2u Mathematics Marathon v1.0 (2 Viewers)

[insert-username]

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]

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

Spoiler

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

 

[Mountain.Dew]

SoulSearcher, please feel free to post up the next question. give a hard one this time, eh?

 

[SoulSearcher]

SoulSearcher, please feel free to post up the next question. give a hard one this time, eh?

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

 

[Riviet]

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

Spoiler

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

Next Question:

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

 

[Mountain.Dew]

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:

Spoiler

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]

k=2log₂7

log₃7=log₂7/log₂3=k/[2log₂3]

log₉6=m, 9^m=6=3^2m, 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

 

[Riviet]

Next question:

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

Spoiler

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

 

[pLuvia]

Spoiler

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

Spoiler

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

Posting a question very soon

 

[pLuvia]

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

 

[Riviet]

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

Spoiler

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 e^2x from first principles

 

[pLuvia]

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]

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

 

[SoulSearcher]

Next Question:
Differentiate e^2x from first principles

Spoiler

let f(x) = e^2x

therefore
f(x+h) = e^2(x+h)
= e^2x+2h
= e^2x * e^2h

by definition,

f'(x) = lim (h -> 0) { f(x+h) - f(x) } / h
f'(x) = lim (h -> 0) { e^2(x+h) - e^2x } / h
f'(x) = lim (h -> 0) { e^2x * e^2h - e^2x } / h
f'(x) = lim (h -> 0) { e^2x( e^2h - 1) } / h
f'(x) = e^2x lim (h -> 0) { e^2h - 1 } / h .............. (1)

now we have to investigate the value of lim (h -> 0) { e^2h - 1 } / h as h gets closer to 0

h = 0.1, lim (h -> 0) { e^2h - 1 } / h = 2.2140 (all rounded to 4 dec. pl.)
h = 0.01, lim (h -> 0) { e^2h - 1 } / h = 2.0201
h = 0.001, lim (h -> 0) { e^2h - 1 } / h = 2.0020
h = 0.0001, lim (h -> 0) { e^2h - 1 } / h = 2.0002
h = -0.1, lim (h -> 0) { e^2h - 1 } / h = 1.8127
h = -0.01, lim (h -> 0) { e^2h - 1 } / h = 1.9801
h = -0.001, lim (h -> 0) { e^2h - 1 } / h = 1.9980
h = -0.0001, lim (h -> 0) { e^2h - 1 } / h = 1.9998

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

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

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

 

[pLuvia]

hmm, that's really long

Feel free to post the next question SoulSearcher

 

[SoulSearcher]

Next question:

Find the volume generated when the curve y = e^x + e^-x between x = -1 and x = 1 is rotated around the x-axis

 

[pLuvia]

Spoiler

V =
1
pi ∫ (e^x+e^-x) dx
-1

= pi [ e^x-e^-x ]¹_-1
= 4.7 pi u³

 

[Mountain.Dew]

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.

anwers correct to 3 dp

 

[sando]

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]

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.
anwers correct to 3 dp

hehe, I believe this is a 3-unit question but I'll do it anyway

Spoiler

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) = ⁿC_r p^rq^n-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.