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STEP I:

1: Solution by Unbounded

2: Solution by Unbounded

3: Solution by nota bene

4: Solution by SimonM

5: Solution by nota bene

6: Solution by Unbounded

7: Solution by sonofdot

8: Solution by Unbounded

9: Solution by cliverlong

10: Solution by Dadeyemi

11:Solution by Unbounded

12: Solution by Robbie10538

13: Solution by Unbounded

14: Solution by brianeverit

STEP II:

1: Solution by sonofdot

2: Solution by dadeyemi

3: Solution by sonofdot

4: Solution by dadeyemi

5: Solution by dadeyemi

6: Solution by welshenglish

7: Solution by Glutamic Acid

8: Solution by Glutamic Acid

9: Solution by tommm

10: Solution by tommm

11: Solution by tommm

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

STEP III:

1: Solution by SimonM

2: Solution by Dadeyemi

3: Solution by Daniel Freedman

4: Solution by Dadeyemi

5: Solution by Dadeyemi

6: Solution by Elongar

7: Solution by Dadeyemi

8: Solution by Dadeyemi

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

Solutions written by TSR members:

1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

1: Solution by Unbounded

2: Solution by Unbounded

3: Solution by nota bene

4: Solution by SimonM

5: Solution by nota bene

6: Solution by Unbounded

7: Solution by sonofdot

8: Solution by Unbounded

9: Solution by cliverlong

10: Solution by Dadeyemi

11:Solution by Unbounded

12: Solution by Robbie10538

13: Solution by Unbounded

14: Solution by brianeverit

STEP II:

1: Solution by sonofdot

2: Solution by dadeyemi

3: Solution by sonofdot

4: Solution by dadeyemi

5: Solution by dadeyemi

6: Solution by welshenglish

7: Solution by Glutamic Acid

8: Solution by Glutamic Acid

9: Solution by tommm

10: Solution by tommm

11: Solution by tommm

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

STEP III:

1: Solution by SimonM

2: Solution by Dadeyemi

3: Solution by Daniel Freedman

4: Solution by Dadeyemi

5: Solution by Dadeyemi

6: Solution by Elongar

7: Solution by Dadeyemi

8: Solution by Dadeyemi

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

Solutions written by TSR members:

1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

(edited 13 years ago)

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STEP III, Question 1

a)

b)

a)

Spoiler

b)

Spoiler

Some more;

Did these quite a while ago I'm afraid some may be partial solutions.

Did these quite a while ago I'm afraid some may be partial solutions.

STEP I 2002 Question 7

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

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

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

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

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

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}

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

(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 $y = ae^{-(x+1)}$ 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 $y = be^{x-1}$ 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.

$\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = e^x - 1 \Rightarrow \ln y = e^x - x + k \Rightarrow y = Ae^{e^x - x}$ for x > 0.

Given that y = e^e when x = 1, e^e = Ae^(e - 1) so A = e; $y = e^{e^x - x + 1}$

$\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = 1 - e^x \Rightarrow \ln y = x - e^x + c \Rightarrow y = Be^{x - e^x}$ for x < 0.

At x = 0, y = $e^{1 - 0 + 1} = e^2$ using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so

$e^2 = Be^{-1} \Rightarrow B = e^3$

$y = e^{e^x - x + 1} \text{for} x > 0$

$y = e^{3 + x - e^x} \text{for} x < 0$

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

(ii) $\displaystyle \lim_{x \to - \infty} e^{3 + x - e^x} e^{-x} = \lim_{x \to -\infty} e^{3 - e^x} = e^3$.

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 $y = ae^{-(x+1)}$ 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 $y = be^{x-1}$ 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.

$\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = e^x - 1 \Rightarrow \ln y = e^x - x + k \Rightarrow y = Ae^{e^x - x}$ for x > 0.

Given that y = e^e when x = 1, e^e = Ae^(e - 1) so A = e; $y = e^{e^x - x + 1}$

$\dfrac{1}{y} \dfrac{\text{d}y}{\text{d}x} = 1 - e^x \Rightarrow \ln y = x - e^x + c \Rightarrow y = Be^{x - e^x}$ for x < 0.

At x = 0, y = $e^{1 - 0 + 1} = e^2$ using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so

$e^2 = Be^{-1} \Rightarrow B = e^3$

$y = e^{e^x - x + 1} \text{for} x > 0$

$y = e^{3 + x - e^x} \text{for} x < 0$

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

(ii) $\displaystyle \lim_{x \to - \infty} e^{3 + x - e^x} e^{-x} = \lim_{x \to -\infty} e^{3 - e^x} = e^3$.

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

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

$\displaystyle =\frac{1}{2}\left[-\cot^2\theta\right]_{\pi/6}^{\pi/4} =$

$\displaystyle = \frac{1}{2}\left(\frac{\sqrt{3}/2}{1/2} - 1\right) = \frac{\sqrt{3}}{2} - \frac{1}{2}$ as required.

