STEP maths I, II, III 1990 solutions Watch

insparato
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#41
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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.
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Speleo
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#42
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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 :/
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Dystopia
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I may as well write up my solution to STEP III, Q1.

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By expanding (\cos\theta + i\sin\theta)^{9} and dividing the imaginary part by the real part you get:

\tan9\theta = \frac{s^{9} - 36s^{7}c^{2} + 126s^{5}c^{4} - 84s^{3}c^{6} + 9sc^{8}}{c^{9} - 36c^{7}s^{2} + 126c^{5}s^{4} - 84c^{3}s^{6} + 9cs^{8}}

Where s = \sin\theta, c = \cos\theta.

Dividing by \cos^{9}\theta, you get:

\tan9\theta = \frac{t^{9} - 36t^{7} + 126t^{5} - 84t^{3} + 9t}{1 - 36t^{2} + 126t^{4} - 84t^{6} + 9t^{8}}

Where t = \tan\theta.

By expanding the brackets, we see this is equivalent to \frac{t(t^{2} - 2)(t^{6} - 33t^{4} + 27t^{2} - 3)}{(3t^{2} - 1)(3t^{6} - 27t^{4} + 33t^{2} - 1)}.

The roots of this equation are given by the roots of the numerator. The roots are \tan\frac{k\pi}{9}, \; k = 0, \; 1, \; 2, \; \cdots, \; 8.

Since \tan\frac{k\pi}{9} = - \tan\frac{(9 - k)\pi}{9} and using the difference of two squares along with the known values for \tan0, \; \tan\frac{\pi}{3}, \; \tan\frac{2\pi}{3}, we get the roots as t(t^{2} - 3)(t^{2} - \tan^{2}\frac{\pi}{9})(t^{2} - \tan^{2}\frac{2\pi}{9})(t^{2} - \tan^{2}\frac{4\pi}{9}).

This is because, for example (t - \tan\frac{\pi}{9})(t - \tan\frac{8\pi}{9}) = (t - \tan\frac{\pi}{9})(t + \tan\frac{\pi}{9}) = (t^{2} - \tan^{2}\frac{\pi}{9}).

So t^{6} - 33t^{4} + 27t^{2} - 3 can be factorised into (t^{2} - \tan^{2}\frac{\pi}{9})(t^{2} - \tan^{2}\frac{2\pi}{9})(t^{2} - \tan^{2}\frac{4\pi}{9}). Examining the constant on both we get -(\tan\frac{\pi}{9}\tan\frac{2\pi  }{9}\tan\frac{4\pi}{9})^{2} = - 3 \Rightarrow \tan\frac{\pi}{9}\tan\frac{2\pi}  {9}\tan\frac{4\pi}{9} = \sqrt{3}.

Setting \alpha = \tan^{2}\frac{\pi}{9}, \; \beta = \tan^{2}\frac{2\pi}{9}, \; \gamma = \tan^{2}\frac{4\pi}{9}, we also have:

\alpha + \beta + \gamma = 33, \; \alpha\beta + \beta\gamma + \gamma\alpha = 27

\alpha^{3} + \beta^{3} + \gamma^{3} = (\alpha + \beta + \gamma)^{3} - 3(\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) + 3\alpha\beta\gamma

\Rightarrow \tan^{6}\frac{\pi}{9} + \tan^{6}\frac{2\pi}{9} + \tan^{6}\frac{4\pi}{9} = 33^{3} - (3 \times 33 \times 27) + 9 = 33273.
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*bobo*
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#44
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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)
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cuddy
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#45
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thanks for the rep speleo! Im 1 point off getting my 3rd gem now
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DFranklin
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#46
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(Original post by Speleo)
1987 III/6

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

at 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).
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Speleo
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Thanks, I'll try that now.
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DFranklin
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(Original post by 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.
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Speleo
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#49
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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

Your new method seems to work great (haven't completed it yet but I've got to the 2nd order de in x). Thanks
I wouldn't be at all surprised if something very similar to this came up in STEP III this year.
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Rabite
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Ack, I ruined the ending for myself by accident
Should have known better than to come here!
Oh well, least I know how to do it.
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Simba
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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.
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Simba
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#52
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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.
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Simba
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#53
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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.
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*bobo*
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#54
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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)
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*bobo*
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#55
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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
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Dystopia
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STEP III, Q13.

