VCE Methods Question Thread!

[polar]

 

why does solving give the answer?

yeah, what darklight said

if your question meant how did I know that solving gives the minimum and not a maximum - the function is a parabola with a minimum, when you square root the function, it doesn't change the fact it has a minimum (note: its minimum wasn't on the x-axis)

[shaiga95]

thanks I understand now

[PsychoM]

Hey i have a ques

Find the coefficient of y^4 in the expansion of ( y + 3)^3 (2 − y)^5.
The MQ textbook is pretty confusing. Is there another faster way to approach this?

[Limista]

Hey i have a ques

Find the coefficient of y^4 in the expansion of ( y + 3)^3 (2 − y)^5.
The MQ textbook is pretty confusing. Is there another faster way to approach this?

Did the MQ textbook expand (y+3)^3 using the (a + b)^3 rule and expand (2 - y)^5 using the binomial theorem, before multiplying all the terms together?

To be honest, I don't think you'd be very likely to get this question on the non-calc exam or SACs, since the binomial theorem isn't strictly part of the methods study design anymore. You could, however, get it as a calc allowed question.

I'm not aware of a faster way to solve the above problem.

[polar]

I forgot you could just expand it lol

here's another way, but it's a bit longer
first, write as and use the facts that when you expand these
-the powers add up
-the coefficients multiply

now, you want all the terms, using index laws, possible combinations to get to 4 are (remembering that the maximum of the first is 3 and the second is 5) 3+1, 2+2, 2+2, 1+3 and 0+4. hence, using the binomial theorem



hence, the coefficient of the term is

[Jaswinder]

yeap just use binomial theorem and look for terms which multiply to get y^4 terms, i got -170 (-80y^4+720y^4-1080y^4+270y^4)

[PsychoM]

Did the MQ textbook expand (y+3)^3 using the (a + b)^3 rule and expand (2 - y)^5 using the binomial theorem, before multiplying all the terms together?

To be honest, I don't think you'd be very likely to get this question on the non-calc exam or SACs, since the binomial theorem isn't strictly part of the methods study design anymore. You could, however, get it as a calc allowed question.

I'm not aware of a faster way to solve the above problem.

they use binomial theorem for both (y+3)^3 (2-y)^5 which seems a bit risky if I suddnely plug in a wrong number.
Okay thanks. But that's my only first chapter holiday hmw  and I have just started only.

I forgot you could just expand it lol

here's another way, but it's a bit longer
first, write as and use the facts that when you expand these
-the powers add up
-the coefficients multiply

now, you want all the terms, using index laws, possible combinations to get to 4 are (remembering that the maximum of the first is 3 and the second is 5) 3+1, 2+2, 2+2, 1+3 and 0+4. hence, using the binomial theorem



hence, the coefficient of the term is

Thanks polar. Great explanations! I totally get it now

yeap just use binomial theorem and look for terms which multiply to get y^4 terms, i got -170 (-80y^4+720y^4-1080y^4+270y^4)

Thanks Jaswinder. I get it now

[~T]

Question 11b) in the Chapter 1 review of the Essentials textbook...
"Use the geometric results CY = CX and AX = AZ to find an expression for r in terms of x"


CA is the hypotenuse of a right angle triangle ABC, with a point X somewhere along it. AB is x cm long, CB is 3 cm long.

In the previous part, we find that AZ = (x-r) and CY = (3-r)... Hence AX = (x-r) and XC = (3-r), and AC = (3 + x - 2r). Now we just use pythag as



















Awesome... Except, the answers say only minus, instead of plus/minus. Which totally makes sense looking at the picture, if x=0 then r=0 or r=3, and it totally couldn't be 3. My question is, where did I go wrong in the maths??? Why can't I have plus/minus?

Thank you

[Scooby]

Can someone explain the difference between even and odd functions?

[Bad Student]

In an even function, f(-x)=f(x).
In an odd function, f(-x)=-f(x).

[Planck's constant]

Question 11b) in the Chapter 1 review of the Essentials textbook...
"Use the geometric results CY = CX and AX = AZ to find an expression for r in terms of x"


CA is the hypotenuse of a right angle triangle ABC, with a point X somewhere along it. AB is x cm long, CB is 3 cm long.

In the previous part, we find that AZ = (x-r) and CY = (3-r)... Hence AX = (x-r) and XC = (3-r), and AC = (3 + x - 2r). Now we just use pythag as



















Awesome... Except, the answers say only minus, instead of plus/minus. Which totally makes sense looking at the picture, if x=0 then r=0 or r=3, and it totally couldn't be 3. My question is, where did I go wrong in the maths??? Why can't I have plus/minus?

Thank you

You done nothing wrong but you have to apply some geometric constraints.
Easiest way to see it is to multiply your final equation x 2.

LHS is now 2r

Looking at the diagram, ask yourself the question, how big can 2r get? Clearly no bigger than x, and you can see this by rotating AC in your diagram anticlockwise until it's almost parallel to BC, and it can't go any further else no triangle.

Therefore the RHS of your equation (after the x2 multiplication) must be < x, and the rest follows.

[BubbleWrapMan]

Awesome... Except, the answers say only minus, instead of plus/minus. Which totally makes sense looking at the picture, if x=0 then r=0 or r=3, and it totally couldn't be 3. My question is, where did I go wrong in the maths??? Why can't I have plus/minus?

Thank you

In geometric situations like these you'd have to infer from the context whether you can have plus or minus. For a plus or minus situation, you'd have to ask yourself whether r is a function of x - that is, should each value of x give a unique value of r? If that's the case, you can only have either plus or minus, and you'd have to decide from given information which of the two options is correct. Which is basically what you did.

Another way would be to use the fact that 3 + x - 2r > 0 (since it is a length), and this means that r < (x+3)/2. Since only the minus option satisfies this, it must be the correct one.

[~T]

Thank you both Calvin Climb and Planck's constant! I could see how it wouldn't work with plus, but I wasn't aware you simply choose one option because the other doesn't make sense. I figured there must be a mathematical way, as there is always a mathematical way! And, Calvin Climb, you provided with r < (x+3)/2.

Thank you again

[vashappenin]

Hey can I please get some help on these two questions? Thanks

[~T]

For the inverse, because it is a 3x3 matrix, you aren't expected to find it without technology. So type the matrix into your calculator and raise it to -1 to find the inverse.

For the inverse to not exist, the determinant must equal 0. You can either find the determinant using your calculator and solve it to equal 0, or you can have a look at the elements in the inverse matrix you found before. Ask yourself, when will these elements be undefined? Ie, for what values of a are the elements dividing by zero?

You'll find there will be two values. For the next question, the way I would do it is to solve the equations with your calculator using both values of a that we found. One will give you the result "false", meaning the equations have no solution, and one will give you an answer in terms of another parameter, meaning there are infinite solutions.