Pyoro
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Index of STEP solutions; 2010 STEP papers.
Thanks to miml for originally running this thread.
4 solutions left to fill.

STEP I
1 – Solution by jasbirsingh
2 – Solution by gcseeeman
3 – Solution by matt2k8
4 – Solution by miml
5 – Solution by miml
6 – Solution by jj193
7 – Solution by jj193
8 – Solution by miml
9 – Solution by matt2k8
10 – Solution by matt2k8
11 – Solution by brianeverit
12 – Solution by curious_lemma
13 – Solution by matt2k8, Get me off the £\?%!^@ computer

STEP II
1 – Solution by Unbounded, Salavant
2 – Solution by Unbounded, Farhan.Hanif93
3 – Solution by Unbounded, matt2k8
4 – Solution by Unbounded, matt2k8
5 – Solution by matt2k8
6 – Solution by jj193
7 – Solution by matt2k8
8 – Solution by ziedj
9 –
10 –
11 – Solution by gcseeeman
12 – Solution by curious_lemma
13 – Solution by metaltron

STEP III
1 – Solution by Unbounded
2 – Solution by matt2k8
3 – Solution by Unbounded
4 – Solution by Unbounded; correction by Dalek1099
5 – Solution by Get me off the £\?%!^@ computer
6 – Solution by Get me off the £\?%!^@ computer
7 – Solution by matt2k8
8 – Solution by matt2k8
9 – Solution by AnonyMatt
10 – Solution by Pyoro
11 –
12 – Solution by Pyoro
13 –
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ziedj
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SHOTGUN STEP II QUESTION 4, WAS THE SEXIEST (easiest) QUESTION
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boromir9111
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The second and third "here" don't open fully?
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miml
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STEP I - Q5
The binomial expansion required is

\displaystyle (1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r

i)
Let x = 1,

\displaystyle 2^n = \sum_{r=0}^n \binom{n}{r}


ii)
Differentiate both sides to get
\displaystyle n(1+x)^{n-1} = \sum_{r=0}^n r\binom{n}{r}x^{r-1}

Let x=1

\displaystyle n2^{n-1} = \sum_{r=0}^n r\binom{n}{r}


iii)
Integrate both sides over 0,1

\displaystyle \int_0^1(1+x)^ndx = \left [ \frac{(1+x)^{n+1}}{n+1} \right ]^1_0 = \frac{1}{n+1}(2^{n+1}-1)

and

\displaystyle \int_0^1 \sum_{r=0}^n \binom{n}{r}x^r dx = \frac{1}{r+1}\sum_{r=0}^n \binom{n}{r} (This is easier to see by writing out the sum explicitly)


iv)

Differentiate both sides twice

\displaystyle \frac{d^2}{dx^2}(1+x)^n = n(n-1)(1+x)^{n-2}

and

\displaystyle\frac{d^2}{dx^2}\sum_{r=0}^n \binom{n}{r}x^r = \sum_{r=0}^n r(r-1)\binom{n}{r}x^{r-2}

Let x=1,

\displaystyle n(n-1)2^{n-2} = \sum_{r=0}^n r(r-1)\binom{n}{r}

Adding this to the result in ii) gives

\displaystyle  n(n-1)2^{n-2}+ n2^{n-1}   = \sum_{r=0}^n r(r-1)\binom{n}{r} + \sum_{r=0}^n r\binom{n}{r}

\displaystyle  n2^{n-2}(n-1+2) = \sum_{r=0}^n (r^2-r+r)\binom{n}{r}

\displaystyle  n(n+1)2^{n-2} = \sum_{r=0}^n r^2\binom{n}{r}


STEP I - Q4

i)
 \displaystyle I = \int \frac{1}{\sqrt{x(x+1)}}dx

 \displaystyle x = \frac{1}{t^2 - 1} \implies \frac {dx}{dt} = \frac{-2t}{(t^2-1)^2}

 \displaystyle \therefore I = \int \frac{-2t}{(t^2-1)^2} \frac{1}{\sqrt{\frac{1}{t^2-1}(\frac{1}{t^2-1}+1)}}dt

