STEP maths I, II, III 1990 solutions

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Hmm im sensing some kind of substitution. So you have them both in terms of two variables but i havent a bloody clue what to substitute.

Reply 41

Speleo

They can be manipulated to:
[xe^(2t)]' = 5ye^(2t)
[ye^(-2t)]' = (2cost - x)e^(-2t)

So by integrating the second and substituting you can remove y, but I'm not sure the equations you get are useable :/

Reply 42

Dystopia

I may as well write up my solution to STEP III, Q1.

Spoiler

Reply 43

*bobo*

STEP III
11)
I= (2/3)Ma^2 + 2Ma^2 + 12Ma^2= (44/3)Ma^2
angular momentum= 6aMV(2^0.5)
using L=I(angular velo.)
initial angular velocity= 9(2^0.5)V/(22a)

KE= 0.5I(angular velo.)^2=(22Ma^2/3)((2x81)V^2/(22^2 a^2))= 54MV^2/22
=(27/11)MV^2
GPE at highest point= 7Mg.2(2^0.5)a= 14Mag(2^0.5)
(27/11)MV^2 > 14Mag(2^0.5)

V^2 > (154/27)ag(2^0.5)

Reply 44

cuddy

thanks for the rep speleo! Im 1 point off getting my 3rd gem now :biggrin:

Reply 45

DFranklin

Speleo

1987 III/6

\frac{dx}{dt} + 2x - 5y = 0

\frac{dy}{dt} + x - 2y = 2\cos t

t=0; x=0, \frac{dy}{dt}=0

Can anyone get me started?

Rewrite it as the equation

\dot{\bf v} + A{\bf v} = {\bf f}

, where

{\bf v} = (x, y)

and

{\bf f} = (0, 2\cos t)

. You should be able to do something useful with the eigenvalues / eigenvectors of A. (You might need to do a little 'fiddling' to get it to work, but once you have the idea, it should be OK).

Reply 46

Speleo

Thanks, I'll try that now.

Reply 47

DFranklin

Speleo

Thanks, I'll try that now.

Sorry, but having tried it, I don't think it works very well. (The matrix doesn't have real eigenvectors, so it's going to get real painful, real fast).

So, differentiate the first equation to get

x''+2x'-5y' = 0

, then use both the original equations to write y' in terms of x, x' and cos t. You then have a 2nd order linear equation for x.

Reply 48

Speleo

Haha, yeah I just did that and found out about the imaginary eigenvalues/complex eigenvectors, don't worry I didn't spend long trying to get that approach to work :wink:

Your new method seems to work great (haven't completed it yet but I've got to the 2nd order de in x). Thanks :smile:

I wouldn't be at all surprised if something very similar to this came up in STEP III this year.

Reply 49

Rabite

Ack, I ruined the ending for myself by accident :frown:

Should have known better than to come here!
Oh well, least I know how to do it.

Reply 50

Simba

II/4:

Line segments:

1: 1 line segment.
2: 4 line segments.
3: 9 line segments.
4: 16 line segments.

So it is reasonable to suppose that n lines are split into n^2 line segments.

Each new line that we introduce splits every existing line into one more piece, and is itself intersected into as many parts as there are lines after it has been added.

So the number of line segments for n lines is 1 + (1 + 2) + (2 + 3) + ... + (n - 1 + n) = 1 + 3 + 5 + ... + 2n - 1.

\sum_{k=1}^n (2k - 1) = n(n+1) - n = n^2

QED.

Planes:

Each new line cuts the existing n - 1 lines in n - 1 places, creating n new regions.

One region exists to begin with.

So the number of planes for n lines is 1 + 1 + 2 + 3 + 4 + ... + n

1 +

\sum_{k=1}^n k = n(n+1)/2 = (n^2+n)/2 = (n^2 + n + 2)/2

as required.

QED.

Reply 51

Simba

II/6:

i) c(b - 2) >= 0 (obviously true since b and c are >= 2).

