couple of questions. (1 Viewer)

[sammeh]

right i broke my brain trying to answer these in my trial. so i thought they might be interesting to see other approaches.

a) prove (cos x) - 1 + (x^2/2) > 0

i stared that this for a good minute or so then said "buggrit" and moved on.

b) <i>refer to attachment for diagram. pardon my crappy drawing, but its clear enough to make out.</i>

a special chess board is made such that it has five rows and eight columns. 3 squares are marked upon it, as in the diagram, x y and m.

i: if a player can only move in the directions shown upon the board, one square at a time, how many ways are there to get from x to y?

ii: how many ways are there to get from x to y if the player <u>must</u> pass through m.

i think i got the 1st part of this one, but not the second.


cheers people.

 

[ngai]

a) when x = 0, cos0 = 1, 1 = 1, 0^2/2 = 0, so LHS = 1-1+0 = 0
0 > 0?...i don't think so

b) X to Y
he must move 2 right, 5 up
and he can only move right and up
so from 7 moves to make, choose 2 to be "right"
therefore 7C2 = wateva, u press the buttons

ii) X to M: similarly, 4C1
M to Y: similarly, 3C1
so 4C1 * 3C1 = 12

 

[Rorix]

its supposed to be >= 0, use calculus

 

[Archman]

well you dunt really hafta use calculus straight away, not that theres anything wrong with it. but you can also
let x = 2y
cos2y - 1 + (2y)^2/2
= (1 - 2sin^2y) - 1 + 2y^2
= 2(y^2 - sin^2y)
well u can use calculus now if you want to, but im sure u've proven |y| > |siny| many times

 

[sammeh]

hrmm.

really wasnt thinking well during exam i suppose. now i feel even stupider than i did when i walked out of the room ^^

thanks all.

 

[CrashOveride]

it says find locus of z if
w = (z-i)/(z-2) is purely imag

I know how to do it algebraically but i was wanting also to do it graphically.

For the upper half of the circle it says argw = pi/2 and for lower half argw = -pi/2

Ive attached my poor diagram. How can i show this geometrically? That would solve my dillemma.

 

[Archman]

well i reckon you still need to do this algebra bit
argw= arg((z-i)/(z-2)) = arg(z-i) - arg(z-2)
now since w is purely imaginary, argw is pi/2 or -pi/2
so we know that the lines from P(z) to the points (0,1) and (2,0) are perpendicular, in other word the angles between them is always pi/2, so it must be a circle with diameter (0,1) and (2,0)