STEP maths I, II, III 1991 solutions

Rabite

Hmm...
I can do the first bit and the last bit.

But not the middle bit (unless we resort to complex numbers, but that's foul play right?)

I don't see anything wrong with using complex numbers. Alternatively, note that it's the sum of the "x-components" of the sides of a 17-sided polygon.

Reply 61

gyrase

3

Rabite

Hmm...
I can do the first bit and the last bit.

But not the middle bit (unless we resort to complex numbers, but that's foul play right?)

I think the summation isn't possible without using complex numbers so it would be fine to use them ... correct me if I'm wrong somebody.

Do please post your working as something is better than nothing! :tsr2:

Reply 62

nota bene

15

Rabite

But not the middle bit (unless we resort to complex numbers, but that's foul play right?)

Hmm, why would it be? It is surely not inaccurate, or am I missing something :s-smilie:

Probably not the way it is supposed to be done, but can't see what would make it invalid.

(I've started on III/7 but seem not to see what to do with c) and I've not got all steps of a) to work yet, I'll try to look at it tomorrow again)

Reply 63

DFranklin

18

nota bene

(I've started on III/7 but seem not to see what to do with c) and I've not got all steps of a) to work yet, I'll try to look at it tomorrow again)

Spoiler

Reply 64

Rabite

8

Okay. I'll start with the last bit, since I quite like that bit.
(attached)
I'll edit the other parts in as I copy them into LaTeX.

Oh, I missed out a little bit. At the end, with the quadratic, I'm not sure how to justify the sign when you solve the quadratic...
Franklin, you're up! :p:

First bit here too. The files are in reverse order.

Speaking of III/7,
we are asked to deduce

\frac{1}{2}\ln{x}<\sqrt{x} \text{ for } x\geq 1

(which is quite easily done) but this inequality is far too weak to deduce that

\frac{\ln{x}}{x} \rightarrow 0 \text{ as } x \rightarrow \infty

which we were asked to show subsequently. I can see that we're probably meant to deduce

\frac{\ln{x}}{x}< x^{\frac{1}{x}}

from the result then proceed from here, but again, this inequality is far too weak. 2 steps of l'hopital rule on the other hand would suffice. Maybe someone can enlighten me :smile:

Thanks

Reply 66

nota bene

15

I used l'hopital, couldn't see a simple way otherwise...

Reply 67

DFranklin

18

khaixiang

Speaking of III/7,
we are asked to deduce

\frac{1}{2}\ln{x}<\sqrt{x} \text{ for } x\geq 1

(which is quite easily done) but this inequality is far too weak to deduce that

\frac{\ln{x}}{x} \rightarrow 0 \text{ as } x \rightarrow \infty

which we were asked to show subsequently. I can see that we're probably meant to deduce

\frac{\ln{x}}{x}< x^{\frac{1}{x}}

from the result then proceed from here, but again, this inequality is far too weak. 2 steps of l'hopital rule on the other hand would suffice. Maybe someone can enlighten me :smile:

Thanks

What's wrong with

\ln(x) < 2 \sqrt{x} \implies \frac{\ln{x}}{x} < 2 \frac{\sqrt{x}}{x} = \frac{2}{\sqrt{x}} \to 0

as

x \to \infty

?

Reply 68

khaixiang

5

DFranklin

What's wrong with

\ln(x) < 2 \sqrt{x} \implies \frac{\ln{x}}{x} < 2 \frac{\sqrt{x}}{x} = \frac{2}{\sqrt{x}} \to 0

as

x \to \infty

?

Yes, you're right :redface:

the answer is just 2 steps away!..... but then, how about the lower limit? I suppose it don't matter that much in this lengthy question?

Reply 69

DFranklin

18

khaixiang

Yes, you're right :redface:

the answer is just 2 steps away!..... but then, how about the lower limit? I suppose it don't matter that much in this lengthy question?

Exactly what lower limit did you have in mind?

Reply 70

generalebriety

16

ln x and x are both positive so lnx / x can't go negative... :smile:

Reply 71

generalebriety

16

nota bene

I'm attempting I/3 I'll see if I sort it out...

Okay, I'll post what I have now (probably should scan a sketch tomorrow as well), but it is not finished, and likely contains a few mistakes

Spoiler

I'm tired of this one, I'll take it up tomorrow again and try to sort it out.

Is this question finished / getting finished? :smile:

I'm pretty convinced I've managed to catch up with all the posts now - if everyone could check my first post and let me know if I've missed anything (with appropriate post number references :wink:

) I'd be grateful. :smile:

Oh, and if you answer a question and it doesn't appear with 24 hours or so, assume I've forgotten about you and nudge me in the first direction. :p:

Reply 72

nota bene

15

generalebriety

Is this question finished / getting finished? :smile:

I'm pretty convinced I've managed to catch up with all the posts now - if everyone could check my first post and let me know if I've missed anything (with appropriate post number references :wink:

) I'd be grateful. :smile:

Oh, and if you answer a question and it doesn't appear with 24 hours or so, assume I've forgotten about you and nudge me in the first direction. :p:

It certainly isn't finished. I think I said somewhere that you should remove it from the OP and leave it free to do for anyone; I might return to it, but I'm a bit fed up with that question :p:

Reply 73

Rabite

8

I put up the rest of that trig question for whoever wanted it - Decota was it?
I have a tiny quibble about choosing which sign for the ± (from a quadratic) at the end, though - anyone want to take a look at it? DFranklin? You know everything it seems! Care to help? :p:

(On another note, I got an anonymous neg rep in this forum with the reason "Ahh"... bleh. :frown:

)

Reply 74

Speleo

7

EDIT: wow it took me almost an hour to do that, when I started only Decota's post was there :tongue:

Anyway, even though it's redundant now (grr), II/4

y = cos[p] + cos[2p]
2y^2 = 2cos^2[p] + 2cos^2[2p] + 4cos[p]cos[2p]
= 1 + cos[2p] + 1 + cos[4p] + 2cos[3p] + 2cos[p]
= cos[4p] + 2cos[3p] + cos[2p] + 2cos[p] + 2

since cos(pi + x) = cos(pi - x) [see cos graph], cos[6pi/5] = cos[4pi/4] and cos[2pi/5] = cos[8pi/5], or cos[3p] = cos[2p], cos[p] = cos[4p].

