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.
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 .
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.
Last edited by mrbunton; 17 Feb 2018 at 11:28 AM.
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.
Last edited by mrbunton; 17 Feb 2018 at 3:15 PM.
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
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 .
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?
Last edited by fan96; 18 Feb 2018 at 1:32 AM.
Since is an even function and is an odd function,
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If , then
Which is the general solution for
Last edited by fan96; 18 Feb 2018 at 12:27 PM.
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.
Last edited by altSwift; 19 Feb 2018 at 10:35 PM.
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
probability: how many 3 digits numbers exist where the middle digit is the average of the first and last digit.
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
yup
Show that the polynomial (where is an integer such that ) has exactly one real zero where .
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
Last edited by mrbunton; 24 Feb 2018 at 10:32 AM.
Last edited by altSwift; 25 Feb 2018 at 7:24 PM.
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
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.
Last edited by mrbunton; 27 Mar 2018 at 6:57 PM.
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