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HSC Mathematics Marathon (1 Viewer)

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That's an interesting question...
Expanding that would give, 1-(sum roots)-(sum of two roots at a time) -......+(product of roots)
I think it would eventually give 1+(n-1)= n.
Lol this is because I saw that sum of non real roots = -1 and product of roots = 1
 

jyu

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The points O, I, Z and P on the Argand Plane represent the complex numbers
0, 1, z and z + 1 respectively, where z = cosθ + i sinθ is any complex number of
modulus 1, with 0 <θ <π .
(i) Explain why OIPZ is a rhombus.
(ii) Show that (z-1)/(z+1) is purely imaginary.
(iii) Find the modulus of z + 1 in terms of θ .
 

jyu

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wow! The graphics & mathematical typeset is awesome.
Anyone can do it. Use equation editor (left arrow) and draw (right arrow) in microsoft office word.

 
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jyu

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Find the gradient of a tangent line to the graph of
at x=1.
 

Drongoski

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Find the gradient of a tangent line to the graph of
at x=1.
Should be staightforward - using implicit differentiation.

Subst x-1 into expression, you end up with y2 - 2y +1 = 0 ==> y = 1 also.

By implicit diffn:

15x4-3x2-2x3y' - 6x2y -3x2y' - 6xy + 4yy'+y' = 0

Rearranging: y' = {15x4-3x2-6x2y-6xy} / {2x3 3x2-4y-1}

Subst. x=1, y=1 to find the gradient.

Can someone please pose another question please.
 
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Drongoski

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I end up with dy/dx=0/0, the gradient is indeterminable.
You are correct. You need to try L'Hospital's Rule then. Differentiate both numerator and denominator wrt x and resubstitute. Hopefully you don't get a 0/0 again.t
 
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^ in any first semester uni maths course, and I think it was in the four unit syllabus ( along time ago, I have seen it asked in some really old 4unit hsc exams papers, like from 1970's, I remember I have seen someone post up a ton of old 4unit exams )
 

Drongoski

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I've never heard of "L'Hospital's Rule" :-(
1) You can Wikipedia it

2) Have you differentiated the numerator and denominator and re-substituted? I'm too lazy doing the mathematical typesetting.

3) Of course L'Hospital's Rule is not in HSC maths.
 
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@Drongoski

im pretty sure you get

dy/dx = - ( 15x^4 -3x^2 -6(x^2)y -6xy ) / ( 1 + 4y -3x^2 - 2x^3 ) and the point of interest is (1,1)

mmm this is a strange one, you cannot differentiate top and bottom ( and if so , with respect to what, x or y?? ) , because that means will get factors of "dy/dx" on the top from the product rules.

EDIT: ohh dont worry, I figured it out, because then we can sub anywhere we see a dy/dx in the next fraction ( obtained by via La Hopitals ) by the massive expression above for dy/dx
 
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Drongoski

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You have to differentiate again(implicitly) the top of the originally derived dy/dx and the bottom separately. You then re-substitute (x,y) @ (1,1). See what you get,
 

jyu

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1) You can Wikipedia it

2) Have you differentiated the numerator and denominator and re-substituted? I'm too lazy doing the mathematical typesetting.

3) Of course L'Hospita;s Rule is not in HSC maths.
You don't need to use l'Hospital's Rule. However, it gives you a quick answer if you know how to use it.

The question remains open.

May be get help from wolframalpha.

When you finish with this question, do the following one.

 
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hscishard

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Yea..no way I'm typing everything back into wolfram to re-check. I probably made a mistake whilst entering the whole thing
BTW, you sound like you've finished 4unit.
 
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