STEP 2013 Solutions

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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

:
$\int_0^1 x^{n-1}(1-x)^n dx = \int_1^0 (1-u)^{n-1} u^n (-1) du = \int_0^1 (1-x)^{n-1} x^n dx$ , 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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Q7 – STEP II

Part i):

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x=1, y=0 is a nice obvious solution.

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

and

:

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

Part iii):

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Now, b is a prime number meaning that

has four possible factors – 1, b,

and

. Let’s examine each of these in turn:

,

This solves to get

which is not possible as a and b are both positive integers.

,

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

,

This solves for

which is not possible as a and b are both positive integers, yet b squared is greater than b implying that a = 0.

,

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

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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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

as the horizontal velocity of the particle and

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

be the time taken to reach B, and

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

:

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

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

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

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

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

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

,

and

, respectively. Now, the first collision:

Conservation of Linear Momentum:

Newton’s Law of Restitution:

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

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

. Plugging this in then immediately yields the speeds of the other two particles.
Now, for a third collision we require

:

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

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

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:

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

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

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 $e < 3 – 2\sqrt{2}$

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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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mimx

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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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happens after HH of TT, HT of TH would give $A \geq 1$

happens after HT of TH:

Now, $p \neq q$ so $p-q \neq 0$ then:

13 (ii)

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A=2 happens after HTT or THH.

S=2 happens after HHT or TTH.

A=3 happens after THTT or HTHH.

S=3 happens after HHHT or TTTH.

13 (iii)

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A=2n happens after (n*(HT))T or (n*(TH))H.

S=2n happens after (n*(HH))T or (n*(TT))H.

$=pq^{2n}+qp^{2n}$
Then we note pq>0 and

and $p^{n-1}-q^{n-1}$ are not equal to 0 since and have the same sign since $p-q \neq 0, \text{and}\: n-1>0$ .
$pq(p^n-q^n)(p^{n-1}-q^{n-1})>0$
$(pq^{2n}+qp^{2n})-(p(pq)^n+q(pq)^n)>0$

A=2n+1 happens after (n*(HT))HH or (n*(TH))TT.

S=2n+1 happens after (n*(HH))HT or (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 $p-q \neq 0, \text{and}\: n-1>0$ . ( $p>q \rightarrow p^m>q^m$ when m>0)
$pq(p^{n+1}-q^{n+1})(p^{n-1}-q^{n-1})>0$
$(pq^{2n+1}+qp^{2n+1})-(p^2(pq)^n+q^2(pq)^n)>0$

----
[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

with

then:
$a=-\frac{1}{3}(p+q+r)$
$b=\frac{1}{3}(pq+qr+rp)$

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.

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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

, 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 $\alpha > 0$ 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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(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

, 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

.

Therefore:

(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

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

Which implies that:

This then follows that

Therefore

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

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

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

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

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 $f(x)>\frac{27}{4}$ we can conclude that $x<-1, -1<x<\frac{1}{2}, \frac{1}{2}<x<2, x>2$ 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

and

which by solving the quadratic we see that x=-2 or 3.

Finally, using the relations

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

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 $f(x)>\frac{343}{36}$ , we use our diagram as an aid and you see that $x<-2, -\frac{1}{2}<x<\frac{1}{3}, \frac{2}{3}<x<\frac{3}{2}, x>3$ where $x\not= 0,1$

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MW24595

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Question 8

i- Let

be the area of any rectangle with sides parallel to the co-ordinate axes, with one side on

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

be the value of x for which

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,

is the maximum value of

so,

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

on the RHS. To see this, let

be the antiderivative of

.

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

is a decreasing function, we have, for all $0 \le x< x_0$ ,

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

Further, for $x_0 \le x < t$ , we have,

$f(x) > f(t) \Rightarrow \displaystyle \int_{x_0}^t f(x)\ dx >\displaystyle \int_{x_0}^t f(t)\ dx \Rightarrow \displaystyle \int_{x_0}^t f(x)- f(t)\ dx >0$

Therefore, we clearly have,

$\displaystyle \int^{x_0}_0 (f(x)-f(t))\ dx + \displaystyle \int^t_{x_0} (f(x)-f(t))\ dx > x_0 (f(x_0) - f(t)) \Rightarrow \displaystyle \int^t_0 (f(x)-f(t))\ dx > A_0(t)$

Now, observe that,

$\displaystyle \int^t_0 (f(x)-f(t))\ dx \beginaligned = \displaystyle \int^t_0 f(x)\ dx - t(f(t)) \beginaligned = t g(t) - t f(t) \beginaligned = -t(f(t) - g(t)) \beginaligned = -t(tg'(t)) = -t^2 g'(t) \Rightarrow -t^2 g'(t) >A_0 (t)$

iii-

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

Now, if order to find

, we simply plug these into the condition for

, 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,

$-t^2 g_0(t) > A_0(t) \Rightarrow ln(1+t) - \frac{t}{1+t} > 1- \frac{2}{ \sqrt {1+t}} + \frac{1}{1+t} \Rightarrow ln (1+t) > 2- \frac{2}{ \sqrt {1+t}} \Rightarrow ln \sqrt{1+t} > 1- \frac{1}{ \sqrt {1+t}}$

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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

with gradient

has the equation

and the unit circle has equation

.

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

-coordinate $\displaystyle \frac{ab^2}{1 + b^2}$ , and from that we see that its

-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

is fixed, the locus of

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,

.

The above equation shows that, since

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

is fixed, so the parametric equations are

and

.

Letting $b = \tan(t)$ , for $-\frac{\pi}{2} < t < \frac{\pi}{2}$ , 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
Views: 2084
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Second part (b)

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Now, both

and

vary, but $0 < b \leq 1$ .

Using

, we can let $a = \cot(t)$ , in the previous part, and notice that $0 < t \leq \frac{\pi}{4}$ .

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
Views: 1894
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When drawing it, notice that

.

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Brister

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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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How do you figure out the direction for friction? Sorry if you explained it already, but it is hard to read your attachments.

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