HSC 2016 MX1 Marathon (archive) (7 Viewers)

InteGrand · Feb 28, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
Sum to Product and Product to Sum

Lol, realised it just as I was waiting for this page to load.

davidgoes4wce · Feb 29, 2016

Re: HSC 2016 3U Marathon

Was going through this question from Grove textbook this morning on Binomial Therorem

$$ If they ask for a specific term like $\binom{n}{1}$=n$

Now my thinking was initially to x=1 into the x-value on both the LHS and RHS, which would have eliminated the 'x' term. From the example shown, they have put x=1 into the LHS and not the RHS. Before substituting x=0 into the last step on the RHS. I'm just trying to figure out a better way of remembering the last few steps.

$I guess my question would be what happens if they asked for $\binom{n}{2}$ \ =n $ for \ this \ same \ sort \ of \ question?$

KingOfActing · Feb 29, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$I guess my question would be what happens if they asked for $\binom{n}{2}$ \ =n $ for \ this \ same \ sort \ of \ question?$

$From the definition of the binomial coefficient: $\binom{n}{2} =n\\\frac{n!}{2!(n-2)!} = n\\\frac{n(n-1)(n-2)!}{2(n-2)!} = n\\n(n-1) = 2n\\n = 0 $ or $3\\$Whether you include 0 or not is up to debate and context of the question (i.e. is the binomial term even defined for $n = 0$?)$

Speed6 · Feb 29, 2016

Re: HSC 2016 3U Marathon

Isn't this the process?

InteGrand · Feb 29, 2016

Re: HSC 2016 3U Marathon

Speed6 said:
Isn't this the process?

$\noindent Not entirely sure what you did here with the $a$ etc., but to show that $\binom{n}{0}=1$ for non-negative integers $n$, recall that $0!=1$, so $\binom{n}{0} =\frac{n!}{0!(n-0)!} =\frac{1}{0!} = 1$.$

Speed6 · Feb 29, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent Not entirely sure what you did here with the $a$ etc., but to show that $\binom{n}{0}=1$ for non-negative integers $n$, recall that $0!=1$, so $\binom{n}{0} =\frac{n!}{0!(n-0)!} =\frac{1}{0!} = 1$.$

Some manipulation got me there, don't know if it's mathematically valid though.

davidgoes4wce · Mar 4, 2016

Re: HSC 2016 3U Marathon

Was just going through Margaret Grove's Year 12 HSC book , the second last question in E 10.5:

$$ If $ (1+x)^{n}= \sum_{r=0}^{n} \tbinom{n}{r} \ x^r$

$$ Show by integration that $ \sum _{r=0}^{n} \frac{(-1)^r}{r+1} \tbinom{n}{r}=\frac{1}{n+1}$

The first part of the Solutions is as follows:

$\int{(1+x)^{n}} \ dx = \int{\sum_{r=0}^{n} \tbinom{n}{r} \ x^r} dx \ \textcircled{1}$

$\frac{(1+x)^{n+1}}{n+1} + C_{1} = \sum_{r=0}^{n} \tbinom{n}{r} \frac{x^{r+1}}{r+1} \ +C_{2} \ \textcircled{2}$

$\therefore \frac{(1+x)^{n+1}}{n+1}= \sum_{r=0}^{n} \frac{x^{r+1}}{r+1}+C \ \textcircled{3}$

$$ I guess my question is could we jump from step $ \textcircled{1} to \textcircled{3} ?$

Drsoccerball · Mar 4, 2016

Re: HSC 2016 3U Marathon

yes

davidgoes4wce · Mar 10, 2016

Re: HSC 2016 3U Marathon

$Fill in the missing word$

$The diagonals of a rhombus$ ___________________ $the vertex angles through which they pass.$

KingOfActing · Mar 10, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$Fill in the missing word$

$The diagonals of a rhombus$ ___________________ $the vertex angles through which they pass.$

Bisect

$The curves $y_1 = \arcsin{x}$ and $ y_2 = \arccos{x} $ meet at a point $ P(\frac{1}{\sqrt{2}},\frac{\pi}{4}) $. The point S is the x-intercept of $y_2$. Find the area enclosed by the curves, OS and the arc PS where O is the origin.$

qwert73 · Mar 10, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
Bisect

$The curves $y_1 = \arcsin{x}$ and $ y_2 = \arccos{x} $ meet at a point $ P(\frac{1}{\sqrt{2}},\frac{\pi}{4}) $. The point S is the x-intercept of $y_2$. Find the are enclosed by the curves, OS and the arc PS where O is the origin.$

0_0 wtf does that mean??????

