HSC 2015 MX1 Marathon (archive) (2 Viewers)

InteGrand · Jan 30, 2015

Re: HSC 2015 3U Marathon

$$Show that $ \sum_{k=0}^{n}k\binom{n}{k}p^k (1-p)^{n-k}=np.$

braintic · Jan 30, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$$Show that $ \sum_{k=0}^{n}k\binom{n}{k}p^k (1-p)^{n-k}=np.$

Not sure, but does this question assume that 0 <= p <= 1 ?
(I can see it easily for those values, but I haven't really considered values outside that range yet.)

InteGrand · Jan 30, 2015

Re: HSC 2015 3U Marathon

braintic said:
Not sure, but does this question assume that 0 <= p <= 1 ?
(I can see it easily for those values, but I haven't really considered values outside that range yet.)

In practice, p would be between 0 and 1, but the result seems to hold true for all other real p too.

braintic · Jan 30, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
In practice, p would be between 0 and 1, but the result seems to hold true for all other real p too.

Just checking .... because it is a LONG LONG time since I've done this, but is this the expected value of a binomial distribution?

InteGrand · Jan 30, 2015

Re: HSC 2015 3U Marathon

braintic said:
Just checking .... because it is a LONG LONG time since I've done this, but is this the expected value of a binomial distribution?

It is indeed.

FrankXie · Jan 30, 2015

Re: HSC 2015 3U Marathon

braintic said:
Just checking .... because it is a LONG LONG time since I've done this, but is this the expected value of a binomial distribution?

InteGrand said:
It is indeed.

but only binomial expansion is needed to solve this problem. I've had one method in mind.

P.S. if you can prove it is true for 0<=p<=1, that actually means it is true for all real values of p. Because both LHS and RHS are polynomials of p of degree less than or equal to n, and if they are equal to each other at more than n points then they must be identical.

FrankXie · Jan 31, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$$Show that $ \sum_{k=0}^{n}k\binom{n}{k}p^k (1-p)^{n-k}=np.$

$k{{n}\choose{k}}=\frac{k\cdot n!}{k!(n-k)!}=\frac{n\cdot(n-1)!}{(k-1)!(n-k)!}=n{{n-1}\choose{k-1}}$
$\therefore\quad LHS=\sum_{k=1}^{n}k{{n}\choose{k}}p^k(1-p)^{n-k}=n\sum_{k=1}^{n}{{n-1}\choose{k-1}}p^k(1-p)^{n-k}=np\sum_{k=1}^{n}{{n-1}\choose{k-1}}p^{k-1}(1-p)^{(n-1)-(k-1)} $(by shifting the index)$ =np\sum_{k=0}^{n-1}{{n-1}\choose{k}}p^{k}(1-p)^{(n-1)-k}=np[p+(1-p)]^{n-1}=np\cdot1=RHS$

InteGrand · Feb 1, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
$k{{n}\choose{k}}=\frac{k\cdot n!}{k!(n-k)!}=\frac{n\cdot(n-1)!}{(k-1)!(n-k)!}=n{{n-1}\choose{k-1}}$
$\therefore\quad LHS=\sum_{k=1}^{n}k{{n}\choose{k}}p^k(1-p)^{n-k}=n\sum_{k=1}^{n}{{n-1}\choose{k-1}}p^k(1-p)^{n-k}=np\sum_{k=1}^{n}{{n-1}\choose{k-1}}p^{k-1}(1-p)^{(n-1)-(k-1)} $(by shifting the index)$ =np\sum_{k=0}^{n-1}{{n-1}\choose{k}}p^{k}(1-p)^{(n-1)-k}=np[p+(1-p)]^{n-1}=np\cdot1=RHS$

Nice. Here was my attempt:

$(x+y)^n = \sum_{k=0}^{n}\binom{n}{k}x^k y^{n-k}$ (by the binomial theorem)

$\Rightarrow \frac{\partial }{\partial x}[(x+y)^n] = \frac{\partial }{\partial x}\left (\sum_{k=0}^{n}\binom{n}{k}x^k y^{n-k} \right )$

$\Rightarrow n(x+y)^{n-1}= \sum_{k=0}^{n}\frac{\partial }{\partial x}\binom{n}{k}x^k y^{n-k}$ (derivative of a sum is sum of derivatives on RHS)

$\Rightarrow n(x+y)^{n-1}= \sum_{k=0}^{n}k\binom{n}{k}x^{k-1} y^{n-k}$ .

Set $x=p$ and $y = 1-p$ , so x + y = 1, then

$n\cdot1^{n-1}= \sum_{k=0}^{n}k\binom{n}{k}p^{k-1} (1-p)^{n-k}$ .

