HSC 2016 MX1 Marathon (archive) (3 Viewers)

DatAtarLyfe · Apr 24, 2016

Re: HSC 2016 3U Marathon

porcupinetree said:
Unles I'm misreading something, the answer should be equal to 4a = 12 because a equals 3 in this case.

Yep that was my answer. However the answers they provided said the answer was 6. Generally i can tell if the answers are wrong or if im wrong, but i didnt know what they did so i couldnt tell

leehuan · Apr 24, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
View attachment 33145
Lee your way makes sense, and its what i thought of as well when the answers went for a amplitude approach. But idk why the answers did 2a when clearly it takes 4 times the amplitude to reach the starting position again

Edit: Part iii

The distance between both the endpoints is 2a. So if they doubled 2a, they got 4a.

The amplitude is just the distance from an endpoint to the equilibrium

DatAtarLyfe · Apr 24, 2016

HSC 2016 3U Marathon

leehuan said:
The distance between the endpoints is 2a. So if they doubled 2a, they got 4a.

The amplitude is just the distance from an endpoint to the equilibrium

Lee i know how to get the answer. Im confused as to what they did.
They didnt do 4a, they just did 2a

Edit: they didnt double 2a, they doubled a

InteGrand · Apr 24, 2016

Re: HSC 2016 3U Marathon

It's possible that the question intended to ask what's the distance travelled from one end to the other? Anyway, as asked (distance in a full cycle), the answer has to be 4A, where A is the amplitude.

leehuan · Apr 24, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
Lee i know how to get the answer. Im confused as to what they did.
They didnt do 4a, they just did 2a

Edit: they didnt double 2a, they doubled a

But I thought you said they found the distance between the endpoints and then doubled it?

If they only found the distance between the endpoints and left it there then they're wrong and you have to refer to InteGrand's suggested alternate question

DatAtarLyfe · Apr 24, 2016

HSC 2016 3U Marathon

Nup, i said they only found the distance between the endpoints (which is equivalent to doubling the amp). I doubled the distance between the endpoints (which is equivalent to quadrupling the amp)

K then, ill assume they are wrong. Smh, got me so upset, stupid answers

Paradoxica · Apr 24, 2016

Re: HSC 2016 3U Marathon

$$\noindent a) $f(x) \equiv \frac{1-x}{1+x}$. Find the range of $f$ for:$ \\ \indent $i) $x \geq 0 \\ \indent $ii) $-1<x<0 \\ \indent $iii) $x<-1 \\ $b) State the domain for which $\cos^{-1}{\left(\frac{1-x}{1+x}\right)}$ exists.$ \\ $c) By using an appropriate substitution for $x$ or otherwise, prove $\\ \indent \cos^{-1}{\left(\frac{1-x}{1+x}\right)} \equiv 2\tan^{-1}{\sqrt{x}}\\ $d) By differentiating $2 x \tan^{-1}{\sqrt{x}}$ with respect to $x$, evaluate $\int \cos^{-1}{\left(\frac{1-x}{1+x}\right)}$d$x$

davidgoes4wce · Apr 26, 2016

Re: HSC 2016 3U Marathon

Circle Geometry Question

Answer is D bit unsure about the explanations.

porcupinetree · Apr 26, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Circle Geometry Question

Answer is D bit unsure about the explanations.

Use Pythagoras' theorem to find half the length of the chord, then double your answer.

Speed6 · Apr 27, 2016

HSC 2016 3U Marathon

Someone please help

si2136 · Apr 27, 2016

Re: HSC 2016 3U Marathon

View attachment 33157

OP is the radius of the circle, which is 14.9 metres.

OP = OQ

Let X be the intersection between Chord PQ.

OX = 5.1 metres (given)

Triangle OQX is congruent to Triangle OPX.

Therefore, (14.9)^2 = (5.1)^2 + (x)^2

Therefore x = 14.

14 is half of the chord. Therefore the chord PQ is 28 metres.

Drongoski · Apr 27, 2016

Re: HSC 2016 3U Marathon

Speed6 said:
Someone please help

Maybe I can do this one.

$\int \frac {ln x}{2x}dx = \frac {1}{2}\int \frac {lnx}{x} dx = \frac {1}{2}\int ln x d lnx = \frac {1}{2} \times \frac {(ln x)^2}{2} + C = \frac {(ln x)^2}{4} + C$

This is equivalent to doing the substitution:

Let u = ln x

.: du = (1/x)dx

$\therefore \int \frac {ln x}{2x}dx = \frac {1}{2}\int lnx \times \frac {1}{x} dx = \frac {1}{2}\int u du = \frac {1}{2} \times \frac {u^2}{2} + C = \frac {(ln x)^2}{4} + C$

Speed6 · Apr 27, 2016

Re: HSC 2016 3U Marathon

Drongoski said:
Maybe I can do this one.

