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Reply 1
STEP III, Question 1

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STEP III, Question 1

a)

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b)

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Reply 3
1ex1-e^{-x} is strictly increasing, but doesn't tend to infinity
Reply 4
STEP II 2002 Question 3

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Reply 6
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.
Reply 7
STEP II 2002 Question 1

Part One

Part Two

Part Three

(edited 11 years ago)
Reply 8
STEP I 2002 Question 7

I=0acosxsinx+cosxdx\displaystyle I = \int_0^a \frac{\cos x}{\sin x + \cos x} \, dx and J=0asinxsinx+cosxdx\displaystyle J = \int_0^a \frac{\sin x}{\sin x + \cos x} \, dx with 0a<3π40 \leq a < \frac{3\pi}{4}

(Note that in the interval 0x<3π40 \leq x < \frac{3\pi}{4}, sinx+cosx0\sin x + \cos x \not= 0)

Unparseable latex formula:

\begin{array}{rl}[br]I+J & \displaystyle = \int_0^a \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \\ \br \\[br]& \displaystyle = \int_0^a 1 \, dx \\ \br \\[br]& \displaystyle = \left[ x \right]_0^a = a \end{array}



Unparseable latex formula:

\begin{array}{rl}[br]I-J & \displaystyle = \int_0^a \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \\ \br \\[br]& \displaystyle = \left[ \ln (\sin x + \cos x) \right]_0^a \\ \br \\[br]& \displaystyle = \ln (\sin a + \cos a)\end{array}



(I+J)+(IJ)=a+ln(sina+cosa)2I=a+ln(sina+cosa)\displaystyle (I+J)+(I-J) = a + \ln (\sin a + \cos a) \Rightarrow \boxed{2I = a + \ln (\sin a + \cos a)}

(i)

(ii)

II/8:

For x < 0, dy/dx = -y ==> 1/y dy/dx = -1 ==> ln y = -x +c ==> y = Ae^(-x); as y = a when x = -1, a = Ae so A = a/e; therefore y=ae(x+1)y = ae^{-(x+1)} for x < 0.

For > 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x + c ==> y = Be^x; as y = b when x = 1, b = Be so B = b/e therefore y=bex1y = be^{x-1} for x > 0.

If a = b, there is no jump.

Note that e^x - 1 is positive if x > 0, and negative if x < 0.

1ydydx=ex1lny=exx+ky=Aeexx\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = e^x - 1 \Rightarrow \ln y = e^x - x + k \Rightarrow y = Ae^{e^x - x} for x > 0.
Given that y = e^e when x = 1, e^e = Ae^(e - 1) so A = e; y=eexx+1y = e^{e^x - x + 1}

1ydydx=1exlny=xex+cy=Bexex\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = 1 - e^x \Rightarrow \ln y = x - e^x + c \Rightarrow y = Be^{x - e^x} for x < 0.

At x = 0, y = e10+1=e2e^{1 - 0 + 1} = e^2 using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so
e2=Be1B=e3e^2 = Be^{-1} \Rightarrow B = e^3

y=eexx+1forx>0y = e^{e^x - x + 1} \text{for} x > 0
y=e3+xexforx<0y = e^{3 + x - e^x} \text{for} x < 0


(i) limx+exp(exx+1)exp(ex)=limx+exp(1x)=0\displaystyle \lim_{x \to + \infty} \exp(e^x - x + 1) \exp(-e^x) = \lim_{x \to + \infty} \exp(1 - x) = 0

(ii) limxe3+xexex=limxe3ex=e3\displaystyle \lim_{x \to - \infty} e^{3 + x - e^x} e^{-x} = \lim_{x \to -\infty} e^{3 - e^x} = e^3.
II/7: (Scary scary vectors.)

??????????

I'm not checking if anyone posted the solutions, cause I completed these not long ago.

