VCE Specialist 3/4 Question Thread!

[allstar]

can someone please help me on this question?
Forces 8N, 16N and 10N act on a particle in equilibrium. Sketch a triangle of forces to represent the three forces. How do I know which side of the triangle is what force value?
and then the angle between the forces 8 and 16?

[Robert123]

can someone please help me on this question?
Forces 8N, 16N and 10N act on a particle in equilibrium. Sketch a triangle of forces to represent the three forces. How do I know which side of the triangle is what force value?
and then the angle between the forces 8 and 16?

Presuming he question does not contain any additional information, it would not matter which side you decide each force to be on so long as the lengths are relative to the magnitude of the force.
To find the angle between the forces, use the cosine rule which is found in the mensuration section of your formula sheet.


When sketching argument graphs on the complex plane, are you meant to use an open circle at the origin?

[tiff_tiff]

Let p be a polynomial function with integer coefficients. If −5, 3 and 1− 2i are roots of p, then the
minimum degree of the polynomial is:
A. 2
B. 3
C. 4
D. 5
E. 6

please and thank you

[Bestie]

this question was from MAV
Can someone pelase go through each option and explain why C is the ans?

thank you!

[nhmn0301]

Let p be a polynomial function with integer coefficients. If −5, 3 and 1− 2i are roots of p, then the
minimum degree of the polynomial is:
A. 2
B. 3
C. 4
D. 5
E. 6

please and thank you

Since you have real coefficients, you can also have 1 + 2i as a root. So 4 is what I think. Hence C.

this question was from MAV
Can someone pelase go through each option and explain why C is the ans?

thank you!

It might take a while to go through every single answer but I'll try to skim and show you what I think. (* stands for "dot" in this explanation)
A is true because. If you draw a line from point B parallel with AC and in the same direction, it will create an angle in "tail-to-tail". Since this is a right triangle, angle BCA = 90 - theta. Also because the lines are parallel, the angle in between AC and BC is equal to 90-theta.
AC * BC = |AC|.|BC| cos (90 - theta) = |AC|.|BC| sin (theta)           
B is of course true.
For C, if you do a bit of calculation ( I guess we can work out from the diagram but I just personally don't like looking at vectors digram). 
(AC - AB) * AB = (OC - OA -OB + OA ) * AB = (BC) * AB . And you know if you extend AB to create a tail-to-tail angle with BC, that angle is 90 thus its dot product should equal 0. That's why C is false.
For D and E, you can do the same thing like what I did above.

[Robert123]

Let p be a polynomial function with integer coefficients. If −5, 3 and 1− 2i are roots of p, then the
minimum degree of the polynomial is:
A. 2
B. 3
C. 4
D. 5
E. 6

please and thank you

The answer would be 4. One thing you need to keep in mind is the 'integer' coefficient part compare to just real. If the 3 was instead something like '1+sqrt(2)', you will need to have a minimum degree of 5 with one root being the conjugate of '1+sqrt(2)' ie '1-sqrt(2)' so that the square roots cancel out when expanding leaving only integer coefficients

[tiff_tiff]

thank you everyone

but the ans says 5 not 4?

[keltingmeith]

thank you everyone

but the ans says 5 not 4?

What's their reasoning behind it?

[tiff_tiff]

please see the attached

[keltingmeith]

please see the attached

I'm guessing they meant to ask something else, more along the lines of "3 is a quadratic root of p", but they didn't, so their answer is wrong. With the wording they've given, the correct answer is 4.

What paper did you get this from?

[Robert123]

please see the attached

I think you copied the question out wrongly on here. Did it say 3 was a root or sqrt(3) is a root? If it said sqrt (3) was a root then refer to my previous explanation on why there is 5 roots (ie, due to integer coefficient you need the conjugate root of sqrt(3) to cancel out when expanding

[nhmn0301]

Hi there, just wondering how much time do you guys spend on Multiple choice ?

[lzxnl]

I took less than 30 minutes normally for multi choice

[M_BONG]

For decimal place errors (eg. if question asks to give to closest metre) and you give exact, is it capped at 1 mark lost per paper for VCAA? or is it every time you make the mistake? Same with tildas for vectors, is it one mark each time?

[M_BONG]

Ok still confused about general solutions...

This is a complex number, hence Arg Z is between - pi and pi

When I am solving for:
cos (npi/12) = 0

This is what I do,

npi/12 = pi/2 + 2k pi     k element of Z

rearranging:

npi = 6pi + 24 k pi

n = 6 + 24k.

Answer is 6 +12 k?


Isn't general solutions just getting the two angles which equal something and adding two pi to it? Or is it different for complex numbers because of the restriction on the argument?