SimonM
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(Updated as far as #112. All done) SimonM - 26.04.2009


It looks as if the mark schemes for the older STEP papers have been lost in the ether so to speak. So as it was suggested how about a thread for the papers for each year where we can give a mark scheme to all the questions on the paper that anyone can contribute. Also I would be happy to latex the final mark scheme once they're done too if theres enough interest in it!

STEP I:
1: Solution bySpeleo
2: Solution by Speleo
3: Solution by ukgea
4: Solution by DFranklin
5: Solution by DFranklin
6: Solution by DFranklin
7: Solution by DFranklin
8: Solution by Speleo
9: Solution by DFranklin
10: Solution by Speleo, Alternative by DFranklin
11: Solution by Unbounded
12: Solution by ukgea
13: Solution by DFranklin
14: Solution by DFranklin


STEP II:
1: Solution by Trangulor
2: Solution by DFranklin
3: Solution by ukgea
4: Solution by Datr
5: Solution by ukgea
6: Solution by khaixiang
7: Solution by khaixiang
8: Solution by ukgea
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by ukgea
12: Solution by Danniella
13: Solution by brianeverit
14: Solution by brianeverit


STEP III:
1: Solution by dvs
2: Solution by dvs
3: Solution by Speleo
4: Solution by dvs
5: Solution by dvs
6: Solution by ukgea
7: Solution by dvs
8: Solution by Farhan.Hanif93 (earlier slightly flawed solution by ukgea)
9: Solution by DFranklin
10: Solution by DFranklin
11: Solution by DFranklin, Alternative by dr_98_98
12: Solution by DFranklin
13: Solution by DFranklin
14: Solution by DFranklin


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
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Speleo
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I am currently typing out STEP I Question 2.

STEP I Question 1

Total number of combinations =
Number using 0 and 1 +
Number using 0 and 2 +
...
Number using 0 and 9 +
Number using 1 and 2 +
...
Number using 8 and 9.

Now, all of these numbers are going to be 5 digits long except for 100,000. You can also see that 0 is going to be tricky because we need to insure that we don't get 4 digits numbers, e.g. 01011
So, put the combinations with 0 to one side for a minute.

There are clearly the same number of combinations of 1 and 2 as of 3 and 4 or of 4 and 7 or indeed any pair (excluding ones with zero).
So our result for (1,2) will hold for all the other pairs, of which there are:
(1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,8)(1,9)
(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9) [remember (2,1) is already counted as (1,2)]
...
(7,8)(7,9)
(8,9)
This is clearly 1 + 2 + ... + 8 = 8*9/2 = 36 pairs.

Now, call our random pair (a,b).
We need the number of combinations with one a, and with 2 a's, and with 3 a's, and with 4a's. 1a and 5a's are not allowed, as the number must contain 5 digits.
The number of ways with one a is the number of ways you can pick one number out of the 5, i.e. 5C1 = 5.
Similarly, the number of ways with two a's is the number of ways you can pick two numbers out of the 5, i.e. 5C2 = 10.
You will end up with 5C1 + ... + 5C4 = 30.
Now, multiply 30 by 36 and we have 1080 combinations excluding the ones with zeroes.

Now, ignore for a minute the number 100,000, and we again have symmetry, the number with (0,1) = the number with (0,2) = ... = the number with (0,9).

So, consider (0,a).
The first digit must be a, or the number will end up less than 10,000. There are then 4 numbers left, of which there can be 1, 2, 3 or 4 0's. By a similar argument to that above, the number of combinations is 4C1 + 4C2 + 4C3 + 4C4 = 4 + 6 + 4 + 1 = 15.
So 15*9 = 135 more combinations to add.
1080 + 135 = 1215.

But, we are still left with 100,000, which is another number which works, so we add one on to get:

1216
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ukgea
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STEP I Question 3

(i) True. Consider the equation
\ln a \cdot \ln b = \ln b \cdot \ln a

Exponentiating both sides, we get

e^{\ln a \cdot \ln b} = e^{\ln b \cdot \ln a}

\left(e^{\ln a}\right)^{\ln b} = \left(e^{\ln b}\right)^{\ln b}

a^{\ln b} = b^{\ln a}

which was to be proven.

