HSC 2016 MX1 Marathon (archive) (1 Viewer)

sazkim · Oct 6, 2016

Re: HSC 2016 3U Marathon

Hi, this is a question from Chapter 10: Polynomials of TERRY LEE 3U

"Prove that the curves 6y=x^3+3x^2-9x-27 and 3y=x^3-3x^2 +9x-27 have only a point of intersection and show that the two curves are tangential to each other."

So, by equating the two I got x^3-9x^2+27x-27=0 which can be factorised to (x-3)^3=0

This would prove that there is only one point of intersection at (3,0)

And for the two curves being tangent to each other..
The solution says because x=3 is a triple root, the two curves are tangent to each other
But I do not understand the significance the triple root has which make the two curves tangents..

Thank you!!

Drongoski · Oct 6, 2016

Re: HSC 2016 3U Marathon

If the 2 curves do not intersect at all - there will be no root.

If they intersect at 3 points there will be 3 distinct roots.

If the 3 roots are identical, as in this case, it means the 2 curves do not cross each other, but do meet at 1 point at x=3; i.e. they touch each other.

davidgoes4wce · Oct 8, 2016

Re: HSC 2016 3U Marathon

This was from Success One HSC 2013 Paper solutions which I didn't quite get for Binomials:

$\textcircled{1} \ x^{2n-k} (x+2)^{2n-k} = \binom {2n-k}{0} 2^{2n-k}x^{2n-k} +\binom {2n-k}{1}2^{2n-k-1}x^{2n-k+1}+\binom{2n-k}{2}2^{2n-k-2}x^{2n-k+2}+.........+\binom {2n-k}{2n-k}2^0 x^{4n-2k}$

$\textcircled{2} \ So (1+x^2+2x)^{2n}= \binom {2n}{0} [\binom {2n}{0} 2^{2n}x^{2n}+\binom {2n}{1}2^{2n-1}x^{2n+1}+.....]$

$+ \binom{2n}{1}[{\binom{2n-1}{0}2^{2n-1}x^{2n-1}+\binom{2n-1}{1}2^{2n-2}{x^{2n}+....]$

$+ \binom{2n}{2}[{\binom{2n-2}{0}2^{2n-2}x^{2n-2}+\binom{2n-2}{1}2^{2n-3}{x^{2n-1}+\binom{2n-2}{2}2^{2n-4}{x^{2n}....]$

$I struggling to find the link between $ \textcircled{1} and \textcircled{2}. $ In Line 2 working out , I dont understand where the terms came from in the expansion.$

davidgoes4wce · Oct 8, 2016

Re: HSC 2016 3U Marathon

I dont get the factorising and all.

KingOfActing · Oct 8, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
I dont get the factorising and all.

I'll use sigma notation because I think it'll make it more clear. Line 2 is this:

$(1+x^2+2x)^{2n} = \sum_{k = 0}^{2n}{{2n}\choose k}(x^2+2x)^{2n-k} = \sum_{k = 0}^{2n}{{2n}\choose k}(x(x+2))^{2n-k}= \sum_{k = 0}^{2n}{{2n}\choose k}x^{2n-k}(x+2)^{2n-k}$

Can you see how line one is then used?

wu345 · Oct 16, 2016

Re: HSC 2016 3U Marathon

quick question -> If tangents have to intersect outside the parabola, then do normals have to intersect inside the parabola?

InteGrand · Oct 16, 2016

Re: HSC 2016 3U Marathon

wu345 said:
quick question -> If tangents have to intersect outside the parabola, then do normals have to intersect inside the parabola?

