HSC 2015 Maths Marathon (archive) (1 Viewer)

[InteGrand]

Re: HSC 2015 2U Marathon

Omfg....

Might be harder than last year, but I doubt too hard.

 

[teridax]

Re: HSC 2015 2U Marathon

Might be harder than last year, but I doubt too hard.

that is still a bad omen haha

nah, but 2u maths isn't too bad if you did your practice (it's mostly memorising methods and applying accordingly)

 

[InteGrand]

Re: HSC 2015 2U Marathon

that is still a bad omen haha

nah, but 2u maths isn't too bad if you did your practice (it's mostly memorising methods and applying accordingly)

Well it shouldn't be easier than last year because last year's 2U was on the easier side wasn't it?

 

[milkytea99]

Re: HSC 2015 2U Marathon

Well it shouldn't be easier than last year because last year's 2U was on the easier side wasn't it?

I just hope that it won't be as hard as the 2013 paper

 

[teridax]

Re: HSC 2015 2U Marathon

I just hope that it won't be as hard as the 2013 paper

nek minut...

 

[milkytea99]

Re: HSC 2015 2U Marathon

nek minut...

Wtf....

 

[teridax]

Re: HSC 2015 2U Marathon

Wtf....

only trolling lmfao

but srsly, the paper shouldn't be too bad - provided that you studied diligently

 

[milkytea99]

Re: HSC 2015 2U Marathon

only trolling lmfao

but srsly, the paper shouldn't be too bad - provided that you studied diligently

I hope so haha because I actually work harder in maths than eco lol. I was too complacent for eco and was fked up by it yesterday

 

[astroman]

Re: HSC 2015 2U Marathon

How do I find the area of the shaded region?

 

[rand_althor]

Re: HSC 2015 2U Marathon

How do I find the area of the shaded region?

Split it into two areas from x=0 to x=1 (point of intersection), and x=1 to x=2. The first part is the area under 4x-x², while the second part is the area under 4-x²:
.

If you evaluate , you get 0.

 

[Flop21]

Re: HSC 2015 2U Marathon

Split it into two areas, one per function:
.

If you find , the area you are finding is

I don't understand why you add them and use 1, instead of 2.

 

[Drsoccerball]

Re: HSC 2015 2U Marathon

I don't understand why you add them and use 1, instead of 2.

Its because the area under the graph are both equal to each other. You could use 2 but would you want to integrate with a negative domain ?
EDIT: you can use it doesnt matter

 

[InteGrand]

Re: HSC 2015 2U Marathon

How do I find the area of the shaded region?

 

[Flop21]

Re: HSC 2015 2U Marathon

Its because the area under the graph are both equal to each other. You could use 2 but would you want to integrate with a negative domain ?
EDIT: you can use it doesnt matter

Oh I just got it! Thanks all.

Another question tho:

Why can't I use the area of the triangle under the line to minus from the area under the curve???

Say instead of doing f(parabola) - f(line), I do f(parabola) - 1/2ab (triangle area). But I get the wrong answer doing the latter.

 

[rand_althor]

Re: HSC 2015 2U Marathon

Why can't I use the area of the triangle under the line to minus from the area under the curve???

Say instead of doing f(parabola) - f(line), I do f(parabola) - 1/2ab (triangle area). But I get the wrong answer doing the latter.

If you're using O, A, and (4,0) to make your triangle, it is not a triangle as a side would be curved - the one from A to (4,0).

 

[Flop21]

Re: HSC 2015 2U Marathon

If you're using O, A, and (4,0) to make your triangle, it is not a triangle as a side would be curved - the one from A to (4,0).

I meant like this:

Using integration, from 3 to 0, f(parabola) - area of triangle. Shouldn't this work?

 

[kawaiipotato]

Re: HSC 2015 2U Marathon

I meant like this:

Using integration, from 3 to 0, f(parabola) - area of triangle. Shouldn't this work?

It will work (but the bounds are 0 to 3 or else it'll give a negative for the integral)

 

[InteGrand]

Re: HSC 2015 2U Marathon

Oh I just got it! Thanks all.

Another question tho:

Say instead of doing f(parabola) - f(line), I do f(parabola) - 1/2ab (triangle area).

These two methods are actually exactly the same. Integrating that line over the right bounds just gives us that triangle's area. If you're getting a wrong answer doing the latter, you might have made a silly mistake somewhere.

 

[nisak]

Re: HSC 2015 2U Marathon

I meant like this:

Using integration, from 3 to 0, f(parabola) - area of triangle. Shouldn't this work?

you're making it too complicated with the triangle, there is no need..
its just the integration of the parabola minus the straight line from 0 to the x-coordinate of A

 

[Mr_Kap]

Re: HSC 2015 2U Marathon

This is one of the 8 marks i got wrong in the 2013 paper: Why is answer D not B?