STEP 2013 Solutions Watch

Slowbro93
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Unofficial STEP Solutions 2013

Here is a thread of all of the STEP Solutions (I, II and III) that have been created by TSR users. I will update (or the relevant mod) when new solutions are posted

STEP I (Sat 25th June 2013)

Question 1: cpdavis
Question 2: mikelbird
Question 3: FJacob
Question 4: Pyoro, cpdavis
Question 5: mikelbird
Question 6: mikelbird
Question 7: Mastermind2
Question 8: metaltron
Question 9: DJMayes
Question 10: DJMayes
Question 11: DJMayes
Question 12: HCIO
Question 13: DFranklin

STEP II (Sat 19th June 2013)

Question 1: mikelbird
Question 2: Smaug123 Alternative to part ii - Fuzzy12345
Question 3: mikelbird Part ii without considering turning points: Nebula
Question 4: jack.hadamard
Question 5: cpdavis
Question 6: metaltron
Question 7: DJMayes
Question 8:MW24595
Question 9: LShirley95
Question 10: DJMayes
Question 11: DJMayes
Question 12: Smaug123
Question 13: Nebula

STEP III (Sat 26th June 2013)

Question 1: Pyoro
Question 2: borealis72
Question 3: jack.hadamard
Question 4: IrrationalNumber, Blutooth
Question 5: jack.hadamard
Question 6: jack.hadamard
Question 7: DJMayes
Question 8: jack.hadamard
Question 9: Brammer
Question 10: DJMayes
Question 11: bananarama2
Question 12: Blutooth, ukdragon37 - only part iic, MAD Phil
Question 13: DFranklin

STEP I Paper: http://www.thestudentroom.co.uk/show...postcount=1205
STEP II Paper: http://www.thestudentroom.co.uk/show...&postcount=671
STEP III Paper: http://www.thestudentroom.co.uk/show...postcount=1384
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LShirley95
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You wouldn't happen to have the full paper would you?
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Slowbro93
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(Original post by LShirley95)
You wouldn't happen to have the full paper would you?
Yep, thanks Jack! http://www.thestudentroom.co.uk/show...&postcount=672
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Smaug123
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Question 2:
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I_n = \int_0^1 x^n (1-x)^n dx
i) Substitute u = 1-x:
\int_0^1 x^{n-1}(1-x)^n dx = \int_1^0 (1-u)^{n-1} u^n (-1) du = \int_0^1 (1-x)^{n-1} x^n dx, as required.
Now,  2 \int_0^1 x^{n-1} (1-x)^n dx = \int_0^1 x^{n-1} (1-x)^n dx + \int_0^1 x^{n-1}(1-x)^n dx = \int_0^1 x^{n-1} (1-x)^n dx + \int_0^1 (1-x)^{n-1} x^n dx by the first part; this is  \int_0^1 x^{n-1}(1-x)^n + (1-x)^{n-1}x^n dx = \int_0^1 x^{n-1}(1-x)^{n-1}(1-x+x)dx = I_{n-1} as required.
We then integrate by parts: u=x^n, \frac{dv}{dx} = (1-x)^n, \frac{du}{dx} = n x^{n-1}, v = -\frac{(1-x)^{n+1}}{n+1}, so I_n = [x^n \cdot \frac{(1-x)^{n+1}}{n+1}]_0^1 - \int_0^1 -\frac{(1-x)^{n+1}}{n+1} \cdot n x^{n-1} dx, which can easily be seen to reduce to the correct formula.

Since:
x^{n-1}(1-x)^{n+1} = x^{n-1}(1-x)^n(1-x) = x^{n-1}(1-x)^n-x^n(1-x)^n
then:
\int_0^1 x^{n-1}(1-x)^{n+1} dx = \int_0^1 x^{n-1}(1-x)^n dx - \int_0^1 x^n(1-x)^n dx = \frac{I_{n-1}}{2} -I_n
Hence:
I_n = \frac{n}{n+1}(\frac{I_{n-1}}{2} -I_n ) and the result follows by rearrangement.

ii) We proceed inductively: it is easy to verify that I_1 = \frac{(1!)^2}{(2 \times 1 + 1)!}. Then I_{n+1} = \frac{n+1}{2(2n+3)} I_n = \frac{n+1}{2(2n+3} \cdot \frac{(n!)^2}{(2n+1)!} = \frac{(n+1) (2n+2) (n!)^2}{2(2n+3)!} = \frac{(n+1)!^2}{(2n+3)!} as required.

iii) I_{\frac12} = \int_0^1 x^{\frac12}(1-x)^{\frac12} dx. Substitute x = \sin^2(\theta), \frac{dx}{d\theta} = \sin(2\theta), so \int_0^1 x^{\frac12}(1-x)^{\frac12} dx = \int_0^{\frac{\pi}{2}} \sin(\theta) \cos(\theta) \cdot \sin(2\theta) d\theta = \frac12 \int_0^{\frac{\pi}{2}} (\sin(2\theta))^2 d\theta; then we use \cos(4\theta) = 1-2 \sin^2(2\theta) to obtain I_{\frac12} = \frac{\pi}{8}.
Then we use the first part: I_{3/2} = \frac{3/2}{2(2 \cdot \frac{3}{2} + 1)} I_{1/2}, which can easily be seen to be \frac{3 \pi}{128}.
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Fuzzy12345
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Alternatively to Q2 ii
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We have:

\displaystyle I_{n} = \frac{n}{2(2n+1)} I_{n-1} so it follows that \displaystyle I_{n-1} = \frac{n-1}{2(2n-1)} I_{n-2} and so forth, thus:

\displaystyle I_{n} = \frac{n}{2(2n+1)} \cdot \frac{n-1}{2(2n-1)} \cdots \frac{2}{2 \cdot 5} I_{1} and I_{1} = \dfrac{1}{3 \cdot 2} hence:

\displaystyle I_{n} = \frac{n}{2(2n+1)} \cdot \frac{n-1}{2(2n-1)} \cdots \frac{2}{2 \cdot 5} \cdot \frac{1}{2 \cdot 3} = \frac{n!}{2^{n} \cdot \prod_{r=1}^{n} (2r+1)}

\displaystyle \prod_{r=1}^{n} (2r+1) = 3 \cdot 5 \cdot 7 \cdots (2n+1) = \frac{(2n+1)!}{2^{n} \cdot n!}

so:

I_{n} = \dfrac{(n!)^{2}}{(2n+1)!}
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DJMayes
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Q7 – STEP II

Part i):
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x=1, y=0 is a nice obvious solution.

 (3p+4q)^2-2(2p+3q)^2 = 1

 (9p^2+12pq+16q^2)-(8p^2+12pq+18q^2)=1

 p^2-2q^2 = 1

Hence the second solution reduces to the first and is therefore a solution if the first is a solution. Using this, we obtain x=3, y=2 as another solution.


Part ii):
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If x is odd and y is even, then write  x=2m+1 and  y = 2n :

 (2m+1)^2-2(2n)^2=1

 4m^2+4m+1-8n^2=1

 4m^2+4m = 8n^2

 n^2 = \frac{1}{2}m(m+1)



Part iii):

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 b^3=c^4-a^2

 b^3 = (c^2-a)(c^2+a)

Now, b is a prime number meaning that  b^3 has four possible factors – 1, b,  b^2 and  b^3 . Let’s examine each of these in turn:

 c^2-a = b^3 ,  c^2+a = 1

This solves to get  2a = 1-b^3 which is not possible as a and b are both positive integers.

 c^2-a = 1 ,  c^2+a = b^3

 c^2 = \frac{1}{2}(b^3+1) ,  a = \frac{1}{2}(b^3-1)

 c^2-a = b^2 ,  c^2+a = b

This solves for  2a = b –b^2 which is not possible as a and b are both positive integers, yet b squared is greater than b implying that a = 0.

 c^2-a = b ,  c^2+a = b^2

 c^2 = \frac{1}{2}b(b+1) ,  ] a = \frac{1}{2}b(b-1)

Now, our final part – we need to find integer solutions of these equations for a, b and c where b is not prime. The easiest place to start is with b and c. From our final solution, we are after a square number which, when doubled, gives us the product of two consecutive numbers. Trying the obvious ones (1-10) and doubling we find that if c = 6, 2c^2 = 72 = 8x9, giving b=8 as a non prime solution, which is what we want. Finally, plugging in b to the equation with a yields a = 28, so a possible set of integer solutions is a=28, b=8, c=6.
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DJMayes
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Q10 - STEP II:

Part i):

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Let U be the speed of projection of the particle, x and y be the horizontal and vertical displacements of the particle, with  V_h as the horizontal velocity of the particle and  V_v as the vertical velocity of the particle. Then, we can say:

 V_h = ucos \alpha ,  V_v = usin\alpha - gt

 tan \phi = \frac{V_v }{V_h } = \frac{usin\alpha - gt}{ucos \alpha }

 = tan \alpha - \frac{gt}{ucos \alpha }

Similarly:

 x = ucos \alpha t ,  y = u sin \alpha t - \frac{1}{2} gt^2

 tan \theta = \frac{y}{x} = \frac{u sin \alpha t - \frac{1}{2} gt^2 }{ucos \alpha t }

 = tan \alpha - \frac{gt}{2ucos \alpha }

 2 tan \theta = tan \alpha +  tan \alpha - \frac{gt}{ucos \alpha }

 2 tan \theta = tan \alpha + tan \phi , as required.



Part ii):

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 2tan\frac{1}{2} \alpha =  tan \alpha +tan\frac-{1}{2} \alpha

 3tan\frac{1}{2} \alpha = tan \alpha

 3tan\frac{1}{2} \alpha = \dfrac{2tan\frac{1}{2} \alpha}{1-tan^2\frac{1}{2} \alpha }

Now, we can divide across by tan of a half alpha, as it's not 0 (Not fired horizontally and multiply through to get:

 3 - 3tan^2\frac{1}{2} \alpha = 2

 3tan^2\frac{1}{2} \alpha = 1

Which solves to give alpha = 60, with the positive root taken as it's an acute angle of elevation.



Part iii):

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Let  t_1 be the time taken to reach B, and  t_2 be the time taken to reach C. Let's rewind:

 tan \phi = tan \alpha - \frac{gt}{ucos \alpha }

 tan \theta =  tan \alpha - \frac{gt}{2ucos \alpha }

Letting phi and theta be the respective angles at B and C, we get:

  \frac{gt_1 }{ucos \alpha } = 2( tan \alpha - tan\frac{1}{2} \alpha )

 \frac{gt_2 }{ucos \alpha } = tan \alpha + tan\frac{1}{2} \alpha

If the particle reaches C first, then  t_2 < t_1 :

 tan \alpha + tan\frac{1}{2} \alpha < 2( tan \alpha - tan\frac{1}{2} \alpha )

 3tan\frac{1}{2} \alpha <  tan \alpha

 3tan\frac{1}{2} \alpha(1-tan^2\frac{1}{2} \alpha ) < 2tan\frac{1}{2} \alpha

 3 - 3tan^2\frac{1}{2} \alpha < 2

 3tan^2\frac{1}{2} \alpha > 1

And this then gives alpha > 60 degrees, which means it reaches B first.