Expanding: bc - 2c >= 0.

Rearranging: bc >= 2c.

QED.

c >= 2.

Add c to both sides: 2c >= 2 + c.

QED.

ii) a(d + 1) <= c(b - 1) (true since d + 1 <= c and a <= b - 1).

Expanding: ad + a <= bc - c.

Rearranging: bc - ad >= a + c.

QED.

iii) a/b < p < c/d implies that a < bp and dp < c.

a(dp + 1) <= c(bp - 1) (true since dp + 1 <= c and a <= bp - 1).

Expanding: adp + a <= bcp - c.

Rearranging: p(bc - ad) >= a + c.

QED.

iv) a(dp + 1) <= c(bp - 1) (true since aq <= (bp - 1) and (dp + 1) <= cq, multiplying gives aq(dp + 1) <= cq(bp - 1), and cancelling gives this.)

Expanding: adp + a <= bcp - c.

Rearranging: p >= (a + c)/(bc - ad).

QED.

qbc - qad >= b + d

d(qa + 1) <= b(qc - 1) (true since qc - 1 >= dp and bp >= qa + 1, multiplying gives bp(qc - 1) >= dp(qa + 1), and cancelling gives this).

Expanding: qbc - qad >= b + d

Rearranging: q >= (b + d)/(bc - ad).

QED.

p >= 17, q >= 19.

17/19 is the only fraction with a denominator less than 20 which is between 8/9 and 9/10.

Reply 52

Simba

III/2:

i) OA = m(a^3 i + a^2 j + ak)

BC = b^3 i + b^2 j + bk + n[(c^3 - b^3)i + (c^2 - b^2)j + (c - b)k]

For an intersection:

b^3 + n(c^3 - b^3) = a^3. --- (1)
b^2 + n(c^2 - b^2) = a^2. --- (2)
b + n(c - b) = a. --- (3)

Sub (3) in (2):

b^2 + n(c^2 - b^2) = [b + n(c - b)]^2 = b^2 + 2bn(c - b) + n^2.(c - b)^2.

c^2 - b^2 = 2b(c - b) + n(c - b)^2.

(c + b)(c - b) - 2b(c - b) - n(c - b)^2 = 0.

(c - b)[c + b - 2b - n(c - b)] = 0.

(c - b)[c - b - n(c - b)] = 0.

(c - b)(c - b)(1 - n) = 0.

But c =/= b, so n = 1 (if they intersect).

Sub in (3):

b + c - b = a.

c = a.

But c =/= a, so the lines don't intersect.

QED.

ii) cos(AOB) = a.b/|a||b| = [(ab)^3 + (ab)^2 + ab]/root(a^6 + a^4 + a^2).root(b^6 + b^4 + b^2)

= [(ab)^3 + (ab)^2 + ab]/ab.root[(a^4 + a^2 + 1)(b^4 + b^2 + 1)]

= [(ab)^2 + ab + 1]/root[(a^4 + a^2 + 1)(b^4 + b^2 + 1)].

= 3/root(a^4.b^4 + a^4.b^2 + a^4 + a^2.b^4 + a^2.b^2 + a^2 + b^4 + b^2 + 1)

= 3/root(3 + a^2 + a^4 + b^2 + a^2 + b^4 + b^2)

= 3/root(a^4 + b^4 + 2a^2 + 2b^2 + 3)

Since a and b are variable, we can make a (or b) as large as we like, so the denominator can approach infinity. So cos(AOB) can be made arbitrarily close to 0. Values of a and b very close to 1 (but not equal to 1 since a and b are distinct) give cos(AOB) arbitrarily close to 1. It is not possible for a sum of squares (+3) to be negative, so cos(AOB) > 0, outruling pi/2 < AOB < pi.

Hence 0 < cos(AOB) < 1, and so 0 < AOB < pi/2.

QED.