(ii) False. Consider the case \theta = 0 Then

 \mathrm{LHS} = \cos (\sin 0) = \cos 0 = 1
 \mathrm{RHS} = \sin (\cos 0) = \sin 1

RHS clearly is not 1, so this provides a counterexample for the statement.

(iii) False.

A polynomial must have a finite degree. Let the degree of P(theta) be n. Consider the nth-degree term of P(theta). Since the polynomial has degree n, the coefficient of this term cannot be zero. So, as theta becomes very large, this term grows faster than all the other terms, and thus the polynomial will tend to either positive or negative infinity, and thus further than 10^-6 away from the interval [-1, 1] which is the range of the cos(theta) function.

The only case when the highest-degree term of a polynomial is indeed zero, is for the trivial polynomial P(theta) = 0 for all theta. But this does not satisfy the requirement either, for say theta = 0,

\mathrm{LHS} = | 0 - \cos 0| = 1
which is obviously larger than 10^-6.

Therefore, there does not exist any polynomial satisfying the requirement.

(iv)True. The difference LHS - RHS is

x^4 + 3 + x^{-4} - 5

x^4 + x^{-4} - 2

\left(x^2 - x^{-2}\right)^2

which is larger than or equal to zero, since it's a square, and so LHS >= RHS, which was to be proven.
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Trangulor
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a) Given (n-3)^3+n^3=(n+3)^3 and n is an integer

Asked to show i) n is even, and ii) n^2 is a factor of 54, and that iii) there are no integer values of n that satisfy the above equation

i) If n is odd, then (n-3)^3 and (n+3)^3 are even, whilst n^3 is odd. Even and odd on LHS add up to odd, but RHS is even. This is a contradiction, so n must be even. If n is even, you have odd+even equalling odd, which is true

Alternatively, expand the brackets to give n^3-9n^2+27n-27+n^3=n^3+9n^2+27n+27
n^3-18n^2-54=0
n^3=2(9n^2+27)
If n is an integer, then RHS is even, so LHS is even. If n^3 is even, then n is also even

ii) from above, {n^2}(n-18)=54
If n is an integer, then n^2 is a factor of 54

iii) Since {n^2}(n-18)=54
n^2>0, 54>0
so n-18>0 or n>18
Thus n^2>324
If n is a positive integer greater than 18, {n^2}(n-18)>324 so cannot equal 54, so there are no integer values of n which satisfy the above equation

b) Given (n-6)^3+n^3=(n+6)^3 and n is an integer

Asked to show i) n is even, and that ii) there are no integer values of n that satisfy the above equation

i) Similar thing to part a). If n is odd, all three terms are odd, and since the LHS' odd+odd gives even, and RHS is odd, there is a contradiction. If n is even, all the terms are even, so it works.

Also, by expanding, you get n^3-18n^2+108n-216+n^3=n^3+18n^2+108n+216
n^3-36n^2-432=0
n^3=2(18n^2+216)
If n is an integer, then RHS is even, so LHS is even. If n^3 is even, then n is also even

ii) {n^2}(n-36)=432
n^2>0, 432>0
so n-36>0 or n>36
Thus n^2>1296
If n is a positive integer greater than 36, {n^2}(n-36)>1296 so cannot equal 432, so there are no integer values of n which satisfy the above equation

Thanks to khaixang for spotting my dodgy arithmetic
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Speleo
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I'm not doing this in latex, so you can copy-paste this. Please tell me if you want me to write it out in latex.

STEP I Question 2

The argument of f needs to be t, so clearly:
t = (x^2 + 1)^(1/2) + x
dt/dx = x(x^2 + 1)^(-1/2) + 1
dt/dx = [x + (x^2 + 1)^(1/2)](x^2 + 1)^(-1/2)
dt/dx . (x^2 + 1)^(1/2)/[x + (x^2 + 1)^(1/2)] = 1

The integral becomes:
INT (x^2 + 1)^(1/2)/[x + (x^2 + 1)^(1/2)] f(t) dt
= INT (t-x)/t f(t) dt
= INT (1 - x/t) f(t) dt
= 1/2 INT (2 - 2x/t) f(t) dt

Now, 1 - 2x/t = [t - 2x]/t
= [(x^2 + 1)^(1/2) - x]/[(x^2 + 1)^(1/2) + x]
= [(x^2 + 1)^(1/2) - x].[(x^2 + 1)^(1/2) + x]/[(x^2 + 1)^(1/2) + x]^2
= 1/t^2

= 1/2 INT [1 + t^(-2)]f(t) dt

When x = 0, t = (0 + 1)^(1/2) + 0 = 1
As x->infinity, both terms of t go to infinity so t-> infinity.
Therefore, the limits are 1 and infinity, as required.