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Q11 – STEP II
Part i):
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Let A, B and C be the three particles (with mass m), and let the speed of the particles after respective collisions be denoted by  V_a ,  V_b and  V_c , respectively. Now, the first collision:

Conservation of Linear Momentum:

 mu = m V_a + m V_b

 u = V_a + V_b

Newton’s Law of Restitution:

 e = \dfrac{ V_b - V_a}{u}

 eu = V_b - V_a

Adding together, we get:

 V_b = \frac{1}{2}u(1+e)

And subtracting we get:

 V_a = \frac{1}{2}u(1-e) (Which is the speed we were after!)

Next, we need to look at the second collision, between B and C. You could plug this all in again, but the second collisions is actually identical to the first but with u replaced by  V_b . Plugging this in then immediately yields the speeds of the other two particles.
Now, for a third collision we require  V_a > V_b :

 \frac{1}{2}u(1-e) > \frac{1}{4}u(1-e^2)

Divide across by u and 1-e (Fine as neither are negative or 0) and multiply by 4:

 2 > 1+e

Which then gives the inequality e<1 which we are given, so another collision must occur.


Part ii):

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Firstly, we need to consider the third collision. Let Va and Vb be the velocities after the collision, similarly to before:

Conservation of linear momentum:

 \frac{1}{2}mu(1-e)+ \frac{1}{4}mu(1-e^2) = mV_a + mV_b

 \frac{1}{2}u(1-e)+ \frac{1}{4}u(1-e^2) = V_a + V_b

Newtons Law of restitution:

 e[\frac{1}{2}u(1-e) - \frac{1}{4}u(1-e^2)] = V_b – V_a

We are only interested in the velocity of B, so add them together:

 2V_b = \frac{1}{2}u(1-e)+ \frac{1}{4}u(1-e^2) +\frac{1}{2}eu(1-e) - \frac{1}{4}eu(1-e^2)

 2V_b = \frac{1}{2}u(1-e)(1+e)  +\frac{1}{4}u(1-e)^2(1+e)

Don’t try expand or simplify this yet. For a fourth collision, we require Vb > Vc so 2Vb > Vc:

 \frac{1}{2}u(1-e)(1+e)  +\frac{1}{4}u(1-e)^2(1+e) &gt; \frac{1}{2}u(1+e)^2

Now, divide across by 1+e and u and multiply by 4, which are all fine due to non-zero and non-negativity:

 2(1-e) + (1-e)^2 &gt; 2(1+e)

 2 – 2e +1-2e+e^2 &gt; 2+2e
 e^2 -6e+1 &gt; 0

This quadratic has roots of  3 \pm 2\sqrt{2} , and looking at the graph of the quadratic (And knowing e cannot be greater than one) we get the inequality  e &lt; 3 – 2\sqrt{2}

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LShirley95
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Got Q9 written with diagrams, unfortunately don't have a scanner so this is the best I can do:
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In case anyone wants a higher res version of the (II) paper to work from:

http://www.thestudentroom.co.uk/show...&postcount=829
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13 (i)
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A=1 happens after HH of TT, HT of TH would give A \geq 1
P(A=1)=P(HH)+P(TT)=p^2+q^2
S=1 happens after HT of TH:
P(S=1)=P(HT)+P(TH)=pq+qp=2pq

Now, p \neq q so p-q \neq 0 then:
(p-q)^2 &gt; 0
p^2 + q^2 - 2pq &gt; 0
p^2 + q^2 &gt; 2pq
P(A=1) &gt; P(S=1)

13 (ii)
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A=2 happens after HTT or THH.
P(A=2)=P(HTT)+P(THH)
=pq^2+qp^2
S=2 happens after HHT or TTH.
P(S=2)=P(HHT)+P(TTH)
=pq^2+qp^2
=P(A=2)

A=3 happens after THTT or HTHH.
P(A=3)=P(THTT)+P(HTHH)
=pq^3+qp^3
S=3 happens after HHHT or TTTH.
P(S=3)=P(HHHT)+P(TTTH)
=pq^3+qp^3
=P(A=3)

13 (iii)
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A=2n happens after (n*(HT))T or (n*(TH))H.
P(A=2n)=P((n^*(HT))T)+P((n^*(TH)  )H)
=p(pq)^n+q(pq)^n
S=2n happens after (n*(HH))T or (n*(TT))H.
P(S=2n)=P((n^*(HH))T)+P((n^*(TT)  )H)
=pq^{2n}+qp^{2n}
Then we note pq>0 and p^n-q^n and p^{n-1}-q^{n-1} are not equal to 0 since and have the same sign since p-q \neq 0, \text{and}\: n-1&gt;0.
pq(p^n-q^n)(p^{n-1}-q^{n-1})&gt;0
(pq^{2n}+qp^{2n})-(p(pq)^n+q(pq)^n)&gt;0
P(S=2n)-P(A=2n)&gt;0
P(S=2n)&gt;P(A=2n)