$\noindent No. Consider the parabola $x^2 = 4y \iff y = \frac{x^2}{4}$. The normal at a point $\left(2t,t^2\right)$ is $y = t^2 -\frac{1}{t}\left(x -2t\right)$. So the normal at the point on the parabola where $t = 1$ (the point $(2,1)$) is $y = 1 -(x-2)\iff y = -x+3$. The normal where $t = -4$ (point $(-8,16)$) is $y = 16 +\frac{1}{4}\left(x + 8\right)$. Solve this simultaneously to find that the intersection point of these two normals is $\left(x_{0},y_{0}\right) = (-12,15)$. Since $x_0 ^2 = 144 > 4\times 15 = 4y_0 \Rightarrow y_{0} < \frac{x_0^2}{4}$, this point is \emph{outside} (under) the parabola.$

InteGrand · Oct 16, 2016

Re: HSC 2016 3U Marathon

(Note that due to the convexity of a parabola, its tangents will never go 'inside' it, so there's no way two tangents could intersect somewhere inside the parabola. With normals though, they go both in and outside the parabola, so a priori they could well meet outside, and in fact can as the above example shows.)

Trebla · Oct 16, 2016

Re: HSC 2016 3U Marathon

Here is a simpler version of the Newton's Method question in the BOS trials.

$$Consider a function $g(x)$ which is continuous in $a\leq x\leq b$ defined by$\\\\g(x)=f(x)-\dfrac{(f(b)-f(a))(x-a)}{b-a}\\\\$where $f(x)$ is some continuous function in that same domain.$\\\\$(i)$\quad $Show that $g(a)=g(b)\\\\$(ii)$\quad $Explain why there must exist an $x=x_0$ such that $g'(x_0)=0\\\\$(iii)$\quad $Hence show that $f'(x_0)=\dfrac{f(b)-f(a)}{b-a}$

$$Let $A(x)$ be some function such that\\\\$A(x)=x-\dfrac{f(x)}{f'(x)}$\\\\Define Newton's method of approximating roots by a sequence $x_{k}$ where $x_{k+1}=A(x_k)$ and let $\alpha$ be a root of $f(x)$ where $f'(\alpha)\neq 0.\\\\$(iv)$\quad $Show that there exists some $c_k$ in the domain such that$\\\\|x_{k+1}-\alpha|=|x_k-\alpha|\dfrac{|f(c_k)f''(c_k)|}{[f'(c_k)]^2}\\\\\\$(v)$\quad $Hence show that if the function is such that $|f(x)f''(x)|<[f'(x)]^2$ in the domain then $x_{k+1}\rightarrow \alpha$ as $k \rightarrow \infty$

Paradoxica · Oct 16, 2016

Re: HSC 2016 3U Marathon

Trebla said:
Here is a simpler version of the Newton's Method question in the BOS trials.

$$Consider a function $g(x)$ which is continuous in $a\leq x\leq b$ defined by$\\\\g(x)=f(x)-\dfrac{(f(b)-f(a))(x-a)}{b-a}\\\\$where $f(x)$ is some continuous function in that same domain.$\\\\$(i)$\quad $Show that $g(a)=g(b)\\\\$(ii)$\quad $Explain why there must exist an $x=x_0$ such that $g'(x_0)=0\\\\$(iii)$\quad $Hence show that $f'(x_0)=\dfrac{f(b)-f(a)}{b-a}$

$$Let $A(x)$ be some function such that\\\\$A(x)=x-\dfrac{f(x)}{f'(x)}$\\\\Define Newton's method of approximating roots by a sequence $x_{k}$ where $x_{k+1}=A(x_k)$ and let $\alpha$ be a root of $f(x)$ where $f'(\alpha)\neq 0.\\\\$(iv)$\quad $Show that there exists some $c_k$ in the domain such that$\\\\|x_{k+1}-\alpha|=|x_k-\alpha|\dfrac{|f(c_k)f''(c_k)|}{[f'(c_k)]^2}\\\\\\$(v)$\quad $Hence show that if the function is such that $|f(x)f''(x)|<[f'(x)]^2$ in the domain then $x_{k+1}\rightarrow \alpha$ as $k \rightarrow \infty$

this triggers me

duxxx · Oct 17, 2016

Re: HSC 2016 3U Marathon

The tidal heights over the next five days for the Yangtze River mouth in Shanghai are shown in screenshot below.