Now, let f(x) = x^(-3)

The integral becomes:
(1/2) INT (1 + t^-2)(t^-3)dt
= (1/2) INT t^-3 + t^-5 dt
= (1/2) [-(t^-2)/2 - (t^-4)/4]
Limits 1 and infinity:
= (1/2) [- 0 - 0] - (1/2)[-1/2 - 1/4]
= (1/2)(3/4)

= 3/8
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insparato
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(Original post by Speleo)
I'm not doing this in latex, so you can copy-paste this. Please tell me if you want me to write it out in latex.

STEP I Question 2

The argument of f needs to be t, so clearly:
t = (x^2 + 1)^(1/2) + x
dt/dx = x(x^2 + 1)^(-1/2) + 1
dt/dx = [x + (x^2 + 1)^(1/2)](x^2 + 1)^(-1/2)
dt/dx . (x^2 + 1)^(1/2)/[x + (x^2 + 1)^(1/2)] = 1

The integral becomes:
INT (x^2 + 1)^(1/2)/[x + (x^2 + 1)^(1/2)] f(t) dt
= INT (t-x)/t f(t) dt
= INT (1 - x/t) f(t) dt
= 1/2 INT (2 - 2x/t) f(t) dt

Now, 1 - 2x/t = [t - 2x]/t
= [(x^2 + 1)^(1/2) - x]/[(x^2 + 1)^(1/2) + x]
= [(x^2 + 1)^(1/2) - x].[(x^2 + 1)^(1/2) + x]/[(x^2 + 1)^(1/2) + x]^2
= 1/t^2

= 1/2 INT [1 + t^(-2)]f(t) dt

When x = 0, t = (0 + 1)^(1/2) + 0 = 1
As x->infinity, both terms of t go to infinity so t-> infinity.
Therefore, the limits are 1 and infinity, as required.


Now, let f(x) = x^(-3)

The integral becomes:
(1/2) INT (1 + t^-2)(t^-3)dt
= (1/2) INT t^-3 + t^-5 dt
= (1/2) [-(t^-2)/2 - (t^-4)/4]
Limits 1 and infinity:
= (1/2) [- 0 - 0] - (1/2)[-1/2 - 1/4]
= (1/2)(3/4)

= 3/8
If people want it ill make a latex document for each paper.
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DFranklin
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I'll try and go through Paper I, Q5,6 tonight.
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datr
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STEP II Q4

\displaystyle I_n = \int_0^\pi (\frac{\pi}{2} - x)sin(nx+\frac{x}{2})cosec(\frac{x}{2}) \,dx

\displaystyle I_n - I_{n-1} = \int_0^\pi (\frac{\pi}{2} - x)cosec(\frac{x}{2})(sin(nx+\frac{x}{2})-sin((n-1)x+\frac{x}{2})) \,dx

\displaystyle I_n - I_{n-1} = \int_0^\pi (\frac{\pi}{2} - x)cosec(\frac{x}{2})(sin(nx+\frac{x}{2})-sin(nx-\frac{x}{2})) \,dx

\displaystyle sin(nx+\frac{x}{2})-sin(nx-\frac{x}{2}) = sin(nx)cos(\frac{x}{2}) + sin(\frac{x}{2})cos(nx) - sin(nx)cos(\frac{x}{2}) + sin(\frac{x}{2})cos(nx) = 2sin(\frac{x}{2})cos(nx)

\displaystyle I_n - I_{n-1} = 2\int_0^\pi (\frac{\pi}{2} - x)cos(nx) \, dx

\displaystyle I_n - I_{n-1} = 2\left(\left[\frac{(\frac{\pi}{2}-x)sin(nx)}{n}\right]_0^\pi + \frac{1}{n}\int_0^\pi sin(nx) \, dx\right)

\displaystyle I_n - I_{n-1} = -\frac{2}{n^2}[cos(nx)]_0^\pi = \frac{4}{n^2}

\displaystyle I_n = \sum_{r=1}^{n} \frac{4}{r^2}

NB As you can see the algebra gets fiddly around the middle, I think it's correct but I'd be grateful to anyone who can confirm i haven't done anything stupid.