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Consider a general point P which makes an angle \theta with the upwards vertical.

The forces acting on the particle are its weight, acting downwards with magnitude mg, and the reaction force R acting towards O. Let the radius of sphere be r.

R + mg\cos\theta = a = \frac{mv^{2}}{r}

R = \frac{mv^{2}}{r} - mg\cos\theta

The particle leaves the surface when R = 0.

\frac{mv^{2}}{r} = mg\cos\theta \Leftrightarrow \frac{v^{2}}{r} = g\cos\theta

v^{2} = rg\cos\theta

Since the velocity is tangential to the circle, the resolved velocity vector makes an angle of \theta to the horizontal.

Therefore, the velocity and position vectors are:

\textbf{v} = (v\cos\theta)\textbf{i} + (v\sin\theta - gt)\textbf{j}

\textbf{r} = (v\cos\theta t)\textbf{i} + t(v\sin\theta - \frac{1}{2}gt)\textbf{j}

Let the collision occur at some point C.

We're trying to find the angle which \theta has to exceed, so we can consider C at the same height as O, and find the resulting value of \theta; then \theta has to be larger than this for the collision to be below O.

The horizontal distance to C is r(1 + \sin\theta); the horizontal velocity is a constant v\cos\theta. Therefore the collision occurs at t = \frac{r(1 + \sin\theta)}{v\cos\theta}.

Inserting this value of t into the equation for the height:

\textbf{r}_{j} = \frac{r(1 + \sin\theta)}{v\cos\theta} \left(v\sin\theta - \frac{rg(1 + \sin\theta)}{2v\cos\theta}\right  )

 = \frac{r(1 + \sin\theta)(2v^{2}\sin\theta\cos  \theta - rg(1 + \sin\theta))}{2v^{2}\cos^{2}\the  ta}

 = \frac{r(1 + \sin\theta)(2rg\sin\theta\cos^{2  }\theta - rg(1 + \sin\theta))}{2rg\cos^{3}\theta} = \frac{r(1 + \sin\theta)(2\sin\theta\cos^{2}\  theta - (1 + \sin\theta))}{2\cos^{3}\theta}

The change in height is -r\cos\theta. Equation the two you get:

(1 + \sin\theta)(2\sin\theta(1 - \sin^{2}\theta) - 1 - \sin\theta) = -2( 1 - \sin^{2}\theta)^{2}

When you multiply this all out and simplify, you end up with 2\sin^{3}\theta + 3\sin^{2}\theta - 1 = 0; the only real solution to this is \sin\theta = \frac{1}{2}, so when the collision is at the same height as O, \theta = \frac{\pi}{6}.

For C to be below O, \theta &gt; \frac{\pi}{6}.

Alternate Method

Energy is conserved, so \frac{1}{2}mv^{2} + mgr(1 + \cos\theta) = \frac{1}{2}mV^{2} + mgr

Where v = velocity at P, V = velocity at C.

This simplifies to V^{2} = v^{2} + 2gr(1 + \cos\theta) - 2gr = gr\cos\theta + 2gr\cos\theta = 3gr\cos\theta.

Using the velocity vector, we find V^{2} = 3gr\cos\theta = (v\cos\theta)^{2} + \left(v\sin\theta - \frac{gr(1 + \sin\theta)}{v\cos\theta}\right)  ^{2}

When this is simplified you end up with the same cubic equation in terms of \sin\theta.

---

In an actual exam you would perhaps want to prove that as \theta increases, the collision becomes lower down, but this is heavily implied by the question so I didn't bother.

Someone might also want to check it, because my mechanics is somewhat rusty.
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ukgea
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#57
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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.
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Dystopia
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STEP I, Q2.

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If \omega =
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
e^{\frac{2i\pi}{3}
, then \omega is a complex cube root of 1 (by DeMoivre's), so \omega^{3} - 1 = 0 \Rightarrow (\omega - 1)(\omega^{2} + \omega + 1) = 0 \Rightarrow \omega^{2} + \omega + 1 = 0 as \omega is complex.

