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2u Mathematics Marathon v1.0 (1 Viewer)

skyrockets1530

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DeanM said:
cant resist the simpsons...

sub x=2, 3, 4 into ln(x-1) to get the following...
1/3 { 0 + (4x0.6931) + 1.0986 }
~1.2904

someone else can post a question.. im off to bed.. for real this time..
Ok, i'll post a question

Serena invests 8000 at the begginning of each year into her supperannuation account which pays 10% pa compounded half-yearly, after 10 years, serena reduces her investment to 5000$, find the amount in the supper account after 20 years.
 

skepticality

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Question 85

There are two points on a wall of a building, A and B. The two points, A and B are 200meters apart. A pole has the perpendicular distance to the building 50m and point A is placed such that the line drawn from A to the pole is perpendicular to the wall. A person standing has the perpendicular distance to the building 80m and point B is placed such that the line drawn from B to the person is perpendicular to the wall. The two perpendicular lines drawn to the wall are 200m apart. The person is sqrt[40900] meters from the pole. The person is asked to reach the pole as fast as he can, ie with minimal distance. However he must touch the wall first. Find the point on the wall the person touches, by expressing it as the distance from B.

theres actually a dumbass(proper) way and a cheapass way
 
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skyrockets1530

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icycloud said:
Alright then DeanM, here goes:
Question 84
Find the equation of the tangent to the curve y = exln(x) at x = 2.
y = exln(x)
y' = exlnx + ex/x
m = e2ln2 + e2/2
y-(e2ln2) = (e2ln2 + e2/2)(x-2)
y = (e2ln2 + e2/2)(x-2) - e2ln2
xe2ln2 + xe2/2 - 2e2ln2 - e2 - y - e2ln2 = 0
thats too long
already posted a question
 
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yabby

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icycloud said:
Alright then DeanM, here goes:
Question 84
Find the equation of the tangent to the curve y = exln(x) at x = 2.
y=e^xlnx
using product rule
dy/dx= e^x.1/x + e^x.lnx
sub x=2
dy/dx= e^2.1/2 +e^2.ln2

When x=2
y= e^2ln2

Sub into formula y-y1=m(x-x1)
y - (e^2.ln2) = (1/2.e^2+e^2.ln2)(x-2)
y= (1/2.e^2+e^2.ln2)(x-2) + (e^2.ln2)
Put that in the calc, and there u go

Umm there was 2 other questions..so i wont add one to make it even more confusing
 

skyrockets1530

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yabby said:
lol sky,doh! u did it way fsater then me..anyways how do u do those square signs..?
haha, for those 2 things just do, sup in brackets and for the ones below, like 2 just do sub in brackets, and close the tags with the /, just click 'reply with quote' and you'll see em
 

skepticality

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skepticality said:
Question 85

There are two points on a wall of a building, A and B. The two points, A and B are 200meters apart. A pole has the perpendicular distance to the building 50m and point A is placed such that the line drawn from A to the pole is perpendicular to the wall. A person standing has the perpendicular distance to the building 80m and point B is placed such that the line drawn from B to the person is perpendicular to the wall. The two perpendicular lines drawn to the wall are 200m apart. The person is sqrt[40900] meters from the pole. The person is asked to reach the pole as fast as he can, ie with minimal distance. However he must touch the wall first. Find the point on the wall the person touches, by expressing it as the distance from B.

theres actually a dumbass(proper) way and a cheapass way
Answer is..
123
for those interested
 

skyrockets1530

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I'll answer my own question...
Serena invests 8000 at the begginning of each year into her supperannuation account which pays 10% pa compounded half-yearly, after 10 years, serena reduces her investment to 5000$, find the amount in the supper account after 20 years.

First do the first ten years
deposit is made once a year and interest is paid twice yearly
A1 = 8000(1.05)
A2 = [8000(1.05)]1.05
= 8000(1.052)
A3 = 8000(1.053) + 8000(1.05)
A4 = 8000(1.054) + 8000(1.052)
A4 = 8000(1.054) + 8000(1.052)
An = 8000(1.05n + 1.05n-2 + 1.05n-4 + ....)
A20 = 8000(1.0520 + 1.0518 + 1.0516 + .... + 1.052)
A20 = 8000[(1.052(1.102510 - 1)) / .1025]
NOTE: the rate is 1.1025 as the series goes 1.052 + 1.054...
A20 = 142,264.2513
Now for the second 10 years
A21 = [142,264.2513 + 5000]1.05
A22 = 147,264.2513(1.052)
A23 = 147,264.2513(1.053) + 5000(1.05)
A24 = 147,264.2513(1.054) + 5000(1.052)
...
A40 = 147,264.2513(1.0520) + 5000[(1.052(1.10259 - 1)) / .1025]
Note the second series of interest only has 9 terms
A40 = $466,384.57 total value of super account
That should be all correct, please let me know if theres an error

Next question:
given that d/dx = ex2 = 2xex2 , evaluate the integral of xex2 between 1 and 0
 

rsingh

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Question: Find the integral of 1/x between 1 and 0.

Thanks.
 

skepticality

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skyrockets1530 said:
Next question:
given that d/dx = ex2 = 2xex2 , evaluate the integral of xex2 between 1 and 0
d/dx = ex2 = 2xex2
therefore 1/2 [integral] boundary 0..1 2xex2 = [ex2]10
=2[ex2]10
=2(e-1)

Next Question (do it guys..)

There are two points on a wall of a building, A and B. The two points, A and B are 200meters apart. A pole has the perpendicular distance to the building 50m and point A is placed such that the line drawn from A to the pole is perpendicular to the wall. A person standing has the perpendicular distance to the building 80m and point B is placed such that the line drawn from B to the person is perpendicular to the wall. The two perpendicular lines drawn to the wall are 200m apart. The person is sqrt[40900] meters from the pole. The person is asked to reach the pole as fast as he can, ie with minimal distance. However he must touch the wall first. Find the point on the wall the person touches, by expressing it as the distance from B.

theres actually a dumbass(proper) way and a cheapass way
 

rama_v

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d/dx (ex2) = 2x ex2
.: ∫xex2 dx
= (1/2) ex2
= (1/2)(e1 - e0)
= (1/2)(e - 1)

Next question:
A car, of initial value $15000 depreciates at a rate of 3% p.a.
Calculate the car's worth after five years.
 
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Twiggyy

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question:
A car, of initial value $15000 depreciates at a rate of 3% p.a.
Calculate the car's worth after five years.

15000(1-0.97)^5 = $12881


next q -

What values of k are always positive?
x^2 - 2kx + 6k
 
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icycloud

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Twiggyy said:
What values of k are always positive?
x^2 - 2kx + 6k
Positive definite: a > 0 and Discriminant < 0

a = 1 > 0

Discriminant = (-2k)^2 - 4(6k)
= 4k^2 - 24k
< 0

4k^2 - 24k < 0
k^2-6k < 0
k(k-6) < 0

Therefore, 0 < k < 6 #

There's already an unanswered question above, post #174.
 

Kutay

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lie_stella said:
Next Question

Solve ln(7x-12) = ln x
well because they got the same base they cancel out ---> log laws there for it becomes 7x - 12 = x and x = 2

Hope thats right, correct me if i am wrong

Next Question:

The point P(x,y) lies on the line y = 2x - 9 and is equidistant from A(0,1) and B(2,3) . Find the Equation of the Locus P

Answer:
ROOT {(x-0)2 + (y-1)2} = ROOT {(x-2)2 + (y-3)2}
Then jus square both sides and u you in up getting 4y +4x -12 = 0 by putting to one side
 
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