A=2n+1 happens after (n*(HT))HH or (n*(TH))TT.
P(A=2n+1)=P((n^*(HT))HH)+P((n^*(  TH))TT)
=p^2(pq)^n+q^2(pq)^n
S=2n+1 happens after (n*(HH))HT or (n*(TT))TH.
P(S=2n+1)=P((n^*(HH))HT)+P((n^*(  TT))TH)
=pq^{2n+1}+qp^{2n+1}
Then we note pq>0 and p^{n+1}-q^{n+1} and p^{n-1}-q^{n-1} are not equal to 0 since and have the same sign since p-q \neq 0, \text{and}\: n-1&gt;0. (p&gt;q \rightarrow p^m&gt;q^m when m>0)
pq(p^{n+1}-q^{n+1})(p^{n-1}-q^{n-1})&gt;0
(pq^{2n+1}+qp^{2n+1})-(p^2(pq)^n+q^2(pq)^n)&gt;0
P(S=2n+1)-P(A=2n+1)&gt;0
P(S=2n+1)&gt;P(A=2n+1)

----
[note: as you may expect, both the odd and even cases have the same result]



Alternate to 3ii without considering turning points:
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The cubic can be factorised as (x-p)(x-q)(x-r) with p,q,r&gt;0
then:
a=-\frac{1}{3}(p+q+r)
b=\frac{1}{3}(pq+qr+rp)
c=-pqr
from which a<0, b>0, c<0 immediately follow.
a^2-b=\frac{1}{9}(p+q+r)^2-\frac{1}{3}(pq+qr+rp)
=\frac{1}{18}(2p^2+2q^2+2r^2-2pq-2qr-2rp)
=\frac{1}{18}((p-q)^2+(q-r)^2+(r-p)^2)
which is positive since is is a sum of 3 non-zero squares times a positive number.
a^2-b&gt;0
a^2&gt;b
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Smaug123
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Q12:

i)
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\mathbb{E}(X) = \sum_{i=0}^{\infty} \mathbb{P}(X=i) i, since the Poisson distribution takes nonnegative integer values.
Now, \mathbb{P}(X=i) is 0 for even i, and e^{-\lambda} \frac{\lambda^i}{i!} otherwise, so the required expectation is \sum_{i=1, i odd}^{\infty} i e^{-\lambda} \frac{\lambda^i}{i!} = \lambda e^{-\lambda} \sum_{i=1, i odd}^{\infty} \frac{\lambda^{i-1}}{(i-1)!} = \lambda e^{-\lambda} \sum_{i=0, i even}^{\infty} \frac{\lambda^{i}}{i!} = \lambda e^{-\lambda} \alpha.
And since X+Y = U, and \mathbb{E}(U) = \lambda from the expectation of the Poisson distribution, and since the expectation of the sum is the sum of the expectations, we have \mathbb{E}(Y) = \lambda - \lambda e^{-\lambda} \alpha = \lambda e^{-\lambda} (e^{\lambda} - \alpha) = \lambda e^{-\lambda} \beta.


ii)
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The variance of X is "the mean of the squares - the square of the mean". The mean is \lambda e^{-\lambda} \alpha, which squares to \lambda^2 e^{-2 \lambda} \alpha^2. Note that \alpha+\beta = e^{\lambda}, and hence the second term in the expression for the variance is precisely "the square of the mean".
So we want "the mean of the squares" to be \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}.
Note that the denominator is e^{\lambda}.
Now, \mathbb{E}(X^2) = \sum_{i=0, i odd}^{\infty}i^2 e^{-\lambda} \frac{\lambda^i}{i!} = \lambda e^{-\lambda} \sum_{i=0, i odd}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!}.
We can see that a factor \lambda e^{-\lambda} = \frac{\lambda}{\alpha+\beta} has appeared, so we just need to show that \alpha + \lambda \beta = \sum_{i=0, i odd}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!}.
But this is certainly true: \sum_{i=0, i odd}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!} = \sum_{i=0, i even}^{\infty}(i+1) \frac{\lambda^i}{i!} = \sum_{i=0, i even}^{\infty}\frac{\lambda^i}{(  i-1)!} + \sum_{i=0, i even}^{\infty} \frac{\lambda^i}{i!} = \alpha+\lambda \beta as required.


Expression for Var(Y):
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We repeat the process: \mathbb{E}(Y) = \lambda e^{-\lambda} \beta, so \mathbb{E}(Y)^2 = \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2}.
Then \displaystyle \mathbb{E}(Y^2) = \sum_{i=0, i even}^{\infty} i^2 e^{-\lambda} \frac{\lambda^i}{i!}, which is e^{-\lambda} \lambda \sum_{i=2, i even}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!}.
This is e^{-\lambda} \lambda \sum_{i=1, i odd}^{\infty}(i+1) \frac{\lambda^i}{i!} = e^{-\lambda} \lambda \sum_{i=1, i odd}^{\infty} \frac{\lambda^i}{(i-1)!} + e^{-\lambda} \lambda \sum_{i=1, i odd}^{\infty} \frac{\lambda^i}{i!} = e^{-\lambda} \lambda (\lambda \alpha + \beta) = \frac{\lambda^2 \alpha + \lambda \beta}{\alpha + \beta}.
Hence the variance is \frac{\lambda^2 \alpha + \lambda \beta}{\alpha + \beta} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2}.