  \displaystyle =\int \frac{-2t}{(t^2-1)^2} \frac{1}{\sqrt{\frac{1}{t^2-1}(\frac{t^2}{t^2-1})}}dt

 \displaystyle =\int \frac{-2t}{(t^2-1)^2} \frac{1}{\sqrt{\frac{t^2}{(t^2-1)^2}}}dt

 \displaystyle =\int \frac{-2t(t^2-1)}{t(t^2 - 1)^2}dt

 \displaystyle  =-2 \int \frac{1}{t^2 - 1} dt

\displaystyle  = -\ln\left | \frac{t-1}{t+1} \right | + c

\displaystyle  =\ln \left |\frac{\sqrt{\frac{x+1}{x}}+1}{\sqrt{\frac{x+1}{x}}+1} \right | + k

 \displaystyle = \ln \left |\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \right| + k

 \displaystyle = \ln \left| {(\sqrt{x+1}+\sqrt{x})^2}\right | + k

 \displaystyle =2\ln{(\sqrt{x+1}+\sqrt{x})} + k


ii)

 \displaystyle V = \pi \int_{\frac{1}{8}}^{\frac{9}{16}} \left( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+1}}\right)^2dx

 \displaystyle= \pi \int_\frac{1}{8}^\frac{9}{16} \frac{1}{x} - \frac{2}{\sqrt{x(x+1)}}+\frac{1}{x+1}dx

 \displaystyle=\pi \left [ \ln{x} - 4\ln{(\sqrt{x}+\sqrt{x+1})} + \ln{(x+1)} \right ]^{\frac{9}{16}}_{\frac{1}{8}}

 \displaystyle=\pi \left( \ln{\frac{9}{16}} - 4\ln (\frac{3}{4}+\frac{5}{4}) + \ln{\frac{25}{16}} \right ) - \pi \left( \ln{\frac{1}{8}} - 4\ln (\frac{1}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}) + \ln{\frac{9}{8}} \right )

 \displaystyle=\pi \left( \ln{\frac{9}{16}} - \ln{16} + \ln{\frac{25}{16}} - \ln{\frac{1}{8}} + 4\ln{\sqrt{2}} - \ln{\frac{9}{8}}\right )

 \displaystyle=\pi \left( \ln{\frac{9}{16}} + \ln{\frac{1}{16}} + \ln{\frac{25}{16}} + \ln{8} + \ln{4} + \ln{\frac{8}{9}} \right )

 \displaystyle=\pi \left( \ln{\frac{9 \times25 \times 8\times4 \times 8}{16 \times 16 \times 16\times 9}} \right )

 \displaystyle=\pi \ln{\frac{25}{16}}

 \displaystyle =2\pi\ln{\frac{5}{4}}


STEP I - Q8
i)
 a^3 = 9c^3 - 3b^3 = 3(3c^3 - b^3)

So we have \displaystyle 3|a^3 \Leftrightarrow 3|a

Now let  a = 3p

Our equation becomes

 27p^3 + 3b^3 = 9c^3 \Leftrightarrow b^3 = 3(c^3-9p^3)

And so  3|b^3 \Leftrightarrow 3|b

Now let b=3q
Our equation becomes

 27p^3 + 3(27)q^3 = 9c^3 \Leftrightarrow 3(p^3+3q^3) = c^3

And so  3|c^3 \Leftrightarrow 3|c

Therefore each of a,b, c must be divisible by 3.
We can write

 a = 3p ,  b = 3q ,  c = 3r

Our equation becomes

 27p^3 + 3(27)q^3 = 9(27)r^3 \Leftrightarrow p^3 + 3q^3 = 9r^3

Therefore each of p,q,r must also be divisible by 3.
We can carry on the previous argument repeatedly wlog
Therefore the numbers a,b,c must contain infinitely many factors of 3.
And so, a=b=c=0 QED


ii)
Consider quartic residues modulo 5
 0^4 \equiv 0(\mod{5})
 1^4 \equiv 1(\mod{5})
 2^4 \equiv 1(\mod{5})
 3^4(=81) \equiv 1(\mod{5})
 4^4(=256) \equiv 1(\mod{5})
(All numbers are congruent 0,1,2,3,4 modulo 5 so this is sufficient)

