HSC 2014 Maths Marathon (archive) (3 Viewers)

[Trebla]

Post any questions within the scope and level of Mathematics (2 unit). Once a question is posted, it needs to be answered before the next question is raised.

I'll start you guys off

 

[photastic]

Re: HSC 2014 2U Marathon

Post any questions within the scope and level of Mathematics (2 unit). Once a question is posted, it needs to be answered before the next question is raised.

I'll start you guys off

Ceebs latex

Next question

Grammar error for part ii cos i speak no english

 

[Rhinoz8142]

Re: HSC 2014 2U Marathon

i) I am not sure if I am correct but I think
L = 5
5 = 150 - e^t
145 = -e^t
In(145) = t
t= 4.976733742
so I think t= 5

I think is wrong, could you correct me ?

ii) i am not sure

 

[photastic]

Re: HSC 2014 2U Marathon

i) I am not sure if I am correct but I think
L = 5
5 = 150 - e^t
145 = -e^t
In(145) = t
t= 4.976733742
so I think t= 5

I think is wrong, could you correct me ?

ii) i am not sure

Why is L=5, you should first find the equation for P, then equate with the given equation for L, so u are w r o n g

 

[Rhinoz8142]

Re: HSC 2014 2U Marathon

Why is L=5, you should first find the equation for P, then equate with the given equation for L, the rest should be all good.

Ahhh, okay...I did it all wrong

 

[integral95]

Re: HSC 2014 2U Marathon

Hint

 

[trungduong12]

Re: HSC 2014 2U Marathon

Hint

How ??
I dont seem to get this question at all

 

[hi im trash]

Re: HSC 2014 2U Marathon

Ceebs latex

Next question

Grammar error for part ii cos i speak no english

for i) t=0 p=5
lions = 150-e^t

dt/dp = 2/p (rearranged)
t = 2ln(p)+c
c = -2ln5
t=2ln(p/5)
therefore p=5e^(t/2)

5e^(t/2) = 150-e^t
set e^(t/2) as "a"
5a = 150-a^2
a^2 +5a -150
therefore a = 10, -15
e(t/2) = 10, -15
but we cannot use the negative 15 because we cannot "ln" a negative number
therefore t = 2ln10
for the second one, i dont really wanna do it D: i dont know how to draw graphs on the forum.

hoping my answer is right )

edit: im not sure why P = 5e^->(t/10) though

 

[Sy123]

Re: HSC 2014 2U Marathon

 

[dunjaaa]

Re: HSC 2014 2U Marathon

(i) 1/2ln(17)
(ii) Using trapezoidal rule with 5 function values;
1/2ln(17)≈1/2(4/17 + 1 + 7/5)
ln(17)≈224/85
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Question:
At the beginning of each year Danny invests in his superannuation. His account pays 5.6% p.a. annually compounding interest. Danny's first investment was $1000 and each of following year his investment was 20% more than the previous year's. Danny continued his investment for 30 years. That is, until the end of the year following his 30th deposit. What percentage of the total value of the account at the end of the 30 years was interest? Express your answer correct to the nearest whole percentage.

 

[Rhinoz8142]

Re: HSC 2014 2U Marathon

Let u= x^2 +1
du/dx = 2x
du = 2x dx
x=4
4^2 +1
= 17
x= 0
0^2 +1
= 1
1/2( 17,1{1/u du )
1/2 (17,1[In u])
1/2 ([In 17] - [In 0])
Ans = 1/2(In17)

ii
I cant remember how to do the traps rule need to revise that.

 

[photastic]

Re: HSC 2014 2U Marathon

Let u= x^2 +1
du/dx = 2x
du = 2x dx
x=4
4^2 +1
= 17
x= 0
0^2 +1
= 1
1/2( 17,1{1/u du )
1/2 (17,1[In u])
1/2 ([In 17] - [In 0])
Ans = 1/2(In17)

ii
I cant remember how to do the traps rule need to revise that.

This is not a 3u marathon.

 

[Carrotsticks]

Re: HSC 2014 2U Marathon

Grammar error for part ii cos i speak no english

This question is fallacious because P = 0 always.

 

[photastic]

Re: HSC 2014 2U Marathon

This question is fallacious because P = 0 always.

how?

 

[Carrotsticks]

Re: HSC 2014 2U Marathon

how?

Because Game of Thrones

 

[Sy123]

Re: HSC 2014 2U Marathon



(If you're a 4U student, please refrain from answering questions in this thread, instead try to give questions)

 

[aDimitri]

Re: HSC 2014 2U Marathon

Let u= x^2 +1
du/dx = 2x
du = 2x dx
x=4
4^2 +1
= 17
x= 0
0^2 +1
= 1
1/2( 17,1{1/u du )
1/2 (17,1[In u])
1/2 ([In 17] - [In 0])
Ans = 1/2(In17)

ii
I cant remember how to do the traps rule need to revise that.

substitution is a 3U method. i'm actually not sure how to do this question using 2U methods.

 

[RealiseNothing]

Re: HSC 2014 2U Marathon

substitution is a 3U method. i'm actually not sure how to do this question using 2U methods.

The top is just the derivative of the bottom, therefore log function.

 

[aDimitri]

Re: HSC 2014 2U Marathon

The top is just the derivative of the bottom, therefore log function.

oh that's 2U?
well then.

 

[photastic]

Re: HSC 2014 2U Marathon

substitution is a 3U method. i'm actually not sure how to do this question using 2U methods.

Tsk tsk tsk