HSC 2015 MX1 Marathon (archive) (2 Viewers)

braintic · Sep 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Probability of one family having 2 boys:
$\binom{4}{2}(\frac{1}{2})^2(\frac{1}{2})^2 =\frac{3}{8}$
Probability of exactly two familys having 2 boys:
$\binom{4}{2}(\frac{3}{8})^2(\frac{5}{8})^2 =\frac{675}{2048}$

I know you understand what you have said, but just to clarify the wording for anyone reading it:

The first calculation is the probability of one particular family having two boys, not the probability of exactly one of the four families having two boys.

sharoooooo · Sep 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Probability of one family having 2 boys:
$\binom{4}{2}(\frac{1}{2})^2(\frac{1}{2})^2 =\frac{3}{8}$
Probability of exactly two familys having 2 boys:
$\binom{4}{2}(\frac{3}{8})^2(\frac{5}{8})^2 =\frac{675}{2048}$

thanks, but i dont understand why its 3/8 and 5/8 ?

braintic · Sep 7, 2015

Re: HSC 2015 3U Marathon

sharoooooo said:
thanks, but i dont understand why its 3/8 and 5/8 ?

The answer in the first calculation is being used in the second.

3/8 is the probability of a specific family having 2 boys and 2 girls, so 5/8 is the probability that they don't.

leehuan · Sep 8, 2015

Re: HSC 2015 3U Marathon

$\\ $1. What is the value of a revolution angle? \\ 2. Hence, explain why the angle sum of a polygon is $ \pi \left( n-2 \right) $ where $n$ is the number of sides of the polygon. \\ 3. [Challenge for 3U, standard for 4U] Verify this result using mathematical induction.$

rand_althor · Sep 9, 2015

Re: HSC 2015 3U Marathon

$$Show that $(\cos{t})^{2n+1}=\cos{t}(1-\sin^2{t})^n$. Hence, using the binomial theorem to expand $(1-\sin^2{t})^n$ where $n$ is a positive integer, show that $\int^{\frac{\pi}{2}}_0 (\cos{t})^{2n+1} \mathrm{d}t=\sum_{r=0}^{n}(-1)^r \cdot \frac{1}{2r+1} \cdot \binom{n}{r}$.$

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

Was not too sure about the step going from the 2nd last line to the last line. Anyone can give me an explanation as to how they go about it?

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

I can derive the 'n' value of 8 but it is through a much longer process of using natural logs.

InteGrand · Sep 10, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Was not too sure about the step going from the 2nd last line to the last line. Anyone can give me an explanation as to how they go about it?

$$It is because$

$$\frac{(1+x)^7}{x^7}=\left(\frac{1+x}{x}\right)^7$ (index law: $\frac{a^r}{b^r} = \left(\frac{a}{b}\right)^r$)$

$and then split the fraction.$

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Show that $(\cos{t})^{2n+1}=\cos{t}(1-\sin^2{t})^n$. Hence, using the binomial theorem to expand $(1-\sin^2{t})^n$ where $n$ is a positive integer, show that $\int^{\frac{\pi}{2}}_0 (\cos{t})^{2n+1} \mathrm{d}t=\sum_{r=0}^{n}(-1)^r \cdot \frac{1}{2r+1} \cdot \binom{n}{r}$.$

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

Will try to upload a LaTeX version as my writing is a bit messy

Drsoccerball · Sep 10, 2015

Re: HSC 2015 3U Marathon

I think it would be more wise to make a substitution of u=sinx before expanding...

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

LaTeX version of my solution

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
I think it would be more wise to make a substitution of u=sinx before expanding...

your right I needed to say

" Let u= sin x"

Drsoccerball · Sep 10, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
your right I needed to say

" Let u= sin x"

No you didnt understand what i was trying to say... You expanded the trig expression and the integrated it after instead you should first let u =sinx so the expansion is easier and the substitution is also easier.

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
No you didnt understand what i was trying to say... You expanded the trig expression and the integrated it after instead you should first let u =sinx so the expansion is easier and the substitution is also easier.

OK I followed a method in the John Kinny-Lewis text which is the harder way and I understood it pretty much step by step doing it his way.

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

I also think in terms of difficulty the Bored of Studies Trails is equivalent of difficulty to the John Kinny-Lewis HSC Exam papers book. I've only seen a few of the BoS trials papers so I can't comment thoroughly about which is the more difficult.

davidgoes4wce · Sep 10, 2015

Re: HSC 2015 3U Marathon

I'm assuming the composers of the BoS trials come from this forum? If anyone knows, who created those papers on a yearly basis? Obviously, they must be experienced and know the content inside out.

rand_althor · Sep 10, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
LaTeX version of my solution

Pretty much what I did. You have skipped steps, but I'll assume you did it correctly.

Next question:

$$Using the binomial theorem, prove that $ r\binom{n}{r}=n\binom{n-1}{r-1}$.$

Drsoccerball · Sep 10, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Pretty much what I did. You have skipped steps, but I'll assume you did it correctly.

Next question:

$$Using the binomial theorem, prove that $ r\binom{n}{r}=n\binom{n-1}{r-1}$.$

$(x+1)^n = \binom{n}{0} +\binom{n}{1}x+...+ \binom{n}{n}x^n$

$n(x+1)^{n-1}=\binom{n}{1} +2\binom{n}{2}x+...+r\binom{n}{r}x^{r-1}...+n \binom{n}{n}x^{n-1}$

Compare coeefiecients of $x^{r-1}$

$r\binom{n}{r} =n\binom{n-1}{r-1}$

rand_althor · Sep 10, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$(x+1)^n = \binom{n}{0} +\binom{n}{1}x+...+ \binom{n}{n}x^n$

$n(x+1)^{n-1}=\binom{n}{1} +2\binom{n}{2}x+...+r\binom{n}{r}x^{r-1}...+n \binom{n}{n}x^{n-1}$

Compare coeefiecients of $x^{r-1}$

$r\binom{n}{r} =n\binom{n-1}{r-1}$

Which expansions are you comparing the coefficients of?

HSC 2015 MX1 Marathon (archive) (2 Viewers)

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)