Therefore we must have
 p^4 \equiv 0(\mod{5}) or  p^4 \equiv 1(\mod{5})
and
 2q^4 \equiv 0(\mod{5}) or  2q^4 \equiv 2(\mod{5})

Also  5r^4 \equiv 0(\mod{5}) \forall r \in \mathbb{Z}

We must have p^4 + 2q^4 \equiv 0(\mod{5}) . Considering possible pairings of residues (0,0),(0,2),(1,0),(1,2), we must have both p and q divisible by 5.

We can write p=5a and  q = 5b
Our equation becomes

 5^4(a^4) + 2(5^4)b^4 = 5r^4 \Leftrightarrow 5^3(a^4+2b^4) = r^4

And so,  5|r^4 \Leftrightarrow 5|r

We can write  p=5a, q=5b,  r=5c

Our equation becomes
5^4(a^4) + 2(5^4)b^4 = 5^5(c^4) \Leftrightarrow a^4+2b^4 = 5c^4

Therefore each of a,b,c must also be divisible by 5.
We can carry on the previous argument repeatedly wlog
Therefore the numbers p,q,r must contain infinitely many factors of 5.
And so, p=q=r=0 QED
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miml
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(Original post by boromir9111)
The second and third "here" don't open fully?
They work for me, but I'll upload it onto mediafire.
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boromir9111
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(Original post by miml)
They work for me, but I'll upload it onto mediafire.
They work now. Thanks!
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Salavant
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Can I shotgun II 1 please? Second-easiest question on paper (4 being the easiest <3) so shouldn't involve much pain to type up.
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Siddd
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I can do Step I question 1 :P
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I/12

Hopefully I've made it clear enough, couldn't get latex working.

Spoiler:
Show
E(X) = 1*P(X=1) + 2*P(X=2) + 3*P(X=3) + ... = sum of r*P(X=r), where r ranges over the positive integers.

The given sum is equal to (P(X=1) + P(X=2) + P(X=3) + ...) + (P(X=2) + P(X=3) + ...) + (P(X=3) + ...) + ... which simplifies to E(X).

------------------------

Notice that P(X=>4) = 1 - P(X<=3), and that P(X=1) equals zero.

For X to equal 2, we can either have MD, or DM, which has a total probability of 2pq.

For X to equal 3, we can either have MMD, or DDM, which has a total probability of p^2q + pq^2 = pq(p+q) = pq(1) = pq.

Therefore P(X=>4) = 1 - (pq + 2pq) = 1 - 3pq = 1 - 3p(1-p)
= 1 - 3p + 3p^2 = (1 - 3p +3p^2 -p^3) + p^3 = (1-p)^3 + p^3 = p^3 + q^3

------------------------

Now notice that for n > 1, for X to take some value n, we must have DDD....M (n-1 Ds and 1 M) or MMM...D (n-1 Ms and 1 D). Therefore, P(X=n) = qp^(n-1) + pq^(n-1). If we sum this, using the formula for a GP, we find that P(X=>n) = p^(n-1) + q^(n-1).

Using the formula for E(X) which we derived earlier, we see that

E(X) = SUM[p^(r-1) + q^(r-1)] - 1, where r ranges over the positive integers.

(the -1 is to take account for the fact that the formula for P(X=>n) gives P(X=>1) as 2, when it should in fact be 1).

