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2u Mathematics Marathon v1.0 (4 Viewers)

insert-username

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It's being fixed. The board isn't handling the special symbols particularly well at the moment due to the upgrade, but the admins are on it. :)

Next question:

For the parabola x2 + 2x + 28y - 111 = 0, find the coordinates of its vertex and focus, and the equations of its directrix and axis. What is the maximum value of this parabola?


I_F
 

SoulSearcher

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insert-username said:
It's being fixed. The board isn't handling the special symbols particularly well at the moment due to the upgrade, but the admins are on it. :)

Next question:

For the parabola x2 + 2x + 28y - 111 = 0, find the coordinates of its vertex and focus, and the equations of its directrix and axis. What is the maximum value of this parabola?


I_F
for this question we have to arrange the equation into the form (x-h)2 = 4a(y-k), where h is the x-value and k is the y-value of the vertex

x2 + 2x + 28y - 111 = 0
x2 + 2x = 111 - 28y
x2 + 2x + 1 = 112 - 28y
(x+1)2 = - 28 (y-4)
(x+1)2 = 4 * -7(y-4)

therefore the vertex of the parabola is ( -1, 4 )
the focus of the parabola is ( -1, 4 - 7 ) = ( -1, -3 ) as the focus is equal to ( h, k + a )
the directrix of the parabola is y = 4 - (-7) = 11
the axis of the parabola is x = -1

the general equation for a concave down parabola with vertex (h,k) is (x-h)2 = -4a(y-k). since the equation (x+1)2 = 4 * -7(y-4) fits this equation, the parabola is concave down.
therefore the maximum value of the parabola is the y-value of the vertex, which is y = 4
 

Riviet

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SoulSearcher said:
i wouldnt have any idea if this is hard but here it goes.

next question:
Find the smallest distance between the graphs of y = x2 - 4x + 12 and y = 2x + 1
y = x2 - 4x + 12

dy/dx = 2x-4

2x-y+1=0 is closest to y = x2 - 4x + 12 at the point where m=2

.: 2x-4=2

x=3
y=9

.: distance= |(ax+by+c)/sqrt(a2+b2)|

= |(6-9+1)/sqrt5|

= 2/sqrt5 units

This is a fun one :lol:

Next Question:

If log96=m and log249=k, express log37 in terms of k and m.
 

Mountain.Dew

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Riviet said:


Next Question:

If log96=m and log249=k, express log37 in terms of k and m.
this one took me some time, but here it is:

1st, make k and m in terms of logs in base 3:

m = log96 = log36/log39 = log36/2
m = log3(3*2) / 2 = log33/2 + log32 / 2 = 1/2 + log32/2

m = 1/2 + log32/2....(1)

k = log249 = log349 / log32 = 2log37/log32

k = 2log37/log32....(2)

2nd, in (2), make log32 and sub (2) into (1). then, make log37 the subject to get ur answer in terms of m and k.

i'll let u guys do the algebra hehehehe :)
 

Slidey

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k=2log27

log37=log27/log23=k/[2log23]

log96=m, 9m=6=32m, 2m.log23=log26,
2m=log26/log23=1/log23 + 1

.'. 2m-1=1/log23, so log37=k/(2m-1)

Next question:

Find the values of x for which (log10x)(log10x2) + log10x3 - 5=0
 
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Riviet

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Slide Rule said:
Next question:

Find the values of x for which (log10x)(log10x2) + log10x3 - 5=0
let u=log10x

Then 2u2+3u-5=0

(2u+5)(u-1)=0

u=-5/2 or u=1

log10x=-5/2 or log10x=1

.: x=10, 10-2.5

New Question:

(a) Show that (1-cos h)/h2 = 1/(1+cos h) . (sin h/h)2

(b) Using part (a) and

lim (sin x)/x = 1, show by first principles that d/dx(sin x) = cos x, given that sin (a+b)=sin a cos b+cos a sin b.
x->0
 
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pLuvia

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Riviet said:
let u=log10x

Then 2u2+3u-5=0

(2u+5)(u-1)=0

u=-5/2 or u=1

log10x=-5/2 or log10x=1

.: x=10, 10-2.5

New Question:

(a) Show that (1-cos h)/h2 = 1/(1+cos h) . (sin h/h)2

(b) Using part (a) and

lim (sin x)/x = 1, show by first principles that d/dx(sin x) = cos x, given that sin (a+b)=sin a cos b+cos a sin b.
x->0
a)
(1-cos h)/h2 = 1/(1+cos h) . (sin h/h)2

RHS = 1/(1+cos h) . (sin2 h/h2)
= sin2 / h2(1 + cos h)

LHS = (1-cos h) / h2 . (1+cos h)/(1+cos h)
= 1-cos2 h / h2(1+cos h)
= sin2 h / h2(1+cos h)
= RHS as req'd

b)
d/dx sinx = lim (h-->0) (sin(x+h)-sinx) / h
= lim (h-->0) [2cos(x+h/2)sin h/2] / h [ Using the 4u rule sinA-sinB = 2cos(1/2{A+B})sin(1/2{A-B}) ]
= lim (h-->0) cos(x+h/2) x lim (h-->0) [sin h/2] / h/2
= cosx x 1

.: d/dx sinx = cosx

:D:D


Posting a question very soon
 
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pLuvia

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Next Question

Find the area bounded by the curve y = 1+sinx, the x axis and the ordinates x = 0 and x = 3pi/2 is rotated about the x-axis, find the volume of the solid generated.

