VCE Methods Question Thread!

[ashoni]

3. The factorisation of 2b^3+a, where a = kb and k is an even whole number. What can it be written as?

The answer is:
2 (b+cuberoot(a/2))(b^2-5b+(cuberoot(a/2))^2

Confused guys... working out would be great!

[ashoni]

Hmm, another question... lol fairly long this one

7). The new personnel manager at the museum determines that the number of people P (in thousands) that visit the museum each month is given by P(x) = -0.25x^3+ax^2+21.25x+b where x represents the end of the given month. For example x=1 represents the end of January.
(a) Find the values of a and b if 42,000 people pass through the doors in January(i.e (1,42)) and 132,000 visited during May.
(b) Find the number that visited in August.
(c) Find the range of the function if x is an element of (0,12]
(d) During which month did the attendance drop dramatically?
(e) If there are fewer than 100,000 visitors during a month then management cuts back the number of employees rostered on. When does this occur?

Help with this question would be amazing! Sorry about how long it is though...

[Hutchoo]

a) Use simultaneous equations:
(x,y) -> Sub in (1,42) (i.e x = 1, y =42) for equation 1.
Sub in (5,132) for equation 2.

That is :
42 = -0.25(1)^3+a(1)^2+21.25(1)+b      ----(1)
132 = -0.25(5)^3+a(5)^2+21.25(5)+b       ----- (2)


Should get a = 3/2, b = 39/2.

b) Number that visited in August. P(August) = 
P(August) = 315/2 = 157.5 = 158 people.


c) P(x) =  -0.25x^3+(3/2)x^2+21.25x+(39/2), Domain restriction from (0,12].
Sub in x = 0, y = 39/2.
Sub in x = 12 therefore y = 117/2.

You can also see this via the graph:


I are le sleepy now. If I've made any mistakes, my bad, if not, then yay.
Continue the rest tmr


Edit: 1.8k posts =D

[naveed76]

Hi can anybody help me with these questions please :

A real estate investor bought a lakeside parcel of land and plans to enclose a portion of it to be used as a recreation area. He has 600 meters of fencing materials to enclose the rectangular area with one side openning to the lake. Let x represent the situation.

1.  draw the detailed diagram to represent this situation.
2. possible values that x can take?
3. express width of the rectangle in terms of x
4. find an expression for the area of the rectangle in terms x?
5. what is the area of the rectangle if x=15 meters
6 if the farmer wants an area of 30,000 m, use the quadratic formula to find the dimensions of the rectangular paddock created.
7. if the investor wants the width to be twice the length, find the dimensions and the final area of the rectangular area formed

Thanks.

[#1procrastinator]

Oh yes, forgot difference of two squares...so in general, you'd root the side with the variable? hypothetically if have something you can't factor

i wouldn't remember it like that, there's no set way of solving inequalities, sometimes you just play around, sometimes you need a bit of wishful thinking, but in this case, factorisation was needed.

How do you derive the combination formula? I'm having a hard time trying to understand intuitively why it works

see attachments, taken from Art and Craft of problem solving, combinatorics chapter.

Thanks, I'll think about it

Big respect for the attachments too TT   

[CommanderElahi]

Hi, I really need help on this extended question, if some one could help I'd be really grateful.

For
f: R+ ->R, f(x) 2e to the power -x  and g(-infinity,2) -> R, g(x)= 1/x-2

 1. State the range of f and g
fin inverse of f and g and state the domains

2.Find g o f and sketch graph

3. Find (g o f) inverse-1

4.Sketch the graph y= (g o f)to the power -1 x


Thanks homes.

[Mr. Study]

Disclaimer: Some bits of this question, I was unsure about. If bits and pieces are wrong, Please point it out! I will learn aswell.

f: R+ ->R, f(x) 2e to the power -x  and g(-infinity,2) -> R, g(x)= 1/x-2

1. State the range of f and g
fin inverse of f and g and state the domains

Lets look at a visual of f: R⁺ ->R, f(x)=2e^-x. I am not too sure about this domain but I do believe its really all positve x-values

.

That is with the domain restricted, as it is R⁺, meaning positive values of x, hence why their is no graph on the left side of the y-axis.

Now we can see and find the range of f(x), which is the y-values for which this function exists. Hence, it is [2,0). We are not including zero as if we did, the function could not exist. It should be a closed circle, (I think), as we can include 2 and the function will still exist.

Now to find the inverse of this function, we can do this:

x=2e^-y
x/2=e^-y
log_e(x/2)=-y
y=-log_e(x/2)

I 'switched' the x and y values around and made y the subject.
Just incase, if you wanted the range of the inverse function, it will be the domain of the original function. So it will be R⁺->R. Also, If you wanted the domain of the inverse function, it will be the range of the original function, so (2,0).

Now for g(x)
Visual Representation:



If you noticed the original equation, it was just a hyperbola, translated 2 units in the positive x-direction. However, the domain was restricted to (-infinity, 2). Hence why the 'top right corner' of the parabola 'is gone'.

