(Updated as far as #57)

STEP II:

1: Solution by Zacken

2: Solution by IrrationalRoot

3: Soluions by Zacken

4: Solution by Zacken

5: Solution by Zacken

6: Solution by Zacken

7: Solution by B_9710

8: Solution by Farhan.Hanif93

9: Solution by A-cute-Angle

10: Solution by Zacken

11: Solution by The-Spartan

12: Solution by Farhan.Hanif3

13: Solution by Farhan.Hanif3

STEP III:

1: Solution by IrrationalRoot

2: Solution by Farhan.Hanif93

3: Solution by Zacken

4: Solution by EricPiphany

5: Solution by IrrationalRoot

6: Solution by Farhan.Hanif93

7: Solution by Farhan.Hanif93

8: Solution by IrrationalRoot

9: Solution by Farhan.Hanif93

10: Solution by Farhan.Hanif93

11: Solution by Farhan.Hanif93

12: Solution by Zacken

13: Solution by Farhan.Hanif3

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 - ...

STEP II:

1: Solution by Zacken

2: Solution by IrrationalRoot

3: Soluions by Zacken

4: Solution by Zacken

5: Solution by Zacken

6: Solution by Zacken

7: Solution by B_9710

8: Solution by Farhan.Hanif93

9: Solution by A-cute-Angle

10: Solution by Zacken

11: Solution by The-Spartan

12: Solution by Farhan.Hanif3

13: Solution by Farhan.Hanif3

STEP III:

1: Solution by IrrationalRoot

2: Solution by Farhan.Hanif93

3: Solution by Zacken

4: Solution by EricPiphany

5: Solution by IrrationalRoot

6: Solution by Farhan.Hanif93

7: Solution by Farhan.Hanif93

8: Solution by IrrationalRoot

9: Solution by Farhan.Hanif93

10: Solution by Farhan.Hanif93

11: Solution by Farhan.Hanif93

12: Solution by Zacken

13: Solution by Farhan.Hanif3

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 7 years ago)

Scroll to see replies

STEP III, Question 1

p and q

st and s+t

p+q=0

(edited 7 years ago)

Original post by Zacken

(Updated as far as #5)

STEP II:

1: Solution by Zacken

2:

3:

4: Solution by Zacken

5: Solution by Zacken

6:

7:

8:

9:

10:

11:

12:

13:

STEP III:

1: Solution by IrrationalRoot

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

13:

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 - ...

STEP II:

1: Solution by Zacken

2:

3:

4: Solution by Zacken

5: Solution by Zacken

6:

7:

8:

9:

10:

11:

12:

13:

STEP III:

1: Solution by IrrationalRoot

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

13:

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 - ...

I'll take ii q11 for now. Will probably focus on the applied seeing as no-one else seems to like it :-P

Posted from TSR Mobile

STEP III, Question 5

$yz+zx+xy=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\dfrac{(1)^2-2}{2}=-\dfrac{1}{2}$

as required.

$x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=x^2(y+z)+y^2(z+x)+z^2(x+y)= x^2(1-x)+y^2(1-y)+z^2(1-z)=x^2+y^2+z^2-(x^3+y^3+z^3)=2-3=-1$

as required.

$-\frac{1}{2}=-\frac{1}{2} \times 1=(yz+zx+xy)(x+y+z)=xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz=3xyz+x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=3xyz-1[br][br]\Rightarrow xyz=\dfrac{1-\frac{1}{2}}{3}=\dfrac{1}{6}$

as required.

We require

$S_{n+1}=aS_n+bS_{n-1}+cS_{n-2} [br][br]\Leftrightarrow x^{n+1}+y^{n+1}+z^{n+1}=ax^n+ay^n+az^n+bx^{n-1}+by^{n-1}+bz^{n-1}+cx^{n-2}+cy^{n-2}+cz^{n-2} [br][br]\Leftrightarrow x^{n-2}(x^3-ax^2-bx-c)+y^{n-2}(y^3-ay^2-by-c)+z^{n-2}(z^3-az^2-bz-c)=0$.

It therefore suffices to find $a$, $b$ and $c$ such that the cubic $m^3-am^2-bm-c=0$ for $m=x$,$y$,$z$.