Do non-zero \lambda exist?
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Note that Var(X)+Var(Y) = \frac{\lambda \alpha+\lambda^2 \beta}{\alpha+\beta} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} + \frac{\lambda^2 \alpha + \lambda \beta}{\alpha + \beta} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} = \frac{\alpha (\lambda + \lambda^2) + \beta (\lambda + \lambda^2)}{\alpha + \beta} - \frac{\lambda^2}{(\alpha+\beta)^  2}(\alpha^2+\beta^2) = \lambda + \lambda^2 - \lambda^2 \frac{\alpha^2 + \beta^2}{(\alpha+\beta)^2},
while Var(X+Y) = Var(U) = \lambda.
So we require \lambda = \lambda+\lambda^2 - \lambda^2 \frac{\alpha^2+\beta^2}{(\alpha+  \beta)^2}, and hence \lambda = 0 or \frac{\alpha^2 + \beta^2}{(\alpha+\beta)^2} = 1 and so 2 \alpha \beta = 0; this is clearly impossible, since \alpha &gt; 0 and \beta = 0 only for \lambda = 0 (by considering -\frac{\lambda}{1!} = \frac{\lambda^3}{3!} + \frac{\lambda^5}{5!} \dots, but dividing through by \lambda gives a sum of squares on the RHS but a negative on the LHS). Hence there are no such \lambda, since we do not define a Poisson variable for \lambda = 0.
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Question 6:

Part i)

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\displaystyle  u_{n+2} = 1 + \frac{1}{u_{n+1}} = 1 + \frac{1}{1+\frac{1}{u_n}} = 1 + \frac{u_n}{u_n +1}

Similarly:

 \displaystyle u_n = 1 + \frac{u_{n-2}}{u_{n-2}+1}

 \displaystyle u_{n+2} - u_n = \frac{u_n}{u_n +1} - \frac{u_{n-2}}{u_{n-2}+1} =  \frac{u_n(u_{n-2} + 1)-u_{n-2}(u_n+1)}{(1+u_n)(1+u_{n-2})} = \frac{u_n - u_{n-2}}{(1+u_n)(1+u_{n-2})}



ii)

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When n=1, u_1 = 1 which satisfies the inequality. Inductive step:

 \displaystyle 1 \leq u_k \leq 2

Now:

 \displaystyle u_{k+1} = 1 + \frac{1}{u_k}

 \displaystyle u_k = \frac{1}{u_{k+1}-1}

Using our inductive step:

 \displaystyle 1 \leq \frac{1}{u_{k+1}-1} \leq 2

 \displaystyle 1 \geq u_{k+1} - 1 \geq \frac {1}{2}

 2 \geq u_{k+1} \geq \frac{3}{2} \geq 1

Hence, if u_k satisfies the inequality so does u_(k+1) and since n=1 satisfies the inequality, by induction, so do all natural numbers n.


iii)

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The two required series converge. This can be seen by considering:

 \displaystyle  \frac{u_n - u_{n-2}}{(1+u_n)(1+u_{n-2})}

From this since the bottom of the fraction is always positive, both the numerator and the LHS have the same sign. It can be shown therefore by computing a few terms that the series with odd n is increasing and the series with even n is decreasing. Hence, the series of odd n is increasing and is bounded above by 2. Also, the negative of the even n sequence is bounded above by -1 and is strictly increasing. It follows that the two series converge.

As the the two required series converge:

 \displaystyle u_{n+2} - u_n \rightarrow 0

Therefore we wish to solve:

 \displaystyle \frac{u_n - u_{n-2}}{(1+u_n)(1+u_{n-2})} = 0

Also:

 \displaystyle u_n = 1 + \frac{u_{n-2}}{u_{n-2}+1}

 \displaystyle u_{n-2}(u_n - 2) = 1 - u_n

 \displaystyle u_{n-2} = \frac{1-u_n}{u_n-2}

We also have:

 \displaystyle u_n - u_{n-2} = 0

so:

 \displaystyle u_n - \frac{1-u_n}{u_n-2} = 0

 \displaystyle u_n^2 - u_n - 1 =0

 \displaystyle u_n = \frac{1 \pm \sqrt5}{2}

We can discard the negative solution since all terms in the series are positive and between 1 and 2 inclusive. Hence both series converge to the above positive fraction. And since both series converge to this value, u_n converges to:

 \displaystyle \frac{1 + \sqrt5}{2}

Changing u_1 to 3 does not change this answer.

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Zorgz
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(Original post by Smaug123)
Q12:

i)
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\mathbb{E}(X) = \sum_{i=0}^{\infty} \mathbb{P}(X=i) i, since the Poisson distribution takes nonnegative integer values.
Now, \mathbb{P}(X=i) is 0 for even i, and e^{-\lambda} \frac{\lambda^i}{i!} otherwise, so the required expectation is \sum_{i=1, i odd}^{\infty} i e^{-\lambda} \frac{\lambda^i}{i!} = \lambda e^{-\lambda} \sum_{i=1, i odd}^{\infty} \frac{\lambda^{i-1}}{(i-1)!} = \lambda e^{-\lambda} \sum_{i=0, i even}^{\infty} \frac{\lambda^{i}}{i!} = \lambda e^{-\lambda} \alpha.
And since X+Y = U, and \mathbb{E}(U) = \lambda from the expectation of the Poisson distribution, and since the expectation of the sum is the sum of the expectations, we have \mathbb{E}(Y) = \lambda - \lambda e^{-\lambda} \alpha = \lambda e^{-\lambda} (e^{\lambda} - \alpha) = \lambda e^{-\lambda} \beta.