2y^2 = 3cos[p] + 3cos[2p] + 2
= 3y + 2

2y^2 = 3y + 2
2y^2 - 3y - 2 = 0
2y^2 - 4y + y - 2 = 0
2y(y-2) + 1(y-2) = 0
(2y+1)(y-2) = 0
A quick look at the cos graph shows that both cos[p] and cos[2p] are less than 1, so y = -1/2.

---

SUM = Re[e^(2pi/17) + e^(4pi/17) + ... + e^(34pi/17)]
= Re[sum of 17th roots of unity] = Re[0] = 0.

---

Clearly from the answer we want to get to: 2z^2 + z - 2 = 0
2z^2 = 2cos^2[p] + 2cos^2[2p] + 2cos^2[4p] + 2cos^2[8p] + 4cos[p]cos[2p] + 4cos[p]cos[4p] + 4cos[p]cos[8p] + 4cos[2p]cos[4p] + 4cos[2p]cos[8p] + 4cos[4p]cos[8p]
= 1 + cos[2p] + 1 + cos[4p] + 1 + cos[8p] + 1 + cos[16p] + 2cos[3p] + 2cos[p] + 2cos[5p] + 2cos[3p] + 2cos[9p] + 2cos[7p] + 2cos[6p] + 2cos[2p] + 2cos[10p] + 2cos[6p] + 2cos[12p] + 2cos[4p]

By symmetry around pi, cos[8p] = cos[9p], cos[7p] = cos[10p], cos[6p] = cos[11p], cos[5p] = cos[12p], cos[4p] = cos[13p], cos[3p] = cos[14p], cos[2p] = cos[15p] and cos[p] = cos[16p]

2z^2 = 4 + 2cos[p] + 3cos[2p] + 4cos[3p] + 3cos[4p] + 2cos[5p] + 4cos[6p] + 2cos[7p] + cos[8p] + 2cos[9p] + 2cos[10p] + 2cos[12p] + cos[16p]
2z^2 + z - 2 = 2 + 3cos[p] + 4cos[2p] + 4cos[3p] + 4cos[4p] + 2cos[5p] + 4cos[6p] + 2cos[7p] + 2cos[8p] + 2cos[9p] + 2cos[10p] + 2cos[12p] + cos[16p]
= 2cos[0p] + 2cos[p] + 2cos[2p] + 2cos[3p] + 2cos[4p] + 2cos[5p] + 2cos[6p] + 2cos[7p] + 2cos[8p] + 2cos[9p] + 2cos[10p] + 2cos[11p] + 2cos[12p] + 2cos[13p] + 2cos[14p] + 2cos[15p] + 2cos[16p]
= twice the sum we found earlier = 2*0 = 0.

2z^2 + z - 2 = 0
z = [-1+-sqrt(17)]/4
-1-sqrt(17)/4 < -1
From the cos graph you can see that 3 parts of z are above zero, only one below, so the sum cannot be less than -1.

Therefore, z = -[1-sqrt(17)]/4

Reply 75

khaixiang

5

Rabite

I put up the rest of that trig question for whoever wanted it - Decota was it?
I have a tiny quibble about choosing which sign for the ± (from a quadratic) at the end, though - anyone want to take a look at it? DFranklin? You know everything it seems! Care to help? :p:

(On another note, I got an anonymous neg rep in this forum with the reason "Ahh"... bleh. :frown:

)

again? this is the second time right? looks like you got yourself a secret admirer.

Reply 76

DFranklin

18

Rabite

Oh, I missed out a little bit. At the end, with the quadratic, I'm not sure how to justify the sign when you solve the quadratic...
Franklin, you're up! :p:

Actually, I didn't even think to justify it when I just looked at this!

But it's not too bad; we have

4z = -1\pm\sqrt{17}

, so it will be enough to convince ourselves that z is positive.

\cos \theta + \cos 8\theta = 2 \cos \frac{9\theta}{2} \cos \frac{7 \theta}{2} = 2 \cos \frac{9\pi}{17}\cos \frac{7\pi}{17} = -2 \cos \frac{8\pi}{17}\cos \frac{7\pi}{17}

. Since

\cos \frac{8 \pi}{17} > 0

and

0 < \cos \frac{7\pi}{17} < 1

, deduce

\cos \theta + \cos 8\theta > -2 \cos \frac{8\pi}{17} = -2 \cos 4\theta

So

\cos \theta + \cos 2 \theta + \cos 4 \theta + \cos 8 \theta > \cos 2 \theta + \cos 4 \theta - 2 \cos 4\theta = \cos 2\theta -\cos 4\theta > 0

.

Speleo

From the cos graph you can see that 3 parts of z are above zero, only one below, so the sum cannot be less than -1.

I don't think this is good enough without more justification. You might have the 3 "+ve" parts so much smaller than the one -ve bit that the -ve bit dominates. Edit: of course it is good enough. My mistake!