{mod comment}: The observant will notice that the value of [\cos(nx)]_0^\pi depends on whether or not n is even. Before making yet another post pointing this out, please look at the last few posts in this thread and decide whether your post actually adds anything new. Thanks.
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ukgea
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STEP II Question 3.

First part: Proof by induction.

Claim:
\displaystyle S_N = \frac{1}{2}\left(\frac{3}{2} + \frac{1}{N+1} - \frac{5}{N+2}\right)
for all N.

For N = 1, the claim holds:
\displaystyle S_N = \frac{1}{1\cdot 2 \cdot 3} = \frac{1}{6}

\displaystyle = \frac{1}{2}\left(\frac{3}{2} + \frac{1}{N+1} - \frac{5}{N+2}\right)

\displaystyle = \frac{1}{2}\left(\frac{3}{2} + \frac{1}{2} - \frac{5}{3}\right)

\displaystyle = \frac{1}{2}\left(\frac{9}{6} + \frac{3}{6} - \frac{10}{6}\right)

\displaystyle = \frac{1}{2}\cdot\frac{2}{6}

\displaystyle = \frac{1}{6}

\displaystyle = S_N.

Now assume the claim holds for some N = k, i.e.

\displaystyle S_k = \frac{1}{2}\left(\frac{3}{2} + \frac{1}{k+1} - \frac{5}{k+2}\right).

We shall prove that it will then follow that it holds for N = k + 1.

First, we note that by adding

\displaystyle \frac{2(k+1) - 1}{(k+1)((k+1)+1)((k+1)+2)}

to S_k we get S_{k+1}, i.e.

\displaystyle S_{k+1} - S_k = \frac{2(k+1) - 1}{(k+1)((k+1)+1)((k+1)+2)}

Secondly, we note that the difference

\displaystyle \frac{1}{2}\left(\frac{3}{2} + \frac{1}{(k+1)+1} - \frac{5}{(k+1)+2}\right) - \frac{1}{2}\left(\frac{3}{2} + \frac{1}{k+1} - \frac{5}{k + 2}\right)

 \displaystyle = \frac{1}{2}\left(\frac{1}{(k+1)+1} - \frac{5}{(k+1)+2} - \frac{1}{k+1} + \frac{5}{k+2}\right)

 \displaystyle =\frac{1}{2} \left(\frac{1}{k+2} - \frac{5}{k+3} - \frac{1}{k+1} + \frac{5}{k+2}\right)

 \displaystyle =\frac{1}{2}\left(\frac{6}{k+2} - \frac{5}{k+3} - \frac{1}{k+1}\right)

 \displaystyle =\frac{1}{2}\left(\frac{6(k+1)(k+3)}{(k+1)(k+2)(k+3)} - \frac{5(k+1)(k+2)}{(k+1)(k+2)(k+3)} - \frac{(k+2)(k+3)}{(k+1)(k+2)(k+3)}\right)

\displaystyle = \frac{1}{2} \left(\frac{6(k+1)(k+3) - 5(k+1)(k+2) - (k+2)(k+3)}{(k+1)(k+2)(k+3)}\right)

\displaystyle = \frac{1}{2} \cdot\frac{6k^2 + 24k + 18 - 5k^2 - 15k - 10 - k^2 - 5k - 6}{(k+1)(k+2)(k+3)}

\displaystyle = \frac{1}{2} \cdot\frac{4k - 2}{(k+1)(k+2)(k+3)}

\displaystyle = \frac{2k - 1}{(k+1)(k+2)(k+3)}.

But this was exactly the difference S_{k+1} - S_k, i.e.