Multiplying through by p yields $np= \sum_{k=0}^{n}k\binom{n}{k}p^{k} (1-p)^{n-k}$ (since $p\sum_{k=0}^{n}f(k)=\sum_{k=0}^{n}p\cdot f(k)$ and $p\cdot p^{k-1}=p^k$ ) . $\blacksquare$

Drsoccerball · Feb 2, 2015

Re: HSC 2015 3U Marathon

$\int x\sqrt{1-x}dx \text{using} u^2=1-x$

photastic · Feb 2, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$\int x\sqrt{1-x}dx \text{using} u^2=1-x$

FrankXie · Feb 2, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION
$The region enclosed between $ y=2x-x^2 $ and the $ x $ axis is rotated about the $ y $ axis. Find the volume of the solid formed.$

LostInHSC · Feb 2, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
NEXT QUESTION
$The region enclosed between $ y=2x-x^2 $ and the $ x $ axis is rotated about the $ y $ axis. Find the volume of the solid formed.$

Is the answer found by working out

$V=\pi \int_{0}^{2} \, [(x)(x-2)]^{2} \, \, dx$

FrankXie · Feb 2, 2015

Re: HSC 2015 3U Marathon

LostInHSC said:
Is the answer found by working out

$V=\pi \int_{0}^{2} \, [(x)(x-2)]^{2} \, \, dx$

incorrect. your calculation is true for the volume of solid generated by rotating that region about x axis, but not y axis.

LostInHSC · Feb 3, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
incorrect. your calculation is true for the volume of solid generated by rotating that region about x axis, but not y axis.

Here is my attempt:

$\\ y=2x-x^{2} \\ y=-(x^{2}-2x) \\ y=-[(x-1^{2}-1)] \,\,\,\,\,($by completing the square$) \\ y=-(x-1)^{2}+1 \\ -y+1=(x-1)^{2}\\ 1\pm\sqrt{-y+1}=x \rightarrow\,\,\,\, $We will need both branches.$ \\\\\\ $Region A: (Using branch 1)$ \\ x=1-\sqrt{-y+1} \\ x^{2}= 2-2\sqrt{-y+1} -y \\ \therefore V=\pi\int_{0}^{1} (2-2\sqrt{-y+1} -y)\,\,dy \\ \\ \\ $Region B: (Using branch 2)$ \\x=1+\sqrt{-y+1}\\ x^{2}=2+2\sqrt{-y+1}-y \\ V=\pi\int_{0}^{1} (2+2\sqrt{-y+1}-y) \,\,\, dy \\\\\\ \therefore \,\,\,$Volume of solid formed can be found by Volume of Region B - Volume of Region A.$ \\$i.e:$ \,\,\,\,\, V= \pi\int_{0}^{1} (2+2\sqrt{-y+1}-y) \,\,\, dy - \pi\int_{0}^{1} (2-2\sqrt{-y+1} -y)\,\,dy$

FrankXie · Feb 3, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION

$$ For what values of $ k, |(3\sin\theta-\cos\theta)+k(\cos\theta-2\sin\theta)|\leq1 $ is true for all real $ \theta ?$

LostInHSC · Feb 3, 2015

Re: HSC 2015 3U Marathon

New Question

$$Differentiate$ \,\,\,\, y=cos^{-1}(\frac{1-x^{2}}{1+x^{2}})$

leehuan · Feb 4, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
NEXT QUESTION

$$ For what values of $ k, |(3\sin\theta-\cos\theta)+k(\cos\theta-2\sin\theta)|\leq1 $ is true for all real $ \theta ?$

http://i1164.photobucket.com/albums/q566/Rui_Tong/IMG_1522_zpsve7bxkqa.jpg
Working may be invalid, however GeoGebra does agree with my final answer.

InteGrand · Feb 4, 2015

Re: HSC 2015 3U Marathon

LostInHSC said:
New Question

$$Differentiate$ \,\,\,\, y=cos^{-1}(\frac{1-x^{2}}{1+x^{2}})$

$y=\cos ^{-1}\left ( \frac{1-x^2}{1+x^2} \right ), \text{ for }-1\leq \frac{1-x^2}{1+x^2} \leq 1$

$\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1-x^2}{1+x^2} \right )=\frac{-2x(1+x^2)-(1-x^2)\cdot2x}{(1+x^2)^2}$

$=\frac{-4x}{(1+x^2)^2}$ .

$\therefore y^\prime = \frac{-\left ( \frac{-4x}{(1+x^2)^2} \right )}{\sqrt{1-\left ( \frac{1-x^2}{1+x^2} \right )^2}}\text{ for}-1< \frac{1-x^2}{1+x^2} < 1$ (using the chain rule)

$=\frac{4x}{(1+x^2)^2 \sqrt{\frac{(1+x^2)^2 - (1-x^2)^2}{(1+x^2)^2}}}$

$=\frac{4x}{(1+x^2)\sqrt{4x^2}}, \text{ as } (1+x^2)^2-(1-x^2)^2 =(1+x^2 + 1-x^2)(1+x^2 -1 + x^2)=2\times 2x^2 = 4x^2$

$=\frac{2x}{|x|(1+x^2)}.$

InteGrand · Feb 4, 2015

Re: HSC 2015 3U Marathon

Find the angle between the curves $y = x^2$ and $y = (x-3)^2$ .

photastic · Feb 4, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Find the angle between the curves $y = x^2$ and $y = (x-3)^2$ .

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