STEP maths I, II, III 1991 solutions

I have an alternative solution for post 7 part b (please add freely if you think it is useful).

Notice that

1+x+x^2+x^3=(1+x)(1+x^2)

Reply 23

nota bene

15

First part:

\sin^2\theta+sin^2\phi+sin^2\psi+2\sin\theta\sin \phi \sin\psi\\

=\sin\theta[\cos(\phi+\psi)+2\sin\phi\sin\psi]+\sin^2\phi+\sin^2\psi\\

=\sin\theta[\cos\phi\cos\psi+\sin\phi\sin\psi]+\sin^2\phi+\sin^2\psi\\

=\cos(\phi+\psi)\cos(\phi-\psi)+\sin^2\phi+\sin^2\psi\\

=\cos^2\phi\cos^2\psi-\sin^2\phi\sin^2\psi+\sin^2\phi+\sin^2\psi\\

=(1-\sin^2\phi)(1-\sin^2\psi)-\sin^2\phi\sin^2\psi+\sin^2\phi+\sin^2\psi\\

=1

Second part:

Letting

\theta=\phi=\pi/5

and

\psi=\pi/10

, using the first part we have:

2\sin^2\frac{\pi}{5}+\sin^2\frac{\pi}{10}+2\sin^2 \frac{\pi}{5}\sin\frac{\pi}{10}-1=0\\

\Rightarrow8\sin^2\frac{\pi}{10}\cos^2\frac{\pi}{10}+\sin^2\frac{\pi}{10}+8\sin^3\frac{\pi}{10}\cos^2\frac{\pi}{10}-1=0\\

\Rightarrow8\sin^2\frac{\pi}{10}-8\sin^4\frac{\pi}{10}+\sin^2\frac{\pi}{10}+8\sin^3\frac{\pi}{10}-8\sin^5\frac{\pi}{10}-1=0\\

\Rightarrow(8\sin^3\frac{\pi}{10}+\sin^2\frac{\pi}{10}-1)+\sin^2\frac{\pi}{10}(1-\sin^2\frac{\pi}{10}-\sin^3\frac{\pi}{10})=0

\Rightarrow(\sin^3\frac{\pi}{10}+\sin^2\frac{\pi}{10}-1)\cos^2\frac{\pi}{10}=0\\

\Rightarrow\sin^3\frac{\pi}{10}+\sin^2\frac{\pi}{10}-1=0

justinsh

Sorry for misprints, it seems LaTex does funny things on this forum. If someone could tell me how to fix the errors, I would be glad to do so.

spaces sometimes does wonders... (didn't quote you just so you can look at the code)

Reply 24

generalebriety

16

justinsh

I have an alternative solution for post 7 part b (please add freely if you think it is useful).

Notice that

1+x+x^2+x^3=(1+x)(1+x^2)

Nice thought. I'll gladly add it if you do the rest of the working because I'm certainly not doing any more Maclaurin today. :wink:

Reply 25

I may have done mistakes but here it goes:

\ln(1+x+x^2+x^3)\\

=\ln(1+x)+\ln(1+x^2)\\

=(x-\frac{1}{2}x^2+\frac{1}{3}x^3+...+\frac{(-1)^{n+1}}{n}x^n)+(x^2-\frac{1}{2}x^4+\frac{1}{3}x^6+...+\frac{(-1)^{n+1}}{n}x^{2n})\\

=x+\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{3}{4}x^4+...+\left[\frac{(-1)^{n+1}}{n}-\frac{1}{2n}\right]x^{2n}+\frac{x^{2n+1}}{2n+1}

Reply 26

STEP II question 3

First part:

Multiplying be xyz everywhere yelds:

x^2yz+xz=xy^2z+xy=xyz^2+yz

(*)

Using (*) three times:

x^2yz+xz=xy^2z+xy

\Rightarrow xyz(x-y)=x(y-z)

(1)

xy^2z+xy=xyz^2+yz

\Rightarrow xyz(y-z)=y(z-x)

(2)

xyz^2+yz=x^2yz+xz

\Rightarrow xyz(z-x)=z(x-y)

(3)

Multiplying (1), (2) and (3) yelds:

(xyz)^3(x-y)(y-z)(z-x)=xyz(y-z)(z-x)(x-y)

\Rightarrow x^2y^2z^2=1

Second part:

(x+\frac{1}{y})^3\\

=(x+\frac{1}{y})(y+\frac{1}{z})(z+\frac{1}{x})\\

=xyz+\frac{1}{xyz}+(x+\frac{1}{y})+(y+\frac{1}{z})+(z+\frac{1}{x})\\

=xyz+\frac{1}{xyz}+3(x+\frac{1}{y})

=\pm2+3(x+\frac{1}{y})

since

xyz=-1

(from first part)

First case: (notation

X=x+\frac{1}{y}

)

X^3=2+3X

\Rightarrow(X+1)^2(X-2)=0

\Rightarrow X\in\{-1;2\}

Second case:

X^3=-2+3X

\Rightarrow(X-1)^2(X+2)=0

\Rightarrow X\in\{1;-2\}

I don't know how to get rid of the degenerate solutions... (help!)

