(Updated as far as #109) SimonM - 23.03.2009

If you don't know of this usual past paper site, just ask like...anyone.

Oh also, if you solved a question using a different method, you can post up your solution too so that people might see other ways of looking at the problem.

One last thing, if you post a partially completed solution, make sure to tell us which bits you did/didn't do. Also, remember to tell us clearly which question it is you're posting. Also, I think there's a bit of STEP 1987 here too...

~~~~~~~~~~~~~~~

STEP I (Mathematics):

1: Solution by ukgea

2: Solution by Dystopia

3: Solution by brianeverit

4: Solution by Dystopia

5: Solutions by Nota and Square

6: Solution by ukgea

7: Solution by SimonM

8: Solution by Dystopia

9: Solution by Dystopia

10: Solution by *bobo*

11: Solution by brianeverit

12: Solution by Dystopia

13: Solution by brianeverit

14: Solution by Nota

15: Solution by SimonM

16: Solution by Nota

STEP II (Further Mathematics Paper A):

1: Solution by Nota

2: Solution by Dystopia

3: Solution by brianeverit

4: Solution by Simba

5: Rant/explanation and solution by ukgea

6: Solution by Simba

7: Solution by SimonM

8: Mostly done by *bobo*, middle bit by DFranklin and last part by Rabite

9: Solution by SimonM

10: Solution by nota bene

11: Solution by *bobo*

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

15: Solution by SimonM

16: Solution by brianeverit

STEP III (Further Mathematics Paper B):

1: Solution by Dystopia

2: Solution by Simba

3: Solution by ukgea

4: Solution by SimonM

5: Solution by Speleo

6: Solution by Dystopia

7: Solution by SimonM

8: Dodgy solution and disclaimer by Dystopia

9: Solution by Speleo

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by Dystopia

14: Solution by brianeverit

15: Solution by SimonM

16: Solution by SimonM

~~~~~~~~~~~~~~~

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

Okay, I made the new thread, since everyone seemed to be ruining General's one.

Pull up a chair and join the cool table.

For those who don't come here often, a short introduction:

Solutions for earlier STEPs are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved; if you see any mistakes in solutions posted, please point them out.

(Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.)

You may be wondering why STEP I has hyperbolic functions and differential equations.

The "usual" past paper website may have the papers mislabelled; please check the front of the paper, not the filename, to see which paper you're doing.

(Mathematics = I, further mathematics A = II, further mathematics B = III.)

Pull up a chair and join the cool table.

For those who don't come here often, a short introduction:

Solutions for earlier STEPs are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved; if you see any mistakes in solutions posted, please point them out.

(Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.)

You may be wondering why STEP I has hyperbolic functions and differential equations.

The "usual" past paper website may have the papers mislabelled; please check the front of the paper, not the filename, to see which paper you're doing.

(Mathematics = I, further mathematics A = II, further mathematics B = III.)

If you don't know of this usual past paper site, just ask like...anyone.

Oh also, if you solved a question using a different method, you can post up your solution too so that people might see other ways of looking at the problem.

One last thing, if you post a partially completed solution, make sure to tell us which bits you did/didn't do. Also, remember to tell us clearly which question it is you're posting. Also, I think there's a bit of STEP 1987 here too...

~~~~~~~~~~~~~~~

STEP I (Mathematics):

1: Solution by ukgea

2: Solution by Dystopia

3: Solution by brianeverit

4: Solution by Dystopia

5: Solutions by Nota and Square

6: Solution by ukgea

7: Solution by SimonM

8: Solution by Dystopia

9: Solution by Dystopia

10: Solution by *bobo*

11: Solution by brianeverit

12: Solution by Dystopia

13: Solution by brianeverit

14: Solution by Nota

15: Solution by SimonM

16: Solution by Nota

STEP II (Further Mathematics Paper A):

1: Solution by Nota

2: Solution by Dystopia

3: Solution by brianeverit

4: Solution by Simba

5: Rant/explanation and solution by ukgea

6: Solution by Simba

7: Solution by SimonM

8: Mostly done by *bobo*, middle bit by DFranklin and last part by Rabite

9: Solution by SimonM

10: Solution by nota bene

11: Solution by *bobo*

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

15: Solution by SimonM

16: Solution by brianeverit

STEP III (Further Mathematics Paper B):

1: Solution by Dystopia

2: Solution by Simba

3: Solution by ukgea

4: Solution by SimonM

5: Solution by Speleo

6: Solution by Dystopia

7: Solution by SimonM

8: Dodgy solution and disclaimer by Dystopia

9: Solution by Speleo

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by Dystopia

14: Solution by brianeverit

15: Solution by SimonM

16: Solution by SimonM

