nacho's MM Qs

[nacho]

okay, just need some indication of whether my technique is correct:
Solved Q's are put in quotes.

Q. Apply the binomial theorem to find the exact value of 1.001^5. Give the exact final answer as a decimal only and show all the working details (Zero mark is ggiven to a solution without using the binomial theorem)

Nacho's notation guide : (5,1) = 5.C.1

(5,0) (1)^5 + (5,1)(1)^4(0.001)^1 + (5,2)(1)^3(0.001)^2 + (5,3)(1)^2(0.001)^3 + (5,4)(1)^1(0.001)^4 + (5,5)(1)^0(0.001)^5

And then, a painful expansion? is there a shortcut to this, it seems dreadfully long...     = SOLVED

Just put 1/1000 instead of 0.001 for a simple expansion T_T

Also, i have a question which requries me to input a matrix, but i don't know how to use latex to create matrix.. so here's a rough indication of what it is:




Q2. where a and b are real constants and b>a>0. Find all the possible values of x such that A^-1 exists.

My workings lead me to find that not equal to -a, a, b or -b , although I had no clue what i was doing, and the answer/workings seemed too general for a 5 mark question..
what i did was:

det(A) = x^3(1 - bx^-1) - a^2(x - b)

x^3 - bx^2 - a^2x + a^2b
x^2(x - b) - a^2(x + b)
(x + a) . (x - a) . (x - b) . (x + b)   =/ 0 (does not = 0)

therefore x is not equal to -a, a, b or -b
R\{-a, a, b or -b}

[david10d]

Pretty sure the first question is incorrect as 'He' probably wants you to use the binomial theorem formula and to not expand.

Thanks for reminding me that it can be +ve or -ve, just not equal to zero. If that's so, can someone confirm for me this answer?

x=to all real numbers other than 2 [is the notation R\{2}? lol]

[pi]

Q. Apply the binomial theorem to find the exact value of 1.001^5. Give the exact final answer as a decimal only and show all the working details (Zero mark is ggiven to a solution without using the binomial theorem)

Nacho's notation guide : (5,1) = 5.C.1

(5,0) (1)^5 + (5,1)(1)^4(0.001)^1 + (5,2)(1)^3(0.001)^2 + (5,3)(1)^2(0.001)^3 + (5,4)(1)^1(0.001)^4 + (5,5)(1)^0(0.001)^5

And then, a painful expansion? is there a shortcut to this, it seems dreadfully long...

I can't think of a more pointless question...



EDIT: 1.9k^th post 

[nacho]

Pretty sure the first question is incorrect as 'He' probably wants you to use the binomial theorem formula and to not expand.

Thanks for reminding me that it can be +ve or -ve, just not equal to zero. If that's so, can someone confirm for me this answer?

x=to all real numbers other than 2 [is the notation R\{2}? lol]

lol you don't just plug in a number and assume it's correct, 'He' does not roll that way.
(epic pun)

[Andiio]

Haha "He"

I wasn't sure if I got the first q right, I had that as well but it was 5 marks so was like :|

Second question was quite easy though, doesn't take long at all!

[nacho]

Second question was quite easy though, doesn't take long at all!

If you weren't being sarcastic, I would have retreated into a shell and sobbed. (yuo were being sarcastic right?)

[david10d]

Pretty sure the first question is incorrect as 'He' probably wants you to use the binomial theorem formula and to not expand.

Thanks for reminding me that it can be +ve or -ve, just not equal to zero. If that's so, can someone confirm for me this answer?

x=to all real numbers other than 2 [is the notation R\{2}? lol]

lol you don't just plug in a number and assume it's correct, 'He' does not roll that way.
(epic pun)

mm thanks for point that out too ._. it couldn't have been that simple.. plus it's incorrect anyways.

 x^3 - bx^2 - a^2x + a^2b
 x^2(x - b) - a^2(x + b)
 (x + a) . (x - a) . (x - b) . (x + b)   =/ 0 (does not = 0)

my algebra skills are quite rusty, but i'm almost sure that it can't be (x^2-a^2) unless they're both under (x+b) or (x-b) [AGAIN could be wrong, nothing can be certain for me on this forum]

maybe you don't have to expand.. x^3(1-b/x) - a^2(x-b)

which leaves:

x=R\{b} (or not 0 too?)

which is essentially the same as my first answer but i didn't substitute 'b' back in

lol looking back it seems WAY too simple. there must be a trick some where or more answers

[Andiio]

Second question was quite easy though, doesn't take long at all!

If you weren't being sarcastic, I would have retreated into a shell and sobbed. (yuo were being sarcastic right?)

o_O well I'm going to redo it later tonight, but I wrote like 5/10^3 and stuff instead of spamming 0's (Not being cocky or anything btw!)

