VCE Methods Question Thread!

[LPadlan]

Use the ration root theorem to help factorise each of the following:
12x^3+20x^2-x-6

[Sine]

Zeroes are the same as roots (basically x- intercepts)

So "simply" it means that a potential factor will always be p/q where p is a factor of the final coefficient and q is a factor of the first coefficient (given the terms go from x³ to x⁰ = 1 (so disappears leaving a constant term in your case -6)


So let's list out the potential zeroes/roots
p is a factor of -6 which means it can be  ±1,  ±2,  ±3,  ±6

q is a factor of 12 which means it can be  ±1,  ±2,  ±3,  ±6,  ±12

p/q =  ±1,  ±1/2,  ±1/3,  ±1/6,  ±1/12 (when p = ±1)
          ±2/1,  ±2/2,  ±2/3,  ±2/6,  ±2/12  (when p = ±2)
          ±3/1,  ±3/2,  ±3/3,  ±3/6,  ±3/12   (when p = ±3)
         ±6/1,  ±6/2,  ±6/3,  ±6/6,  ±6/12   (when p = ±6)

Since I have done it systematically I have got some repeats - let's find all the possible unique solutions now       

p/q =  ±1,  ±2,  ±3,  ±6,  ±1/2,  ±3/2,  ±1/3,  ±2/3,  ±1/6,  ±1/12,  ±1/4

p/q = ±1/12,  ±1/6,  ±1/4,  ±1/3,  ±1/2,  ±2/3, ±1,  ±3/2, ±2,  ±3,  ±6 (just ordered them not necessary)

Now these are the solutions we must "guess" - tedious i know

Now let's define the expression you were given

f(x) = 12x³+20x²-x-6
So say we start checking these solutions and first find that f(1/2) = 0 This means (x - 1/2) is a factor so from there we can utilise long division by dividing the original cubic by this factor to find a quadratic factor of the cubic. Remember we need to check solutions until we find one that is f(p/q) = 0.

If we did this correctly we end up with

Factorising this quadratic correctly yields

So finally we get


And finally state our answer as required - remember we defined f(x)
So

In it's factorised form is


I doubt this would ever come up on an exam and it never has from the past 20 anyway!

[geminii]

I'm havnig trouble with this question...

f(x) = (x+1)^2(x-1)
Find the values for which f(x) = p has
i. one
ii. two
iii. three solutions, where p is a real number.

[Sine]

I'm havnig trouble with this question...

f(x) = (x+1)^2(x-1)
Find the values for which f(x) = p has
i. one
ii. two
iii. three solutions, where p is a real number.

Graph it on CAS or by hand to give a better understanding of my solution - whilst labelling turning points


Let's use CAS to find the deririvite

Now let's find the x-values of the turning points


he y-values of the turning points


Now imagine a horizontal line (parallel to the x-axis) cutting the graph if it cuts the graph once it has 1 solutions, cuts the graph twice it has two solutions and thrice - 3 solutions.
For 1 solution it will be above the turning point on the left or below the turning point on the right and hence cutting the graph once
thus p E (∞, -32/27) U (0, ∞ )
Two solutions will be when it cuts the graph hitting the turning point
p = 0, -32/27
Three solution cutting the graph between the y-values of the turning points
p E (-32/27, 0)

[Max Kawasakii]

I'm havnig trouble with this question...

f(x) = (x+1)^2(x-1)
Find the values for which f(x) = p has
i. one
ii. two
iii. three solutions, where p is a real number.

It's asking when the line y=x intersects the cubic once, twice and three times, essentially asking you for what values of P will satisfy this.

To answer this you essentially need to know the turning points of the graph which for your graph are (-1,0) and (1/3,-32/27)

f(x)=p intersects the line once, think to yourself if I drew a horizontal line through the Y axis, where would it only intersect the Graph once?
 Answer is; (-Infinity,-32/27)  U (0,Infinity) = when P < -32/27 or P > 0

f(x)=p intersects twice, same as above, when does a horizontal line (P) intersect the cubic twice, this happens when the the line P toches the turning points. so when P = 0 or P = -32/27

f(x)=p intersects 3 times when the horizontal line P intersects the the cubic between your turning points, so f(x)=P has 3 solutions when P < 0 or P > -32/27 or when -32/27 < P < 0

i) P < -32/27 or P > 0
ii) P = 0 or P = -32/27
iii) -32/27 < P < 0

I don't know how to use a lot of features on here but hopefully this helps somewhat.

[Max Kawasakii]

Find the derivative of (3x^2 - 2x)^3 with respect to x.

I can do this on CAS with ease but the question is actually from a short answer (Tech free) test, I can't seem to get the answer which is

= 6x^2(3x-2)^2(3x-1)

I keep on getting 6(3x^2-2x)^2(3x-1), thanks in advance for any help.

[Sine]

Find the derivative of (3x^2 - 2x)^3 with respect to x.

I can do this on CAS with ease but the question is actually from a short answer (Tech free) test, I can't seem to get the answer which is

= 6x^2(3x-2)^2(3x-1)

I keep on getting 6(3x^2-2x)^2(3x-1), thanks in advance for any help.

you need to use the chain rule - or colloquially differentiate the outside and differentiate the inside
let








but we just made up what y = so

Check CAS for equivalence! 

[geminii]

Thanks Max and Sine! I knew the coordinates of the turning point and I had already drawn the graph, I just didn't know how to use this to find the answer. Thanks again!

[TheCommando]

In f(x)=cos(x-pie/3)
Is f(0)=1/2 as -pie/3 is in the fourth quadrant? why is it positive

[Max Kawasakii]

Check CAS for equivalence! 

Wow thanks for that! I was getting your answer via chain rule as it turns out, could you please show me how our answer;



is equivalent to the back of the book answer of;

[Max Kawasakii]

In f(x)=cos(x-pie/3)
Is f(0)=1/2 as -pie/3 is in the fourth quadrant? why is it positive

Do you know CAST

Quadrant 4; Cos is Positive
Quadrant 1: All are positive
Quadrant 2; Sin Is positive
Quadrant 3, Tan is positive.

Is this what you are asking? There is a mathematical way to remember this as well, I find CAST more simple in most circumstances.

[TheCommando]

Yes i do
So its positive cause its in the  4th quadrant as the minus in front of pie/3 makes it go anticlockwise

How do i silve the following equation for theta with a dmain of [0,2pie]
For sin(2theta)=-.08

And what is f(0) for sin2(x-pie/3)

Mod Edit [Aaron]: Double post merged. Please edit your posts in future unless you can justify a double post.

[zhen]

Wow thanks for that! I was getting your answer via chain rule as it turns out, could you please show me how our answer;



is equivalent to the back of the book answer of;

This is how I would simplify it so that it's equivalent to the book answer.

[Sine]

Wow thanks for that! I was getting your answer via chain rule as it turns out, could you please show me how our answer;



is equivalent to the back of the book answer of;

Take a factor of 2 out from the 6x-2

Factor out an x from the quadratic

Square the quadratic and x separately

Factor out the x squared term


beaten by zhen but I have a few intermedaite steps which may be useful! 

[MattBro]

Hi,
What is the simplest method to use to sketch sine and cosine graphs with translations?
Thanks