HSC 2016 MX1 Marathon (archive) (1 Viewer)

leehuan · Dec 7, 2015

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Not sure what has happened but I have tried accessing your link but can't read it from 2 computers (a uni) and my mac book (which I use regularly)

Question's already done, but for your reference LaTeX wasn't working.

davidgoes4wce · Dec 11, 2015

Re: HSC 2016 3U Marathon

the answer is y=38 degrees. Anyone can explain the step by step with reasoning?

InteGrand · Dec 11, 2015

Re: HSC 2016 3U Marathon

davidgoes4wce said:
the answer is y=38 degrees. Anyone can explain the step by step with reasoning?

Use the following facts:

• We can find angle CBE using the fact that adjacent angles on a straight line add up to 180 degrees (that is, are supplementary).

• Then we can find angle BCE isn't angle sum of triangle BCE.

• But angle DCE equals angle BCE as we are told that CE bisects angle BCD. So we now have angle DCE too.

• Angle DCE + y deg = 95 deg (exterior angle of triangle DCE), so we have y = 95 – angle DCE, and we found angle DCE above.

davidgoes4wce · Dec 14, 2015

Re: HSC 2016 3U Marathon

Not sure if anyone here uses the Extension 1 (Year 11) 2/3 Unit text but there is a mistake on Q2a page 182, the answer should be : -16/(2x+3)^5 (the derivative of y=2/(2x+3)^4)

Not -24/(2x+3)^5 as stated in the answer.

InteGrand · Dec 14, 2015

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Not sure if anyone here uses the Extension 1 (Year 11) 2/3 Unit text but there is a mistake on Q2a page 182, the answer should be : -16/(2x+3)^5 (the derivative of y=2/(2x+3)^4)

Not -24/(2x+3)^5 as stated in the answer.

Which textbook is this (author)?

davidgoes4wce · Dec 14, 2015

Re: HSC 2016 3U Marathon

Its by Dr Max Shur

leehuan · Dec 14, 2015

Re: HSC 2016 3U Marathon

I think that textbook is more frequently used for the purpose of study. The contents of it are quite simplistic.
--------------------------------------
So why are you putting errors here instead of making your own thread?

davidgoes4wce · Dec 14, 2015

Re: HSC 2016 3U Marathon

Here was the question:

davidgoes4wce · Dec 14, 2015

Re: HSC 2016 3U Marathon

OK I won't post any more mistakes on this thread any more. I thought you meant Grove specifically.

leehuan · Dec 14, 2015

Re: HSC 2016 3U Marathon

Grove got repetitive cause it's Grove. Worst textbook ever.

Dude just so you know though every one of us knows your intention is pure. Nobody's gonna yell at you for your own thread. It's just a bit derailing this one
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Yeah, looks like that textbook writer derped and thought that the x was on the 3.

davidgoes4wce · Dec 17, 2015

Re: HSC 2016 3U Marathon

Can't seem to do this one, the solution is x=-0.5 and x=1

Parvee · Dec 17, 2015

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Can't seem to do this one, the solution is x=-0.5 and x=1

The one you posted has complex solutions

I think you meant it with a +
But yeah just multiply everything by x^2 and solve 2x^2-x-1=0

davidgoes4wce · Dec 17, 2015

Re: HSC 2016 3U Marathon

Parvee said:
The one you posted has complex solutions

I think you meant it with a +
But yeah just multiply everything by x^2 and solve 2x^2-x-1=0

I got it from Dr Max Shurs book from page 50 Q1a:

So far have noticed quite a few mistakes from the text. You are right there should be a positive in front of 1/x^2, as I also had a complex solution.

leehuan · Dec 17, 2015

Re: HSC 2016 3U Marathon

http://www.wolframalpha.com/input/?i=1/x-1/x^2=2

What a joke for a 3U text. Yeah, that's a mistake.

