.
I did use this whilst searching for my roots, and clearly -1 doesn't work, -2 does, -3 and -4 fail after quick mental calculation, -6 is a hit. Then I just solved the quadratic. I suspected all way along that I was supposed to do something using more of the previous result, but as I knew that bashing it out by guessing roots would be quick I didn't think more about it. (I did these two questions last week, so just typed them up now)

DFranklin

I think this is your problem:

Indeed, corrected.
Thanks!

Reply 25

steppity-step

Dadeyemi

Some more;

Did these quite a while ago I'm afraid some may be partial solutions.

RE Q3: You lost an a squared very early on and so came to get a minimum on the x-axis. In fact the minimum occurs when

x = \frac{a^2 b}{a^2 - 1}

, and at this point f(x)>0. Also, your axes are very inappropriately placed seeing as b>0. Thankfully you then gave up, because if you had carried on you would have got into quite a mess!

For anyone looking for a solution to this question I got:

c^2 < b(a^2-1)

gives 0 solutions

c^2 > ba^2

gives 1 solution

ba^2 \geq c^2 > b(a^2-1)

gives 2 solutions

c^2 = b(a^2-1)

also gives 1 solution

For part (i) note that we have the case where

c^2 = b(a^2-1)

, and require

x = \frac{a^2 b}{a^2 - 1} = \frac{9}{4}

. This is the only solution.

For part (ii) it isn't quite as neat, we have the case where

ba^2 \geq c^2 > b(a^2-1)

so we're looking for two solutions. One is x=4 which you see quickly on inspection, for the other we have to trawl through some algebra to get x=49/16. Even if the first part of the question isn't to taste I think part two is worth doing to check you know what to do to transform the equation into something managable, namely a standard quadratic.

Reply 26

maltodextrin

Dadeyemi

Some more;

Did these quite a while ago I'm afraid some may be partial solutions.

I think you're missing an a^2, i get the minimum to be (b(a^2)/(a^2 - 1), (b(a^2 - 1))^1/2. To finish the question off i get:

By looking at the graph it is clear that:
Conditions for exactly one solution are c = b(a^2)/(a^2 - 1) or c > a(b^1/2)
For exactly two solutions b(a^2 - 1) < c^2 < (a^2)b

i) Squaring once gives 8x - 14 = 8(x - 2)^1/2, squaring again gives 16(x^2) - 72x + 81 = 0, by the quadratic formula, x = 9/4

ii) Squaring once 8x - 22 = 10(x - 3)^1/2 squaring again 16(x^2) - 113x + 196 = 0 by the quadratic formula x = 4 or x = 49/16

Seems too easy so i've probably gone wrong. I can't see how the first bit helps with the second, it tells you how many solutions the equations have but you find that out for yourself when you solve them. Any feedback welcome.

Reply 27

maltodextrin

toasted-lion

RE Q3: You lost an a squared very early on and so came to get a minimum on the x-axis. In fact the minimum occurs when

x = \frac{a^2 b}{a^2 - 1}

c^2 < b(a^2-1)

gives 0 solutions

c^2 > ba^2

gives 1 solution

ba^2 \geq c^2 > b(a^2-1)

gives 2 solutions

c^2 = b(a^2-1)

also gives 1 solution

For part (i) note that we have the case where

c^2 = b(a^2-1)

, and require

x = \frac{a^2 b}{a^2 - 1} = \frac{9}{4}

. This is the only solution.

For part (ii) it isn't quite as neat, we have the case where

ba^2 \geq c^2 > b(a^2-1)

I left this page open overnight and didn't refresh, well i guess 2 solutions is better than none lol!

Reply 28

maltodextrin

SimonM

STEP III, Question 1

Spoiler

I might be missing something obvious but how did you determine that (lna)^2/a and (lna)/a tend to zero as a tends to infinity? I knew that (lna)/a tended to zero but proved it using l'hopital because I thought they'd want to see proof, i didn't know that (lna)^2/a did though so also ended up showing it using l'hopital.

Reply 29

steppity-step

maltodextrin

I left this page open overnight and didn't refresh, well i guess 2 solutions is better than none lol!

