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Help with some questions (1 Viewer)

MC Squidge

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i) The complex numbers 2+1, -1-2i and z represent points A, B and C on the Argand Diagram. Find One Possible value of z such that Triangle ABC is right angled at C and |CB|=2|CA|

ii)

iii) Given x^2-y^2+xy=5
a)find dy/dx and hence points on the curve whose tangents are vertical or horizontal
b) Discuss the behaviour of the curve for large values of x
c) Sketch the curve
 
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alakazimmy

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ii)

Prove true for n=1
So (1 + sqrt(2))<sup>1</sup> = 1 + 1*sqrt(2)

So p<sub>n</sub> = 1, and q<sub>n</sub> = 1
Hence, it is true for n=1, as p<sub>n</sub> and q<sub>n</sub> are integers

Assume true for n = k.
That is (1 + sqrt(2))<sup>k</sup> = p<sub>k</sub> + q<sub>k</sub>*sqrt(2)

Now, we need to prove true for n= k + 1
That is, prove that: (1 + sqrt(2))<sup>k+1</sup> = p<sub>k+1</sub> + q<sub>k+1</sub>*sqrt(2)

Using our assumption, (1 + sqrt(2))<sup>k+1</sup> = (1 + sqrt(2))<sup>k</sup> (1 + sqrt(2))
= (p<sub>k</sub> + q<sub>k</sub>*sqrt(2)) (1 + sqrt(2))
= p<sub>k</sub> + q<sub>k</sub>*sqrt(2) + p<sub>k</sub>*sqrt(2) + 2q<sub>k</sub>
= (p<sub>k</sub> + 2q<sub>k</sub>) + (q<sub>k</sub> + p<sub>k</sub>)*sqrt(2)
= p<sub>k+1</sub> + q<sub>k+1</sub>*sqrt(2)

where p<sub>k+1</sub> = p<sub>k</sub> + 2q<sub>k</sub>, which is an integer
and q<sub>k+1</sub> = q<sub>k</sub> + p<sub>k</sub>, which is an integer

Therefore, by the principles of M.I ... bla bla bla, it's true for all integer values of n which are larger than 1.
 
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youngminii

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iii) Using impleeeecit differentiation,
2x - 2y(dy/dx) + x(dy/dx) + y = 0
2x + y = (dy/dx)(2y - x)
dy/dx = (2x + y)/(2y - x)
I may have made a mistake, 'cause I'm pretty rusty with implicit.
So the points on the curve where the tangent is horizontal is dy/dx = 0 and vertical is undefined.
For horizontal: 0 = (2x + y)/(2y - x)
2x + y = 0
y = -2x
Subbing y -> Original equation
x^2 - 4x^2 - 2x^2 = 5
-5x^2 = 5
x^2 = -1
Oh dear. It seems that it's never horizontal? I may be horribly wrong lol.
For vertical, undefined means denominator = 0.
Therefore, 2y - x = 0
x = 2y
Subbing x -> Original equation
4y^2 - y^2 + 2y^2 = 5
5y^2 = 5
y = +-1
When y = 1, x = 2 and -3
When y = -1, x = 3 and -2
And so tangents are never horizontal and vertical at (2, 1), (-3, 1), (3, -1) and (-2, -1)
 
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Trebla

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The points (1, -3) and (-1, 3) do not have vertical tangents because they do not satisfy the condition x = 2y. This condition as well as the original equation must be satisfied in order to have an undefined gradient.
 

youngminii

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Ooh I never knew that :D My bad. Lol should've just subbed in y = +-1 into x = 2y
 

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