So E(X) = 1/(1-p) + 1/(1-q) -1 = 1/p + 1/q -1 = (p+q)/pq -1 = 1/pq -1

Now, since pq = p(1-p) = p - p^2 = 0.25 - (p - 0.5)^2, we can see that pq has a maximum value of 0.25, and so 1/pq has a minimum value of 4. From this it follows that E(X) => 3.
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Salavant
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OK, here we go, II 1, LaTeX isn't fantastic but that's because I'm still learning how to use LyX. (I really hope this is right, but I used a grapher, looks right, circle osculates nicely). Question 11 on II I also remember being easy but only if you can draw a diagram... hard to do on computer. I'd tackle a III, but would have to start from scratch on questions and it's a little late.
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Farhan.Hanif93
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(Original post by miml)
...
STEP II - Q2
cos(3x)
= \cos (2x+x)
= \cos (2x)\cos (x) - \sin (2x) \sin (x)
=  2\cos ^3(x) - \cos (x) - 2(1 - \cos ^2(x))\cos (x)
= 4\cos ^3(x) -3\cos (x) as required.

sin(3x)
= \sin (2x+x)
= \sin (2x)\cos (x) + \cos (2x)\sin (x)
= 2\sin (x) \cos ^2(x) + \sin (x) - 2\sin ^3(x)
= 3\sin (x) - 4\sin ^3(x) as required.

(i)
I(\alpha ) = \displaystyle\int ^{\alpha}_0 7\sin (x) - 8\sin ^3(x)dx = \displaystyle\int ^{\alpha}_0 2\sin (3x) + \sin (x) dx
= \left[-\frac{2}{3}\cos (3x) - \cos (x)\right]^{\alpha}_0

= -\frac{2}{3}\cos (3\alpha) - \cos (\alpha) + \frac{5}{3}

=-\frac{2}{3}(4\cos ^3(\alpha) - 3\cos (\alpha)) - \cos (\alpha) + \frac{5}{3}

Let c=\cos (\alpha):

I(\alpha) = -\frac{8}{3}c^3 + c + \frac{5}{3} as required.

Note that c=1 \implies I(\alpha) = 0.

(ii)
Considering the proper value of I(\beta):

If c=\cos (\beta) = -\frac{1}{6} \implies I(\beta) =\frac{8}{3} \times  \frac{1}{216} - \frac{1}{6} + \frac{5}{3} = \frac{245}{162}

Considering the value of I(\beta) using Eustace's method:

I(\beta) = \displaystyle\int ^{\beta}_0 7\sin (x) - 8\sin ^3(x) dx

=\left[\frac{7}{2}\sin ^2(x) - 2\sin ^4(x)\right]^{\beta}_0

 = \frac{7}{2}\sin ^2(\beta) - 2\sin ^4(\beta)

Note that we're considering where \cos (\beta) = -\frac{1}{6} \implies \sin (\beta) = \frac{\sqrt{35}}{6}:

 I(\beta) = \frac{7}{2}\times \frac{35}{36} - 2\left(\frac{35}{36}\right)^2 = \frac{245}{162} as required.

Considering all values of \alpha for which Eustace's method gives the right answer (Let c=\cos (\alpha)):
\implies \frac{7}{2}\sin ^2(\alpha) - 2\sin^4(\alpha)= -\frac{8}{3}\cos ^3(\alpha) + \cos (\alpha) + \frac{5}{3}

\implies \frac{7}{2}(1-c^2) - 2(1-2c^2+c^4) = -\frac{8}{3}c^3 + c + \frac{5}{3}

\implies 12c^4 - 16c^3 - 3c^2 + 6c +1 = 0

\implies (6c+1)(2c+1)(c-1)^2=0

\implies \cos (\alpha)  = -\frac{1}{6}, -\frac{1}{2}, 1

\implies \alpha = [2n\pi, (2n+1)\pi \pm \frac{\pi}{3}, (2n+1)\pi \pm \cos ^{-1}\left(\frac{1}{6}\right)] \forall n\in \mathbb{Z}.
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Farhan.Hanif93
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SHOTGUN STEP I - Q6 for tomorrow.
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jasbirsingh
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(Original post by miml)
...
STEP I - Q1

Spoiler:
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5x^2 + 2y^2 -6xy+4x-4y \equiv a(x-y+2)^2 + b(cx+y)^2+d

\textrm{Expanding } a(x-y+2)^2 + b(cx+y)^2+d
=a(x-y+2)(x-y+2)+b(cx+y)(cx+y) +d
=a(x^2-2yx+4x-4y+y^2+4)+b(c^2x^2+2cyx+y^2)+d
=(a+bc^2)x^2+(a+b)y^2+(2bc-2a)yx+4ax-4ay+(4a+d)