NB: sin2x = (1-cos2x) / 2
 
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Riviet

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pLuvia said:
Next Question

Find the area bounded by the curve y = 1+sinx, the x axis and the ordinates x = 0 and x = 3pi/2 is rotated about the x-axis, find the volume of the solid generated.

NB: sin2x = (1-cos2x) / 2
3π/2
∫ π(1+sinx)2 dx =
0
3π/2
∫ π[1+2sinx+sin2x] dx =
0
3π/2
∫ π[1+2sinx+1/2-(cos2x)/2] dx = [x-2cosx+x/2-(sin2x)/4] from 3π/2->0
0

= π{[3π/2-0+3π/4-0]+2}

=π(9π+8)/4 u3

Next Question:
Differentiate e2x from first principles
 
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pLuvia

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Error here Riviet.

For Volumes V = ∫ yπ dx
You forgot the π.

And it's +2 not -2 since you have to [3π/2-0+3π/4-0]- (-2)

Answer being π (9π/4 + 2) u3
 

Riviet

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Argh, thanks for spotting all those, a bit rusty with the volumes. The errors are fixed and I made the answer one fraction. :)
 

SoulSearcher

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Riviet said:
Next Question:
Differentiate e2x from first principles
let f(x) = e2x

therefore
f(x+h) = e2(x+h)
= e2x+2h
= e2x * e2h

by definition,

f'(x) = lim (h -> 0) { f(x+h) - f(x) } / h
f'(x) = lim (h -> 0) { e2(x+h) - e2x } / h
f'(x) = lim (h -> 0) { e2x * e2h - e2x } / h
f'(x) = lim (h -> 0) { e2x( e2h - 1) } / h
f'(x) = e2x lim (h -> 0) { e2h - 1 } / h .............. (1)

now we have to investigate the value of lim (h -> 0) { e2h - 1 } / h as h gets closer to 0

h = 0.1, lim (h -> 0) { e2h - 1 } / h = 2.2140 (all rounded to 4 dec. pl.)
h = 0.01, lim (h -> 0) { e2h - 1 } / h = 2.0201
h = 0.001, lim (h -> 0) { e2h - 1 } / h = 2.0020
h = 0.0001, lim (h -> 0) { e2h - 1 } / h = 2.0002
h = -0.1, lim (h -> 0) { e2h - 1 } / h = 1.8127
h = -0.01, lim (h -> 0) { e2h - 1 } / h = 1.9801
h = -0.001, lim (h -> 0) { e2h - 1 } / h = 1.9980
h = -0.0001, lim (h -> 0) { e2h - 1 } / h = 1.9998

as can be seen, as h gets closer to 0, the value of lim (h -> 0) { e2h - 1 } / h approaches a limit of value 2. Therefore the value of lim (h -> 0) { e2h - 1 } / h is given as 2 and thus can be substituted into equation 1

f'(x) = e2x lim (h -> 0) { e2h - 1 } / h
f'(x) = e2x * 2 (substituting 2 for lim (h -> 0) { e2h - 1 } / h)
f'(x) = 2e2x

This is the way I learnt it, if there is an easier way please feel free to post it up
 
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pLuvia

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hmm, that's really long :p

Feel free to post the next question SoulSearcher
 

Mountain.Dew

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heres one:

a random sample of 10 people is made. assuming either sez is equally likely, find the probability that

(i) there is an equal number of each sex
(ii) more chics than blokes. :p

anwers correct to 3 dp
 

sando

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mmm.... considering that i havnt done probability since yr 9. i'll pass on this question

but i have a question ready for all u to answer. i'll just wait for someone to answer mountain dews ?
 

SoulSearcher

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Mountain.Dew said:
heres one:
a random sample of 10 people is made. assuming either sez is equally likely, find the probability that
(i) there is an equal number of each sex
(ii) more chics than blokes. :p
anwers correct to 3 dp
hehe, I believe this is a 3-unit question but I'll do it anyway
since the chance of picking each sex is equally likely, let the chance of picking a female equal 0.5 = p and the chance of picking a male equal 0.5 = q, and let the chance of picking a female as a success.
using the binomial probability formula, where P (X=r) = nCr prqn-r, where p = probability of success and q = probability of failure in a single eventand r = 10
a) equal number of each sex
P (X=5) = 10C5(0.5)5(0.5)10-5

P (X=5) = 10C5(0.5)5(0.5)5

P (X=5) = 10C5(0.5)10

P (X=5) = 252 * (0.5)10

P (X=5) = 0.246 to 3 dp
b) more chics then blokes
P (X>5) = P (X=6) + P (X=7) + P (X=8) + P (X=9) + P (X=10)
= { 1 - P (X=5) } / 2
= { 1 - 0.246 } / 2
= 0.754 / 2
= 0.377 to 3 dp.
 

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