HSC 2017-2018 Maths Marathon (1 Viewer)

boredofstudiesuser1 · Feb 8, 2017

Re: HSC 2017 Maths (Advanced) Marathon

davidgoes4wce said:
I was also looking to compare (b) v (f)....................earlier I mentioned (b) v (d)

Yeah, I just assumed so

since d had no angle marked or hypotenuse

davidgoes4wce · Feb 8, 2017

Re: HSC 2017 Maths (Advanced) Marathon

boredofstudiesuser1 said:
Yeah, I just assumed so since d had no angle marked or hypotenuse

I learnt something new I guess, I always had the assumption that with congruency tests, that if there was a right angle in there, it was always RHS.

Not sure if the majority of HSC and prelim students knew that in the back of their mind. (And I'm not afraid to say this was from a Year 9 Maths textbook

)

19KANguy · Feb 11, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Show that x^3 - 8 has only one linear factor over the real number field,

What is the real number field?

Mahan1 · Feb 17, 2017

Re: HSC 2017 Maths (Advanced) Marathon

19KANguy said:
Show that x^3 - 8 has only one linear factor over the real number field,

What is the real number field?

To solve it in a different way, we can look at the roots, since $x^{3}-8$ is a cubic polynomial then there are three roots, $\alpha, \beta, \gamma$ .Obviously one of the roots is 2,so let $\alpha = 2$
then sum and product of the roots imply
$\beta+ \gamma = -2, \beta \gamma = 4$
then
$( \beta+ \gamma)^{2} = \beta \gamma = 4$
which implies
$\beta^{2}+\gamma^{2} = -4$
and no real number satisfies above equality, hence there is only one linear factor

dragon658 · Feb 19, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Hey guys,
I've got little to no idea about how to solve the logarithmic equation questions I've come across in the past hsc papers.

Can someone help me solve: 2lnx=ln(5+4x) AND 2log(base5)^3=log(base5)^x-log(base5)^6

1729 · Feb 19, 2017

Re: HSC 2017 Maths (Advanced) Marathon

dragon658 said:
Hey guys,
I've got little to no idea about how to solve the logarithmic equation questions I've come across in the past hsc papers.

Can someone help me solve: 2lnx=ln(5+4x) AND 2log(base5)^3=log(base5)^x-log(base5)^6

$$\noindent \textbf{QUESTION 1} \\ $ \begin{align*} 2\ln{x} &= \ln{(5+4x)} \\ \ln{x^2} &= \ln{(5+4x)} \\ x^2 &= 5 + 4x \\ x^2 - 4x - 5 &= 0 \\ (x-5)(x+1) &= 0 \\ x &= -1, 5 \\ x \neq -1 \textrm{ } (x > 0), \textrm{ } x &= 5 \end{align*} \\\\\\ $\textbf{QUESTION 2} \\ $ \begin{align*} 2\log_5{3} &= \log_5{x} - \log_5{6} \\ \log_5{9} &= \log_5{\frac{x}{6}} \\ 9 &= \frac{x}{6} \\ x &= 54 \end{align*}$

Mathew587 · Feb 19, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Herooo, Pls help with this

1729 · Feb 19, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Mathew587 said:
Herooo, Pls help with this
View attachment 33803

$$\noindent $T_n = S_n - S_{n-1}.$ Your answer is $T_{16} = 249 - 233 = 16$

envlam · Feb 19, 2017

Re: HSC 2017 Maths (Advanced) Marathon

$\begin{align*}T_{50}&=S_{50}-S_{49}\\&= 249-233\\&= 16\end{align*}$

edit: My bad, didn't see there was an answer already

Mathew587 · Feb 20, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Oh you can do that? My teachers a bit weird and would want us to actually write the whole solution out. How could you prove Tn = Sn - Sn-1?? ^^D

1729 · Feb 20, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Mathew587 said:
Oh you can do that? My teachers a bit weird and would want us to actually write the whole solution out. How could you prove Tn = Sn - Sn-1?? ^^D

$$\noindent $\begin{align*} S_n &= T_1 + T_2 +T_3 + ... + T_{n-1} + T_n \\ S_{n-1} &= T_1 + T_2 + T_3 + ... + T_{n-1} \end{align*} \\ $As can be seen, $S_n - S_{n-1} = T_n$

dan964 · Mar 20, 2017

Re: HSC 2017 Maths (Advanced) Marathon

Just a reminder this is a marathon thread, not for posting textbook errata. I have removed those posts and put into a separate thread
http://community.boredofstudies.org...l-maths-questions-textbook-errata-2017-a.html

davidgoes4wce · Apr 1, 2017

Re: HSC 2017 Maths (Advanced) Marathon

$$ Given x= $ \frac{6}{5} $ and $ y=1+\frac{1}{x}$

$Express y as a simple fraction$

Mahan1 · Apr 5, 2017

Re: HSC 2017 Maths (Advanced) Marathon

davidgoes4wce said:
$$ Given x= $ \frac{6}{5} $ and $ y=1+\frac{1}{x}$

$Express y as a simple fraction$

$x=\frac{6}{5} \Rightarrow y = 1+\frac{1}{x} = 1 + \frac{1}{\frac{6}{5}} = 1 + \frac{5}{6} = \frac{11}{6}$

davidgoes4wce · Apr 7, 2017

Re: HSC 2017 Maths Marathon

davidgoes4wce · Apr 7, 2017

Re: HSC 2017 Maths Marathon

OK I got nothing else to do on a Friday night I am going to have an honest go at this question.

davidgoes4wce · Apr 8, 2017

Re: HSC 2017 Maths Marathon

davidgoes4wce · Apr 8, 2017

Re: HSC 2017 Maths Marathon

davidgoes4wce · Apr 8, 2017

Re: HSC 2017 Maths Marathon

davidgoes4wce · Apr 9, 2017

Re: HSC 2017 Maths Marathon

HSC 2017-2018 Maths Marathon (1 Viewer)

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