\textrm{Now, equating the coefficients to find a,b,c,d}


\textrm{Equating the coefficents of } x^2 \textrm{ gives}: a + bc^2 = 5
 \textrm{Equating the coefficents of } y^2 \textrm{ gives}: a+b = 2
 \textrm{Equating the coefficents of } xy \textrm{ gives}: 2bc -2a = -6
 \textrm{Equating the coefficents of } x \textrm{ gives} : 4a = 4
 \textrm{Equating the coefficents of } y \textrm{ gives} : -4a = -4
 \textrm{Equating the constants} : 4a+d=0

\textrm{Solving these equations simultaneously gives}
a=1,b=1,c=-2, d=-4

\textrm{Now, write the second equation in same form as first one.}

6x^2 + 3y^2 -8xy+8x-8y \equiv a(x-y+2)^2 + b(cx+y)^2+d

\textrm{Equate the coefficients to find a,b,c,d}


\textrm{Equating the coefficents of } x^2 \textrm{ gives}: a + bc^2 = 6
 \textrm{Equating the coefficents of } y^2 \textrm{ gives}: a+b = 3
 \textrm{Equating the coefficents of } xy \textrm{ gives}: 2bc -2a = -8
 \textrm{Equating the coefficents of } x \textrm{ gives} : 4a = 8
 \textrm{Equating the coefficents of } y \textrm{ gives} : -4a = -8
 \textrm{Equating the constants} : 4a+d=0


\textrm{Solving these equations simultaneously gives}
a=2,b=1,c=-2, d=-8

5x^2 + 2y^2 -6xy+4x-4y \equiv (x-y+2)^2 + (-2x+y)^2-4
6x^2 + 3y^2 -8xy+8x-8y \equiv 2(x-y+2)^2 + (-2x+y)^2-8

5x^2 + 2y^2 -6xy+4x-4y = 9
(x-y+2)^2 + (-2x+y)^2-4 =9
(x-y+2)^2 + (-2x+y)^2 =13 ... (i)

x^2 + 3y^2 -8xy+8x-8y = 9
 2(x-y+2)^2 + (-2x+y)^2-8 =14
2(x-y+2)^2 + (-2x+y)^2 =22 ...(ii)

\textrm{Subtracting (ii) from (i) gives}
(x-y+2)^2 = 9
\textrm{So} x-y+2 = 3 ....(1) [latex]

[latex]\textrm{So} x-y+2 = -3 ....(2) [latex]



[latex]2(i) - (ii) /textrm{gives}
(-2x+y)^2 = 4
-2x+y = 2 ... (3)
-2x+y = -2 ...(4)

\textrm{Solving (1) and (3) gives x=-3 and y=-4}
\textrm{Solving (1) and (4) gives x=1 and y=0}
\textrm{Solving (2) and (3) gives x=3 and y=8}
\textrm{Solving (2) and (4) gives x=7 and y=12}

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gcseeeman
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I/2

y = \left(\dfrac{x-a}{x-b}\right)e^x
Let u = \dfrac{x-a}{x-b} \Rightarrow \frac{du}{dx} = \dfrac{a-b}{(b-x)^2}
Let v = e^x \Rightarrow \frac{dv}{dx} = e^x

\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} =  \dfrac{e^x\left[ x^2 - (a+b)x + (a + ab - b)\right]}{(x-b)^2}

Stationary points when dy/dx = 0. e^x > 0 for all x, so we need the quadratic to = 0 i.e. statisfy the condition for 2 real roots:

\alpha = 1
\beta = -(a+b)
\gamma = (a + ab - b)

2 distinct real roots if \beta^2 &gt; 4 \alpha \gamma  \Rightarrow (a+b)^2 &gt; 4 (a + ab -b) \Rightarrow a^2 + b^2 - 2ab &gt; 4(a-b) \Rightarrow (a-b)^2 &gt; 4 (a-b)

This inequality holds if a-b &lt; 0 or  a - b &gt; 4 as required.
i)
As above, stationary points when
x^2 - (a+b)x + (a + ab - b) = 0

Subbing in a and b gives:

x^2 - \frac{x}{2} -\frac{1}{2} = 0
(x - \frac{1}{4})^2  = \frac{9}{16}

x = \frac{1}{4} \pm \sqrt{\frac{9}{16}} = 1 or  -0.5 (hmm probably would have been easier to factorise)

Asymptote when x - b = 0 ==> x = 1/2 so these are on either side, as required.