What was the answer again?

I think i'm wrong for Q2 though, it seems much too simple T_T

[david10d]

Second question was quite easy though, doesn't take long at all!

If you weren't being sarcastic, I would have retreated into a shell and sobbed. (yuo were being sarcastic right?)

o_O well I'm going to redo it later tonight, but I wrote like 5/10^3 and stuff instead of spamming 0's (Not being cocky or anything btw!)

What was the answer again?

I think i'm wrong for Q2 though, it seems much too simple T_T

The answer is 1.00501.

By the way which spesh class are you in? Haha. Did you end up figuring out how to do the last question (the one with the spammed cos) using compound formulas?

[Andiio]

Second question was quite easy though, doesn't take long at all!

If you weren't being sarcastic, I would have retreated into a shell and sobbed. (yuo were being sarcastic right?)

o_O well I'm going to redo it later tonight, but I wrote like 5/10^3 and stuff instead of spamming 0's (Not being cocky or anything btw!)

What was the answer again?

I think i'm wrong for Q2 though, it seems much too simple T_T

The answer is 1.00501.

By the way which spesh class are you in? Haha. Did you end up figuring out how to do the last question (the one with the spammed cos) using compound formulas?

Sweet I think that's what I got Like I said i'm going to redo them later! Only just got home haha

Sat mornings! Not yet, I didnt spend much time on it - did you do it?

[david10d]

Second question was quite easy though, doesn't take long at all!

If you weren't being sarcastic, I would have retreated into a shell and sobbed. (yuo were being sarcastic right?)

o_O well I'm going to redo it later tonight, but I wrote like 5/10^3 and stuff instead of spamming 0's (Not being cocky or anything btw!)

What was the answer again?

I think i'm wrong for Q2 though, it seems much too simple T_T

The answer is 1.00501.

By the way which spesh class are you in? Haha. Did you end up figuring out how to do the last question (the one with the spammed cos) using compound formulas?

Sweet I think that's what I got Like I said i'm going to redo them later! Only just got home haha

Sat mornings! Not yet, I didnt spend much time on it - did you do it?

Nope, couldn't be bothered with that one after doing all his other questions >_<

Too long ~_~

Could you also explain how you did the binomial question using the correct method when you're up to it? Pretty keen on knowing the steps.

[xZero]

Q2.
, this equation will have a linear factors so you look at the last number , this tells me that the linear factors will be something along the line of .
Using null factor theorem, you sub or into the equation and see which one equals to 0, and does so is a factor.

Since we have 1 factor, we can use long division or calculator to find the other, which is , you can then simply it down to

so

[luffy]

Q2.
, this equation will have a linear factors so you look at the last number , this tells me that the linear factors will be something along the line of .
Using null factor theorem, you sub or into the equation and see which one equals to 0, and does so is a factor.

Since we have 1 factor, we can use long division or calculator to find the other, which is , you can then simply it down to

so

Using null factor theorem for this question is not the best approach to take, simply because if one of the factors was (x - b/2), it would be hard to think of the right term to apply.

Therefore, as Nacho did, I would suggest using "grouping two and two", which would clearly work in this question. The only issue was Nacho's working out, which contained a minor error.
(I don't know how to use Latex yet unfortunately... Its on my "to do" list)
A^-1 exists if det(A) does not equal 0.
Thus:

x^3(1 - bx^-1) - a^2 (x - b) does not equal 0.
multiplying one of the x's in the x^3, gives:

x^2( x - b) - a^2 ( x - b) =/ 0
(x^2 - a^2)( x - b) =/ 0 (grouping two and two).
(x + a) ( x - a) (x -b) does not equal 0.
Thus, x is not equal to a, -a or b.
i.e. x is an element of R\{a,-a, b}.

Problem solved.

(Also, I might not have shown sufficient working out, since its a 5-mark question. My teacher always taught us to draw a graph for inequalities, and to "solve the equality" and then work it through for an inequality. Any of these might be required to obtain all the marks for this question)

[nacho]

oh crap..silly mistake when i factorised.

from
i should get

and then ofcourse:


EDIT: DAMN YOU LUFFY

[david10d]

Q2.
, this equation will have a linear factors so you look at the last number , this tells me that the linear factors will be something along the line of .
Using null factor theorem, you sub or into the equation and see which one equals to 0, and does so is a factor.

Since we have 1 factor, we can use long division or calculator to find the other, which is , you can then simply it down to

so

Awesome! I kept looking back at it to find another answer but couldn't figure out how to find it. You'd have to use long division coz this question is supposed to be done without a calculator.

Once again, good stuff....

cool forum code[/sarcasm]

mm could also be done using above methods.