~~~~~~~~~~~~~~~

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

Scroll to see replies

I am

Edit: Okay I'll start typing up II/1

Prove that both $x^4-2x^3+x^2$ and $x^2-8x+17$ are non-negative for all real x.

Factorise $x^4-2x^3+x^2$ to $(x^2-1)^2$, which clearly is positive.

We can verify that $x^2-8x+17$ is positive for all real x by rewriting it as $(x-4)^2+1$, which clearly is positive.

The equation $x^4-2x^3+x^2-8x+17=0$ has clearly no real solutions for $x\le0$ as we have earlier shown that $x^4-2x^3+x^2$ is always positive, and -8(-x)+17 is always positive since two negative numbers multiplied is positive...

For the interval x>2 we can consider $x^4-2x^3$ as we have earlier shown that$x^2-8x+17$ always is positive for all real x. Factorise it to get $x^3(x-2)$. For positive x $x^3$ is always positive, and for numbers over 2$(x-2)$ is always postitive (and when x=2 the whole equation is still positive, as we have the x^2 etc. terms as well.)

Now consider the interval $0<x\le2$. Factorise $x^4-2x^3+x^2$ to $(x^2-1)^2$, which clearly is positive. For 17-8x where x is less than 2 it is obvious that it will never be negative.

Prove that the equation $x^4-x^3+x^2-4x+4=0$ has no real roots. First, consider the interval x<1, where we can use the factorisation into $4(1-x)+x^2((x-\frac{1}{2})^2+\frac{3}{4})$. Now all squared things are clearly positive, as is the constant, and for the chosen interval 4(1-x) is positive as well.

Consider x=1, here everything stays the same, except the 4(1-x) part that disappears, and the expression as a whole is still positive.

Now consider x>1 - factorise into$x^3(x-1)$ (which is positive as the cube of a positive number is positive, and x-1 for x>1 is positive) and $(x-2)^2$, which obviously is positive as it is an even power.

Therefore , for all x the equation $x^4-x^3+x^2-4x+4=0$ has no real solutions (as all y-values are above the x-axis).

Likely I have a few typos and reversed inequalities in there, point out if you see any

Edit: Okay I'll start typing up II/1

Prove that both $x^4-2x^3+x^2$ and $x^2-8x+17$ are non-negative for all real x.

Factorise $x^4-2x^3+x^2$ to $(x^2-1)^2$, which clearly is positive.

We can verify that $x^2-8x+17$ is positive for all real x by rewriting it as $(x-4)^2+1$, which clearly is positive.

The equation $x^4-2x^3+x^2-8x+17=0$ has clearly no real solutions for $x\le0$ as we have earlier shown that $x^4-2x^3+x^2$ is always positive, and -8(-x)+17 is always positive since two negative numbers multiplied is positive...

For the interval x>2 we can consider $x^4-2x^3$ as we have earlier shown that$x^2-8x+17$ always is positive for all real x. Factorise it to get $x^3(x-2)$. For positive x $x^3$ is always positive, and for numbers over 2$(x-2)$ is always postitive (and when x=2 the whole equation is still positive, as we have the x^2 etc. terms as well.)

Now consider the interval $0<x\le2$. Factorise $x^4-2x^3+x^2$ to $(x^2-1)^2$, which clearly is positive. For 17-8x where x is less than 2 it is obvious that it will never be negative.