EDIT: sorry khaixiang, nice solution though!

Reply 27

khaixiang

5

Ahhh.. I wanted to do II/3! Since we're using completely different methods, I suppose this is still worth posting. And there're many more ways to do this question (I've seen at least 3 more different ways)

\\x+\frac{1}{y}=y+\frac{1}{z}\\ y+\frac{1}{z}=z+\frac{1}{x}\\ z+\frac{1}{y}=z+\frac{1}{x}

Rewrite the above equations:

\\x-y=\frac{1}{z}-\frac{1}{y}\\ y-z=\frac{1}{x}-\frac{1}{z}\\ x-z=\frac{1}{x}-\frac{1}{y}

\\x-y=\frac{y-z}{yz}\\ y-z=\frac{z-x}{xz}\\ x-z=\frac{y-z}{xy}

Multiply them together:

\\(x-y)(y-z)(x-z)=\frac{(x-y)(y-z)(x-z)}{x^2y^2z^2}\\ xyz=\pm 1

Now let:

\\x+\frac{1}{y}=y+\frac{1}{z}=m

(1)

\\x+\frac{1}{y}=m\\ xy+1=my

(2)

\\y+\frac{1}{z}=m\\ \text{ If } xyz=1\\ \text{ Then }y+xy=m

(3)

\\y+\frac{1}{z}=m\\ \text{ If } xyz=-1\\ \text{ Then }y-xy=m

(1)-(2) and (1)+(3) gives

m=\pm1

Reply 28

STEP II question 4 (under construction)

First part:

2y^2=3y+2

\Leftrightarrow y\in\{\frac{-1}{2},2\}

Because y is composed of two cosines smaller than 1,

y=\frac{-1}{2}

.

Second part:

[br]\displaystyle\sum_{k=1}^{16}\cos{k\theta}=\sum_{k=1}^{16}\Re e^{ik\theta}=\Re\left(\sum_{k=1}^{16}e^{ik\theta} \right) = \Re\left( \frac{1-e^{i17\theta}}{1-e^{i\theta}} \right)=\Re \left(\frac{1-1}{1-e^{i\theta}}\right)=0

Third part:

Considering the fact that the inverse of a complex number of modulus 1 is its conjugate:

S=2(\cos\theta+\cos2\theta+\cos4\theta+\cos8\theta)

=e^{i\theta}+e^{i2\theta}+e^{i4\theta}+e^{ i 8 \theta }+e^{i9\theta}+e^{i13\theta}+e^{i15\theta}+e^{i16 \theta }

We then compute

S^2

realising that the square of the terms equals S and simplying the rest using the second part. We eventually find that

S^2-S+4=0

.

A drawing then let us determine the sign of the correct expression.

Nvm.
XD

III/5

All derivatives with respect to t.

r'' + 4r = 5sin3t

CF -> m^2 + 4 = 0
m = +-2i
r = Acos2t + Bsin2t

PI -> r = psin3t + qcos3t
r'' = -9psin3t - 9qcos3t
-5psin3t - 5qcos3t = 5sin3t
p = -1, q = 0

r = Acos2t + Bsin2t - sin3t
r' = -2Asin2t + 2Bcos2t - 3cos3t

t = pi/2, r = 1
1 = -A + 1
A = 0

t = pi/2, r' = -2
-2 = -2B
B = 1

r = sin2t - sin3t

r(pi/5) = sin2pi/5 - sin(3pi/5) = sin(2pi/5) - sin(pi - 3pi/5) = 0
r(3pi/5) = sin(6pi/5) - sin(9pi/5) = -sin(pi/5) - sin(-pi/5) = 0
Since the function is clearly continuous, and the curve starts and finishes at the origin, it is a closed loop.

