Note the denominator in the first line
That looks wrong. However they later multiplied by instead, so it seems two mistakes have cancelled out?
I don't think there are any further restrictions.
The only conditions given to you are
and
ruse-solution.png
Did they make a mistake on line two, with x(x-a)?? [When I multiplied out, i got x(x+a), ending up with the locus as a circle with centre (-a/2, b/2) instead of (a/2, b/2)]
also, is there supposed to be a restriction on the locus? like excluding (a,0) (because z =/ a on the denominator)
Thanks!
Note the denominator in the first line
That looks wrong. However they later multiplied by instead, so it seems two mistakes have cancelled out?
I don't think there are any further restrictions.
The only conditions given to you are
and
Last edited by fan96; 6 Feb 2018 at 5:53 PM.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
in the very first line, its works out as
Last edited by darkk_blu; 6 Feb 2018 at 6:08 PM.
I can't believe that question was also in my 4U test today, but I really shouldn't have been. I stuffed up and put instead, lol, lost a mark.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
The graph of a locus represents all the possible points that satisfies a given equation.
You have to exclude (a,0), as it lies on the circle.
Otherwise, you are saying that z=a satisfies Re(z-ib/z-a)=0, which is false.
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