$\displaystyle x = \sin{2\theta} \implies \theta = \frac{1}{2}\arcsin{x}$

and $\implies \mathrm{d}x =$

$\displaystyle = 2\cos{2\theta}\mathrm{d}\theta$

$\displaystyle \implies \int_{\sqrt{3}/2}^1 \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x =$

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

$\displaystyle = \sqrt{3}-1-\int_{\pi/6}^{\pi/4}2\,\mathrm{d}\theta = \sqrt{3} - 1 - \frac{\pi}{6}$, as required.

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

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

$\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-xy^2}{\frac{1}{x}-\sqrt{\frac{1}{x^2}-1}} \,\mathrm{d}x =$

$\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-x^2y^2}{1-\sqrt{1-x^2}} \,\mathrm{d}x =$

$\displaystyle = \int_{\sqrt{3}{2}}^{1} \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x = \sqrt{3}-1-\frac{\pi}{6}$

2002 II Question 1

Unparseable latex formula:

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

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

$\displaystyle =\frac{1}{2}\left[-\cot^2\theta\right]_{\pi/6}^{\pi/4} =$

$\displaystyle = \frac{1}{2}\left(\frac{\sqrt{3}/2}{1/2} - 1\right) = \frac{\sqrt{3}}{2} - \frac{1}{2}$ as required.

$\displaystyle x = \sin{2\theta} \implies \theta = \frac{1}{2}\arcsin{x}$

and $\implies \mathrm{d}x =$

$\displaystyle = 2\cos{2\theta}\mathrm{d}\theta$

$\displaystyle \implies \int_{\sqrt{3}/2}^1 \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x =$

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

$\displaystyle = \sqrt{3}-1-\int_{\pi/6}^{\pi/4}2\,\mathrm{d}\theta = \sqrt{3} - 1 - \frac{\pi}{6}$, as required.

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

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

$\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-xy^2}{\frac{1}{x}-\sqrt{\frac{1}{x^2}-1}} \,\mathrm{d}x =$

$\displaystyle = \int_{1}^{\sqrt{3}{2}} \frac{-x^2y^2}{1-\sqrt{1-x^2}} \,\mathrm{d}x =$

$\displaystyle = \int_{\sqrt{3}{2}}^{1} \frac{1}{1-\sqrt{1-x^2}}\,\mathrm{d}x = \sqrt{3}-1-\frac{\pi}{6}$

2002 II Question 11

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

Total moment

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

We know that total moment $\displaystyle = W\overline{x}$, so $\displaystyle\overline{x} = \frac{\alpha l}{\alpha+1}$, 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 $\displaystyle \overline{x_A} = l - \overline{x} = \frac{l}{\alpha+1}$ to be the distance of c.o.m. from A. Angle $\displaystyle\theta$ going from the horizontal to the line AB.

Equating the horizontal forces: $\displaystyle F_A = R_B$. Noting that $\displaystyle F_A = \mu R_A$, we get $\displaystyle\mu R_a = r_B$. Noting that $\displaystyle f_B=\mu R_B$, we get $\displaystyle F_B = \mu^2 R_A$

Equating the vertical forces: $\displaystyle W = R_A+F_B = R_A(1+\mu^2)$.

Equating the moments: $\displaystyle W\overline{x_A}\cos\theta = R_Bl\sin\theta + F_Bl\cos\theta \implies R_Bl\tan\theta = W\overline{x_A} - F_Bl = \mu R_al\tan\theta = R_A(1+\mu^2)\overline{x_A} - \mu^2R_Al$

$\displaystyle \implies \mu l\tan\theta = (1+\mu^2)\overline{x_A} - \mu^2l = (1+\mu^2)\frac{l}{\alpha+1}-\mu^2l$

$\displaystyle \implies \mu\tan\theta = \frac{1+\mu^2}{\alpha+1} - \mu^2 = \frac{1-\alpha\mu^2}{\alpha+1}$

$\displaystyle \implies \tan\theta = \frac{1-\alpha\mu^2}{(1_\alpha)\mu}$ as required.

Not sure if I'm right, but I'll give it a go:

$\displaystyle \alpha = \frac{3}{2}$, and $\displaystyle \theta = \frac{\pi}{4} \implies (1+\alpha)\mu = 1-\alpha\mu^2 \implies 3\mu^2+5\mu-2 = 0 \implies \mu = \frac{2}{6} = \frac{1}{3}$

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

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

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

We know that total moment $\displaystyle = W\overline{x}$, so $\displaystyle\overline{x} = \frac{\alpha l}{\alpha+1}$, 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 $\displaystyle \overline{x_A} = l - \overline{x} = \frac{l}{\alpha+1}$ to be the distance of c.o.m. from A. Angle $\displaystyle\theta$ going from the horizontal to the line AB.

Equating the horizontal forces: $\displaystyle F_A = R_B$. Noting that $\displaystyle F_A = \mu R_A$, we get $\displaystyle\mu R_a = r_B$. Noting that $\displaystyle f_B=\mu R_B$, we get $\displaystyle F_B = \mu^2 R_A$

Equating the vertical forces: $\displaystyle W = R_A+F_B = R_A(1+\mu^2)$.