By all means.

porcupinetree · Apr 27, 2016

Re: HSC 2016 3U Marathon

Speed6 said:
Someone please help

Speed6 · Apr 27, 2016

Re: HSC 2016 3U Marathon

Cheers

davidgoes4wce · Apr 29, 2016

Re: HSC 2016 3U Marathon

hedgehog_7 said:
Two particles A and B are moving along a horizontal line with their distances X(A) and X(B) to the right of the origin O at time t given by
X(A) = 4te^(-t) and X(B) = -4(t^2)(e^-t). The particles are joined by a piece of elastic, whose midpoint M has position X(M) at time t.

When are A and B furthest apart?
Ans: t = 0.5(1+root 5)

ive tried adding X(A) and X(B) and then differentiating it to find the max tp and i dont seem to get the answer. Thx

$Cambridge 3B Q 14 (a)$

$$ I had a slightly different style in answering this question. $ With \ x_{m}= \frac{x_{a}+x_{b}}{2}= \frac{4t e^{-t}+-4t^2 e^{-t}}{2} =2e^{-t}(t-t^2)$

$\frac{dx_{m}}{dt}=(t-t^2)(-2e^{-t})+(2e^{-t}(1-2t))=-2e^{-t}t+2e^{-t}t^2+2e^{-t}-4e^{-t}t$

$Which simplifies down to v= 2e^{-t}(t^2-3t+1)$

$Red represents $ x_{a}, $ Blue represents $ x_{b}, $ Violet represents $ x_{m}, $ and orange represent the velocity functions.$

$By letting $ x_{m}=0, $ we can solve for the time returning back to the origin $ 0= 2te^{-t}(1-t), $ as shown graphically by the violet function in the graph$

$$ t=1 , then substituting this back into the speed function one can solve for speed.$ |v(1)|=|2e^{-1}[(-1)^2-3(1)+1]|=|\frac{-2}{e}|=\frac{2}{e}$

davidgoes4wce · Apr 29, 2016

Re: HSC 2016 3U Marathon

I have decided to finish the rest of Cambridge Year 12 3 Unit Ex 3B Q 14 (b) (c) and (d) from my point of view.

$Q 14 (b) $ By setting v=0 $ , 0=t^2-3t+1$

$t=\frac{3 \pm \sqrt{9-4(1)(1)}}{2(1)}=\frac{3\pm \sqrt{5}}{2}$

$$ By observation from the graphs above, the graph is furthest right when $ t=\frac{3-\sqrt{5}}{2}$

$$ The function is furthest left when $ t=\frac{3+\sqrt{5}}{2}$

$Q 14 (c) As \ t \to \infty, x_{a} \to 0$

$As \ t \to \infty, x_{b} \to 0$

$As \ t \to \infty, x_{m} \to 0$

$Q 14 (d) x_{d}=x_{a}-x_{b} =4te^{-t}-(-4t^2e^{-t})$

$=4te^{-t}+4t^2e^{-t}$

$= 4te^{-t} (1+t)$

$\frac{dx_{d}}{dt}=(1+t)[e^{-t}(4)+4t(-e^{-t})]+4te^{-t}$

$= (1+t)[4e^{-t}-4te^{-t}]+4te^{-t}$

$=4e^{-t}-4te^{-t}+4te^{-t}-4t^2e^{-t}+4te^{-t}$

$=4e^{-t}+4te^{-t}-4t^2e^{-t} =4e^{-t}(1+t-t^2)$

$Set \frac{dx_{d}}{dt}=0$

$0=4e^{-t}(1+t-t^2)$

$Can also be written as : t^2-t-1=0$

$$ By using the quadratic roots formula: $ t= \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$= \frac{1 \pm \sqrt{5}}{2}$

$t in this case cannot be $ t=\frac{1-\sqrt{5}}{2} $ since time cannot be a negative quantity.$

$\therefore $ the solution for t must be $ t=\frac{1+\sqrt{5}}{2}$

davidgoes4wce · Apr 29, 2016

Re: HSC 2016 3U Marathon

The graph below shows the maximum distance apart which occurs roughly at t=1.618, it basically is the graph of

$x_{m}=x_{a}-x_{b}$

leehuan · Apr 30, 2016

Re: HSC 2016 3U Marathon

Instructive question.

$The statement $\frac{d}{dx}\sin^2{x}=\sin{2x} $ should not come as a surprise. Why?$

DatAtarLyfe · Apr 30, 2016

HSC 2016 3U Marathon

leehuan said:
Instructive question.

$The statement $\frac{d}{dx}\sin^2{x}=\sin{2x} $ should not come as a surprise. Why?$

Because expecting the unexpected makes the unexpected expected

Q.E.D

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