VCE Specialist 3/4 Question Thread!

[keltingmeith]

Alternatively, use an addition formula. sin (x-y)=sin(x)cos(y)-sin(y)cos(x)

[IndefatigableLover]

Chain rule. Let u=x-y

Just checking my working (if I set it out properly):



So u=x-y



(do to implicit differentiation of u=x-y).

So the final differentiate form for the left hand side is:



EDIT: Just saw your post EulerFan101!
Might try using that method and see how comfortable I am with it

EDIT 2: Just tried your method EulerFan101 and it's nice but the amount of mistakes I made (careless) before I arrived at the final answer hurt my head >.>

[IndefatigableLover]

Has anyone used Specialist Maths Dimensions (written by Dr. G) before?
Is it worth going through some of the questions in the textbook for extra revision because I have heard that the textbook does hold some difficult questions..

[pi]

Yeah, I thought it had some decent questions. The CAS tips are notably ancient though haha.

[IndefatigableLover]

Yeah, I thought it had some decent questions. The CAS tips are notably ancient though haha.

Nice! I guess I'll do some Revision Questions/Application Tasks from there then!
And yes the CAS tips are pretty ancient LOL (TI-84 & TI-89)

[Eiffel]

my brains just gone bit retarded.

when drawing/sketcing rays such as Arg ( z -(2+i)) < pi/4, i know theres an open startpoint at 2,1. and the ray inclined at pi/4 radians with respect to the x axis, btu when shading why do we shade all the way to -pi and stop there? (or in this case -pi in respect to the inclination - which is -pi + 1 unit vertically if that makes sense)?

also, we do we need to shade inside the ray?

[pi]

Think about the definition of Arg as compared to arg

[Eiffel]

i was thinking that, but also thought it may have been do to something else. just confirming you are talking about the (-pi, ..

[Eiffel]

Quadrilateral ABCD is circumscribed by a circle. The tangent to the circle at B meets the extension of AD at the point E. given that angle BCA = 30 degrees, and angle ACD = 70 degrees, find the magnitude of angle BEA

[knightrider]

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This is

(Image removed from quote.)

is the upper angle, is the lower angle. refers to the angle formed by adding those two angles together.

So, the letter in the middle tells you at which point the angle forms. The first and last letter tell you how the angle extends out.

Make sense?

Thankyou so much eulerfan101

It makes sense.

[alchemy]

There's a question which asks to find all solutions of the equation z^3 - (2-i)z^2 + z -2 + i , given that 2-i is a solution.

I straight away wrote 2+i down as the first solution, because of the conjugate factor theorem stating so, before proceeding to find other solutions. However, it doesn't have 2+i given in the answers as one of the solutions...
How come?

[cooldude123]

There's a question which asks to find all solutions of the equation z^3 - (2-i)z^2 + z -2 + i , given that 2-i is a solution.

I straight away wrote 2+i down as the first solution, because of the conjugate factor theorem stating so, before proceeding to find other solutions. However, it doesn't have 2+i given in the answers as one of the solutions...
How come?

Note that the conjugate root theorem which states that 2+i is a factor only applies when all coefficients of the polynomial are real, and since z^3 - (2-i)z^2 + z -2 + i has coefficients that imaginary, it doesn't apply to this problem.

[alchemy]

Note that the conjugate root theorem which states that 2+i is a factor only applies when all coefficients of the polynomial are real, and since z^3 - (2-i)z^2 + z -2 + i has coefficients that imaginary, it doesn't apply to this problem.

Right. I need to read the finer details of a theorem next time...

[IndefatigableLover]

Pretty simple looking question but it's that 'i' that's messing with me :/
Sketch:

[cooldude123]

Pretty simple looking question but it's that 'i' that's messing with me :/
Sketch:

I think this is how you do it:







And then it should be just a simple sketching in of the sector from to