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HSC 2015 Maths Marathon (archive) (2 Viewers)

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Flop21

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Re: HSC 2015 2U Marathon

First sketch the graph
Since its that little triangle thingy:
What is this "area of a rectangle" thing? Where can I find info how to do this in my cambridge textbook?
 

Flop21

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Re: HSC 2015 2U Marathon

Has anyone got any question on these topics: integration, logs and exponential functions ?


Here's one for someone: Integrate this... (2x+3)^10
 

photastic

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Re: HSC 2015 2U Marathon

Has anyone got any question on these topics: integration, logs and exponential functions ?


Here's one for someone: Integrate this... (2x+3)^10


When is trapezoidal rule for accurate than the Simpson's Rule? Use a diagram to justify your answer.
 

Flop21

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Re: HSC 2015 2U Marathon



When is trapezoidal rule for accurate than the Simpson's Rule? Use a diagram to justify your answer.
The trapezoidal rule is more accurate than The Simpson's Rule when more subintervals are used.

(Can someone else do the diagram).


Another question:

- Differentiate: xe^(2x)
 

matchalolz

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Re: HSC 2015 2U Marathon

The trapezoidal rule is more accurate than The Simpson's Rule when more subintervals are used.

(Can someone else do the diagram).


Another question:

- Differentiate: xe^(2x)
d/dx(xe^[2x])
let u = x
u dash = 1

let v = e^(2x)
v dash = 2e^(2x)

d/dx(xe^[2x]) = e^(2x) + 2xe^(2x)
= e^(2x) * (1+2x)
 

Flop21

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Re: HSC 2015 2U Marathon

d/dx(xe^[2x])
let u = x
u dash = 1

let v = e^(2x)
v dash = 2e^(2x)

d/dx(xe^[2x]) = e^(2x) + 2xe^(2x)
= e^(2x) * (1+2x)
Quick question.

Do we always have to factorise our final answers? Will we loose points if we don't? Will we gain points if we do?
 

Drsoccerball

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Re: HSC 2015 2U Marathon

Quick question.

Do we always have to factorise our final answers? Will we loose points if we don't? Will we gain points if we do?
no you're all gee if you leave it unfactorised version but you Have to simplify fully
 

leehuan

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Re: HSC 2015 2U Marathon

Correct, but there shouldn't have been a need to split PV into 2 parts. It seems you split it then recombined without a purpose.
Probably because my original plan was to base it off an area of a rectangle and a triangle, but when I realised what the definition did I forgot to swap it back.

What is this "area of a rectangle" thing? Where can I find info how to do this in my cambridge textbook?
This isn't really a thing; more the concept of visualising the question.
From the graph of y=e^(2x), it appears that the region bound by the curve and the lines y=1 and x=2, is almost the value of the integral, however not quite.

The integral of e^(2x) from 0 to 1 also includes a rectangle beneath it. This rectangle, that does belong to the value of the integral, does not belong in the area of the question required. All he did was subtract off the area to match the question.

(I tried to use GeoGebra to create a graph, but e^4 is too large (about 60) and if I change the scale, the rectangle will appear far too narrow.)

Next question:


Hint for (ii):
Consider the two points in part (i), i.e. (0,0) and (r,h). What must be the boundaries of the integral?
 

matchalolz

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Re: HSC 2015 2U Marathon

Probably because my original plan was to base it off an area of a rectangle and a triangle, but when I realised what the definition did I forgot to swap it back.



This isn't really a thing; more the concept of visualising the question.
From the graph of y=e^(2x), it appears that the region bound by the curve and the lines y=1 and x=2, is almost the value of the integral, however not quite.

The integral of e^(2x) from 0 to 1 also includes a rectangle beneath it. This rectangle, that does belong to the value of the integral, does not belong in the area of the question required. All he did was subtract off the area to match the question.

(I tried to use GeoGebra to create a graph, but e^4 is too large (about 60) and if I change the scale, the rectangle will appear far too narrow.)

Next question:


Hint for (ii):
Consider the two points in part (i), i.e. (0,0) and (r,h). What must be the boundaries of the integral?
Part i. FInd gradient
m = h/r
y-h = h/r(x-r)
ry - rh = hx - h
ry - hx - rh - h = 0
therefore ry - hx - h(r+1) = 0

I think I've almost got part two, just slipped up somewhere and cant find my mistake
 

InteGrand

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Re: HSC 2015 2U Marathon

Part i. FInd gradient
m = h/r
y-h = h/r(x-r)
ry - rh = hx - h
ry - hx - rh - h = 0
therefore ry - hx - h(r+1) = 0

I think I've almost got part two, just slipped up somewhere and cant find my mistake
The line for (i) is simply .

(A line through the origin is of the form , where is the slope, so once we've found the slope, we can quickly write down its equation without having to use point-slope form.)
 

matchalolz

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Re: HSC 2015 2U Marathon

LOL i was trying to find volume next to x-axis but was using V = pi integrand whatever x^2 dy massive ooops

idk I got πh^5/3r^2, might try again later but gtg atm
 
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leehuan

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Re: HSC 2015 2U Marathon

I had the two mixed up. I actually intended to use (h,r)
 

photastic

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Re: HSC 2015 2U Marathon

Find the positive number for which the sum of the number, triple the number and eight times its reciprocal is a minimum.
 

leehuan

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Re: HSC 2015 2U Marathon

Find the positive number for which the sum of the number, triple the number and eight times its reciprocal is a minimum.
I'm still trying to figure out what you mean here.
 

Ambility

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Re: HSC 2015 2U Marathon

Find the positive number for which the sum of the number, triple the number and eight times its reciprocal is a minimum.
In the following answer, please pretend that the negative square root of two is actually positive on the second to last line of working. For some reason, if I remove the negative, LaTeX screws up. If anyone knows why, let me know.

 

InteGrand

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Re: HSC 2015 2U Marathon

In the following answer, please pretend that the negative square root of two is actually positive on the second to last line of working. For some reason, if I remove the negative, LaTeX screws up. If anyone knows why, let me know.



^Seems to work now (I put a space between the "(" bracket and the backslash).
 
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