2002 II Question 1

Unparseable latex formula:

\displaystyle \int_{\pi/6}^{\pi/4} \frac{1}{1-\cos{\2\theta}} \,\mathrm{d}\theta =



=π/6π/412sin2θdθ=\displaystyle = \int_{\pi/6}^{\pi/4} \frac{1}{2\sin^2\theta} \,\mathrm{d}\theta =
=12[cot2θ]π/6π/4=\displaystyle =\frac{1}{2}\left[-\cot^2\theta\right]_{\pi/6}^{\pi/4} =
=12(3/21/21)=3212\displaystyle = \frac{1}{2}\left(\frac{\sqrt{3}/2}{1/2} - 1\right) = \frac{\sqrt{3}}{2} - \frac{1}{2} as required.

x=sin2θ    θ=12arcsinx\displaystyle x = \sin{2\theta} \implies \theta = \frac{1}{2}\arcsin{x}
and     dx=\implies \mathrm{d}x =
=2cos2θdθ\displaystyle = 2\cos{2\theta}\mathrm{d}\theta

    3/21111x2dx=\displaystyle \implies \int_{\sqrt{3}/2}^1 \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x =
=π/6π/412sin2θsin2θdθ=\displaystyle =\int_{\pi/6}^{\pi/4} \frac{1-2\sin^2\theta}{\sin^2\theta}\,\mathrm{d}\theta =
=31π/6π/42dθ=31π6\displaystyle = \sqrt{3}-1-\int_{\pi/6}^{\pi/4}2\,\mathrm{d}\theta = \sqrt{3} - 1 - \frac{\pi}{6}, as required.

Using x=1y\displaystyle x = \frac{1}{y}, dy=y2dx\displaystyle \mathrm{d}y = -y^2\mathrm{d}x, we obtain:

12/31y(yy212)dy=\displaystyle \int_{1}^{2/\sqrt{3}} \frac{1}{y(y-\sqrt{y^2-1^2})} \,\mathrm{d}y =

=132xy21x1x21dx=\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-xy^2}{\frac{1}{x}-\sqrt{\frac{1}{x^2}-1}} \,\mathrm{d}x =
=132x2y211x2dx=\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-x^2y^2}{1-\sqrt{1-x^2}} \,\mathrm{d}x =
=321111x2dx=31π6\displaystyle = \int_{\sqrt{3}{2}}^{1} \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x = \sqrt{3}-1-\frac{\pi}{6}
2002 II Question 11

Total weight (to check our calculus) =0lαWl1(x/l)α1dx=aWlα[1αxα]0l=Wlαlα=W\displaystyle = \int_0^l \alpha Wl^{-1}(x/l)^{\alpha-1}\,\mathrm{d}x = aWl^{-\alpha}\left[\frac{1}{\alpha}x^\alpha \right]_0^l = Wl^{-\alpha}l^\alpha = W

Total moment
Unparseable latex formula:

\displaystyle = \int_0^l \alpha Wl^{-1}(x/l)^{\alpha-1}x\,\mathrm{d}x = \alphaWl^{-\alpha}\int_0^l x^\alpha \,\mathrm{d}x


=αα+1Wl\displaystyle = \frac{\alpha}{\alpha+1}Wl.

We know that total moment =Wx\displaystyle = W\overline{x}, so x=αlα+1\displaystyle\overline{x} = \frac{\alpha l}{\alpha+1}, as required.

Draw out a diagram: A touching the ground, with R_A going up, F_A going right. B touching the wall, with F_B going up, R_B going left. Somewhere ebtween A and B, we have the centre of mass. We denote xA=lx=lα+1\displaystyle \overline{x_A} = l - \overline{x} = \frac{l}{\alpha+1} to be the distance of c.o.m. from A. Angle θ\displaystyle\theta going from the horizontal to the line AB.

