• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

2u Mathematics Marathon v1.0 (1 Viewer)

insert-username

Wandering the Lacuna
Joined
Jun 6, 2005
Messages
1,226
Location
NSW
Gender
Male
HSC
2006
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

Active Member
Joined
Oct 13, 2005
Messages
6,757
Location
Entangled in the fabric of space-time ...
Gender
Male
HSC
2007
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

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
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

Magician, and Lawyer.
Joined
Nov 6, 2005
Messages
825
Location
Sydney, Australia
Gender
Male
HSC
2005
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

But pieces of what?
Joined
Jun 12, 2004
Messages
6,600
Gender
Male
HSC
2005
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
 
Last edited:

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
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
 
Last edited:
P

pLuvia

Guest
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
 
Last edited by a moderator:
P

pLuvia

Guest
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
 
Last edited by a moderator:

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
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
 
Last edited:
P

pLuvia

Guest
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

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
Argh, thanks for spotting all those, a bit rusty with the volumes. The errors are fixed and I made the answer one fraction. :)
 

SoulSearcher

Active Member
Joined
Oct 13, 2005
Messages
6,757
Location
Entangled in the fabric of space-time ...
Gender
Male
HSC
2007
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
 
P

pLuvia

Guest
hmm, that's really long :p

Feel free to post the next question SoulSearcher
 

Mountain.Dew

Magician, and Lawyer.
Joined
Nov 6, 2005
Messages
825
Location
Sydney, Australia
Gender
Male
HSC
2005
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

HSC IS EVIL
Joined
Oct 31, 2005
Messages
1,123
Gender
Male
HSC
2006
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

Active Member
Joined
Oct 13, 2005
Messages
6,757
Location
Entangled in the fabric of space-time ...
Gender
Male
HSC
2007
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.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top