1 + \omega^{2} = -\omega
|-\omega| = 1
arg(-\omega) = \frac{-\pi}{3}

(1 + \omega)^{n} = (-\omega^{2})^{n}
\binom{n}{0} + \binom{n}{1}\omega + \cdots + \binom{n}{n} = (-1)^{n}\omega^{2n}

(1 + \omega^{2})^{n} = (-\omega)^{n}
\binom{n}{0} + \binom{n}{1}\omega^{2} + \cdots + \binom{n}{n}\omega^{2n} = (-1)^{n}\omega^{n}

(1 + \omega^{3})^{n} + 2^{n}
\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^{n}

Note that if you add the three together, the coefficients of all the binomial coefficients, \binom{n}{k}, except for those where k is a multiple of three, are equal to \omega^{2} + \omega + 1 = 0. When k is a multiple of three, the coefficient is three.

So \binom{n}{0} + \binom{n}{3} + \cdots + \binom{n}{k} = \frac{1}{3}(2^{n} + 2\cos\frac{n\pi}{3})

As (-1)^{n}e^{\frac{i2n\pi}{3}} + (-1)^{n}e^{\frac{i4n\pi}{3}} = 2\cos\frac{n\pi}{3} by the sum to product formulae. I think I showed this a different way when I first did this question, but I can nolonger remember how.

\binom{25}{0} + \binom{25}{3} + \cdots + \binom{25}{24} = \frac{1}{3}(2^{25} + 2\cos\frac{\pi}{3}) = 11,184,811.
(\cos\frac{25\pi}{3} = \cos\frac{\pi}{3})

\binom{24}{2} + \binom{24}{5} + \cdots + \binom{24}{23} = \frac{1}{2}\left(2^{24} - \frac{1}{3}(2^{24} + 2)\right) = \frac{1}{3}(2^{24} - 1) = 5,592,405.
(\cos\frac{24\pi}{3} = 1)

Due to the symmetry of the binomial coefficients: \binom{24}{1} = \binom{24}{23}, \; \binom{24}{2} = \binom{24}{22}, and so on.
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Dystopia
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STEP II, Q2.

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A + B + C + D = \pi



\sin A\sin C + \sin B\sin D = \sin (A + B)\sin (A + D)



\Leftrightarrow -2\sin A\sin C - 2\sin B\sin D = -2\sin (A + B)\sin (A + D)

Using the product to sum and compound angle formulae:

LHS = \cos (A + C) - \cos (A - C) + \cos (B + D) - \cos (B - D)



\cos (A + C) = \cos (\pi - (B + D)) = - \cos (B + D)



\Rightarrow LHS = - \cos (A - C) - \cos (B - D) = - \cos (C - A) - \cos (B - D)

This is because cos(x) is an even function.

- \cos (C - A) = \cos (\pi - (C - A)) = \cos (2A + B + C)



\cos (2A + B + D) - \cos (B - D) = -2\sin (A + B) \sin (A + D) = RHS

For the second part, draw the cyclic quadrilateral with vertices P, Q, R, S, and also the diagonals PR and QS.

Let angle(PRQ) = angle(PSQ) = A, angle(QSR) = angle(QPR) = B, angle(RQS) = angle(RPS) = C, angle(SRP) = angle(SQP) = D. Each pair of angles is equal as they are angles subtended by the same arc.

It is clear that A + B + C + D = \pi, as the angles make up the angles of a triangle.

\frac{PQ}{\sin A} = \frac{QR}{\sin B} = \frac{RS}{\sin C} = \frac{SR}{\sin D} = \frac{PR}{\sin (A + B)} = \frac{QS}{\sin (A + D)} = 2R

This is a result of the (full) Sine Rule. R is the radius of the circle.

Using the result of the first part, you get:

\sin A\sin C + \sin B\sin D = \sin (A + B)\sin (A + D)

\Rightarrow \frac{PQ . RS}{4R^{2}} + \frac{QR . SP}{4R^{2}} = \frac{PR . QS}{4R^{2}}

\Rightarrow PQ . RS + QR . SP = PR . QS
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nota bene
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#60
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#60
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.

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.
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