{1}{3(1+x)} .

\displaystyle y = \frac{1}{3u} \frac{\text{d} u}{\text{d} x} \Rightarrow y^2 = \frac{1}{9u^2} \left(\frac{\text{d} u}{\text{d} x}\right)^2, \quad \frac{\text{d} y}{\text{d} x} = \frac{-1}{3u^2}\frac{\text{d} u}{\text{d} x}  + \frac{1}{3u} \frac{\text{d}^2 u}{\text{d} x^2}

Applying substitution,

\displaystyle \frac{-1}{3u^2}\left(\frac{\text{d} u}{\text{d} x}\right)^2  + \frac{1}{3u} \frac{\text{d}^2 u}{\text{d} x^2} + \frac{1}{3u^2} \left(\frac{\text{d} u}{\text{d} x}\right)^2 + \frac{x}{1+x} \frac{1}{3u} \frac{\text{d} u}{\text{d} x} = \frac{1}{3(1+x)}

\displaystyle \frac{\text{d}^2 u}{\text{d} x^2} + \frac{x}{1+x} \frac{\text{d} u}{\text{d} x} - \frac{1}{1+x} u = 0

which has general solution u = Ax + Be^{-x} ,

i.e.

\displaystyle \frac{1}{3u} = \frac{1}{3(Ax + Be^{-x})} ,

\displaystyle \frac{\text{d} u}{\text{d} x} = A - Be^{-x} ,

\displaystyle y = \frac{A - Be^{-x}}{3(Ax + Be^{-x})} .

y(0) = 0:

\displaystyle 0 = A - B \Rightarrow A = B

\displaystyle \therefore y = \frac{A - Ae^{-x}}{3(Ax + Ae^{-x})} = \frac{1 - e^{-x}}{3(x + e^{-x})}.

(ii)

\displaystyle \frac{\text{d} y}{\text{d} x} + y^2 + \frac{x}{1-x} y = \frac{1}{1-x} .

Applying the substitution \displaystyle y = \frac{1}{u} \frac{\text{d} u}{\text{d} x} gives (detail omitted):

\displaystyle \frac{\text{d}^2u}{\text{d} x^2} + \frac{x}{1-x} \frac{\text{d} u}{\text{d} x} - \frac{1}{1-x} u = 0

to which "obvious" solutions are u = x, u = exp x, and so the general solution is u = Ax + Be^x. This gives:

\displaystyle \frac{1}{u} = \frac{1}{Ax + Be^x} ,\quad \frac{\text{d} u}{\text{d} x} = A + Be^x

\displaystyle \therefore y = \frac{A + Be^x}{Ax + Be^x} .

y(0) = 2:

\displaystyle 2 = \frac{A + B}{B} \Rightarrow A = B

\displaystyle \therefore y = \frac{A + Ae^x}{Ax + Ae^x} = \frac{1 + e^x}{x + e^x} .
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calcium878
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Alternative I/3. As inspirato for the beginning.

(Original post by insparato)
STEP I Question 3
 cos^4\theta - sin^4\theta = (cos^2\theta + sin^2\theta)(cos^2\theta - sin^2\theta) = cos2\theta (*)

 cos^4\theta + sin^4\theta = \frac{1}{4}(cos^22\theta + 2cos2\theta + 1) + \frac{1}{4}(cos^22\theta - 2cos2\theta +1)

 = \frac{1}{2}cos^22\theta + \frac{1}{2} = \frac{1}{2}(1-sin^22\theta) + \frac{1}{2}

 = 1 - \frac{1}{2}sin^22\theta (**)
Now, (*) + (**) gives 2cos^4\theta = cos2\theta + 1 - \frac{1}{2}sin^22\theta so that cos^4\theta = \frac{1}{2}(cos2\theta + 1 - \frac{1}{2}sin^2 2\theta)
And, (*)-(**) gives sin^4\theta = \frac{1}{2}(1 - \frac{1}{2}sin^2 2\theta - cos2\theta)

So, \displaystyle \int_0^{\frac{\pi}{2}} cos^4 \theta \text{d}\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} cos2\theta + 1 - \frac{1}{2}sin^2 2\theta \text{d}\theta

-\frac{1}{2}sin^2 2\theta = \frac{1}{4}cos4 \theta - \frac{1}{4}

So that \displaystyle \frac{1}{2} \int_0^{\frac{\pi}{2}} cos2\theta + \frac{1}{4}cos4\theta + \frac{3}{4} = \frac{1}{2} \left[\frac{1}{2}sin2\theta + \frac{3}{4}\theta + \frac{1}{16}sin4\theta\right]_0^{\frac{\pi}{2}}

= \frac{1}{2}([\frac{3\pi}{8}]-[0]) = \frac{3}{16}\pi

The one with sin^4 \theta is very nearly the same, just a sign change here and there (skipping the intermediate rearrangements):

\displaystyle \int_0^{\frac{\pi}{2}} sin^4 \theta \text{d}\theta = \frac{1}{2} \int_0^{\frac{\pi}{2}} -cos2\theta + \frac{1}{4}cos4\theta + \frac{3}{4} = \frac{1}{2} \left[-\frac{1}{2}sin2\theta + \frac{3}{4}\theta + \frac{1}{16}sin4\theta\right]_0^{\frac{\pi}{2}}

 = \frac{1}{2}([\frac{3\pi}{8}]-[0]) = \frac{3}{16}\pi

Then a traditional mash through the trig for the 6-th power ones, which I won't type up :p:
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insparato
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Yep i did it that way originally on paper. But i thought im going to have to do reduction formula anyways so might as well follow through. Btw I was going to post that but the Latex was a ***** anyways so i couldnt be bothered lol.
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calcium878
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I'm still 4 chapters away from the reduction formula in FP2 ;dry;
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insparato
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Get reading then :p:. I've posted two examples there .
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calcium878
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Yes, sir!
After STEP :ninja:
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eponymous
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Step I Q2

ok, so this is my first time EVER using latex , here goes...




ok, so i cant master latex , see fatuous philomath's post
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JeremyC
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#19
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(Original post by eponymous)
Step I Q2

ok, so this is my first time EVER using latex , here goes...






\tan ( A+B )= \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}= 1



\Rightarrow A+B = arc\tan(1) = \frac{\pi}{4}

now
\arctan \frac{1}{p} + \arctan\frac{1}{q} = \frac{\pi}{4}

so from the first part;

\frac{\frac{1}{p}+\frac{1}{q}}{1-\frac{1}{pq}}= 1

rearranging a bit; \frac{p+q}{pq}= 1-\frac{1}{pq}

pq-p-q-1=0



pq-p-q-1+2=2



(p-1)(q-1)=2

now, 2 = 1\times2 or -1\times-2

so (p-1) = 1 or 2 and (q-1) = 2 or 1

hence p = 2 or 3 and q = 3 or 2

(p-1) and (q-1) cannot be equal to -1, since this would make p or q equal to zero and p and q are non zero integers.
I think you mean...
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eponymous
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(Original post by fatuous_philomath)
I think you mean...
thats it!
thank you so much!!
i really can't get latex right
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