harder 4u cambridge q's i couldn't do (1 Viewer)

[Grey Council]

Q 16, exercise 4.1
This is hard to write out, sorry. i'll do my best, but its better if you go and check yourself. btw, sorry turtle, i couldn't do this. i tried a bit, but I didn't know how to progress. ah well

If P(x) - 1 - x + (x^2)/2! - ... + (-1)^n.((x^n)/(n!))
show that P(x) has no multiple zeros for n>=2

Q 14, exercise 4.2
P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 4. One zero is root 2, snother zero is rational and the sum of the zeros is positive. Factorise P(x) fully over R.

if the second one isn't too hard, tell me. I'll have another go at it tommorow.

 

[turtle_2468]

4.1:
For n=2, obvious.
Suppose true for n=k.
Then consider n=k+1. Consider the graph of P(x). Suppose x=m (m constant) is a multiple zero for n=k+1. (1)
Then that means that x=m is a zero for n=k. (2)
So: From (1), 1+m-m^2/2......+(-1)^k.(m^k/k!)+(-1)^(k+1).(m^(k+1)/(k+1)!)=0 (A)
From (2), 1+m-m^2/2......+(-1)^k.(m^k/k!)=0 (B)
So subtracting, eq. (A)-(B) yields
(-1)^(k+1).(m^(k+1)/(k+1)!)=0 (only term on LHS of (A) not on LHS of (B))
So m=0 is a soln
But then subbing in m=0 we get P(m)=1
so m is not a root *CONTRADICTION THUNDER!!!!**
Hence m is not a root, and we don't have any multiple roots by contradiction. So by PMI this is true.

This was a hard q...
Second one looks quite hard... try guessing/rational root theorem (I THINK there is such a thing.. but then it's 1 ;-) )
bump this post up tomorrow if you haven't figured out 2, and when I get back I'll try get a soln

 

[Giant Lobster]

Irrational conjugate theorem? if all coeffs are rational, and an irrational root exists, then its conjugate is also a root.

I asked for a proof for this a while ago. I think J0n answered me, if i remember correctly.

 

[Grey Council]

shiet, that was a killer. No wonder I didn't get it. lol

I'll try and do question 2 now. Berloody, its good to tuck in ones chin when trying the last cambridge questions. Your gonna get a blow, so might as well get ready for it.

 

[grimreaper]

The second question isnt too hard. Since root 2 is a zero, the remaining zeroes must multiply to give 2 root 2. Just by inspection, and knowing at least one root is rational and the sum of the roots is positive, you can tell the other roots are 2, -1, - root 2. Hahah theres probably some dodgy logic in there but I think I'm right, am I???

 

[Grey Council]

GGGNNNRRRR!! shuddup, you! lol, i figured it out, and now your saying its easy. Just shut up.

heehee, here is my brilliant logic:

P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 4. One zero is root 2, snother zero is rational and the sum of the zeros is positive. Factorise P(x) fully over R.

so the equation is in form:
x⁴ + ax³ + bx² + cx + 4
a > 0

so to start off, the equation factorised looks like this:
(x - sqrt2)Q(x)
now product of roots will be +4. So to get rid of any square root signs, we can let another root be (x + sqrt2), not minus sqrt2 cause it says ONE root is minus sqrt2.

so now it looks like this:
(x - sqrt2)(x + sqrt2)Q(x)
as we know the product of roots gives us 4, so
-2x = 4 (explanation: sqrt2 * -sqrt2 = -2)
x = -2
factors of -2 are: -2 and 1 OR 2 and -1 (we know at least one other root is rational. )
however, zeros will add up to be positive (a is positive, given)

therefore zeros are sqrt2, -sqrt2, 2 and -1

.'. (x-2)(x+1)(x - sqrt2)(x + sqrt2)
and thats QED beep. Q berloody E bloody D.

damn, i am, once again, impressed at my own brilliance. And Grimreaper, don't say how easy it easy, act impressed. Your ruining my day, man, let me be happy. I don't often solve these hard questions.

 

[grimreaper]

You're brilliant Grey Council, I would never have been able to answer that question so elegantly

Anyway, exercises 4.1-4.3 in cambridge are all ok once you know the processes, but 4.4 is very hard..... none of my answers seem to come out right, so look forward to that one

 

[Grey Council]

You serious? I didn't get taught 3u polynomials last year at all, so i didn't do ANY homework. I crammed (as much as you can "cram" for maths) in the yearlies for year 11, so i know the very basics (ie -b/a, c/a etc lol ). 4.3 is revision of 3u mostly, and i'm struggling mate . Today at school, we covered the theory for 4.4, i didn't get a single word. heh, i'll just have to do it myself, hrm.

oh and thank you about the compliment. ^__^

btw, i just did my first trial 4u paper ever. I don't think I have to mention that I got owned. I couldn't even do the complex numbers section. hrmph. It was question 3, and i got half of question 3. q 1 (integration) and 2(graphs) i skipped, 4(conics) i skipped, i did question 5b (growth and decay... WTF thats 2u stuff, whats wrong? Am i doing something wrong? is there new theory for growth and decay in 4u? this is an 1986 paper). question 8(b) looked easy, i got happy, i wrote a few lines and I got stuck. heh, knew it couldn't be that easy.

 

[grimreaper]

Actually, 4.3 is really different from 3u... I dont remember having to make new equations from variations on the roots in 3u(someone correct me if I'm wrong), but its really not difficult at all since there are only a couple of processes that fit every question. Also, I recently found out that the stuff at the end of 4.4 requires binomial theorum, so I'm guessing thats why I couldnt do it hehe (the early questions arent TOO bad....)

Oh and as for doing past papers.... Lets just say I'll leave it a while, all I've done in 4u is complex no.'s and polynomials

 

[Grey Council]

lol i haven't even done polynomials.

and the 3u stuff, i meant questions 1-11. After that you hafta use the modifications of roots to get equation/however you say it in words stuff. hehe

 

[Grey Council]

according to my teacher, we've finished polynomials. wtf, we started it last thursday'ish

 

[:: ck ::]

LOl ur like us

finished the whole freakn chapter and started it last friday or something

 

[Heinz]

Originally posted by :: ryan.cck ::
LOl ur like us

finished the whole freakn chapter and started it last friday or something

I think all maths teachers are like that. In the past 2-3 weeks, weve supposedly covered polynomials, integration and volumes. Went through volumes in one lesson "thats basically it" says my teacher...

 

[KeypadSDM]

Originally posted by Grey Council
So to get rid of any square root signs, we can let another root be (x + sqrt2), not minus sqrt2 cause it says ONE root is minus sqrt2.

It says one, it never says, "Not 2". Tut tut, you can't just assume things

 

[Grey Council]

well then, you can always post up how to do it.

tsk tsk tsk Keypad. Thats bad form. Deflating my puffed up ego, and then not even telling me how to do it.

 

[KeypadSDM]

It's the buchernan approach to learning.

YOU GOT IT WRONG, AND I'M NOT RELEASING MY ANSWERS!

 

[freaking_out]

Originally posted by KeypadSDM
It's the buchernan approach to learning.

YOU GOT IT WRONG, AND I'M NOT RELEASING MY ANSWERS!

LoL- and hence the students looses.

 

[KeypadSDM]

Originally posted by freaking_out
LoL- and hence the students looses.

But since they don't go to taylors, he doesn't mind