Equating the horizontal forces: FA=RB\displaystyle F_A = R_B. Noting that FA=μRA\displaystyle F_A = \mu R_A, we get μRa=rB\displaystyle\mu R_a = r_B. Noting that fB=μRB\displaystyle f_B=\mu R_B, we get FB=μ2RA\displaystyle F_B = \mu^2 R_A
Equating the vertical forces: W=RA+FB=RA(1+μ2)\displaystyle W = R_A+F_B = R_A(1+\mu^2).
Equating the moments: WxAcosθ=RBlsinθ+FBlcosθ    RBltanθ=WxAFBl=μRaltanθ=RA(1+μ2)xAμ2RAl\displaystyle W\overline{x_A}\cos\theta = R_Bl\sin\theta + F_Bl\cos\theta \implies R_Bl\tan\theta = W\overline{x_A} - F_Bl = \mu R_al\tan\theta = R_A(1+\mu^2)\overline{x_A} - \mu^2R_Al
    μltanθ=(1+μ2)xAμ2l=(1+μ2)lα+1μ2l\displaystyle \implies \mu l\tan\theta = (1+\mu^2)\overline{x_A} - \mu^2l = (1+\mu^2)\frac{l}{\alpha+1}-\mu^2l
    μtanθ=1+μ2α+1μ2=1αμ2α+1\displaystyle \implies \mu\tan\theta = \frac{1+\mu^2}{\alpha+1} - \mu^2 = \frac{1-\alpha\mu^2}{\alpha+1}
    tanθ=1αμ2(1α)μ\displaystyle \implies \tan\theta = \frac{1-\alpha\mu^2}{(1_\alpha)\mu} as required.

Not sure if I'm right, but I'll give it a go:
α=32\displaystyle \alpha = \frac{3}{2}, and θ=π4    (1+α)μ=1αμ2    3μ2+5μ2=0    μ=26=13\displaystyle \theta = \frac{\pi}{4} \implies (1+\alpha)\mu = 1-\alpha\mu^2 \implies 3\mu^2+5\mu-2 = 0 \implies \mu = \frac{2}{6} = \frac{1}{3}
2002 II Question 12

P(k of K days, something with probability of q happens)=(Kk)qk(1q)Kk\displaystyle P(\text{k of K days, something with probability of q happens}) = \binom{K}{k}q^k(1-q)^{K-k}
q=q = probability such that more than l coins of the L coins will produce less than m heads

Estimating q:
Unparseable latex formula:

Q \~ B(L,z)

, where z is the probability of a coin producing less than m heads.
We can say that
Unparseable latex formula:

\displaystyle Q' \~ N(Lz,Lz(1-z))

, which gives:
Unparseable latex formula:

\displaystyle q = P(Q>l) \approx 1 - \Phi\left(\frac{(l-\frac{1}{2})-Lz}{\sqrtLz(1-z)}\right) = \Phi\left(\frac{2Lz-2l+1}{2\sqrt{Lz(1-z)}}\right) \approx \Phi\left(\frac{2Lz-2l+1}{2\sqrt{Lz}}\right)

for small z.
We then substitute h=Lz\displaystyle h = Lz to get q=Φ(2h2l+12h)\displaystyle q = \Phi\left(\frac{2h-2l+1}{2\sqrt{h}}\right) as required.

z is the probability that less than m heads are made after M throws.
Z B(M,p)    ZN(Mp,Mp(1p))\displaystyle Z ~ B(M,p) \implies Z \approx N (Mp, Mp(1-p))
z=P(Z<m)=Φ((m12)MpMp(1p))=Φ(2m1Mp2Mp(1p))\displaystyle z = P(Z<m) = \Phi\left(\frac{(m-\frac{1}{2})-Mp}{\sqrt{Mp(1-p)}}\right) = \Phi\left(\frac{2m-1-Mp}{2\sqrt{Mp(1-p)}}\right).

h=Lz=LΦ(2m1Mp2Mp(1p))\displaystyle h = Lz = L\Phi\left(\frac{2m-1-Mp}{2\sqrt{Mp(1-p)}}\right) as required.