Graph (in Wolfram Alpha):
Image
ii)
Similar working for ii) leads to x = 1.5 or 3 (both on the same side of the asymptote at x = 0), and to the graph below:
Graph (in Wolfram Alpha):
Image


II/11

[See Diagram Below]

Resolving Horizontally:

R\cos (\alpha + \theta) = T\cos (\beta - \theta) (*)

Taking Moments about the COM:

2l \times R\sin\alpha = l \times T\sin\beta

\Rightarrow T = \dfrac{2R\sin\alpha}{\sin\beta} (**)

Substituting (**) into (*), cancelling R, and expanding using double angle formulae:

R\cos (\alpha + \theta) = \dfrac{2R\sin\alpha}{\sin\beta} \cos (\beta - \theta)
(\cos\alpha \cos\theta - \sin\alpha \sin\theta) \sin\beta=2(\cos\beta \cos\theta + \sin\beta \sin\theta) \sin\alpha

Factoring out sin and cos alpha e.t.c:

\cos\alpha (\cos\theta \sin\beta) = \sin\alpha (2 \cos\beta \cos\theta + 3 \sin\beta \sin\theta)

\cot\alpha = \dfrac{2 \cos\beta \cos\theta + 3 \sin\beta \sin\theta}{\cos\theta \sin\beta} = \boxed {2 \cot\beta + 3 \tan \theta} as required.

There may be (probably is) a better way to do this last part, but hopefully just plugging in the values given and expanding tan (45 - 30) for tan(15) is at least OK to show it. i.e.

\frac{1}{\tan\alpha} = \frac{2}{\tan 45} + 3 \tan 30 = 2+ \sqrt{3} \Rightarrow \tan\alpha = 2 - \sqrt{3}

\tan 15 = \tan (45 - 30) = \dfrac{\tan 45 - \tan 30}{1 + \tan 45 \tan 30} = \dfrac{3 - \sqrt{3}}{3+\sqrt{3}} = 2 - \sqrt{3}

\therefore \alpha = 15 (as alpha < 180)
Diagram(ish)
Image


Hopefully there are no typos (or mistakes).
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matt2k8
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I'm about to do II- q4
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matt2k8
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II/3

i) using values for F_0, F_1, F_2,F_3 we have
a + b = 0 From F_0 so  b = -a
a\lambda +b \mu = 1 from F_1
a\lambda^2 + b\mu^2 = 1 fom F_2
a\lambda^3 + b\lambda^3 = 2 from F_3

so using the first one we actually have

a(\lambda-\mu) =1
a(\lambda^2 - \mu^2) = 1
a(\lambda^3 - \lambda^3)=2

and the last one factorises to

a(\lambda-\mu)(\lambda^2 + \mu\lambda + \mu^2) = 2
so using the first, \lambda^2 + \mu\lambda + \mu^2 = 2 as required. (*)

Also factorising the second one, then using the first one means that
\lambda + \mu = 1

Substituting this into (*) and some other algebra leads to
 a= \frac{1}{\sqrt5}, b = \frac{-1}{\sqrt5}, \lambda=\frac{1+\sqrt5}{2}, \mu = \frac{1-\sqrt5}{2}

ii) substitute in then use binomial theorem gives F_6 = 8

iii) using sum of GP formula, then some algebra gives the required sum as 1.


