harder 3U Q (1 Viewer)

[mojako]

Can smeone do this pls..?
Note: I'm not really sure what the RHS in part (b) is.

 

[Xayma]

Was there any diagram that came with this? Im intrigued by "lower trapezia"

 

[googleplex]

just drawing trapeziums under the graph getting all those areas < integrating from 1 to n
?

yeh was there any pics or lead ups to this q

 

[nike55]

hehe usually i dont post anymore but im in a good mood from my umat scores
(btw it would be good if someone got latex or some math program for this board like other maths forums do)

anyway

area under y = lnx (from 1 to n) > area under sum of trapeziums from 1 to n
nlnn - n + 1 > 1/2(0+ln2) + 1/2(ln2+ln3) + .... + 1/2 (ln(n-1) + lnn)
1/2lnn + nlnn - n + 1 > lnn! (adding 1/2lnn to each side)
e(1/2lnn + nlnn - n + 1) > n!
e(1/2lnn) * e(nlnn) * e(-n) * e > n!
(n^1/2 * n^n * e)/e(n) > n!
(e * n^(n+1/2)) / e(n) > n! as required

 

[mojako]

Was there any diagram that came with this? Im intrigued by "lower trapezia"

unfortunately there wasnt

I'll read nike's answer.
Thanks!

When I tried to do this question I used the wrong formula for trapezoid rule.... and also I tried to find the area from 1 to (n+1).

 

[mojako]

Is there a term to say if I make "x" becomes "2^x" or "e^x"?
If you dont get what I mean... making "x" becomes "x^2" is called.. raising x to the power of two.

EDIT: I know now... somebody told me.. it's "raising 2 to the x" or "raising e to the x", hehe..

 

[CrashOveride]

http://img65.exs.cx/img65/6325/CIRCLES.jpg

Question 5. Circle Geo aint going so well ^^

 

[mojako]

another Q:
solve 6x^4 - 7x^3 - 28x^2 +35x -10 = 0

I'm sure many, many of you are much, much smarter than me
Just do the circle and polynomial questions by "inspection" like in <a href=http://www.boredofstudies.org/community/showthread.php?t=41677>here</a>

Inspection inspection.. beware.. be careful.. the inspectors are coming.. hide hide hide... cover up any mess.
Now, where are our inspectors??
I need them!!

 

[withoutaface]

by inspection only works if they don't give you the answer

 

[mojako]

by inspection only works if they don't give you the answer

I dont think I give the answer for the polynomial Q

 

[ngai]

solve 6x^4 - 7x^3 - 28x^2 +35x -10 = 0

lol u asked, so here it is
clearly, x = 1/2, 2/3, +- sqrt(5) (by inspection)

 

[mojako]

oh.. my dearest inspector...
anyway I sneaked into your office and found your inspection method
now I'm gonna make the whole community knows

since the coefficients of the polynomial are all integers, the numerator of any rational roots is a factor of the constant term, and the denominator is a factor of the leading coefficient.
BTW do we have to know this?? It's not mentioned in the syllabus, which however has a section (section 7.1) which specifically talks about integer roots being factors of the constant term. I've seen a question like this but I thought it was a bad question.

Now that you secret inspection procedure has been broken, would be very friendly to do the circle question not by inspection?

 

[ngai]

since the coefficients of the polynomial are all integers, the numerator of any rational roots is a factor of the constant term, and the denominator is a factor of the leading coefficient.
BTW do we have to know this?? It's not mentioned in the syllabus, which however has a section (section 7.1) which specifically talks about integer roots being factors of the constant term. I've seen a question like this but I thought it was a bad question.

Now that you secret inspection procedure has been broken, would be very friendly to do the circle question not by inspection?

ROFL i didnt even bother with that
i used maple

hahaha alrite i'll do the circle geo
extend PO to meet circle again at E
BD & OP intersection is F
now, let angle FDP = a
so CDB = a
and so CAB = a
and so PEC = a
and so triangles ECP, FDP similar (equal angles a and P)
so angle DFP = angle ECP = 90

 

[mojako]

Thank you

How do you know how to do this though??
How long did it take you?

 

[Archman]

'eres another soln to the circle geo (by inspection )
draw in tangent at P, call a point to the right of P on the tangent E
let PO meet BD at F
EPA = PCA = DBP so PE and DB are parallel.
now since PE is a tangent, so PE is perpendicular to OP, so DB is perpendicular to OP.

 

[CrashOveride]

You mean E is to the right of P on the tangent...then all works well thanks for that Archman ^_^

 

[mojako]

omg.. inspection..
its such a wonderful and powerful technique..
I wonder why we weren't taught that technique at school

On the serious side, how do these things come to your mind?

 

[Archman]

You mean E is to the right of P on the tangent...then all works well thanks for that Archman ^_^

thx, its corrected.

 

[Rorix]

On the serious side, how do these things come to your mind?

You inspect it.

hence, 'by inspection'.

 

[withoutaface]

omg.. inspection..
its such a wonderful and powerful technique..
I wonder why we weren't taught that technique at school

On the serious side, how do these things come to your mind?

You fuck around with a few things in your head until you find one that works.