HSC 2014 Maths Marathon (archive) (1 Viewer)

mreditor16 · Aug 13, 2014

Re: HSC 2014 2U Marathon

yep very nice!

mreditor16 · Aug 13, 2014

Re: HSC 2014 2U Marathon

haha no worries!

integral95 · Aug 13, 2014

Re: HSC 2014 2U Marathon

harrypotterfan said:
Oh LOL sorry I didn't physically do the question, literally just guessed by looking at the sigma notation :L I've done this question (properly) before though, is this from a CSSA trial? because I don't usually remember questions haha so I must have done it recently

There's a similar question like that in the one of the past 2U HSC paper (don't remember which one haha)

mreditor16 · Aug 13, 2014

Re: HSC 2014 2U Marathon

integral95 said:
There's a similar question like that in the one of the past 2U HSC paper (don't remember which one haha)

...............

mreditor16 said:
next one (From a past 2u trial paper)

upload img

integral95 · Aug 13, 2014

Re: HSC 2014 2U Marathon

mreditor16 said:
...............

my bad

Sy123 · Aug 14, 2014

Re: HSC 2014 2U Marathon

$\\ $(i) Show that$ \ x^n - 1 = (x-1)(1+ x + x^2 + \dots + x^{n-1}) \\ \\ $(ii) Show that if a number is in the form$ \ 18^n - 1 \ $for some integer$ \ n \ $then such a number is divisible by$ \ 17 \\ \\ $(iii) Using the result in part (i), explain why$ \ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$

Nukeboy · Aug 14, 2014

Re: HSC 2014 2U Marathon

Sy123 said:
$\\ $(i) Show that$ \ x^n - 1 = (x-1)(1+ x + x^2 + \dots + x^{n-1}) \\ \\ $(ii) Show that if a number is in the form$ \ 18^n - 1 \ $for some integer$ \ n \ $then such a number is divisible by$ \ 17 \\ \\ $(iii) Using the result in part (i), explain why$ \ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$

Thats how i did part 1. I just divided the two eqs. dont know if it sufficiently proves it though.

Nukeboy · Aug 14, 2014

Re: HSC 2014 2U Marathon

And Part 2. I solved it using induction.

Dont you think part 3 is missing something. Maybe x<1. Because if x>1 then 1/1-x is negative, or am i missing something here.

aDimitri · Aug 14, 2014

Re: HSC 2014 2U Marathon

Nukeboy said:
And Part 2. I solved it using induction.
View attachment 30769
Dont you think part 3 is missing something. Maybe x<1. Because if x>1 then 1/1-x is negative, or am i missing something here.

well i think for part 2 you were supposed to use part 1. its pretty simple. also induction is a 3U technique, this is a 2U thread

Nukeboy · Aug 14, 2014

Re: HSC 2014 2U Marathon

aDimitri said:
well i think for part 2 you were supposed to use part 1. its pretty simple. also induction is a 3U technique, this is a 2U thread

Oh yeah, whoops. That makes more sense.

dunjaaa · Aug 14, 2014

Re: HSC 2014 2U Marathon

Question:

Screen shot 2014-08-14 at 11.27.41 PM.png

hi im trash · Aug 15, 2014

Re: HSC 2014 2U Marathon

answer is 0 (i think).
its tedious i may have made a mistake T_T!!!!!

integral95 · Aug 15, 2014

Re: HSC 2014 2U Marathon

Sy123 said:
$\\ $(i) Show that$ \ x^n - 1 = (x-1)(1+ x + x^2 + \dots + x^{n-1}) \\ \\ $(ii) Show that if a number is in the form$ \ 18^n - 1 \ $for some integer$ \ n \ $then such a number is divisible by$ \ 17 \\ \\ $(iii) Using the result in part (i), explain why$ \ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$

The result in part iii is only true for $\ \ \ -1<x<1$

integral95 · Aug 15, 2014

Re: HSC 2014 2U Marathon

$\int \frac{1}{xlnx} \ \ dx \\ \\ \\ \int tan^2x \ \ dx$

Thank_You · Aug 15, 2014

Re: HSC 2014 2U Marathon

integral95 said:
$\int \frac{1}{xlnx} \ \ dx \\ \\ \\ \int tan^2x \ \ dx$

BTW how do you type up the question like that, or is that just a screenshot ?

1. Let u = lnx
Hence, du = 1/x dx

Hence, Integration = f 1/u du
= lnu +C
= ln(lnx) + C

2. I = f sec^2x - 1 dx
= tanx -x +C

NEXT QUESTION: http://postimg.org/image/id86iq0qb/

photastic · Aug 15, 2014

Re: HSC 2014 2U Marathon

Thank_You said:
BTW how do you type up the question like that, or is that just a screenshot ?

1. Let u = lnx
Hence, du = 1/x dx

Hence, Integration = f 1/u du
= lnu +C
= ln(lnx) + C

2. I = f sec^2x - 1 dx
= tanx -x +C

NEXT QUESTION: http://postimg.org/image/id86iq0qb/

How would you the first one using 2u method only?

integral95 · Aug 15, 2014

Re: HSC 2014 2U Marathon

aarondapho said:
How would you the first one using 2u method only?

$y = b^x \\ ln(y) = xln(b)\\ \frac{ln(y)}{ln(b)}= x \\ \frac{dx}{dy} = \frac{1}{yln(b)} = \frac{1}{b^xln(b)} \\\therefore \frac{dy}{dx}= b^xln(b)$

integral95 · Aug 15, 2014

Re: HSC 2014 2U Marathon

randomly_generated said:
nice easy 2U question:

Bump... lol how did that guy get banned.

integral95 · Aug 15, 2014

Re: HSC 2014 2U Marathon

$i)\ \ $By drawing the graph$ \ \ \ f(x) = \sqrt{1-x^2} $ and considering the area, find$ \\ \int_{0}^{1}\sqrt{1-x^2}\ dx \\ \\ ii) $Find the approximation of$ \int_{0}^{1}\sqrt{1-x^2}\ dx \ $using the Simpson's Method$ \\ iii) $Hence, find the approximate value of \pi$$

dunjaaa · Aug 15, 2014

Re: HSC 2014 2U Marathon

Question:

Screen shot 2014-08-15 at 10.17.08 PM.png

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