Therefore we can take $a=x+y+z=1$, $b=-(yz+zx+xy)=\dfrac{1}{2}$ and $c=xyz=\dfrac{1}{6}$.

$yz+zx+xy=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\dfrac{(1)^2-2}{2}=-\dfrac{1}{2}$

as required.

$x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=x^2(y+z)+y^2(z+x)+z^2(x+y)= x^2(1-x)+y^2(1-y)+z^2(1-z)=x^2+y^2+z^2-(x^3+y^3+z^3)=2-3=-1$

as required.

$-\frac{1}{2}=-\frac{1}{2} \times 1=(yz+zx+xy)(x+y+z)=xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz=3xyz+x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=3xyz-1[br][br]\Rightarrow xyz=\dfrac{1-\frac{1}{2}}{3}=\dfrac{1}{6}$

as required.

We require

$S_{n+1}=aS_n+bS_{n-1}+cS_{n-2} [br][br]\Leftrightarrow x^{n+1}+y^{n+1}+z^{n+1}=ax^n+ay^n+az^n+bx^{n-1}+by^{n-1}+bz^{n-1}+cx^{n-2}+cy^{n-2}+cz^{n-2} [br][br]\Leftrightarrow x^{n-2}(x^3-ax^2-bx-c)+y^{n-2}(y^3-ay^2-by-c)+z^{n-2}(z^3-az^2-bz-c)=0$.

It therefore suffices to find $a$, $b$ and $c$ such that the cubic $m^3-am^2-bm-c=0$ for $m=x$,$y$,$z$.

Therefore we can take $a=x+y+z=1$, $b=-(yz+zx+xy)=\dfrac{1}{2}$ and $c=xyz=\dfrac{1}{6}$.

(edited 7 years ago)

Can I reserve STEP III Q4?

Original post by EricPiphany

Can I reserve STEP III Q4?

I was gonna do that one but I'm pretty new to Latex and so I'd be terribly slow with all those integrals and stuff so go ahead haha.

I'd like to reserve II Q2 (i.e. I'm doing it now) .

EDIT: Meant to say Q2 sorry.

(edited 7 years ago)

STEP III Q4

(i).

Laplace transform of $\displaystyle e^{-bt}f(t)$ is

(ii).

Laplace transform of $\displaystyle f(at)$ is

$\displaystyle \int_0^\infty e^{-st}f(at) \,\mathrm{d}t$, making the substitution $\displaystyle u = at, \mathrm{d}u = a \mathrm{d}t$,

$\displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}u}f(u) \,\mathrm{d}u = \displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}t}f(t) \,\mathrm{d}t = a^{-1}F\left(\frac{s}{a}\right)$.

(iii).

Laplace transform of $\displaystyle f'(t)$ is

$\displaystyle \int_0^\infty e^{-st}f'(t) \,\mathrm{d}t = \left[e^{-st}f(t)\right]_0^\infty + s\int_0^\infty e^{-st}f(t) \,\mathrm{d}t = -f(0)+sF(s)$, using integration by parts.

(iv).

Laplace transform of $\displaystyle \sin(t)$ is

$\displaystyle \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = \left[-e^{-st}\cos(t)\right]_0^\infty -s \int_0^\infty e^{-st}\cos(t) \,\mathrm{d}t =$

$\displaystyle =1 - s \left[e^{-st}\sin(t)\right]_0^\infty -s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = 1- s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t$.

So, $\displaystyle \mathrm{I} = 1-s^2 \mathrm{I}$, so $\displaystyle \mathrm{I}=\frac{1}{1+s^2}$.

Now, $\displaystyle \cos(t)$ is the derivative of $\displaystyle \sin(t)$, so by part (iii), Laplace transform of $\displaystyle \cos(t)$ is $\dfrac{s}{1+s^2}$.

And by part (ii), Laplace transform of $\displaystyle \cos(qt)$ is $\displaystyle q^{-1} \left(\frac{s/q}{1+(s/q)^2}\right) = \frac{s}{s^2+q^2}$.

Thus, by part (i), Laplace transform of $\displaystyle e^{-pt}\cos(qt)$ is $\displaystyle \frac{s+p}{q^2+(s+p)^2}$.

What a mess, feel free to edit and repost.

(i).