Area = (1/2) INT (pi/5 to 3pi/5) r^2 dt
= (1/2) INT sin^2(2t) + sin^2(3t) - 2sin2tsin3t dt
= (1/2) INT (1/2) - (1/2)cos4t + (1/2) - (1/2)cos6t + cos5t - cost dt
= (1/2)[t - (1/8)sin4t - (1/12)sin6t + (1/5)sin5t - sint](pi/5 to 3pi/5)
= (1/2)[3pi/5 - (1/8)sin(12pi/5) - (1/12)sin(18pi/5) + (1/5)sin(15pi/5) - sin(3pi/5)] - (1/2)[pi/5 - (1/8)sin(4pi/5) - (1/12)sin(6pi/5) + (1/5)sin(5pi/5) - sin(pi/5)]
= (1/2)[3pi/5 - (1/8)sin(2pi/5) + (1/12)sin(3pi/5) + 0 - sin(3pi/5)] - (1/2)[pi/5 - (1/8)sin(pi/5) + (1/12)sin(pi/5) + 0 - sin(pi/5)]
= pi/5 - (1/16)sin(2pi/5) - (11/24)sin(3pi/5) + (25/48)sin(pi/5)
= pi/5 - (3/48)sin(2pi/5) - (22/48)sin(2pi/5) + (25/48)sin(pi/5)

= pi/5 + (25/48)[sin(pi/5) - sin(2pi/5)]

Hope there's another straightforward differential equation question on this year's STEP III :smile:

Reply 31

II, 2:
I'm not entirely sure how to finish this one off, so someone else do that.
We're given

3y^2 +3x^2 -10xy +16y - 16x +15=0

Differentiate and divide by 2 - I'll write m for dy/dx, since I hate fractions:

3ym + 3x - 5[y+xm] + 8m -8 = 0

[3y-5x+8]m = 8-3x+5y

m = \frac{8-3x+5y}{3y-5x+8}

Now use

y-y_1 = m(x-x_1)

, the infamous straight line formula, with x1=s, y1=t:

y = mx-ms+t

Consider the ms-t bit, putting it all over the denominator of m:

-ms+t = \frac{-8s+3s^2-5st +3t^2 -5st + 8t}{3t -5s+8}

From our original equation,

3s^2 + 3t^2 -10st = 16s - 16t -15

as (s,t) is on the curve. So -ms+t becomes

\frac{16s-16t-15-8s+8t}{3t-5s+8} = \frac{8s-8t-15}{3t-5s+8}

Now you have to wonder why you're missing a minus sign, and realise that they've factored it out for some reason, so it's the same thing really.
Next bit:
As x, y (or s,t, whatever) get large, we can ignore the linear terms (and the constant of course) in the original equation:
The curve approximates another curve -

3y^2-10xy +3x^2=0

y^2 - \frac{10}{3} xy + x^2 =0

(y-\frac{5}{3} x)^2 +x^2 - \frac{25}{9} x^2 =0

y = \frac{5}{3} x ± \frac{4}{3} x

Those = signs should probably be --> to represent some kind of tending to, but...hm. Anyway.

The above relation gives y = 3x and y=x/3. Putting those in (with s and t) gives the required result.

For the next bit, use dy/dx = tan{angle between x axis}.
And dx/dy = tan{angle between y axis}.

For the axes of the hyperbola, use y=mx+c, and you know that both axes cross through the intersection of the two asymptotes (which is to be found).

I'm not sure how to determine the gradient of the axes (they bisect the asymptotes, but still...)
Anyone want to tell me?

I'm guessing that the gradients are 1 and -1, since tan(m) is probably something to do with averaging the angles, ie, ½(arctan(3) + arctan(1/3)) = 45 (works for any arctan x and 1/x).

Reply 32

Oh crap. You post it then, I'll do away with mine :P
Sorry bout that.

Reply 33

Speleo

7

Am I going crazy or are matrices no longer on the STEP syllabus?
http://www.ocr.org.uk/Data/publications/specifications_syllabuses_and_tutors_handbooks/Specificat98487.pdf

Also, does anyone have the '05 papers? (II and III '06 are available if you go back a couple hundred pages to last June in this forum).

Reply 34

Wow, I can't find Matrices there either.
(Thanks about the 06 papers, by the way...did you want to start threads about those newer ones some time, too?)

Reply 35

generalebriety

16

Decota

GE you might want to check this line.

Oh, ****.

I'll fix it, just not now. Cheers. Thought it was too simple.

Reply 36

Speleo

7

There are full sets of solutions online for 1999 onwards, so I'm not sure there's much point. I think we should continue going back in time for now.

EDIT: It would seem that vectors beyond C4 are also not on the syllabus :biggrin:

Reply 37

generalebriety

16

Speleo: correct, matrices aren't on it any more.
Rabite: it was DFranklin who suggested that. :smile:

Can someone check I've got that bloody question right this time please? :redface:

Reply 38