Equating the moments: $\displaystyle W\overline{x_A}\cos\theta = R_Bl\sin\theta + F_Bl\cos\theta \implies R_Bl\tan\theta = W\overline{x_A} - F_Bl = \mu R_al\tan\theta = R_A(1+\mu^2)\overline{x_A} - \mu^2R_Al$

$\displaystyle \implies \mu l\tan\theta = (1+\mu^2)\overline{x_A} - \mu^2l = (1+\mu^2)\frac{l}{\alpha+1}-\mu^2l$

$\displaystyle \implies \mu\tan\theta = \frac{1+\mu^2}{\alpha+1} - \mu^2 = \frac{1-\alpha\mu^2}{\alpha+1}$

$\displaystyle \implies \tan\theta = \frac{1-\alpha\mu^2}{(1_\alpha)\mu}$ as required.

Not sure if I'm right, but I'll give it a go:

$\displaystyle \alpha = \frac{3}{2}$, and $\displaystyle \theta = \frac{\pi}{4} \implies (1+\alpha)\mu = 1-\alpha\mu^2 \implies 3\mu^2+5\mu-2 = 0 \implies \mu = \frac{2}{6} = \frac{1}{3}$

2002 II Question 12

$\displaystyle P(\text{k of K days, something with probability of q happens}) = \binom{K}{k}q^k(1-q)^{K-k}$

$q =$probability such that more than l coins of the L coins will produce less than m heads

Estimating q:

We can say that

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

z is the probability that less than m heads are made after M throws.

$\displaystyle Z ~ B(M,p) \implies Z \approx N (Mp, Mp(1-p))$

$\displaystyle z = P(Z<m) = \Phi\left(\frac{(m-\frac{1}{2})-Mp}{\sqrt{Mp(1-p)}}\right) = \Phi\left(\frac{2m-1-Mp}{2\sqrt{Mp(1-p)}}\right)$.

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

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

$\displaystyle P(\text{k of K days, something with probability of q happens}) = \binom{K}{k}q^k(1-q)^{K-k}$

$q =$probability such that more than l coins of the L coins will produce less than m heads

Estimating q:

Unparseable latex formula:

, where z is the probability of a coin producing less than m heads.Q \~ B(L,z)

We can say that

Unparseable latex formula:

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

Unparseable latex formula:

for small z.\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)

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

z is the probability that less than m heads are made after M throws.

$\displaystyle Z ~ B(M,p) \implies Z \approx N (Mp, Mp(1-p))$

$\displaystyle z = P(Z<m) = \Phi\left(\frac{(m-\frac{1}{2})-Mp}{\sqrt{Mp(1-p)}}\right) = \Phi\left(\frac{2m-1-Mp}{2\sqrt{Mp(1-p)}}\right)$.

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

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

2002 II Question 13

$F(a) = 0$

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

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

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

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

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

We know $\displaystyle 1 \leq (2-F(x))^2 \leq 4 \implies \frac{1}{4} \leq \frac{1}{(2-F(x))^2} \leq 1 \implies \frac{1}{2} \leq \frac{2}{(2-F(x))^2} \leq 2$

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

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

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

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

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

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

So $\displaystyle Var[Y] \leq 2Var[X] + 2E[X]^2 - E[Y]^2 \leq 2Var[X] + 2E[X]^2 - \frac{1}{4}E[X]^2 =$

$\displaystyle = 2Var[X] + \frac{7}{4}E[X]^2$, as required.

$F(a) = 0$

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

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

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

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

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

We know $\displaystyle 1 \leq (2-F(x))^2 \leq 4 \implies \frac{1}{4} \leq \frac{1}{(2-F(x))^2} \leq 1 \implies \frac{1}{2} \leq \frac{2}{(2-F(x))^2} \leq 2$

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

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

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

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

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

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

So $\displaystyle Var[Y] \leq 2Var[X] + 2E[X]^2 - E[Y]^2 \leq 2Var[X] + 2E[X]^2 - \frac{1}{4}E[X]^2 =$

$\displaystyle = 2Var[X] + \frac{7}{4}E[X]^2$, as required.

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

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

Standard seperating of variables yields:

$\displaystyle \frac{y^4}{(y^2 - 1)^2} (\mathrm{d}y)^4 = 1 . (\mathrm{d}x)^4$

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

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

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

Resulting in:

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

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

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

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

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

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

Resulting in:

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

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

Or simply:

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

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

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

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

$\displaystyle \boxed{c = \pm \frac{1}{2}}$

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

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

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

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

and

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

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

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

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

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

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

I will attach a sketch as soon as I can.

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

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

Standard seperating of variables yields:

$\displaystyle \frac{y^4}{(y^2 - 1)^2} (\mathrm{d}y)^4 = 1 . (\mathrm{d}x)^4$

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

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

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

Resulting in:

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

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

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

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

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

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

Resulting in:

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

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

Or simply:

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

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

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

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

$\displaystyle \boxed{c = \pm \frac{1}{2}}$

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

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

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

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 $y = \pm 1$ satisfies the differential equation, we simply substitute them straight in, yielding,

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

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

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

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

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

I will attach a sketch as soon as I can.

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