Prove that the equation $x^4-x^3+x^2-4x+4=0$ has no real roots. First, consider the interval x<1, where we can use the factorisation into $4(1-x)+x^2((x-\frac{1}{2})^2+\frac{3}{4})$. Now all squared things are clearly positive, as is the constant, and for the chosen interval 4(1-x) is positive as well.

Consider x=1, here everything stays the same, except the 4(1-x) part that disappears, and the expression as a whole is still positive.

Now consider x>1 - factorise into$x^3(x-1)$ (which is positive as the cube of a positive number is positive, and x-1 for x>1 is positive) and $(x-2)^2$, which obviously is positive as it is an even power.

Therefore , for all x the equation $x^4-x^3+x^2-4x+4=0$ has no real solutions (as all y-values are above the x-axis).

Likely I have a few typos and reversed inequalities in there, point out if you see any

Rabite

Not in public.

Ahh...it's not copying the formatting ._.

I don't want to have to type out all those links again!

Ahh...it's not copying the formatting ._.

I don't want to have to type out all those links again!

Was that a hint?

Spoiler

III/9 Easy as $\pi e$

tant = sinhu

v = sechu = 1/coshu = 1/sqrt(1+sinh^2u)

= 1/sqrt(1+tan^2t) = 1/sqrt(sec^2t) = 1/sect = cost

dt/du = (dv/du)/(dv/dt) = (-sechutanhu)/(-sint)

= sechu.sqrt(1-sech^2u)/sqrt(1-cos^2t)

= v.sqrt(1-v^2)/sqrt(1-v^2) = v

dv/du = -sechutanhu, -2dv/du = 2sechutanhu

= 2v.sqrt(1-v^2) = 2costsint = 2sin2t

d2v/du2 = d/du(-sechutanhu) = sechutanh^2u - sech^3u

-coshud2v/du2 = sech^2u - tanh^2u = v^2 - (1-v^2)

= cos^2t - sin^2t = cos2t

du/dt.d2v/dt2 = 1/v.(-cost) = -cost/cost = -1

dv/dt.d2u/dt2 = -sint.d/dt(1/v) = -sint.d/dt(sect) = -sint.sect.tant

= -tan^2t

(du/dt)^2 = 1/v^2 = 1/cos^2t = sec^2t = 1 + tan^2t

Sum of these parts = - 1 - tan^2t + 1 + tan^2t = 0

tant = sinhu

v = sechu = 1/coshu = 1/sqrt(1+sinh^2u)

= 1/sqrt(1+tan^2t) = 1/sqrt(sec^2t) = 1/sect = cost

dt/du = (dv/du)/(dv/dt) = (-sechutanhu)/(-sint)

= sechu.sqrt(1-sech^2u)/sqrt(1-cos^2t)

= v.sqrt(1-v^2)/sqrt(1-v^2) = v

dv/du = -sechutanhu, -2dv/du = 2sechutanhu

= 2v.sqrt(1-v^2) = 2costsint = 2sin2t

d2v/du2 = d/du(-sechutanhu) = sechutanh^2u - sech^3u

-coshud2v/du2 = sech^2u - tanh^2u = v^2 - (1-v^2)

= cos^2t - sin^2t = cos2t

du/dt.d2v/dt2 = 1/v.(-cost) = -cost/cost = -1

dv/dt.d2u/dt2 = -sint.d/dt(1/v) = -sint.d/dt(sect) = -sint.sect.tant

= -tan^2t

(du/dt)^2 = 1/v^2 = 1/cos^2t = sec^2t = 1 + tan^2t

Sum of these parts = - 1 - tan^2t + 1 + tan^2t = 0

III/5

$\begin{pmatrix} n \\ r \end{pmatrix} + \begin{pmatrix} n \\ r-1 \end{pmatrix} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} = \frac{n!(n-r+1)}{r!(n-r+1)!} + \frac{n!r}{r!(n-r+1)!}$

$= \frac{n!(n-r+1) + n!r}{r!(n-r+1)!} = \frac{n!(n+1)}{r!(n-r+1)!} = \frac{(n+1)!}{r!(n+1-r)!} = \begin{pmatrix} n+1 \\ r \end{pmatrix}$

By induction. Case when n=1 is clearly just the product rule:

$(uv)^{(1)} = u^{(1)}v + uv^{(1)}$

Assume true for n=k, then differentiate again to get:

$(uv)^{(k+1)} = u^{(k+1)}v + u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k-1)}v^{(2)} + \begin{pmatrix} k \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$

$= u^{(k+1)}v + (\begin{pmatrix} k \\ 0 \end{pmatrix}+\begin{pmatrix} k \\ 1 \end{pmatrix}) u^{(k)}v^{(1)} + (\begin{pmatrix} k \\ 1 \end{pmatrix}+ \begin{pmatrix} k \\ 2 \end{pmatrix})u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$

$= u^{(k+1)}v + \begin{pmatrix} k+1 \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k+1 \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$

Which is the proposition with n=(k+1), therefore by induction it is true for all natural n.

$y=\arcsin x => \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} => \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^\frac{3}{2}}$

Therefore, the proposition for n=0 reduces to:

$\frac{x}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}} = 0$

which is obviously true.

Now assume that the proposition is true for n=k and differentiate:

$(1-x^2)y^{(k+3)} - 2xy^{(k+2)} - (2k+1)xy^{(k+2)} - (2k+1)y^{(k+1)} - k^2y^{(k+1)} = 0$

$(1-x^2)y^{(k+3)} - (2(k+1)+1)xy^{(k+2)} - (k+1)^2y^{(k+1)} = 0$

Which is the proposition for n=k+1. Therefore, by induction, the proposition is true for all non-negative integer n.

I hate latex, it took me probably 3 times as long to type this up as to actually do it on paper.

$\begin{pmatrix} n \\ r \end{pmatrix} + \begin{pmatrix} n \\ r-1 \end{pmatrix} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} = \frac{n!(n-r+1)}{r!(n-r+1)!} + \frac{n!r}{r!(n-r+1)!}$

$= \frac{n!(n-r+1) + n!r}{r!(n-r+1)!} = \frac{n!(n+1)}{r!(n-r+1)!} = \frac{(n+1)!}{r!(n+1-r)!} = \begin{pmatrix} n+1 \\ r \end{pmatrix}$

By induction. Case when n=1 is clearly just the product rule:

$(uv)^{(1)} = u^{(1)}v + uv^{(1)}$

Assume true for n=k, then differentiate again to get:

$(uv)^{(k+1)} = u^{(k+1)}v + u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k-1)}v^{(2)} + \begin{pmatrix} k \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$

$= u^{(k+1)}v + (\begin{pmatrix} k \\ 0 \end{pmatrix}+\begin{pmatrix} k \\ 1 \end{pmatrix}) u^{(k)}v^{(1)} + (\begin{pmatrix} k \\ 1 \end{pmatrix}+ \begin{pmatrix} k \\ 2 \end{pmatrix})u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$

$= u^{(k+1)}v + \begin{pmatrix} k+1 \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k+1 \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}$

Which is the proposition with n=(k+1), therefore by induction it is true for all natural n.

$y=\arcsin x => \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} => \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^\frac{3}{2}}$

Therefore, the proposition for n=0 reduces to:

$\frac{x}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}} = 0$

which is obviously true.

Now assume that the proposition is true for n=k and differentiate:

$(1-x^2)y^{(k+3)} - 2xy^{(k+2)} - (2k+1)xy^{(k+2)} - (2k+1)y^{(k+1)} - k^2y^{(k+1)} = 0$

$(1-x^2)y^{(k+3)} - (2(k+1)+1)xy^{(k+2)} - (k+1)^2y^{(k+1)} = 0$

Which is the proposition for n=k+1. Therefore, by induction, the proposition is true for all non-negative integer n.

I hate latex, it took me probably 3 times as long to type this up as to actually do it on paper.

I've been doing III/6, I'm OK down to the part where it says 'state the area of the region bounded by E' (thanks to the parameterisation method that I picked up after I think Rabite used it), and I was wondering whether there was an easy way to work that out based on the original curve, whose area is obviously pi, and the transformation.

I knew there was something funny about that!