\displaystyle S_{k+1} - S_k = \frac{1}{2}\left(\frac{3}{2} + \frac{1}{(k+1)+1} - \frac{5}{(k+1)+2}\right) - \frac{1}{2}\left(\frac{3}{2} + \frac{1}{k+1} - \frac{5}{k + 2}\right).

Adding this to the induction hypothesis, we get

\displaystyle S_{k+1}= \frac{1}{2}\left(\frac{3}{2} + \frac{1}{(k+1)+1} - \frac{5}{(k+1)+2}\right)

and hence by induction, the claims holds for all n \geq 1.

As N \rightarrow \infty,

\displaystyle \frac{1}{N+1} \rightarrow 0

\displaystyle \frac{5}{N + 2} \rightarrow 0

and so

\displaystyle S_N \rightarrow \frac{1}{2}\cdot\frac{3}{2} = \frac{3}{4}.

Second part:

\displaystyle \frac{a_n}{a_1} = \frac{a_2}{a_1}\frac{a_3}{a_2}\frac{a_4}{a_3} \cdots \frac{a_n}{a_{n-1}}

\displaystyle = \prod_{k=2}^{n} \frac{a_k}{a_{k-1}}

\displaystyle = \prod_{k=2}^{n} \frac{(k-1)(2k-1)}{(k+2)(2k-3)}

\displaystyle = \frac{\left(\prod_{k=2}^{n} (k - 1)\right)\left(\prod_{k=2}^{n}(2k-1)\right)}{\left(\prod_{k=2}^{n} (k+2)\right)\left(\prod_{k=2}^{n}(2k-3)\right)}

\displaystyle = \frac{\left(\prod_{k=2}^{n} (k - 1)\right)\left(\prod_{k=2}^{n}(2k-1)\right)}{\left(\prod_{k=5}^{n+3}(k-1)\right)\left(\prod_{k=1}^{n-1}(2k-1)\right)}

(\prod_{k=5}^n (k- 1) and \prod_{k=2}^{n-1} (2k-1) cancel out)

\displaystyle = \frac{\left((2-1)(3-1)(4-1)\right)(2n-1)}{\left((n+1-1)(n+2-1)(n+3-1)\right)(2\cdot 1 - 1)}

\displaystyle = \frac{6(2n-1)}{n(n+1)(n+2)}

Then,

\displaystyle \sum_{n=1}^\infty a_n

\displaystyle =a_1 \sum_{n=1}^\infty \frac{a_n}{a_1}

\displaystyle =a_1 \sum_{n=1}^\infty \frac{6(2n-1)}{n(n+1)(n+2)}

\displaystyle =6a_1 \sum_{n=1}^\infty \frac{2n-1}{n(n+1)(n+2)}

(by the first part)

\displaystyle = 6a_1\frac{3}{4}

\displaystyle = 6\cdot\frac{2}{9}\cdot\frac{3}{4}

= 1

(LaTeXing this was just plain torture! :eek: So I probably made some typos up there :p:. And perhaps some silly mistakes as well. Let me know if you find any!)
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Speleo
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STEP I Question 8

(1/r).d/dr(rdv/dr) = -k
d/dr(rdv/dr) = -kr
rdv/dr = -k(r^2/2) + C
dv/dr = -(1/2)kr + C/r
v = -(1/4)kr^2 + Clnr + D

r->0
lnr -> - infinity
r^2 -> 0

Therefore |v| -> infinity as long as C =/= 0.

v = -(1/4)kr^2 + D
When r=a, v=0
0 = -(k/4)a^2 + D
D = (k/4)a^2

v = (k/4)(a^2 - r^2)

F = 2pi INT rv dr
= 2pi INT (k/4)a^2r - (k/4)r^3 dr
= 2pi [(k/8)a^2r^2 - (k/16)r^4]
limits 0 and a.
-> 2kpi[a^4/8 - a^4/16]
= (a^4).kpi/8


NB: I left a minus sign out early on and had to go through changing it right at the end, so there is a small possibility that there is a mistake somewhere, I would appreciate if someone could check it.
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dvs
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STEP III

Q1.
The sketch is manageable. We need the turning points of f(x) for the next bit. So let's solve the equation f'(x) = 0:
f'(x) = 2sinxcosx - 2sinx = 0
=> sinx(cosx - 1) = 0
=> sinx=0 or cosx=1

