HSC 2015 MX1 Marathon (archive) (1 Viewer)

davidgoes4wce · Oct 21, 2015

Re: HSC 2015 3U Marathon

Have learnt this course for nearly 12 months now ( did the prelim for 6 months). My slight weakness is Binomial Theorem and Geometry.

Its not that I haven't practiced enough but I kind of went through 4/5 of the book in Fitzpatrick and did zilch form Margaret Grove in Prelim and HSC texts.

I will make sure I go over Grove and the concepts again in summer.

kawaiipotato · Oct 21, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Simplify

$\log _{b}a * \log _{a^2 }b = \frac{ \ln a}{ \ln b} * \frac{ \ln b}{ \ln a^2 }$

$= \frac{ \ln a}{ \ln b} * \frac{ \ln b}{2 \ln a}$

$= \frac{1}{2}$

$\Rightarrow $A$$

davidgoes4wce · Oct 21, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
$\log _{b}a * \log _{a^2 }b = \frac{ \ln a}{ \ln b} * \frac{ \ln b}{ \ln a^2 }$

$= \frac{ \ln a}{ \ln b} * \frac{ \ln b}{2 \ln a}$

$= \frac{1}{2}$

$\Rightarrow $A$$

I have seen it in books time and time again but never made the effort to memorise that in 1 step. I will do now.

InteGrand · Oct 21, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I have seen it in books time and time again but never made the effort to memorise that in 1 step. I will do now.

$It can be generalised to any log base on the R.H.S.: $\log _b a = \frac{\log _c a}{\log _c b}$ (for any $a,b,c>0$).$

davidgoes4wce · Oct 22, 2015

Re: HSC 2015 3U Marathon

I understand how angle AFG= angle AOG (angle subtended by the same arc AG)

My question with regards to a cyclic quad, must one of the points go through the centre , O, for a cyclic quad to exist?

Carrotsticks · Oct 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I understand how angle AFG= angle AOG (angle subtended by the same arc AG)

My question with regards to a cyclic quad, must one of the points go through the centre , O, for a cyclic quad to exist?

Yes.

Also what paper is this from? Looks like Grammar.

davidgoes4wce · Oct 22, 2015

Re: HSC 2015 3U Marathon

Its from Samir El Hosri study guide

davidgoes4wce · Oct 22, 2015

Re: HSC 2015 3U Marathon

This is the solution for part (i)

I wanted to state my reason as: ("The angle is subtended by the same chord AB"). Would this be good enough to get the 1 mark ?

Crisium · Oct 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I wanted to state my reason as: ("The angle is subtended by the same chord AB"). Would this be good enough to get the 1 mark ?

You have to include the implication of it being subtended by the same chord

It's like saying "co-interior angles in parallel lines" but it excludes whether they're equal, not equal, etc. so be specific and say "co-interior angles in parallel lines are supplementary"

davidgoes4wce · Oct 22, 2015

Re: HSC 2015 3U Marathon

Looking at the question first time, I had no idea where to start. When checking the solutions, I noticed they placed an x^2 term next to the (1+x)^n. When you integrate this coefficient you do see the 1/3 nC0 (which is 1/3) for the first time.

My question is in past HSC or other papers must we be able to spot the integration coefficients by inspection? I for one know the integrated value of x^2 = is x^3/3 and when you substitute a value of x=1 or x=-1 you can determine the sign whether it be positive or negative.

davidgoes4wce · Oct 22, 2015

Re: HSC 2015 3U Marathon

I'm going to get some spaghetti Bolognaise after spending time doing Binomials

dathat · Oct 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I understand how angle AFG= angle AOG (angle subtended by the same arc AG)

My question with regards to a cyclic quad, must one of the points go through the centre , O, for a cyclic quad to exist?

how do you prove the cyclic quad??
as far as i could get is angle AOG=CAB=DAG and i dont know where to from that?

Drsoccerball · Oct 22, 2015

Re: HSC 2015 3U Marathon

dathat said:
how do you prove the cyclic quad??
as far as i could get is angle AOG=CAB=DAG and i dont know where to from that?

Angles at center are two times angle at circumference and angles in triangle and straight line add up to 180 degrees. Therefore cycli(angles in same segment) ceebs doing it look for what I did lel

thomasdo1 · Oct 22, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Looking at the question first time, I had no idea where to start. When checking the solutions, I noticed they placed an x^2 term next to the (1+x)^n. When you integrate this coefficient you do see the 1/3 nC0 (which is 1/3) for the first time.

My question is in past HSC or other papers must we be able to spot the integration coefficients by inspection? I for one know the integrated value of x^2 = is x^3/3 and when you substitute a value of x=1 or x=-1 you can determine the sign whether it be positive or negative.

How did you integrate the LHS, I tried substitution but it got ugly

Drsoccerball · Oct 22, 2015

Re: HSC 2015 3U Marathon

thomasdo1 said:
How did you integrate the LHS, I tried substitution but it got ugly

Let u = (1+x)^n

davidgoes4wce · Oct 23, 2015

Re: HSC 2015 3U Marathon

Not sure how to go about this .

flavurr · Oct 23, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
Not sure how to go about this .

Is the answer possibly x/(x+1) ?

davidgoes4wce · Oct 23, 2015

Re: HSC 2015 3U Marathon

flavurr said:
Is the answer possibly x/(x+1) ?

That is the answer. Any idea on how to explain it?

davidgoes4wce · Oct 23, 2015

Re: HSC 2015 3U Marathon

Going to listen to Omi-Hula Hoop to get more motivated for maths.

flavurr · Oct 23, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
That is the answer. Any idea on how to explain it?

Screen Shot 2015-10-23 at 12.19.11 PM.png

Sorry for how its set out and all. And I dont know if theres a proper ext way of doing it

HSC 2015 MX1 Marathon (archive) (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Retired

Well-Known Member

Well-Known Member

Pew Pew

Well-Known Member

Well-Known Member

New Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

New Member

Well-Known Member

Well-Known Member

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)