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2u Mathematics Marathon v1.0 (1 Viewer)

SoulSearcher

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You sure are bored aren't you pLuvia? :p
Ok then, just a simple probability question to get other people in.

Next Question:

Find the least number of throws of a fair die that will ensure a 75% chance of obtaining at least one six
 

Sober

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SoulSearcher said:
You sure are bored aren't you pLuvia? :p
Ok then, just a simple probability question to get other people in.

Next Question:

Find the least number of throws of a fair die that will ensure a 75% chance of obtaining at least one six
Chances of a 6 having appeared after "n" throws = 1 - (5/6)^n = 0.75

(5/6)^n = 0.25

log base (5/6) of 0.25 = 7.603... rounds up to 8
 

rama_v

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Here's a nice easy question

Next Question:

If f(x) = √x , find f'(x) from first principles.
 

yoakim

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how did you do that super complex question soulsearcher? you must have a super good tutor :D i'm in year 11 as well and yet i have absolutely no idea of how to even do that question, ...as i have not learnt that topic yet :D
 
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pLuvia

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yoakim said:
how did you do that super complex question soulsearcher? you must have a super good tutor :D i'm in year 11 as well and yet i have absolutely no idea of how to even do that question, ...as i have not learnt that topic yet :D
SoulSearcher has already completed the 2u course, accelerated year 12 2u maths last year and I think receive a very high mark

Do be able to do that question, if I think it's the one you are talking about, you must have learnt sequences and series and also that specific question was one of the questions from a CSSA trial
 

SoulSearcher

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yoakim said:
how did you do that super complex question soulsearcher? you must have a super good tutor :D i'm in year 11 as well and yet i have absolutely no idea of how to even do that question, ...as i have not learnt that topic yet :D
pLuvia said:
SoulSearcher has already completed the 2u course, accelerated year 12 2u maths last year and I think receive a very high mark
Yeah, that's pretty much it.
 

SoulSearcher

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rama_v said:
Here's a nice easy question

Next Question:

If f(x) = √x , find f'(x) from first principles.
Yeah, I'll do this one as well, although the solutions taxed from the Cambridge 3 unit book.

lim (u-->x) [f(u) - f(x)] / [u - x]
= lim (u-->x) (√u - √x) / (u - x)
= lim (u-->x) (√u - √x) / (√u - √x)(√u + √x)
= lim (u-->x) 1 / (√u + √x)
= 1 / (√x + √x)
= 1/2√x
 

rama_v

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Yeah. Thats similar to what I had (I think!) :

f'(x) = lim(h->0) [f(x+h- - f(x)] / h

= lim(h->0) [√(x+h) - √x] / h * (√(x-h) + √x)/(√(x-h) + √x)

= lim(h->0) (x+h-x)/h(√(x-h) + √x)

= lim(h->0) h/ h(√(x-h) + √x)

= lim(h->0) 1/(√(x-h) + √x)

= 1/(√x + √x) = 1/(2√x)
 

sando

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NEW QUESTION:

Find the equation of the normal to the curve y = ex at the point where x = 3, in exact form
 
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pLuvia

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I think you meant ex
y=ex

dy/dx=ex

At x=3 y=e3

dy/dx=e3

MNormal=-1/e3

y-e3=-1/e3(x-e3)

(1/e3)x+y+(1-e3)=0
 
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Riviet

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I guess he still hasn't learnt how to use sup and sub tags yet. :p

Feel free to post the next question up when you're ready pLuvia. ;)
 

sando

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Riviet said:
I guess he still hasn't learnt how to use sup and sub tags yet. :p

Feel free to post the next question up when you're ready pLuvia. ;)
not funny lol...

i finally figured out on word.. but it wouldnt let me copy it onto here
 

Riviet

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Nope, not allowed to anymore... :(

But let the marathon continue! (pLuvia's turn to post question)
 
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pLuvia

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Riviet said:
Nope, not allowed to anymore... :(

But let the marathon continue! (pLuvia's turn to post question)
lol it was kind of funny that thread :p

Next Question
A sector of a circle with centre O and radius rcm is bounded by the radii OP and OQ and by the arc PQ. The angle POQ is θ radians.
(i)Given that r and θ vary in such a way that the area of the sector POQ has a constant value of 100cm2, show that θ=200/r2
(ii)Given also that the radius is increasing at a constant rate of 0.5cm/s, find the rate at which the angle POQ is changing when r=10cm
 

word.

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1.
pi * r^2 * @/2pi = 100
r^2 * @ / 2 = 100
r^2 * @ = 200
@ = 200/r^2

2.
d@/dr = -400/r^3
dr/dt = 0.5
d@/dt = ?

d@/dt = dr/dt * d@/dr

at r = 10, d@/dt = 1/2 * -4/10 = -1/5
i.e. @ is decreasing at 1/5 radians per second

Question
Solve for x:
2ln(x) - ln(14x+1) = 0
 

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