ii)
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The variance of X is "the mean of the squares - the square of the mean". The mean is \lambda e^{-\lambda} \alpha, which squares to \lambda^2 e^{-2 \lambda} \alpha^2. Note that \alpha+\beta = e^{\lambda}, and hence the second term in the expression for the variance is precisely "the square of the mean".
So we want "the mean of the squares" to be \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}.
Note that the denominator is e^{\lambda}.
Now, \mathbb{E}(X^2) = \sum_{i=0, i odd}^{\infty}i^2 e^{-\lambda} \frac{\lambda^i}{i!} = \lambda e^{-\lambda} \sum_{i=0, i odd}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!}.
We can see that a factor \lambda e^{-\lambda} = \frac{\lambda}{\alpha+\beta} has appeared, so we just need to show that \alpha + \lambda \beta = \sum_{i=0, i odd}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!}.
But this is certainly true: \sum_{i=0, i odd}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!} = \sum_{i=0, i even}^{\infty}(i+1) \frac{\lambda^i}{i!} = \sum_{i=0, i even}^{\infty}\frac{\lambda^i}{(  i-1)!} + \sum_{i=0, i even}^{\infty} \frac{\lambda^i}{i!} = \alpha+\lambda \beta as required.


Expression for Var(Y):
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We repeat the process: \mathbb{E}(Y) = \lambda e^{-\lambda} \beta, so \mathbb{E}(Y)^2 = \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2}.
Then \displaystyle \mathbb{E}(Y^2) = \sum_{i=0, i even}^{\infty} i^2 e^{-\lambda} \frac{\lambda^i}{i!}, which is e^{-\lambda} \lambda \sum_{i=2, i even}^{\infty} i \frac{\lambda^{i-1}}{(i-1)!}.
This is e^{-\lambda} \lambda \sum_{i=1, i odd}^{\infty}(i+1) \frac{\lambda^i}{i!} = e^{-\lambda} \lambda \sum_{i=1, i odd}^{\infty} \frac{\lambda^i}{(i-1)!} + e^{-\lambda} \lambda \sum_{i=1, i odd}^{\infty} \frac{\lambda^i}{i!} = e^{-\lambda} \lambda (\lambda \alpha + \beta) = \frac{\lambda^2 \alpha + \lambda \beta}{\alpha + \beta}.
Hence the variance is \frac{\lambda^2 \alpha + \lambda \beta}{\alpha + \beta} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2}.


Do non-zero \lambda exist?
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Note that Var(X)+Var(Y) = \frac{\lambda \alpha+\lambda^2 \beta}{\alpha+\beta} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} + \frac{\lambda^2 \alpha + \lambda \beta}{\alpha + \beta} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} = \frac{\alpha (\lambda + \lambda^2) + \beta (\lambda + \lambda^2)}{\alpha + \beta} - \frac{\lambda^2}{(\alpha+\beta)^  2}(\alpha^2+\beta^2) = \lambda + \lambda^2 \frac{\alpha^2 + \beta^2}{(\alpha+\beta)^2},
while Var(X+Y) = Var(U) = \lambda.
So we require \lambda = \lambda + \lambda^2 \frac{\alpha^2+\beta^2}{(\alpha+  \beta)^2}, and hence \lambda = 0 or \frac{\alpha^2 + \beta^2}{(\alpha+\beta)^2} = 0; this is clearly impossible, being a sum of squares, unless \alpha = \beta = 0; but also \alpha is a non-zero sum of squares. Hence there are no such \lambda, since we do not define a Poisson variable for \lambda = 0.

I got very similar but ended up having to show that a^2+b^2/(a+b)^2= 1, which of course isn't possible for non zero values. Not sure but think you might have missed out a lambda squared term when cancelling.
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Smaug123
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(Original post by Zorgz)
Not sure but think you might have missed out a lambda squared term when cancelling.
Thanks for that - I lost a minus sign and a lambdasquared. Fixed that's what you get for trying to do it all in LaTeX without using paper first!
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Slowbro93
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Getting ready to type up a solution to question 5 :yep: Also going to update OP
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Slowbro93
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Question 5:

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Part i


The first part states that f(x)=f(1-x).

Therefore: f'(x)=-f'(1-x) (via the chain rule).

at x= \frac{1}{2} we see that f'(\frac{1}{2})=-f'(\frac{1}{2})

Therfore 2f'(\frac{1}{2})=0

And we see that f'(\frac{1}{2})=0 as required.

We also have the additional information that: f(x)=f(\frac {1}{x}).

Therefore: f'(x)=-\frac{1}{x^2} f'(\frac {1}{x})

At x=-1

f'(-1)=-1(-1)^2 f'(\frac {1}{-1})

Which implies that: f'(-1)=-f'(-1)

This then follows that f'(-1)=-f'(-1)

Therefore f'(-1)=0

At x=2, f'(2)=-\frac{1}{4}f'(\frac{1}{2})

However, using the fact that f'(\frac{1}{2})=0, it then follows that

f'(2)=0 as required.



Part ii
We are given the function:

f(x)=\dfrac{(x^2 -x +1)^3}{(x^2 -x)^2}

It's easy to show that f(x)=f(1-x) and f(x)=f(\frac{1}{x}) by algebra manipulation (I won't show here unless asked).

As we've shown that f(x) satisfies the above conditions, from part i, we know that stationary points would lie at x=-1, 2 and \frac{1}{2}

We also see that f(x) would be asymptotic at x=0 and 1.

Now to find the coordinates of the stationary points, plug the x values into f(x). You only need to do this once and get the other two by using the fact that f(x)=f(1-x)=f(\frac{1}{x}) In this case, you find the stationary points occur at (-1, \frac{27}{4}), (2, \frac{27}{4}), (\frac{1}{2}, \frac{27}{4})

Finally, look at extreme values of x. As there will be an x^2 term if you divide out the polynomial, we see that f(x) will tend towards this term in both directions. Close to the poles, we we that again this x^2 term will mean that we have f(x) going towards positive infinity. (Diagram to be attached shortly).




