HSC 2015 Maths Marathon (archive) (3 Viewers)

FrankXie · Feb 1, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
New question:
$Find the sum of:$ k^r +k^\frac{rk+1}{k} +k^\frac{rk+2}{k}$ for$ k<1. $Express in terms of k and r$

$$ do u mean $ k^r +k^\frac{rk+1}{k} +k^\frac{rk+2}{k}+\cdots$

Drsoccerball · Feb 1, 2015

Re: HSC 2015 2U Marathon

ye

Carrotsticks · Feb 2, 2015

Re: HSC 2015 2U Marathon

Kaido said:
Interesting solve by integrand. Is that legible in an exam? lol

leehuan said:
I don't see why not. The only important thing to notice however is that he used both parametrics and inverse functions, which are parts of the 3U course. (Nonetheless, I haven't heard of 3U knowledge being prohibited in a 2U paper.)

I wouldn't accept this as a proper proof from scratch. Reason being that the meat of the problem is essentially circumvented by using the fact that the derivative of x^3 is 3x^2.

InteGrand · Feb 2, 2015

Re: HSC 2015 2U Marathon

Carrotsticks said:
I wouldn't accept this as a proper proof from scratch. Reason being that the meat of the problem is essentially circumvented by using the fact that the derivative of x^3 is 3x^2.

What if you included the first-principles differentiation of $x^3$ first? I would have put it in my post, but it was done by someone else just two posts above.

Carrotsticks · Feb 2, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
What if you included the first-principles differentiation of $x^3$ first? I would have put it in my post, but it was done by someone else just two posts above.

Yep, then that would of course be a complete proof.

FrankXie · Feb 2, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
$$ do u mean $ k^r +k^\frac{rk+1}{k} +k^\frac{rk+2}{k}+\cdots$

$$ It is a GP with $ a=k^r, 0< $ common ration $=k^{\frac1k}<1. $ So its limiting sum $=\frac{k^r}{1-k^{\frac1k}}.$

braintic · Feb 2, 2015

Re: HSC 2015 2U Marathon

Isn't the point of these marathon threads for HSC students to answer the questions, and for teachers to jump in only when there is an issue with the question, when a student has made an incorrect attempt at the question, or after a couple of days when no-one has been successful.

The higher Ext 2 thread is a bit different because the thinking involved is closer to uni level (and I have jumped in there after a couple of days), but I don't see the point of a teacher jumping in and answering one of these 2 unit questions only half a day after it was asked.

FrankXie · Feb 2, 2015

Re: HSC 2015 2U Marathon

ok then this is a new question

$Evaluate $ \int_{-1}^1\ln\left(x+\sqrt{x^2+1}\right)dx $ using properties of integral.$

FrankXie · Feb 2, 2015

Re: HSC 2015 2U Marathon

braintic said:
Isn't the point of these marathon threads for HSC students to answer the questions, and for teachers to jump in only when there is an issue with the question, when a student has made an incorrect attempt at the question, or after a couple of days when no-one has been successful.

The higher Ext 2 thread is a bit different because the thinking involved is closer to uni level (and I have jumped in there after a couple of days), but I don't see the point of a teacher jumping in and answering one of these 2 unit questions only half a day after it was asked.

To my best understanding, it doesn't matter who tries or solves the question. What matters is everyboday lets the ball rolling, ie, when one question is surely solved, post a new question! The students still have chance to try the question that is already solved because they don't have to read others' solution if they do want to have a go.

leehuan · Feb 6, 2015

Re: HSC 2015 2U Marathon

$Let\quad f\left( x \right) =\ln { \left( x+\sqrt { { x }^{ 2 }+1 } \right) } \\ Then\quad f\left( -x \right) =\ln { \left( -x+\sqrt { \left( -x \right) ^{ 2 }+1 } \right) } \\ =\ln { \left( -x+\sqrt { { x }^{ 2 }+1 } \right) } \\ =-\left( -\ln { \left( \sqrt { { x }^{ 2 }+1 } -x \right) } \right) \\ =-\ln { \left( \frac { 1 }{ \sqrt { { x }^{ 2 }+1 } -x } \right) }$
$=-\ln { \left( \frac { \sqrt { { x }^{ 2 }+1 } +x }{ \left( { x }^{ 2 }+1 \right) -{ x }^{ 2 } } \right) } \quad by\quad rationalising\quad the\quad denominator\\ =-\ln { \left( \frac { x+\sqrt { { x }^{ 2 }+1 } }{ 1 } \right) } \\ =-\ln { \left( x+\sqrt { { x }^{ 2 }+1 } \right) } \\ =-f\left( x \right)$
$Hence\quad \ln { \left( x+\sqrt { { x }^{ 2 }+1 } \right) } \quad represents\quad an\quad odd\quad function\\ \therefore \int _{ -1 }^{ 1 }{ \ln { \left( x+\sqrt { { x }^{ 2 }+1 } \right) } dx } =0\quad \\ as\quad \int _{ -a }^{ a }{ g\left( x \right) dx } =0\quad if\quad g(x)\quad is\quad continuous\quad over\quad -a\le x\le a\quad and\quad is\quad an\quad odd\quad function$

astroman · Feb 8, 2015

Re: HSC 2015 2U Marathon

need help, i keep getting 9 but answer says 10
using 1 subinterval/use trapezoidal rule
$\int_{0}^{2}(x^{3}+1)\\\\h=\frac{2-0}{1}=2\\\\A=1[(9)]$

InteGrand · Feb 8, 2015

Re: HSC 2015 2U Marathon

astroman said:
need help, i keep getting 9 but answer says 10
using 1 subinterval/use trapezoidal rule
$\int_{0}^{2}(x^{3}+1)\\\\h=\frac{2-0}{1}=2\\\\A=1[(9)]$

The function's value is 0³ + 1 = 1 at x = 0.

FrankXie · Mar 9, 2015

Re: HSC 2015 2U Marathon

new question

$$ Find the area enclosed by the curve $ y=\frac{1}{x}, $ the line $ y=2-x, $ the $ x-$axis, and the line $ x=4.$

Drsoccerball · Mar 12, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
new question

$$ Find the area enclosed by the curve $ y=\frac{1}{x}, $ the line $ y=2-x, $ the $ x-$axis, and the line $ x=4.$

$The interception point is at x=1$

$\therefore \pi \int_4^1(4-4x+x^2)-\pi \int_1^4(\frac{1}{x^2})$

$Area= \frac{9\pi}{4}$

FrankXie · Mar 12, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
$The interception point is at x=1$

$\therefore \pi \int_4^1(4-4x+x^2)-\pi \int_1^4(\frac{1}{x^2})$

$Area= \frac{9\pi}{4}$

hint:1. draw a diagram 2. with area no \pi

Drsoccerball · Mar 12, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
hint:1. draw a diagram 2. with area no \pi

i did volume D: ceebs

PhysicsMaths · Mar 12, 2015

Re: HSC 2015 2U Marathon

ln4 +3/2 ?

Drsoccerball · Mar 12, 2015

Re: HSC 2015 2U Marathon

PhysicsMaths said:
ln4 +3/2 ?

ln(4) +9

PhysicsMaths · Mar 12, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
ln(4) +9

How did you do it?
I took the integral of 1/x -2 +x bounded by x= 1 and x =4

FrankXie · Mar 12, 2015

Re: HSC 2015 2U Marathon

sorry but all tries are wrong. you do need to draw a diagram

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