Now the range, It will be (-infinity,0). It cannot exist at zero and you can't really define a function at infinity, Hence why we have round brackets.

Inverse
Same as before, switch x and y over.

x=1/(y-2)
y-2=1/x
y=1/x + 2.

Now the domain of this inverse function will be the range of the original function. Hence, it will be (-infinity,0)

2.Find g o f and sketch graph

This means g(f(x)), So for all values of the function g(x), we will substitute it with f(x).

g o f= 1/ (2e^-x-2)

To sketch it, will should consider if can even exist!
To make sure it exists, the range of f(x) must be a subset of the domain g(x).

So is (2,0) a subset of (-infinity, 2)?

I don't think so... Some check this one!!!

It is!

Hence, we can sketch this function on our graphics calculator... or desmos.



Hopefully thats correct.

3. Find (g o f) inverse-1

Same as before.

x=1/ (2e^-y-2)
2e^-y-2=1/x
2e^-y=1/x + 2
e^-y  = (1/x+2)/2
-y    = log_e((1/x+2)/2)
y= -log_e((1/x+2)/2)

4.Sketch the graph y= (g o f)to the power -1 x

I think this is what you're asking.



I'm not too good with desmos and I have no idea if the bit says ^-1x=-x or it is actually ....-1x

Could you elaborate on this question? I'm getting confused now. I THINK you're asking to sketch the inverse graph but... with that x....

[CommanderElahi]

Mr Study,  you explained it really well, thanks for taking time to help me out. (:
You know your stuff hahah!

[Insa]

Hey guys,

Can anyone help me out on this question?

Find two integer values of p such that the quadratic equation has two rational roots.

Thanks.

[rife168]

Hey guys,

Can anyone help me out on this question?

Find two integer values of p such that the quadratic equation has two rational roots.

Thanks.

I have a feeling that there may be a simpler method to do this, but what the hell... I like this way.



From the quadratic equation:

we can see that for a root to be rational, must be a perfect square.
We can express this in the following way:

So therefore






Now let's see what happens when k=1


so then

That's one pair of integer values but using the same method (trying different values for k) then you can find more solutions
i.e. k=5 gives the solutions

[kamil9876]

Here is how you can find all such solutions, picking up where fletch-j left off:






So really you're just looking for all possible factorizations of into integers. An even quicker way of finding all of them, let , then the equation is . So really you just have to find all factorizations of 24 into two numbers whose difference is even. But of course at least one factor is even, so really you're just looking for all possible factorizations of 24 into two even numbers.

[yawho]

Hey guys,

Can anyone help me out on this question?

Find two integer values of p such that the quadratic equation has two rational roots.

Thanks.

I have a feeling that there may be a simpler method to do this, but what the hell... I like this way.



From the quadratic equation:

we can see that for a root to be rational, must be a perfect square.
We can express this in the following way:

So therefore






Now let's see what happens when k=1


so then

That's one pair of integer values but using the same method (trying different values for k) then you can find more solutions
i.e. k=5 gives the solutions

Can you prove 4 values only or more?

[rife168]

Hey guys,

Can anyone help me out on this question?

Find two integer values of p such that the quadratic equation has two rational roots.

Thanks.

I have a feeling that there may be a simpler method to do this, but what the hell... I like this way.



From the quadratic equation:

we can see that for a root to be rational, must be a perfect square.
We can express this in the following way:

So therefore






Now let's see what happens when k=1


so then

That's one pair of integer values but using the same method (trying different values for k) then you can find more solutions
i.e. k=5 gives the solutions

Can you prove 4 values only or more?

More can be found. See Kamil's post above.

[#1procrastinator]

(hopefully the last I'll ask about quadratic inequalities lol)

When you solve a quadratic inequality like x^2-x-6, you split it into two cases and
then combine the solution sets right? If one of the solution sets (for the above) is
x < -2 and x > 3, why doesn't it make sense to say that the set is (-∞, -2) U (3, ∞)
(aside from the fact that simple inspection will tell you it's wrong)?

----

And here's what kind of led me back to ^

Solve for |2x+6| - |x+3| = |x| for x

I think I've got the gist of how to solve these...
(=> is equal to greater than, lol, sorry don't have my symbols handy)

For |2x+6|
2x + 6 if x ≥ -3
-2x - 6 if x < -3

|x+3|
x+3 if x ≥ -3
-x-3 if x<-3

|x|
x if x ≥ 0
-x if x < 0

So if what I've gathered is right, you're supposed to find an x that fits in the intervals and then
solve the equation? e.g. I can't choose 2x+ 6, x+3 and x because if x can't be
greater than or equal to both 3 and 0

But what about x is less than -3 and x is less than 0? If it's less than -3 then it's also less than 0
but the equation has no solution.

Thanks

[dinosaur93]

Could some help me with differential calculus?

Show working...

1. Find the exact value of h'(2) if ?

2. Find the derivative of .

3. Through quotient rule, find the derivative of ?