# Thread: HSC 2018-2019 MX2 Marathon

1. ## Re: HSC 2018 MX2 Marathon

i) define the graph as

Let the lower bound be such that .

Now,

So for the integral to be zero,

Rearranging,

Using the quadratic formula,

Since ,

So

(One other solution is also )
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ii) The area of the semicircle is , and

The graph doesn't seem to be defined for .

Clearly , so if one of the bounds is , the only bound that will make the integral zero is .  Reply With Quote

2. ## Re: HSC 2018 MX2 Marathon

2^2*pi/2 = 2pi >6 therefore ii) is possible. remember it has radius 2 not 1
also u got the gradient of the second line for i) incorrect, getting the gradient wrong also affected the y-intercept.

noticing k=5 was impressive although.  Reply With Quote

3. ## Re: HSC 2018 MX2 Marathon Originally Posted by mrbunton 2^2*pi/2 = 2pi >6 therefore ii) is possible. remember it has radius 2 not 1
It's a semicircle with diameter from -3 to -1 isn't it? So the diameter has length 2 and radius 1. Originally Posted by mrbunton also u got the gradient of the second line for i) incorrect, getting the gradient wrong also affected the y-intercept.
Oops. The line is .

So the new equation is

And the quadratic is now which gives .  Reply With Quote

4. ## Re: HSC 2018 MX2 Marathon

rip the circle was meant to have radius 2; i changed it now so the circle touches -5 to -1.
edit: i keep writing questions incorrectly; i check next time btw.  Reply With Quote

5. ## Re: HSC 2018 MX2 Marathon  Reply With Quote

6. ## Re: HSC 2018 MX2 Marathon

where

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If is real, then

The smallest value of is .
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If is purely imaginary, then

The smallest value of is .  Reply With Quote

7. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 where

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If is real, then

The smallest value of is .
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If is purely imaginary, then

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

8. ## Re: HSC 2018 MX2 Marathon

The equation does not seem to hold for .

https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

Is there supposed to be an somewhere?  Reply With Quote

9. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 The equation does not seem to hold for .

https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

Is there supposed to be an somewhere?
Yes it should be  Reply With Quote

10. ## Re: HSC 2018 MX2 Marathon

Since is an even function and is an odd function,

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If , then

Which is the general solution for  Reply With Quote

11. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 The equation does not seem to hold for .

https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

Is there supposed to be an somewhere? Originally Posted by jathu123 Yes it should be
My bad, yeah thats the correct equation  Reply With Quote

12. ## 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.  Reply With Quote

13. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 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.  Reply With Quote

14. ## Re: HSC 2018 MX2 Marathon  Reply With Quote

15. ## Re: HSC 2018 MX2 Marathon Originally Posted by altSwift Trivially, .  Reply With Quote

16. ## Re: HSC 2018 MX2 Marathon Originally Posted by sharky564 Trivially, .
Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too  Reply With Quote

17. ## Re: HSC 2018 MX2 Marathon Originally Posted by sharky564 Trivially, . Originally Posted by altSwift Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too  Reply With Quote

18. ## Re: HSC 2018 MX2 Marathon Originally Posted by sharky564 Trivially, .
What made the polynomial so easy to factor?  Reply With Quote

19. ## Re: HSC 2018 MX2 Marathon

probability: how many 3 digits numbers exist where the middle digit is the average of the first and last digit.  Reply With Quote

20. ## 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  Reply With Quote

21. ## Re: HSC 2018 MX2 Marathon

yup  Reply With Quote

22. ## Re: HSC 2018 MX2 Marathon

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

23. ## 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  Reply With Quote

24. ## Re: HSC 2018 MX2 Marathon  Reply With Quote

25. ## 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.  Reply With Quote