K = 7, k = 2, L = 500, l = 4, M = 100, m = 48, p = 0.6. z=Φ(981602100×0.6×0.4)=Φ(37224)Φ(3.7)\displaystyle z = \Phi\left(\frac{98-1-60}{2\sqrt{100\times 0.6\times 0.4}}\right) = \Phi\left(\frac{37}{2\sqrt{24}}\right)\approx\Phi(3.7) , which is a large z, so we are not allowed to approximate as needed.
2002 II Question 13

F(a)=0F(a) = 0
f(b)=1f(b) = 1, a>0a > 0

G(y)=F(y)2F(y)\displaystyle G(y) = \frac{F(y)}{2-F(y)}


F(a)=F(a)2F(a)=02=0\displaystyle F(a) = \frac{F(a)}{2-F(a)} = \frac{0}{2} = 0

G(b)=F(b)2F(b)=121=1\displaystyle G(b) = \frac{F(b)}{2-F(b)} = \frac{1}{2-1} = 1

G(y)=(2F(y))F(y)F(y)(F(y)(2F(y))2=2F(y)(2F(y))20\displaystyle G'(y) = \frac{(2-F(y))F'(y)-F(y)(-F'(y)}{(2-F(y))^2} = \frac{2F'(y)}{(2-F(y))^2} \geq 0 since F(y)0\displaystyle F'(y) \geq 0 since it's a cumulative distribution function.

We know 1(2F(x))24    141(2F(x))21    122(2F(x))22\displaystyle 1 \leq (2-F(x))^2 \leq 4 \implies \frac{1}{4} \leq \frac{1}{(2-F(x))^2} \leq 1 \implies \frac{1}{2} \leq \frac{2}{(2-F(x))^2} \leq 2

We know E[Y]=abxG(x)dx=ab2xF(x)(2F(x))2dx=abxF(x)×2(2F(x))2dx\displaystyle E[Y] = \int_a^b xG'(x)\,\mathrm{d}x = \int_a^b \frac{2xF'(x)}{(2-F(x))^2}\, \mathrm{d}x = \int_a^b xF'(x) \times \frac{2}{(2-F(x))^2}\, \mathrm{d}x

Using the condition found before, we deduce that ab12xF(x)dxabxF(x)×2(2F(x))2dxab2xF(x)dx\displaystyle \int_a^b \frac{1}{2}xF'(x)\, \mathrm{d}x \leq \int_a^b xF'(x) \times \frac{2}{(2-F(x))^2}\, \mathrm{d}x \leq \int_a^b 2xF'(x)\, \mathrm{d}x

    12E[X]E[Y]2E[Y]\displaystyle \implies \frac{1}{2}E[X] \leq E[Y] \leq 2E[Y]

Var[Y]=E[Y2]E[Y]2\displaystyle Var[Y] = E[Y^2] - E[Y]^2.

We know E[Y2]2E[X2]\displaystyle E[Y^2]\leq 2E[X^2], so E[Y2]2E[X2]=2E[X]2+2Var[X]E[Y^2] \leq 2E[X^2] = 2E[X]^2+2Var[X].

We also know that (E[Y])214E[X]2    E[Y]214E[X]2\displaystyle (E[Y])^2 \geq \frac{1}{4}E[X]^2 \implies -E[Y]^2 \leq -\frac{1}{4}E[X]^2

So Var[Y]2Var[X]+2E[X]2E[Y]22Var[X]+2E[X]214E[X]2=\displaystyle Var[Y] \leq 2Var[X] + 2E[X]^2 - E[Y]^2 \leq 2Var[X] + 2E[X]^2 - \frac{1}{4}E[X]^2 =
=2Var[X]+74E[X]2 \displaystyle = 2Var[X] + \frac{7}{4}E[X]^2, as required.
Reply 15
I have some additions to the solutions to some of the STEP III qu's posted by Dadeyemi that would complete them. Would they be wanted?

Also, SimonM (or anybody else), what is the LaTeX for surrounding particular results in boxes, as you sometimes do for answers to parts of questions? That might make my solutions a little more readable...