II/4

i. Let u = a-x, so du = -dx
so
I = \int^0_a \frac{f(a-u)}{f(a-u)+f(u)}(-du)
= \int^a_0 \frac{f(a-x)}{f(x)+f(a-x)}dx
using the facts that \int^a_b = -\int^b_a and that x and u are both dummy variables.

adding these the 2 expressions for I gives 2I = \int^a_0 dx so I=a/2

First integral:
2+x-x^2 = (x+1)(2-x), so ln(2+x-x^2) = ln(x+1) + ln(2-x) = ln(x+1) + ln(1 - (x+1))

So the integral is equivalent to
\int^1_0 \frac{ln(x+1)}{ln(x+1)+ln(1-(x+1))}dx

so we can use the above result with a = 1, f(x) = ln(x+1) and find that the first integral has the value 1/2.

Second integral:
sin(x+\pi/4) = \frac{1}{\sqrt2}(sin x + cos x) using compound angle formulae

and  cos x = sin(\frac{\pi}{2} - x)
so we can write the second integral as

\sqrt2 \int^{\frac{\pi}{2}}_0 \frac{sin x}{sin x + sin(\frac{\pi}{2} - x)}

so we can use the above result with f(x) = sin x and a=\frac{\pi}{2} to find that the integral is equal to \frac{\sqrt2 \pi}{4}.

ii)
Let J = \int^2_{\frac{1}{2}} \frac{sin x}{x(sin x+sin(\frac{1}{x}))} dx

Set x = 1/u, then some re-arranging, and the fact that u and x are just dummy variables to find that also,
J = \int^2_{\frac{1}{2}}\frac{sin \frac{1}{x}}{x(sin x+sin(\frac{1}{x}))} dx

So we can add the 2 expressions for J to find that

2J = \int^2_{\frac{1}{2}} \frac{1}{x} dx
so  J = ln 2
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miml
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(Original post by matt2k8)
I'm about to do II- q4
(Original post by ziedj)
Won't be happy...
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matt2k8
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Woops didn't see that post haha
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ziedj
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(Original post by matt2k8)
Woops didn't see that post haha
You saved me a lot of typing, no worries
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curious_lemma
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II/12

I did this in a hurry so there might be a few mistakes!
Spoiler:
Show


Since f is a probability density function, the integral of f(x) between 0 and 1 will be equal to 1. This means that [INT(a) between 0 and k] + [INT(b) between k and 1] will equal one. Carrying out the integration and rearranging, we find that k = (1-b)/(a-b).

Now k > 0 => (1-b)/(a-b) > 0 => 1-b > 0 => b < 1
and k < 1 = (1-b)/(a-b) < 1 => 1-b < a-b => a > 1

------------------------------

E(X) = [INT(ax) between 0 and k] + [INT(bx) between k and 1]
= (ak^2)/2 + b(1 - k^2)/2

and if we substitute in our expression for k, factorise/expand/cancel factors, we arrive at E(X) = (1 - 2b + ab)/2(a-b) which is what we wanted.

------------------------------

We have have two possible scenarios: either m < k, or m > k. Consider the first: we must have [INT(a) between 0 and k] > 0.5, and so ak < 0.5. If we substitute in for k and rearrange we see that this implies that a+b > 2ab. To find m in this case we must solve

[INT(a) between 0 and m] = 0.5
i.e. am = 0.5, and so m = 1/(2a)

The second possible situation, where m > k, occurs when [INT(a) between 0 and k] < 0.5. Carrying out the integration gives the condition that a+b < 2ab. We can now solve for m as before:

[INT(a) between 0 and k] + [INT(b) between k and m] = 0.5
=> ak + bm - bk = 0.5
=> k(a-b) + bm = 0.5
=> 1 - b + bm = 0.5
=> b(m-1) = -0.5
=> m = 1 - 1/(2b)

-------------------------------

Suppose first that m = 1/(2a). Then, after some rearranging and factorising, E(X) - m = (1 - 2b + ab)/2(a-b) - 1/(2a) = (b(a-1)^2)/2(a-b) > 0. Therefore E(X) > m.

Now suppose that m = 1 - 1/(2b). Then E(X) - m = (1 - 2b + ab)/2(a-b) - 1 + 1/(2b) = (a(b-1)^2)/2(a-b) > 0, and so E(X) > m.

Thus, whatever the value of m, E(X) > m.
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