Screen Shot 2016-10-14 at 2.13.05 PM.png

The heights in the tidal prediction chart show the height above the chart datum which is 7m. For example, a low tide of 0.5m is actually a depth of 7.5m.

a)Use the tidal data to synthesise a model for the depth of water at the Yangtze River mouth for all the data you have captured.
A horizontal translation is required in your model. Use the following information to assist you
The general trigonometric function
f(x) = asinb)x+c) + d
has an amplitude of a, a period of 2pi/b, a horizontal translation of c units to the left and a vertical translation of d units up.

b) Produce a single graph containing the actual data given in the tidal prediction chart and your model. Comment on the strength and limitations of the model.

davidgoes4wce · Oct 25, 2016

Re: HSC 2016 3U Marathon

This question was from the 2013 HSC Exam which got me thinking over the past few days

Q12 d 2013 HSC Exam

$0= \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...+(-1)^n \binom{n}{n}$

I always thought of binomial expansions (as being of the form)

$(1+x)^n$

In the Corneos Publications it has this as the answer:

$(1-x)^n$

$=\sum_{0}^n \binom{n}{r} 1^{n-r} (-x)^r$

$=\binom{n}{0} 1^{n} +\binom{n}{1} 1^{n-1}(-x)+\binom{n}{2} 1^{n-2}(x)^2+...+\binom{n}{n} 1^{n-n}(-x)^n$

^^^ This was the solution typed exactly in the Corneos Publications, I also think they have done an error on that 3rd term it should be (-x)^2 .

Would love more feed back

davidgoes4wce · Oct 25, 2016

Re: HSC 2016 3U Marathon

duxxx said:
The tidal heights over the next five days for the Yangtze River mouth in Shanghai are shown in screenshot below.
View attachment 33514
The heights in the tidal prediction chart show the height above the chart datum which is 7m. For example, a low tide of 0.5m is actually a depth of 7.5m.

a)Use the tidal data to synthesise a model for the depth of water at the Yangtze River mouth for all the data you have captured.
A horizontal translation is required in your model. Use the following information to assist you
The general trigonometric function
f(x) = asinb)x+c) + d
has an amplitude of a, a period of 2pi/b, a horizontal translation of c units to the left and a vertical translation of d units up.

b) Produce a single graph containing the actual data given in the tidal prediction chart and your model. Comment on the strength and limitations of the model.

Do you have an answer for the values for part (a) ? I just want to compare my answers.

leehuan · Oct 25, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
This question was from the 2013 HSC Exam which got me thinking over the past few days

Q12 d 2013 HSC Exam

$0= \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...+(-1)^n \binom{n}{n}$

I always thought of binomial expansions (as being of the form)

$(1+x)^n$

In the Corneos Publications it has this as the answer:

$(1-x)^n$

$=\sum_{0}^n \binom{n}{r} 1^{n-r} (-x)^r$

$=\binom{n}{0} 1^{n} +\binom{n}{1} 1^{n-1}(-x)+\binom{n}{2} 1^{n-2}(x)^2+...+\binom{n}{n} 1^{n-n}(-x)^n$

^^^ This was the solution typed exactly in the Corneos Publications, I also think they have done an error on that 3rd term it should be (-x)^2 .

Would love more feed back

Whilst that transition is dodgy as it skips an important step it's not wrong simply due to the fact that (-x)^2 = x^2

KingOfActing · Oct 25, 2016

Re: HSC 2016 3U Marathon

Could I one line that?

0 = (1-1)^n = nC0 - nC1 + nC2 - ... + (-1)^n*nCn or would the markers not be happy with that?

InteGrand · Oct 25, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
Could I one line that?

0 = (1-1)^n = nC0 - nC1 + nC2 - ... + (-1)^n*nCn or would the markers not be happy with that?

That's a classic method. Should be fine. (But probably safest to write it out with the binomial theorem formula (with the Sigma notation) before it, and/or say you're using the binomial theorem.)

KingOfActing · Oct 25, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
That's a classic method. Should be fine.

Lazy people : 1
Everyone else: ceebs counting

HSC 2016 MX1 Marathon (archive) (1 Viewer)

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