Part iii
From part ii, we can see that the values that satisfy f(x)=\frac{27}{4} are x=-1, \frac{1}{2}, 2. For f(x)&gt;\frac{27}{4} we can conclude that x&lt;-1, -1&lt;x&lt;\frac{1}{2}, \frac{1}{2}&lt;x&lt;2, x&gt;2 where x\not= 0,1

The second part of this asks to find f(x)=\dfrac{343}{36}=\dfrac{7^3}  {6^2}=\dfrac{(x^2 -x +1)^3}{(x^2 -x)^2}

By equating the numerator and denominator, we can see that x^2-x=6 and x^2-x+1=7 which by solving the quadratic we see that x=-2 or 3.

Finally, using the relations f(x)=f(1-x) and f(x)=f(\frac{1}{x}) we also find that x=-\frac{1}{2} and x=\frac{1}{3}. You get another solution for x by using the fact that f(x)=f(1-x) and plugging in x=\frac{1}{3} you find that x=\frac{2}{3}. Finally you get the final solution by using f(x)=f(\frac{1}{x}) and plugging in x=\frac{2}{3} you see the last solution for x is x=\frac{3}{2}. To see when f(x)&gt;\frac{343}{36}, we use our diagram as an aid and you see that x&lt;-2, -\frac{1}{2}&lt;x&lt;\frac{1}{3}, \frac{2}{3}&lt;x&lt;\frac{3}{2}, x&gt;3 where x\not= 0,1


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MW24595
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Question 8

i- Let A(x, t) be the area of any rectangle with sides parallel to the co-ordinate axes, with one side on y= f(t) and another on the y-axis, and with one corner on the curve f(x). It is clear that, the rectangle with the maximum area will be one of this type.

Looking at a rudimentary diagram, it is evident that:

 A(x,t)= x(f(x)-f(t))



\Rightarrow \displaystyle \frac{dA(x,t)}{dx} = (f(x)-f(t)) + x f'(x)

Now, as we're concerned with maximising area, let  x_0 be the value of x for which  A(x,t) is greatest. So,

\displaystyle \frac{dA(x,t)}{dx} = 0



\Rightarrow (f(x_0)-f(t)) + x f'(x_0) = 0



\Rightarrow x_0 f'(x_0)+ f(x_0) = f(t)

By definition, A_0 (t) is the maximum value of A(x,t) so,

 A_0(t) = x_0 (f(x_0)-f(t))

ii- So, we have,

 tg(t) = \displaystyle \int_0^t f(x)\ dx 



\Rightarrow \frac{d(t g(t)}{dt} = \frac{d}{dt} (\displaystyle \int_0^t f(x)\ dx )

Now, notice that on integrating, we have a difference between a function of t and a constant. On differentiating this with respect to t, we just get  f(t) on the RHS. To see this, let h(x) be the antiderivative of  f(x) .

Then,

 tg(t) = h(t) - h(0)



\Rightarrow \frac{d(t g(t))}{dt} = \frac{d(h(t) - h(0))}{dt} = \frac{d h(t)}{dt} = f(t)



\Rightarrow t g'(t) + g(t) = f(t)



\Rightarrow tg'(t) = f(t)- g(t)


Now, consider that:

 \displaystyle \int^t_0 (f(x)-f(t))\ dx = \displaystyle \int^{x_0}_0 (f(x)-f(t))\ dx + \displaystyle \int^t_{x_0} (f(x)-f(t))\ dx

Further, as  f(x) is a decreasing function, we have, for all  0 \le x&lt; x_0 ,

 f(x) &gt; f(x_0)



\Rightarrow \displaystyle \int^{x_0}_0 f(x)\ dx &gt; \displaystyle \int^{x_0}_0 f(x_0)\ dx = f(x_0) \displaystyle \int^{x_0}_0\ dx = x_0 f(x_0)



\Rightarrow \displaystyle \int^{x_0}_0 f(x)\ dx - x_0(f(t)) = x_0 (f(x_0)- f(t))



\Rightarrow \displaystyle \int^{x_0}_0 f(x)-f(t)\ dx = x_0 (f(x_0)- f(t))


Further, for  x_0 \le x &lt; t , we have,

 f(x) &gt; f(t)



\Rightarrow \displaystyle \int_{x_0}^t f(x)\ dx &gt;\displaystyle \int_{x_0}^t f(t)\ dx



\Rightarrow \displaystyle \int_{x_0}^t f(x)- f(t)\ dx &gt;0


Therefore, we clearly have,

\displaystyle \int^{x_0}_0 (f(x)-f(t))\ dx + \displaystyle \int^t_{x_0} (f(x)-f(t))\ dx &gt; x_0 (f(x_0) - f(t))





\Rightarrow \displaystyle \int^t_0 (f(x)-f(t))\ dx &gt; A_0(t)

Now, observe that,

 \displaystyle \int^t_0 (f(x)-f(t))\ dx



\beginaligned =  \displaystyle \int^t_0 f(x)\ dx - t(f(t))



\beginaligned = t g(t) - t f(t)



\beginaligned = -t(f(t) - g(t))



\beginaligned = -t(tg'(t)) = -t^2 g'(t)