Laplace transform of $\displaystyle e^{-bt}f(t)$ is

Unparseable latex formula:

.\displaystyle \int_0^\infty e^{-st}\left(e^{-bt}f(t)\right) \,\mathrm{d}t = \int_0^\infty e^{-(s+b)t}f(t)\right) \,\mathrm{d}t = F(s+b)

(ii).

Laplace transform of $\displaystyle f(at)$ is

$\displaystyle \int_0^\infty e^{-st}f(at) \,\mathrm{d}t$, making the substitution $\displaystyle u = at, \mathrm{d}u = a \mathrm{d}t$,

$\displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}u}f(u) \,\mathrm{d}u = \displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}t}f(t) \,\mathrm{d}t = a^{-1}F\left(\frac{s}{a}\right)$.

(iii).

Laplace transform of $\displaystyle f'(t)$ is

$\displaystyle \int_0^\infty e^{-st}f'(t) \,\mathrm{d}t = \left[e^{-st}f(t)\right]_0^\infty + s\int_0^\infty e^{-st}f(t) \,\mathrm{d}t = -f(0)+sF(s)$, using integration by parts.

(iv).

Laplace transform of $\displaystyle \sin(t)$ is

$\displaystyle \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = \left[-e^{-st}\cos(t)\right]_0^\infty -s \int_0^\infty e^{-st}\cos(t) \,\mathrm{d}t =$

$\displaystyle =1 - s \left[e^{-st}\sin(t)\right]_0^\infty -s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = 1- s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t$.

So, $\displaystyle \mathrm{I} = 1-s^2 \mathrm{I}$, so $\displaystyle \mathrm{I}=\frac{1}{1+s^2}$.

Now, $\displaystyle \cos(t)$ is the derivative of $\displaystyle \sin(t)$, so by part (iii), Laplace transform of $\displaystyle \cos(t)$ is $\dfrac{s}{1+s^2}$.

And by part (ii), Laplace transform of $\displaystyle \cos(qt)$ is $\displaystyle q^{-1} \left(\frac{s/q}{1+(s/q)^2}\right) = \frac{s}{s^2+q^2}$.

Thus, by part (i), Laplace transform of $\displaystyle e^{-pt}\cos(qt)$ is $\displaystyle \frac{s+p}{q^2+(s+p)^2}$.

What a mess, feel free to edit and repost.

(edited 7 years ago)

Original post by StrangeBanana

Doing STEP III, Q5

Done a while ago!!

Original post by IrrationalRoot

Done a while ago!!

hmm

I'm tired

EDIT:

I'll come back tomorrow and see if any are still available

(edited 7 years ago)

STEP II, Question 2

(i)

(ii)

(iii)

(iv)

(edited 7 years ago)

Original post by Zacken

STEP II, Question 4

So it must be the case that$p'(x) = k(x-1)^4(x+1)^4 = k(x^2-1)^4$

So it must be the case that$p'(x) = k(x-1)^4(x+1)^4 = k(x^2-1)^4$

Just a small typo

(edited 7 years ago)

Original post by IrrationalRoot

...

Would have been simpler to note that $\sin (\pi e^x)$ is a bounded function, so that $a^{\sin (\pi e^x)}$ is maximum when $\sin (\pi e^x)$ is 1, giving rise to $a^1$ and $\sin (\pi e^x) = -1$ giving rise to $a^{-1}$.

Original post by EricPiphany

Just a small typo

I'm being incredibly thick, but I can't see it?

Original post by Zacken

I'm being incredibly thick, but I can't see it?

coz I've tried to correct it

Original post by EricPiphany

coz I've tried to correct it

Ahhh, yeah lol, dumb mistake that one. Thanks!

STEP II, Question 3:

$\displaystyle \tan \left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} = \frac{1 - 2\sin \frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} = \frac{1 - \sin x}{\cos x}$.

Which simplifies down to $\sec x - \tan x$.

$\displaystyle \tan \left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} = \frac{1 - 2\sin \frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} = \frac{1 - \sin x}{\cos x}$.

Which simplifies down to $\sec x - \tan x$.

(i)

(ii)

(iii)

(edited 7 years ago)

STEP II Q8

Solution

(edited 7 years ago)

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