It's an ellipse, I'm quite sure. I think you're meant to work out that

"An ellipse centered at the origin can be viewed as the image of the unit circle under a linear map associated with a symmetric matrix A = PDPt, D being a diagonal matrix with the eigenvalues of A, both of which are real positive, along the main diagonal, and P being a real unitary matrix having as columns the eigenvectors of A."

(From Wikipedia)

And the enclosed by an ellipse is pi*ab where a and b are the 'radii' along the y axis and x axis.

It's an ellipse, I'm quite sure. I think you're meant to work out that

"An ellipse centered at the origin can be viewed as the image of the unit circle under a linear map associated with a symmetric matrix A = PDPt, D being a diagonal matrix with the eigenvalues of A, both of which are real positive, along the main diagonal, and P being a real unitary matrix having as columns the eigenvectors of A."

(From Wikipedia)

And the enclosed by an ellipse is pi*ab where a and b are the 'radii' along the y axis and x axis.

I got the first two parts out. For the 3rd

Im stuck with

$2 \displaystyle [y(t)]^2 = \int [y(t)]^4 + 2 \int [x(t)]^2y(t)y'(t)$

The thing is for $\int 2[x(t)]^2y(t)y'(t)$

could you just do

$u = blah$

$du = blah$

$dv = x(t)y(t)$

$v = \int x(t)y(t) = 0$

Therefore $\int [x(t)]^2y(t)y'(t)$ = blah(0) - \int blah(0) = 0

Hmm i havent looked at the spoiler because this very frustrating. Even if i get it out, i doesn't make me a better mathematician for being able to do it. There seems to be loads of ways you can do by parts with these integrals all leading on to different integrals. It just seems thats its luck you pick the right combination of u = .. and dv/dx = ... and maybe a bit of experience in these matters.

Im stuck with

$2 \displaystyle [y(t)]^2 = \int [y(t)]^4 + 2 \int [x(t)]^2y(t)y'(t)$

The thing is for $\int 2[x(t)]^2y(t)y'(t)$

could you just do

$u = blah$

$du = blah$

$dv = x(t)y(t)$

$v = \int x(t)y(t) = 0$

Therefore $\int [x(t)]^2y(t)y'(t)$ = blah(0) - \int blah(0) = 0

Hmm i havent looked at the spoiler because this very frustrating. Even if i get it out, i doesn't make me a better mathematician for being able to do it. There seems to be loads of ways you can do by parts with these integrals all leading on to different integrals. It just seems thats its luck you pick the right combination of u = .. and dv/dx = ... and maybe a bit of experience in these matters.

Speleo

I've been doing III/6, I'm OK down to the part where it says 'state the area of the region bounded by E' (thanks to the parameterisation method that I picked up after I think Rabite used it), and I was wondering whether there was an easy way to work that out based on the original curve, whose area is obviously pi, and the transformation.

insparato

I got the first two parts out. For the 3rd

Im stuck with

$2 \displaystyle [y(t)]^2 = \int [y(t)]^4 + 2 \int [x(t)]^2y(t)y'(t)$

Im stuck with

$2 \displaystyle [y(t)]^2 = \int [y(t)]^4 + 2 \int [x(t)]^2y(t)y'(t)$

Edit: At first I didn't even really bother looking at what you'd done because to be honest, I thought it was a dead end. But I just looked a little harder, and thought, actually, maybe that might work. But it's a bit hard to tell unless you fill in the details a bit. The problem I'm expecting is that you don't know $v(t) = 0$, you only know $\int_0^1 v dt = 0$. So I think v is going to end up in some of your integrals, which looks like a showstopper. But I might be wrong here...

Hmm i havent looked at the spoiler because this very frustrating. Even if i get it out, i doesnt make me a better mathematician for being able to do it. There seems to be loads of ways you can do by parts with these integrals all leading on to different integrals. It just seems thats just luck you pick the right combination of u = .. and dv/dx = ... and maybe abit of experience in these matters.

If it makes you feel better, there's no way I'd have got this question out under exam conditions. (Well, not unless I was following my advice of "better to get one complete question out than 6 with tiny mistakes" ).

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