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STEP 2003 Solutions Thread

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Reply 20
Thank you, I shall remember that :smile:
tommm
STEP II 2003 Q1

Spoiler




For implications I'd use \implies and \iff since your \rightarrow is coming out as a 'tends to' (or \to) on my screen atleast. :wink:
Reply 22
You could us \Rightarrow (note the capital)
8 Horizontal
Define a sequence An=m+Un1 A_n = m+U_n-1 of which all the terms are clearly irrational since U_n is irrational for all n and a rational number + irrational number is always irrational. Yet it converges to m as n n \to \infty

Is this what you mean?
Yes. I doubt it would lose you much here, because it's so trivial to fix it to work for the general case. But the standard interpretation of "show XXX for a given integer n" means "it has to work for any n, but you can consider n as fixed".
Reply 24
8 Horizontal
For implications I'd use \implies and \iff since your \rightarrow is coming out as a 'tends to' (or \to) on my screen atleast. :wink:


Cheers, I'll clean that up.
DFranklin
Yes. I doubt it would lose you much here, because it's so trivial to fix it to work for the general case. But the standard interpretation of "show XXX for a given integer n" means "it has to work for any n, but you can consider n as fixed".


ok thanks :smile:
Reply 26
STEP II 2003 Q2

Spoiler

STEP I question 2

First part:

Spoiler



Second part:

Spoiler

Reply 28
STEP II, Question 7

Spoiler

Reply 29
STEP II 2003 Q4

Spoiler



Upon using the substitution y=Rcosuy = R\cos u, and some manipulation with the cosine double-angle formula, we get

R2(u0.5sin(2u)0arccos(d/R)R^2(u - 0.5\sin(2u)|^{\arccos(d/R)}_0 (N.B. that represents limits of integration)

The difficulty in evaluating this lies in the evaluation of 0.5sin(2arccos(d/R))0.5\sin(2\arccos(d/R)). First, using the sine double angle formula, we get

0.5sin(2arccos(d/R))=sin(arccos(d/R))cos(arccos(d/R))=(d/R)sin(arccos(d/R))0.5\sin(2\arccos(d/R)) = \sin(\arccos(d/R))\cos(\arccos(d/R)) = (d/R)\sin(\arccos(d/R))

By drawing a suitable right-angled triangle, we find that
Unparseable latex formula:

\sin(2 \arccos (d/R)) = \frac{\sqrt{R^2 - d^2}{R}}

and from this we can find the required result.


To find the area in the three listed cases, we note that i) and ii) are both specific versions of iii), therefore we need only find the answer to iii) and i) and ii) follow easily.

From a diagram, we see that we must replace d with the perpendicular distance from the line to the furthest point on the circle from the line. This gives us:

d+D=Rd + D = R, where D is the shortest distance from the origin to the line. The distance from a point (x, y) to the centre is given by

D2=(xa)2+(yb)2D^2 = (x - a)^2 + (y - b)^2

But the point (x, y) lies on the line y = mx + c, therefore

D2=(xa)2+(mx+cb)2D^2 = (x - a)^2 + (mx + c - b)^2

To find the minimum value of D, we must differentiate this expression and set it equal to 0. This gives us

xa+m2x+mcmb=0x - a + m^2x + mc - mb = 0

    x=mbmc+am2+1\implies x = \displaystyle\frac{mb - mc + a}{m^2 + 1}

substituting this back into our expression for D^2, we get

D2=(mbmcm2a)2+(ma+cb)2(m2+1)2D^2 = \displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}

and therefore, to find the new area, we must modify (*) such that

d=R(mbmcm2a)2+(ma+cb)2(m2+1)2d = R - \sqrt{\displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}}

The answer to i) follows by letting m = 0; the answer to ii) follows by letting a = b = 0.

Reply 30
DFranklin
I grant you it's not 100% clear, but I would interpret this part as "given an integer m, give an example of a sequence of irrational numbers that converges to m". (In other words, you have to produce a sequence that works for every m).


Sorry to pester you, but when it says "converges to a given integer m", does that mean that its sum converges to m or does that mean the general term tends to m?
Reply 31
tommm
Sorry to pester you, but when it says "converges to a given integer m", does that mean that its sum converges to m or does that mean the general term tends to m?


If a sequence converges to a value, then the general term converges to that value
If a series or sum converges to a value then the sum of the series converges to that value
I/13
Show that the given probability is true

Let XX and YY be the results of team A and team B respectively.

The probability of each team scoring can be modelled by a binomial distribution:

Unparseable latex formula:

X\tildaB(5,P_A)


Unparseable latex formula:

Y\tildaB(5,P_B)[br]



The situation of neither side winning after the initial 10-shot period occurs when each team has the same number of success, or fails.

Let ii denote the number of fails.

P(X=i)=(5i)PA5i(1PA)i[br]P(Y=i)=(5i)PB5i(1PB)iP(X=i) = \binom{5}{i}P_A^{5-i}(1-P_A)^i[br]P(Y=i) = \binom{5}{i}P_B^{5-i}(1-P_B)^i

Therefore α=(5i)2(1PA)i(1PB)iPA5iPB5i\alpha = \binom{5}{i}^2(1-P_A)^i(1-P_B)^iP_A^{5-i}P_B^{5-i}
as required.


Show that the expected number of shots is given by the stated expression




My LaTeX is probably all wrong and I can't figure out the second part, but there's half a solution (I think), and it's my first one, so woo!
Reply 33
STEP II 2003 Q8

Spoiler

Reply 34
STEP II 2003 Q5

Spoiler

I/14

I feel like I may have messed up the second part, anyone care checking?

i



ii

STEP II Q6

First part:

Spoiler



Second part:

Spoiler

Reply 38
FFFFFFFFFFUUUUUUUUUU I just got halfway through typing up II/6 when I realised it'd been done.
tommm
FFFFFFFFFFUUUUUUUUUU I just got halfway through typing up II/6 when I realised it'd been done.



Sorry :o:

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