InteGrand · Mar 10, 2016

Re: HSC 2016 3U Marathon

qwert73 said:
0_0 wtf does that mean??????

He meant "area" (it was a typo). He just named the curves y1 and y2 so that he could refer to the curves easily.

KingOfActing · Mar 10, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
He meant "area" (it was a typo). He just named the curves y1 and y2 so that he could refer to the curves easily.

Oopsies

Yeah, I meant area.

Graphing it may make it easier to understand.

KINGOM 885 · Mar 12, 2016

Re: HSC 2016 3U Marathon

can someone post a hard integration by substitution question, please

KingOfActing · Mar 12, 2016

Re: HSC 2016 3U Marathon

KINGOM 885 said:
can someone post a hard integration by substitution question, please

I don't think any of those exist.

$$By considering the substitution $ x = 2 + 4\tan^2{\theta} $ or otherwise, find $ \int \frac{\sqrt{x+2}+\sqrt{x-2}}{(\sqrt{x+2}-\sqrt{x-2})(x-2)^2}\,dx$

KINGOM 885 · Mar 13, 2016

Re: HSC 2016 3U Marathon

nice

Paradoxica · Mar 13, 2016

Re: HSC 2016 3U Marathon

KINGOM 885 said:
can someone post a hard integration by substitution question, please

$$\noindent By considering the substitution $y = x^3-\frac{1}{x^2}$, evaluate $\int \frac{x(3x^5 +2)\text{d}x}{x^{10}-2x^5+x^4+1}.$

leehuan · Mar 14, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
I don't think any of those exist.

$$By considering the substitution $ x = 2 + 4\tan^2{\theta} $ or otherwise, find $ \int \frac{\sqrt{x+2}+\sqrt{x-2}}{(\sqrt{x+2}-\sqrt{x-2})(x-2)^2}\,dx$

Have you tried this one of Paradoxica's evil integrals: $\int { \frac { \left( x-1 \right) \sqrt { { x }^{ 4 }+2{ x }^{ 3 }+4{ x }^{ 2 }+2x+1 } }{ { x }^{ 2 }\left( x+1 \right) } dx }$

leehuan · Mar 14, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$$\noindent By considering the substitution $y = x^3-\frac{1}{x^2}$, evaluate $\int \frac{x(3x^5 +2)\text{d}x}{x^{10}-2x^5+x^4+1}.$

Lol I had to make sure a 3U student could do this.

But yeah this one is really easy

leehuan · Mar 14, 2016

Re: HSC 2016 3U Marathon

Trigonometric substitutions were already split off from simplistic substitutions in 4U/3U for the difficulty of having to sub back in. So I'm just going to spoil some of the answer.

$\\ x-2=4\tan ^{ 2 }{ \theta } \Rightarrow dx=8\tan { \theta } \sec ^{ 2 }{ \theta }\, d\theta \\ \int { \frac { \sqrt { x+2 } +\sqrt { x-2 } }{ \left( \sqrt { x+2 } -\sqrt { x-2 } \right) { \left( x-2 \right) }^{ 2 } } dx } \\ =\int { \frac { \sqrt { 4+4\tan ^{ 2 }{ \theta } } +\sqrt { 4\tan ^{ 2 }{ \theta } } }{ \left( \sqrt { 4+4\tan ^{ 2 }{ \theta } } -\sqrt { 4\tan ^{ 2 }{ \theta } } \right) { \left( 4\tan ^{ 2 }{ \theta } \right) }^{ 2 } } \cdot 8\tan { \theta } \sec ^{ 2 }{ \theta } d\theta }$
$\\ =\int { \frac { \left( 2\sec { \theta } +2\tan { \theta } \right) }{ \left( 2\sec { \theta } -2\tan { \theta } \right) \left( 16\tan ^{ 4 }{ \theta } \right) } \cdot 8\tan { \theta } \sec ^{ 2 }{ \theta } d\theta } \\ =\frac { 1 }{ 2 } \int { \frac { \sec ^{ 2 }{ \theta } +2\sec { \theta } \tan { \theta } +\tan ^{ 2 }{ \theta } }{ \left( \sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta } \right) \left( \tan ^{ 3 }{ \theta } \right) } \sec ^{ 2 }{ \theta } d\theta } \\ =\frac { 1 }{ 2 } \int { \frac { 1+2\sin { \theta } +\sin ^{ 2 }{ \theta } }{ \sin ^{ 3 }{ \theta } \cos { \theta } } d\theta }$

And at this point I'm still not sure how this is done using only 3U methods

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