Reply 53

*bobo*

III 11)
total moment of intertia= m(k^2 + r^2)

i) conservation of angular momentum:
m(k^2+a^2)omega=m(k^2+r^2)w
w= omega(k^2 + a^2)/(k^2+r^2)

force acting on bead= mrw^2
dv/dt= v dv/dr= rw^2
=r(omega)^2(k^2+a^2)^2/(k^2+r^2)^2 = rk^2V^2(k^2+a^2)/(a^2(k^2+r^2)^2)

0.5v^2= -k^2V^2(k^2+a^2)/(2a^2(k^2+r^2)) +c
when r=a, v=V
v= dr/dt= -( V^2 +(kV/a)^2- (kV)^2(k^2+a^2)/(a^2(k^2+r^2)))^0.5
= -(omega)r(k^2+a^2)/(k(k^2+r^2)^0.5) (by substittion of given equation)

w= d(theta)/dt= omega(k^2+a^2)/(k^2+r^2)

dr/d(theta)= (dr/dt) /(dtheta/dt)= -r(k^2+r^2)^0.5/k

INT -k/(r(k^2+r^2)^0.5) dr =INT 1 dtheta
u=1/r
-r^2du=dr
INT ((1/k^2) +u^2)^(-0.5) du= theta + a where a is a constant
=> arsinh(ku)= theta+ a
ku= k/r= sinh(theta +a)
=> k= rsinh(theta+a)

Reply 54

*bobo*

II
8)i) (1 and 0)INT x(t)y(t)dt= INT x'(t)x(t) dt= (0.5(x(t))^2)(bet 1 and 0)
=0.5((x(1))^2- (x(0))^2)= 0.5((x(1))^2
^ x(0)= (bet 0 and 0)INT y(t) dt= 0

INT y'' +(y^2 -1)y' +y= y' +y^3/3- y + x=0

y(0)=y(1) and y'(0)=y'(1)

apply y(0) and y(1) to above equation and minus one from the other.
=> x(1)-x(0)=0
(x(0)=0)
=>x(1)=0

ii)(bet 1 and 0) INT x(x''' +((x')^2-1)x''- x') dt=0
u=x(t)
du/dt=x'(t)
dv/dt= x''' +((x')^2-1)x''- x'
v= (x''- x + (x')^3/3- x')

(bet 1 and 0)(x(t)(x''- x + (x')^3/3- x')- INT x'(t)(x''- x + (x')^3/3- x')dt)=0
x(0)=x(1)=0 ,so:
=> (bet 1 and 0)INT x'(x''-1) -(x')^2 + (x')^4/3 dt=0

=>(bet 1 and 0)(0.5(x')^2 - x)= (bet 1 and 0)INT (x')^2- ((x')^4)/3 dt= INT y^2 - (y^4)/3 dt

0.5(x')^2 -x y(0)=y(1) so x'(0)=x'(1) and x(0)=x(1)=0
so LHS integral summation equals 0
=> (bet 1 and 0) INTy^2- (y^4)/3 dt
=>(0->1)INT y^2= (1/3) (0->1)INT y^4 dt

Reply 55

Dystopia

STEP III, Q13.

Spoiler

Reply 56

ukgea

III/3:

i) Given:

(1):

a*a = e

(2):

b*b = e

(3):

a*b*a*b = e

.

From (1) and (2), by mutliplying by

a^{-1}

and

b^{-1}

, respectively, it follows

(4)

a = a^{-1}

(5)

b = b^{-1}

.

Multiplying (3) by

a^{-1}

from the left and

b^{-1}

from the right, we have

b*a = a^{-1}*b^{-1}

and then subbing in (4) and (5) into RHS gives

b*a = a*b

as required.

ii)
Assume that the orders of

c*d

and

d*c

are

n

and

m

, respectively.

Then,

(1)

\displaystyle \underbrace{c*d*c*d*\cdots*c*d}_{n} = e

Multiplying by

c^{-1}

from the left and by

c

from the right gives

\displaystyle \underbrace{d*c*d*c*\cdots*d*c}_{n} = c^{-1}*c

\displaystyle \underbrace{d*c*d*c\cdots*d*c}_n = e

\displaystyle (d*c)^{n} = e

from which it follows that

m|n

.