Meanwhile,

Question 6, STEP III, 2002

y4(dydx)4=(y21)2\displaystyle y^4 \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)^4 = (y^2 - 1)^2

Standard seperating of variables yields:

y4(y21)2(dy)4=1.(dx)4\displaystyle \frac{y^4}{(y^2 - 1)^2} (\mathrm{d}y)^4 = 1 . (\mathrm{d}x)^4

At this point, we would like to take roots, but notice that we must consider seperately the cases y1| y | \le 1, and y>1| y | > 1, or we will be taking square roots of a negative number.

(a) For y>1| y | > 1:

yy21dy=±1.dx\displaystyle \int \frac{y}{\sqrt{y^2 - 1}} \mathrm{d}y = \int \pm 1 . \mathrm{d}x

Resulting in:

y21=±x+c\displaystyle \sqrt{y^2 - 1} = \pm x + c

y2=(±x+c)2+1\displaystyle y^2 = (\pm x + c)^2 + 1

Notice that if x[,]x \in [- \infty, \infty], it is no longer necessary to consider ±x\pm x, and we have:

y2=(x+c)2+1\displaystyle \boxed{y^2 = (x + c)^2 + 1}

(b) For y1| y | \le 1:

y1y2dy=±1.dx\displaystyle \int \frac{y}{\sqrt{1 - y^2}} \mathrm{d}y = \int \pm 1 . \mathrm{d}x

Resulting in:

1y2=±x+c\displaystyle - \sqrt{1 - y^2} = \pm x + c

y2=1(±x+c)2\displaystyle y^2 = 1 - (\pm x + c)^2

Or simply:

y2=1(x+c)2\displaystyle \boxed{y^2 = 1 - (x + c)^2}

Now we consider parts (i) and (ii) of the question:

(i) 32<1\displaystyle \frac{\sqrt{3}}{2} < 1, therefore we use the second formula to determine cc:

Substituting (0,32) \left(0 , \frac{\sqrt{3}}{2} \right) into y2=1(x+c)2\displaystyle y^2 = 1 - (x + c)^2 yields,

c=±12\displaystyle \boxed{c = \pm \frac{1}{2}}

(ii) 52>1\displaystyle \frac{\sqrt{5}}{2} > 1, therefore we use the first formula to determine cc:

Substituting (0,52) \left(0 , \frac{\sqrt{5}}{2} \right) into y2=(x+c)2+1\displaystyle y^2 = (x + c)^2 + 1 yields (as if by magic),

c=±12\displaystyle \boxed{c = \pm \frac{1}{2}}.

Therefore, all the solution curves that pass through one, pass through the other point as well, and we have the following two functions:

Unparseable latex formula:

\displaystyle[br]y^2 = \left\{ \begin{array}{ll}[br]1 - (x + \frac{1}{2})^2 & \mathrm{if} ~| y | \le 1\\[br] & \\[br]1 + (x + \frac{1}{2})^2 & \mathrm{if} ~| y | > 1\\[br]\end{array}[br]



and

Unparseable latex formula:

\displaystyle[br]y^2 = \left\{ \begin{array}{ll}[br]1 - (x - \frac{1}{2})^2 & \mathrm{if} ~| y | \le 1\\[br] & \\[br]1 + (x - \frac{1}{2})^2 & \mathrm{if} ~| y | > 1\\[br]\end{array}[br]



To prove that y=±1y = \pm 1 satisfies the differential equation, we simply substitute them straight in, yielding,

dydx=0\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 0, as desired.

It should be fairly clear that in the case y<1| y | < 1, we have the formula for a circle:

y2+(x+c)2=1\displaystyle y^2 + (x + c)^2 = 1

Specifically, for the two functions, circles with centers (±12,0)\displaystyle \left(\pm \frac{1}{2} , 0 \right), and radius 1.

To sketch the functions for y>1| y | > 1 , we simply consider what happens as x±x \to \pm \infty, and notice that they are symmetric in the xx axis.

I will attach a sketch as soon as I can.
Reply 16
Many thanks :smile:. I'll bear that in mind.
Reply 17
STEP II 2002 Q9

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Reply 18
STEP II 2002 Q10

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Reply 19
STEP II 2002 Q11

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Yay, all three mechanics.

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