At least you know you didn't go wrong now!

Reply 30

steppity-step

maltodextrin

How's this:

\frac {ln(x)}{x}

tends to 0 as x tends to infinity, which we know and seems intuitively obvious (but please post your proof as I've never heard of l'hopital) so, letting

x = \sqrt a

\frac {ln(\sqrt a )}{\sqrt a}

tends to 0 and also

\frac {2ln(\sqrt a )}{\sqrt a} = \frac {ln(\sqrt a ^2 )}{\sqrt a} = \frac {ln( a )}{\sqrt a}

tends to 0, so squaring,

\frac {(ln(a))^2}{a}

tends to 0. That seems ok to me, now your "l'hopital"?

Reply 31

sonofdot

toasted-lion

How's this:

\frac {ln(x)}{x}

tends to 0 as x tends to infinity, which we know and seems intuitively obvious (but please post your proof as I've never heard of l'hopital) so, letting

x = \sqrt a

\frac {ln(\sqrt a )}{\sqrt a}

tends to 0 and also

\frac {2ln(\sqrt a )}{\sqrt a} = \frac {ln(\sqrt a ^2 )}{\sqrt a} = \frac {ln( a )}{\sqrt a}

tends to 0, so squaring,

\frac {(ln(a))^2}{a}

tends to 0. That seems ok to me, now your "l'hopital"?

http://en.wikipedia.org/wiki/L'hopital's_rule

Reply 32

Dadeyemi

STEP III, Question 8
this was a bit laborious to type out

Spoiler

Reply 33

maltodextrin

toasted-lion

How's this:

\frac {ln(x)}{x}

tends to 0 as x tends to infinity, which we know and seems intuitively obvious (but please post your proof as I've never heard of l'hopital) so, letting

x = \sqrt a

\frac {ln(\sqrt a )}{\sqrt a}

tends to 0 and also

\frac {2ln(\sqrt a )}{\sqrt a} = \frac {ln(\sqrt a ^2 )}{\sqrt a} = \frac {ln( a )}{\sqrt a}

tends to 0, so squaring,

\frac {(ln(a))^2}{a}

tends to 0. That seems ok to me, now your "l'hopital"?

That's a nice little proof, looks fine to me. You should definitely check out l'hopitals rule, I use it quite a bit. I'm just worried that they won't let us get away with (lnx)/x tending to zero being obvious so if it comes up i'll probably prove it to stay on the safe side.

Reply 34

steppity-step

maltodextrin

I think we're safe with (lnx)/x tending to 0 being obvious but I think you're right to be cautious with anything more complicated. I checked out l'hopital's rule but didn't really follow the proof, I think I'll see if my maths teacher can help next week. I think I'm starting to get somewhere with STEP now, I can normally find 3 or 4 questions I'm ok with and scribble something down for another 2 or 3, still not totally confident of getting 2 1s though. It's time as well, I just took over an hour on a STEP II question!

Reply 35

Evan247

Sorry, but I suppose in Q3, the minimum value of f(x) should be sqrt[(a^2-1)*b]? because in your inequality x>a^2*(x-1), you should solve it like:
(1-a^2)x>-b*a^2...

Dadeyemi

Some more;

Did these quite a while ago I'm afraid some may be partial solutions.

Reply 36

Dadeyemi

I've removed III q3 from my post it was utter nonsense.

Reply 37

Evan247

But I wanna ask if in TSR there's correct solutions somewhere else?

Dadeyemi

I've removed III q3 from my post it was utter nonsense.

Reply 38

Dadeyemi

Evan247

But I wanna ask if in TSR there's correct solutions somewhere else?

Post #36 and #37

Reply 39

mikru

sonofdot

STEP II 2002 Question 1

Part One

Part Two

Part Three

How did you decide upon your substitution for part three? It's things like this I would have no chance finding in the actual exam. I managed to get the same result as you using the substitution

x = \mathrm{cosec\ } \theta

(which I think followed on from the previous parts of the question better), but I think yours might have reached the result quicker (although I don't really follow your working in the middle).