So the turning points of f(x) have coords:
(0, 3), (pi, -1) and (2pi, 3)

Now let's solve g'(x) = 0. Using the quotient rule then equating the denominator to zero gives us:
(cf(x) + d)(af'(x)) - (cf'(x))(af(x)+b) = 0
f'(x) (acf(x) + ad - acf(x) - bc) = 0
f'(x) (ad - bc) = 0

Since ad != bc, then f'(x) = 0. So the turning points of g have the same x-coords as those of f. Now let's write them out:
(0, (3a+b)/(3c+d)), (pi, (b-c)/(d-c)) and (2pi, (3a+b)/(3c+d))

If d/c < -3, then d/c < 1. So d-c != 0 and d+3c != 0. Since we're working on the closed interval [0, 2pi], then g(x) cannot be larger in magnitude than (3a+b)/(3c+d) or (b-c)/(d-c). Same comment applies if d/c > 1, because then d/c > -3, and so d+3c != 0 too.
--------------


Q2.
(i)
I(b,a) = \small \int_0^1 t^b (1-t)^a dt

Let's use the substitution T=1-t. Then dT/dt = -1, and:
I(b,a) = - \small \int_1^0 (1-T)^b T^a dT
= \small \int_0^1 T^a (1-T)^b dT
= I(a,b)

Because T is just a dummy variable.

(ii)
I(a+1, b) + I(a, b+1) = \small \int_0^1 t^(a+1) (1-t)^b dt + \small \int_0^1 t^a (1-t)^(b+1) dt
= \small \int_0^1 t^(a+1) (1-t)^b + t^a (1-t)^(b+1) dt
= \small \int_0^1 t^a (1-t)^b (t + 1-t) dt
= \small \int_0^1 t^a (1-t)^b dt
= I(a,b)

(iii)
If we apply integration by parts to I(a,b), differentiating t^a and integrating (1-t)^b, we get:
I(a,b) = [((1-t)^b t^(a+1))/(a+1)]{with limits 0 and 1} - \small \int_0^1 (t^(a+1))/(a+1) * (-b (1-t)^(b-1)) dt
= 0 + b/(a+1) \small \int_0^1 t^(a+1) (1-t)^(b-1) dt
= b/(a+1) I(a+1,b-1)

So
(a+1) I(a,b) = b I(a+1,b-1)

Finally, let's calculate I(a,b):
I(a,b) = b/(a+1) I(a+1, b-1)
= b/(a+1) * (b-1)/(a+2) I(a+2, b-2)
= b/(a+1) * (b-1)/(a+2) * (b-2)/(a+3) I(a+3, b-3)
...
= b/(a+1) * (b-a)/(a+2) * ... * (b-b+1)/(a+b) I(a+b, b-b)
= b!/((a+b)!/a!) \small \int_0^1 t^(a+b) dt
= (a! b!)/(a+b)! * 1/(a+b+1)
= (a! b!)/(a+b+1)!
--------------


Hopefully it's all OK. I'll see if I can attempt a few more later on.
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Speleo
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(not in latex but I may well come through and edit latex in later)

STEP I Question 10 - Credit to Lusus Naturae for catching my glaring error

The particle is launched with speed v at angle a above the horizontal.

Time:
's = ut + 1/2 at^2'
0 = vtsina - gt^2/2
gt^2 = 2vtsina
sin^2a = (gt/2v)^2

Range = horizontal velocity * time
r = vtcosa
cos^2a = (r/vt)^2

1 = (gt/2v)^2 + (r/vt)^2
1 = g^2t^2/4v^2 + r^2/v^2t^2
4v^2t^2 = g^2t^4 + 4r^2
g^2t^4 - 4v^2t^2 + 4r^2 = 0

Quadratic in t^2.

t^2 = [4v^2 +- sqrt(16v^4 - 16r^2g^2)]/2g^2
2g^2t^2 = 4v^2 +- sqrt(16v^4 - 16r^2g^2)

(1/2)g^2t^2 = v^2 +- sqrt(v^4 - r^2g^2)

