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HSC 2018-2019 MX2 Marathon (1 Viewer)

iStudent

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Re: HSC 2018 MX2 Marathon

RPQ and PQB are supplementary (parallel lines)

PQB = QSB (angle in alt segment)

hence RPQ + PSB = 180

since opposite angles are supplementary, RSQP is cyclic?

been 4 years hope I am right :p
 

pikachu975

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Re: HSC 2018 MX2 Marathon

Cant seem to figure out how to do this question:
https://imgur.com/a/jRaWrmq
BSQ = PQB (alternate angle theorem)
PQB = 180-RPQ (co-interior angles in a parallelogram are supplementary)
BSQ + RPQ = 180

Hence opposite angles are supplementary so RSQP is a cyclic quad

Hope this reasoning is legit, forgot maths

EDIT: Nvm too slow
 

Drongoski

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Re: HSC 2018 MX2 Marathon

It is easy to prove RSQP is cyclic; so that this can be easily shown in many ways.

e.g.

Let angle AQS = @; .: angle QBS = @; .: angle PRS = @ (PR // QB)

.: angle AQS = angle PRS

.: external angle AQS of quadrilateral RSQP = int opp angle PRS

.: RSQP must be cyclic

QED
 
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fan96

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Re: HSC 2018 MX2 Marathon

a)


b)



c)

For , .

If , then clearly .

Therefore, .

From b) . Since ,



Replacing with , we get with and so



Giving, as required,



d)

From b),

.







From c),

 

fan96

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Re: HSC 2018 MX2 Marathon

yoyoyo
there was a nice question i found a while back and thought i would post it here:


given that e^x can be written as a sum of an odd and even function, find the two functions.
Is the intended answer the Taylor series for ?

Edit: nevermind, I found some others
 
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HeroWise

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Re: HSC 2018 MX2 Marathon

Is the intended answer the Taylor series for ?

Edit: nevermind, I found some others


its Hyperbolic function of sin and cos
so e^x = sinhx+coshx
Its legit like deriving cosx and sinx in complex field


The taylor expansion is just an approximate for e^x for values less than 1 or close to 0 but can get quite close. You would just factorise the even and odd parts of the function. However, that does not mean you cant use it, I tried it with Maclaurin series and it seems to look solid.
 
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fan96

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Re: HSC 2018 MX2 Marathon

The equation of the tangent at the hyperbola is given by



Because the shortest distance between a point and a line is the perpendicular distance, is the intersection between the tangent line and the line perpendicular to the tangent which also passes through the origin.

That line is



Solving these two equations gives



Now,





Hence the locus of is given by the equation .







And similarly,



So,









 
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