But by using the exact same argument, only swapping c and d, we get that

n|m

from which it follows that

n = m

as required.

iii)

Given:

(1)

c^{-1}*b*c = b^r

First part: Induction on s:

Statement:

(2)

c^{-1} * b^s * c = b^{sr}

The case s = 1 is merely the given (1).

Assume (2) is true for s = k, i.e.

c^{-1} * b^{k} * c = b^{kr}

Left multiplying by (1):

c^{-1}*b*c*c^{-1}*b^{k}*c = b^r*b^{kr}

c^{-1}*b^{k+1}*c = b^{(k+1)r}

i.e. the statement is true for s = k + 1 as well, and thus is true by induction for all natural s. By instead using induction backwards, multiplying by

c^{-1} * b^{-1} * c = b^{-r}

(

c^{-1} * b^{-1} * c

is the inverse of

c^{-1} * b * c

), we can show it to be true for all integers s.

Second part:

Known (from the last part): For any integer m,

(1)

c^{-1} * b^m * c = b^{mr}

Now use induction on n:

Statement:

c^{-n} * b^s * c^n = b^{sr^n}

Assume true for n = k:

c^{-k} * b^s * c^k = b^{sr^k}

Now let

m = sr^k

and (using (1)):

c^{-(k+1)}*b^s * c^{k+1}

= c^{-1}*c^{-k} * b^s * c^k * c

= c^{-1}*b^m * c

= b^{mr}

= b^{sr^{k+1}}

which proves the statements is true for n = k + 1 as well, and by induction blah blah..., as required.

Reply 57

Dystopia

STEP I, Q2.

Spoiler

Reply 58

Dystopia

STEP II, Q2.

Spoiler

Reply 59

nota bene

I've attempted this, but I'm crap at graph sketching and I'm not confident about part ii, so where I'm talking crap, just correct it :wink:

.

STEP I Question 5

i)

\displaystyle\int_1^3\frac{1}{6x^2+19x+15}dx= \displaystyle\int\frac{1}{(3x+5)(2x+3)}dx

Do partial fractions on that so we have

\frac{A}{3x+5}+\frac{B}{2x+3}

Comparing coefficients

A(2x+3)+B(3x+5)=1 \newline 2A+3B=0 \newline 3A+5B=1 \newline A=-\frac{3}{2}B \newline -\frac{9}{2}B+5B=1 \text{so,}B=2, A=-3

This leaves us with

\displaystyle\int\frac{2}{(2x+3)}dx + \displaystyle\int -\frac{3}{(3x+5)}dx

This integrates to

\ln|2x+3|-\ln|3x+5|+c=\ln|\frac{2x+3}{3x+5}|

Applying the limits we have:

\ln|\frac{9}{14}|-ln|\frac{5}{8}|=\ln|\frac{36}{35}|

ii)

We have

f(x)=x^{1760}-x^{220}+q

, because it has even powers the x^{1760} and x^{220} will always be positive.This means negative values and positive values of x behave in the same way (so how the function looks to the left of the y-axis is described by a mirroring in the y-axis of how it looks to the right of the y-axis).

Now we can consider three intervals:
i) |x|>1

x^{1760}-x^{220}

will always be positive and graphically the function will be increasing very rapidly. The equation

x^{1760}-x^{220}+q=0

can at most have one solution, for a suitably negative choice of q in this interval.

ii) |x|=1 and |x|=0
At these two points the value of f(x) is entirely dependant on q. If q=0 with these choices of |x| we can have f(x)=0.

iii) |x|<1
Here

x^{1760}-x^{220}

will be negative, and therefore a root will exist if q is positive.

Because the three intervals require different types of values of q we can see that there exists at most of root of the equation f(x)=0.