's = ut + (1/2)at^2'
h = vtsina - gt^2/2

'v = u + at'
0 = vsina - gt
sina = gt/v

h = gt^2 - gt^2/2
h = gt^2/2

So, by the previous result:

h = [v^2 +- sqrt(v^4 - r^2g^2)]/g


NB: I haven't checked this thoroughly, but the first part gets to a stated result and the second part is short and seems to work.
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Speleo
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I've done STEP I question 10 (and 8), not II
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datr
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I've updated my post with STEP II Q4 as well.
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insparato
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(Original post by Speleo)
I've done STEP I question 10 (and 8), not II
Corrected.
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DFranklin
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Step I, Q5:

(i)  Q=4+6i, A = 10+2i, A-Q = 6 - 4i. B-Q will be A-Q rotated by 60 degrees (pi /3) clockwise. So
B-Q = (A-Q)e^{-i\pi/3} = (6-4i)(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}) = (3-2i)(1-i\sqrt{3}) = (3-2\sqrt{3})-(2+3\sqrt{3})i.

So B = Q+(B-Q) = (7 - 2\sqrt{3})+(4-3\sqrt{3})i.

\text{(ii) } Im\left(\frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i}\right) = Im\left(\frac{a(1-i)}{(1+i)(1-i)}+\frac{b(1-2i)}{(1+2i)(1-2i)}+\frac{c(1-3i)}{(1+3i)(1-3i)}\right)\\=

-\left(\frac{a}{1+1}+\frac{2b}{1+4}+\frac{3c}{1+9} \right)
\text{thus } \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} \text{ is real when } \frac{a}{2}+\frac{2b}{5}+\frac{3c}{10} = 0 \text{ i.e. }5a+4b+3c=0.

(Hmm, why did I pick a question with so much opportunity for errors and no result to check against? Plus, I'd forgotten what a pain Latex can be...)
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Speleo
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STEP III Question 3

V_n+1 = (1+c)V_n - d

Let V_n = Ak^n + B

Ak^(n+1) + B = (1+c)(Ak^n + B) - d
kAk^n + B = (1+c)Ak^n + B + Bc - d

Let B = d/c

kAk^n = (1+c)Ak^n
k = (1+c)

V_n = A(1+c)^n + (d/c)
V_0 = A + (d/c)

V_n = (V_0 - d/c)(1+c)^n + (d/c)

c = 0
V_n+1 = V_n - d
Arithmetic progression.
V_n = V_0 - nd

(V_0 - d/c)(1+c)^n + (d/c)
= V_0(1+c)^n - d/c[(1+c)^n - 1]
= V_0(1+c)^n - d/c[1 + nc + n(n-1)c^2/2! + ... - 1]
= V_0(1+c)^n - d/c[nc + n(n-1)c^2/2! + ...]
= V_0(1+c)^n - d[n + n(n-1)c/2! + ...]

lim c->0

= V_0 - nd as required
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dvs
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STEP III

Q4.
If r=cos(t), then r^2 = rcos(t). So:
x^2 + y^2 = x
x^2 - x + y^2 = 0
(x - 1/2)^2 + y^2 = 1/4, which is a circle in the (x,y) plane.

Now the curves r=cos(2nt) look like roses with 2n petals symmetric about the x-axis and y-axis (or the initial line and the half-line t=pi/2).

And the area enclosed by such a curve is
1/2 \small \int_0^{2\pi} cos^2(2nt) dt
= 1/4 \small \int_0^{2\pi} cos(4nt) + 1 dt
= 1/4 [(1/4)sin(4nt) + t]{0,2pi}
= pi/2, which is independent of n.

The curve r=cos(3t) looks also like a rose but with 3 petals, etc. I'm sure this bit is manageable.
--------------


Q5.
It's easy to show that M^2 = -I. (Does this remind you of i = sqrt(-1)? ) Then:
M^3 = -M
M^4 = I
M^5 = M
M^6 = -I
etc.

So,
exp(tM) = I + tM - (t^2/2) I - (t^3/3!) M + (t^4/4!) I + (t^5/5!) M - ...
= I(1 - t^2/2 + t^4/4! - ...) + M(t - t^3/3! + t^5/5! - ...)
= Icos(t) + Msin(t) (Surprise?!)
= \left( \begin{array}{cc} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \\ \end{array} \right)

This is, of course, the rotation matrix. That is, if you multiply a vector by it, that vector is rotated about the origin through an angle t (anticlockwise).

