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2u Mathematics Marathon v1.0 (2 Viewers)

airie

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SoulSearcher said:
Ok:

Two circles of radii 2cm and 3cm touch externally at P. AB is a common tangent. Calculate, in cm2 correct to two decimal places, the area of the region bounded by the tangent and the arcs AP and BP
You can use the calculator, right?

Actually, never mind. I don't have another question ready :p
 

Riviet

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airie said:
You can use the calculator, right?

Actually, never mind. I don't have another question ready :p
Of course you can use the calculator, how else would you do it? :p

You don't have to post up a new question straight after you solved the last one. ;)
 

ianc

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Circle with centre X has a radius of 2cm
Circle with centre Y has a radius of 3cm

XA perpendicular to AB (tangent perp to radius)
Likewise YB perp to AB

Drew line XZ such that ABZX is a rectangle

Need to find angle YXZ:
sin(YXZ) = 1/5
angle YXZ = 0.201 (keep it in rad, store it in calc memory)

Hence angle XYZ = (pi/2)- angleYXZ (store this in memory as well)

Area sector AXP:
A=(1/2)(4)(pi/2 + angleYXZ)

Area sector PYB
A = (1/2)(9)(angleXYZ)

Length XZ = sqrt(24) from pythagoras

Therefore total area of trapezium AXYB
= (1/2)[sqrt(24)][3+2]

Therefore area between two circles and AB
= trapeziumAXYB - sectorAXP - sectorPYB
= 2.504 cm squared
 

Raginsheep

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The only "2u" one I know is that because for some certain x-values, there are more than one y-value which is shown by the horizontal line test.

Actually, the uni one is pretty much the same except a bit more fancy.
 

SoulSearcher

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Ok, I'll just throw in a question here:

A coin is tossed continuously until, for the first time, the same result appears twice in succession. That is, you continue tossing until you toss two heads or two tails in a row.
i) Find the probability that the game ends before the sixth toss of the coin.
ii) Find the probability that an even number of tosses will be required.
 

Sober

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Exphate said:
I think (i) is right, as for two its a random guess :D
(i) 1- (1/2)5
=1-(1/32)
=(31/32)
:. The chance of ending before the sixth roll is 31/32

(ii) 1/2? (nfi!)
Surely i) is actually:

1 - (1/2)4

// Because it has to end before the sixth toss, so it is the complement of the 2nd,3rd,4th and 5th tosses alternating.

As for ii):

It has to be greater than 1/2 as that is the odds that it ends on the second toss alone, I think it is more like:

(1/2) + (1/2)^3 + (1/2)^5 ...

Which is a geometric expression:

first = 1/2
ratio = 1/4

sum = (1/2)(1-(1/4)) = 2/3
 

Riviet

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Next Question:

i) Show that
d {y.[f(y)]2} = 2y.f(y).f'(y) + f(y)2
dy

ii) Deduce that
d {y(lny)2 - 2ylny + 2y} = (lny)2
dy

ii) Hence evaluate
e
∫(lny)2 dy
1
 
Last edited:

Riviet

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Exphate said:
Oh well was i right in the first place?
Unfortunately you weren't. :p

I've changed it so you pretty much know whether or not you're correct in part i) and ii).
 

CharlieB

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Can someone please do this question...

Solve cosx = cos2x (Solve for 0<x<2pi)

How do you go around doing this?
 
P

pLuvia

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cosx=cos2x
cosx=2cos2x-1
2cos2x-cosx-1=0
(2cosx+1)(cos-1)=0
cosx=-1/2 or cosx=1
x=0, 2pi/3, 4pi/3, 2pi
 

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