InteGrand · Dec 18, 2015

Re: HSC 2016 3U Marathon

$\textbf{NEW QUESTION}$

$\noindent Let $\ell_1$ and $\ell_2$ be two lines in the Cartesian plane and suppose that neither line is horizontal. Let the slope of $\ell_1$ and $\ell_2$ be $m_1$ and $m_2$, respectively. Prove that $m_1 m_2 = -1$ if and only if $\ell _1$ is perpendicular to $\ell _2$.$

$\noindent (This fact is commonly taught in early high school but seems to be taken for granted and rarely proven when taught.)$

integral95 · Dec 18, 2015

Re: HSC 2016 3U Marathon

InteGrand said:
$\textbf{NEW QUESTION}$

$\noindent Let $\ell_1$ and $\ell_2$ be two lines in the Cartesian plane and suppose that neither line is horizontal. Let the slope of $\ell_1$ and $\ell_2$ be $m_1$ and $m_2$, respectively. Prove that $m_1 m_2 = -1$ if and only if $\ell _1$ is perpendicular to $\ell _2$.$

$\noindent (This fact is commonly taught in early high school but seems to be taken for granted and rarely proven when taught.)$

using the angle between 2 line formula

$tan \theta = \frac{m_1- m_2}{1+m_1m_2}$

since the lines are perpendicular, then the angle between the lines make a right angle or 90 degrees $\theta = 90$

that makes LHS undefined (tan90), so that would mean RHS is undefined.

RHS is undefined if the denominator is 0 i.e

$1+m_1m_2 = 0 \Rightarrow m_1m_2 =-1$

Lel it's probably not the best.

leehuan · Dec 22, 2015

Re: HSC 2016 3U Marathon

integral95 said:
using the angle between 2 line formula

$tan \theta = \frac{m_1- m_2}{1+m_1m_2}$

since the lines are perpendicular, then the angle between the lines make a right angle or 90 degrees $\theta = 90$

that makes LHS undefined (tan90), so that would mean RHS is undefined.

RHS is undefined if the denominator is 0 i.e

$1+m_1m_2 = 0 \Rightarrow m_1m_2 =-1$

Lel it's probably not the best.

A reference to the limit as x->0 from the right of 1/x could be helpful
-------------------------------------------------------------
NEXT QUESTION:
Let's bring the questions back to the rote for a moment:
$\\ $a) Explain why the statement of the remainder theorem of polynomials holds true. \\ b) If a polynomial $P(x)$ when divided by $(x-3)(x+10)$ yields a remainder of $21x-234$, find the remainder when divided by $(x-3)$

DatAtarLyfe · Dec 22, 2015

Re: HSC 2016 3U Marathon

a) Let P(x) = D(x)Q(x) + R(x), where D(x) is the divisor function, Q(x) is the quotient function and R(x) is the remainder function
Let D(x) = (x - a)
P(x) = (x - a)Q(x) + R(x)
P(a) = (a - a)Q(a) + R(a)
P(a) = R(a)
Hence, when a polynomial P(x) is divided by (x - a), the remainder is P(a)
b) P(x) = (x - 3)(x + 10) + 21x - 234
Using remainder theorem, sub x = 3
P(3) = (3 - 3)(3 + 10) + 21(3) - 234
P(3) = 21(3) - 234 = -171
P(x) = (x - 3)Q(x) + R(x)
using remainder theorem, sub x = 3
P(3) = (3 - 3)Q(3) + R(3) = R(3)
However, P(3) = -171 (from above)
Therefore R(x) = -171

InteGrand · Dec 22, 2015

Re: HSC 2016 3U Marathon

$\noindent \textbf{NEW QUESTION}$

$\noindent Explain the flaw (if any) in the following reasoning.$

$\noindent For a particle moving in a straight line, we know its acceleration is given by $\ddot{x} = v \frac{\mathrm{d}v}{\mathrm{d}x}$. Hence if $v =0$, then $\ddot{x}=0$. So whenever the particle is stationary, its acceleration is 0.$

leehuan · Dec 22, 2015

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
a) Let P(x) = D(x)Q(x) + R(x), where D(x) is the divisor function, Q(x) is the quotient function and R(x) is the remainder function
Let D(x) = (x - a)
P(x) = (x - a)Q(x) + R(x)
P(a) = (a - a)Q(a) + R(a)
P(a) = R(a)
Hence, when a polynomial P(x) is divided by (x - a), the remainder is P(a)
b) P(x) = (x - 3)(x + 10) + 21x - 234
Using remainder theorem, sub x = 3
P(3) = (3 - 3)(3 + 10) + 21(3) - 234
P(3) = 21(3) - 234 = -171
P(x) = (x - 3)Q(x) + R(x)
using remainder theorem, sub x = 3
P(3) = (3 - 3)Q(3) + R(3) = R(3)
However, P(3) = -171 (from above)
Therefore R(x) = -171

a) - Good
b) - Not sure what happened to your working. The logic is that:

P(x) = (x-3)(x+10)Q(x) + (21x-234)

A direct substitution of x=3 was sufficient to establish that the remainder is -171

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