Moving on.

Notice that N^2 = 0. So N^3, N^4, ... are all zero. Thus:
exp(sN) = I + sN
= \left( \begin{array}{cc} 1 & s \\ 0 & 1 \\ \end{array} \right)

So what does this matrix do geometrically? Suppose you have some vector \left( \begin{array}{cc} x \\ y \\ \end{array} \right), this matrix sends it to \left( \begin{array}{cc} x + sy \\ y \\ \end{array} \right). So it's a shear of sorts (parallel to the x-axis).

Finally, if exp(tM) exp(sN) = exp(sN) exp(tM), then (after multiplying these out), we find that sin(t) = 0. So t = kpi. Geometrically, rotation by kpi either brings us back where we started or flips the vector so it points in the antiparallel direction. Therefore, to interchange rotation and shearing, for any type of shear, we require the rotation to not mess with the alignment of the vector.
--------------

(Thanks to khaixiang for pointing out a mistake in the last paragraph!)
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Kolya
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(Original post by Speleo)
(not in latex but I may well come through and edit latex in later)

STEP I Question 10

Range = horizontal velocity * time
r = 2vtcosa
cos^2a = (r/2vt)^2


1 = (gt/2v)^2 + (r/2vt)^2
1 = g^2t^2/4v^2 + r^2/4v^2t^2
4v^2t^2 = g^2t^4 + r^2
g^2t^4 - 4v^2t^2 + r^2 = 0
Small error in bold here Speleo. You seem to have increased it by a factor of 2 for some reason, r = vtcosa instead.

This gets the actual result that they want, which is: 0.5g^2t^2=v^2 \pm \sqrt{v^4-g^2r^2}
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DFranklin
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Paper I, Q6:

a_1 = \cos x, b_1 = 1\\

a_2 = \frac{1}{2}(1 + \cos x) = \cos^2(x/2), b_2 = \sqrt{a_2} = \cos(x/2)\\

a_3 = \frac{1}{2}(\cos(x/2)+\cos^2(x/2)) = \frac{1}{2}(1+\cos(x/2))\cos(x/2)=\cos^2(x/4)\cos(x/2),\\ b_3 = \sqrt{\cos^2(x/4)\cos(x/2)\cos(x/2)} = \cos(x/2)\cos(x/4)

Claim that for n>=3,
\displaystyle a_n = \cos\left(\frac{x}{2^{n-1}}\right)\prod_{i=1}^{n-1}\cos\left(\frac{x}{2^i}\right), b_n = \prod_{i=1}^{n-1}\cos \left(\frac{x}{2^i}\right).

Proof by induction on n: True for n = 3. Suppose true for k=n. Then
\displaystyle a_{k+1} = \frac{1}{2} \left(\cos\left(\frac{x}{2^{k-1}} \right)\prod_{i=1}^{k-1}\cos\left(\frac{x}{2^i} \right) + \prod_{i=1}^{k-1}\cos \left(\frac{x}{2^i} \right) \right)

=\frac{1}{2} \left( 1+\cos\left(\frac{x}{2^{k-1}}\right)\right)\prod_{i=1}^{k-1}\cos \left(\frac{x}{2^i}\right)

=\cos^2\left(\frac{x}{2^{k}} \right) \right)\prod_{i=1}^{k-1}\cos \left(\frac{x}{2^i} \right)

= \cos\left(\frac{x}{2^k}\right)\prod_{i=1}^{k}\cos \left(\frac{x}{2^i}\right)

b_{k+1} = \sqrt{\cos\left(\frac{x}{2^k} \right)\prod_{i=1}^{k}\cos \left(\frac{x}{2^i}\right) \prod_{i=1}^{k-1}\cos \left(\frac{x}{2^i} \right)} \\=\sqrt{\cos^2\left(\frac{x}{2^k}\right)  \prod_{i=1}^{k-1}\cos^2\left(\frac{x}{2^i} \right)} = \prod_{i=1}^{k}\cos \left(\frac{x}{2^i}\right)
Q.E.D. (Well, other than the LaTeX, Quite Easily Done...)
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