\Rightarrow -t^2 g'(t) &gt;A_0 (t)

iii-

 f(x) = \frac{1}{1+x} \Rightarrow f'(x) = \frac{-1}{(1+x)^2}

Now, if order to find x_0 , we simply plug these into the condition for x_0 , namely:

 x_0 f'(x_0)+ f(x_0) = f(t)



\Rightarrow \displaystyle \frac{-x_0}{(1+x_0)^2} + \frac{1}{1+x_0} = \frac{1}{1+t}



\Rightarrow x_0 = -1+ \sqrt {1+t}

Now,

 g(t) = \frac{1}{t} \displaystyle \int_0^t \frac{1}{1+x}\ dx = \frac{1}{t} (ln(1+t))



\Rightarrow tg'(t) = f(t) - g(t) = \frac{1}{1+t} - \frac{1}{t} (ln(1+t))



\Rightarrow -t^2 g'(t) = ln(1+t) - \frac{t}{1+t}

Similarly, we have,

 A_0(t) = x_0 (f(x_0) - f(t))



\beginaligned = \displaystyle (-1+ \sqrt {1+t})(\frac{1}{ \sqrt {1+t}} - \frac{1}{1+t})



\beginaligned \Rightarrow A_0(t) = 1- \frac{2}{ \sqrt {1+t}} + \frac{1}{1+t}

Now, from the previous part's inequality, we get,

 -t^2 g_0(t) &gt; A_0(t)



\Rightarrow ln(1+t) - \frac{t}{1+t} &gt; 1- \frac{2}{ \sqrt {1+t}} + \frac{1}{1+t}



\Rightarrow ln (1+t) &gt; 2- \frac{2}{ \sqrt {1+t}}



\Rightarrow ln \sqrt{1+t} &gt; 1-  \frac{1}{ \sqrt {1+t}}
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jack.hadamard
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I had to practice some drawing on a software, so enjoy.

Question 4


Preliminary stuff

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Name:  circle.png
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The line passing through (a,0) with gradient b has the equation y = b(x - a) and the unit circle has equation x^2 + y^2 = 1.

Hence,

x^2 + [b(x - a)]^2 = 1 \iff (1 + b^2)x^2 - 2ab^2x + (ab)^2 - 1 = 0

implies that

\displaystyle x = \frac{2ab^2 \pm \sqrt{(-2ab^2)^2 - 4(1 + b^2)[(ab)^2 - 1]}}{2(1 + b^2)} = \frac{ab^2 \pm \sqrt{1 + b^2(1 - a^2)}}{1 + b^2}.

It follows that the midpoint has x-coordinate \displaystyle \frac{ab^2}{1 + b^2}, and from that we see that its y-coordinate is

\displaystyle y = b(x - a) = ab\left[\frac{b^2}{1 + b^2} - 1\right] = -\frac{ab}{1 + b^2}.

Therefore, the midpoint is \displaystyle M\left(\frac{ab^2}{1 + b^2}, -\frac{ab}{1 + b^2} \right).


First part

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Since b is fixed, the locus of M is given by the parametric equations

\displaystyle x(a) = a\frac{b^2}{1 + b^2} and \displaystyle y(a) = a \frac{-b}{1 + b^2}.

Clearly, -by = x.

The above equation shows that, since b is fixed, the locus is a straight line.

Now, the length is

\displaystyle \sqrt{[x(1) - x(-1)]^2 + [y(1) - y(-1)]^2} = \sqrt{[x(1) - x(-1)]^2 + \frac{1}{(-b)^2}[x(1) - x(-1)]^2}

which is just \displaystyle \frac{2b^2}{1 + b^2} \sqrt{\frac{1 + b^2}{b^2}} = \frac{2b}{\sqrt{1 + b^2}}.

The sketch looks like the following.

Name:  circle_locus.png
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Second part (a)

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Now a is fixed, so the parametric equations are x(b) and y(b).

Letting b = \tan(t), for -\frac{\pi}{2} &lt; t &lt; \frac{\pi}{2}, we see that

\displaystyle x = a\frac{\tan^2(t)}{1 + \tan^2(t)} = a\sin^2(t) = \frac{a}{2}[1 - \cos(2t)]}
and
\displaystyle y = a\frac{-\tan(t)}{1 + \tan^2(t)} = -a\sin(t)\cos(t) = -\frac{a}{2}\sin(2t)

It follows that (2x - a)^2 + (2y)^2 = a^2 \iff (x - \frac{a}{2})^2 + y^2 = \left(\frac{a}{2}\right)^2.

Bearing the interval in mind, we see that this traces out a complete circle. The sketch below.

Name:  circle_locus_2.png
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Second part (b)

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Now, both a and b vary, but 0 &lt; b \leq 1.

Using ab = 1, we can let a = \cot(t), in the previous part, and notice that 0 &lt; t \leq \frac{\pi}{4}.

Hence,
\displaystyle x = \frac{a}{2}[\sin^2(t)] = \frac{\cot(t)}{2}\sin(2t) = \sin(t)\cos(t) = \frac{1}{2}\sin(2t)
and
\displaystyle y = -a\sin(t)\cos(t) = -\cot(t)\sin(t)\cos(t) = -\cos^2(t).

Therefore, (2x)^2 + (2y + 1)^2 = 1 \iff x^2 + (y + \frac{1}{2})^2 = \left(\frac{1}{2}\right)^2.

Again, bearing the interval we see that it is a quarter of the circle. Sketch below.

Name:  circle_locus_3.png
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When drawing it, notice that y = b(x - a) = bx - 1.
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(Original post by LShirley95)
Got Q9 written with diagrams, unfortunately don't have a scanner so this is the best I can do:
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