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HSC 2018-2019 MX2 Marathon (1 Viewer)

fan96

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Re: HSC 2018 MX2 Marathon

where


_____________________________________________________

If is real, then



The smallest value of is .
_____________________________________________________

If is purely imaginary, then



The smallest value of is .
 

altSwift

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Re: HSC 2018 MX2 Marathon

where


_____________________________________________________

If is real, then



The smallest value of is .
_____________________________________________________

If is purely imaginary, then



The smallest value of is .
Correct!
Here is another (I really like these questions)





 
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fan96

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Re: HSC 2018 MX2 Marathon









Since is an even function and is an odd function,








_____________________________________________________________

If , then



Which is the general solution for



 
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fan96

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Re: HSC 2018 MX2 Marathon

An interesting roots of unity question from the SGS notes.

i) Find the fourth roots of in the form .

ii) Hence write as a product of two quadratic factors with real co-efficients.
 

altSwift

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Re: HSC 2018 MX2 Marathon

An interesting roots of unity question from the SGS notes.

i) Find the fourth roots of in the form .

ii) Hence write as a product of two quadratic factors with real co-efficients.










 

fan96

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Re: HSC 2018 MX2 Marathon

Note that only even numbers can be halved to obtain an integer, and and . Finally, the average of any two digits is never greater than 9 or less than 0.

If the first digit is odd (1, 3, 5, 7, 9), then the last digit must be odd too (1, 3, 5, 7, 9).

If the first digit is even (2, 4, 6, 8), then the last digit must be even or zero (0, 2, 4, 6, 8).

Any two digits can only have one possible average, so the middle digit doesn't matter.

Therefore the number of digits is
 

fan96

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Re: HSC 2018 MX2 Marathon

Show that the polynomial (where is an integer such that ) has exactly one real zero where .
 

mrbunton

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Re: HSC 2018 MX2 Marathon

derive equation to get turning points at -1 and 1
find y values of these turning points:
(-1,2+k) and (1,k-2)
since the poly is odd it will have at least one root.
when k>2; both turning points are above the x-axis; so no extra roots are possible.

proving that r<-1 or similar value:
when letting k=2; that poly is less then 0. the poly can be factored into (x+2)(x-1)^2<0
condition is only true at x<-2 or r<-2
-2<-1 so r<-1
 
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mrbunton

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Re: HSC 2018 MX2 Marathon

i) the line created is bx+ay=0.since you said i) is not the locus of z; and z can be truly anything; then we can observe that bx+ay=0 only yields to purely real values; so it becomes the line y=0 on the imaginary number plane

ii) (x-p)^2+(y-q)^2= k^2
sub y=0 and expand and bring to one side
x^2 -(2p)x +(p^2+ q^2 - k^2) =0
product of roots: z1*z2= p^2 + q^2 - k^2

which is the same thing as we are meant to show. normally one would say the product of the roots are just the real parts of z1 and z2; but there is no imaginary part so condition is still true.
 
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altSwift

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Re: HSC 2018 MX2 Marathon

i) the line created is bx+ay=0.since you said i) is not the locus of z; and z can be truly anything; then we can observe that bx+ay=0 only yields to purely real values; so it becomes the line y=0 on the imaginary number plane

ii) (x-p)^2+(y-q)^2= k^2
sub y=0 and expand and bring to one side
x^2 -(2p)x +(p^2+ q^2 - k^2) =0
product of roots: z1*z2= p^2 + q^2 - k^2

which is the same thing as we are meant to show. normally one would say the product of the roots are just the real parts of z1 and z2; but there is no imaginary part so condition is still true. But in an exam u would of been expected to spend a couple more lines to actually be rigorous but im lazy.
How can you say that the equation is y=0 on the imaginary plane but then sub into an equation that's on the real number plane? Wouldn't you have to sub in bx + ay = 0?
 

mrbunton

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Re: HSC 2018 MX2 Marathon

ya rip me. i misinterpreted what z as a variable point meant. i thought it meant z could be anything and that the line was dependent on what z was; although that doesnt make much sense once you think about it.
 
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