# SAT II Math IIC Questions

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1. Aug 11, 2007

### lLovePhysics

Here are some SAT II Math IIC Questions that I have. Fun!

I'm in the chapter of "Higher-Degree Polynomials" right now and I have some questions concerning some facts:

How do these work?:

1) "The Remainder Theorem- If a polynomial P(x) is divided by x-r (where r is any constant), then the remainder is P(r).

So does this mean if you choose any number, for example, 5 and then divide a polynomial (lets say $$x^{4}+x^{3}+3$$) by x-5, the remainder will always be P(5)??

Is it beneficial to test these kind of things? In order to get these kind of problems, right you really need mathematical acumen right? Should I test these out to "prove" them to myself? I've tried with this one but I ended up messing up on long division =/

Also, if a polynomial has rationa coefficients, then will it always have irrational zeros? and if a polynomial has real coefficients, does it always have complex zeros? I know that both irrational zeros and complex zeros have conjugate pairs.

Are Irrational numbers most often seen as radicals or rationals or both? Thanks for your help in advance!

2. Aug 11, 2007

### lLovePhysics

Wow, I tested the remainder theorem and it really does work. Why is this? Should I just memorize these "facts" or should I delve into them and find out why? I need to take this test in less than 2 months so what do you say?

3. Aug 11, 2007

### learningphysics

Did you do polynomial division to test the remainder theorem? Polynomial division is like arithmetic division...

eg: When you divide 7 by 2... you get 3 remainder 1. That means that 3*2+1 = 7, note that the remainder is always less than the divisor.

same way with polynomial division... suppose when you divide P(x) by Q(x) you get G(x) remainder H(x). That means that G(x)*Q(x) + H(x) = P(x), and also note that the degree of the remainder H(x) is always less than the degree of Q(x)... by degree I mean the highest power x in the polynomial. eg: the degree of $$x^5+3x^3+2$$ is 5.

So suppose Q(x) is of the form x-r, then the remainder H(x) will have a lower degree therefore it is a constant, call it a.

G(x)*(x-r) + a = P(x)

So calculate both sides at x = r... left side becomes a right side becomes P(r), so the remainder a = P(r)

4. Aug 11, 2007

### lLovePhysics

I'll try to figure that out by reading what you've said again.

I have another type of problem that I know how to solve, but I don't know if it is the "best" and a correct way to solve it.

Problem: If x varies as y and z^2, and inversely as the square root of w, what is the effect on y when x is doubled, z is halved, and w is multiplied by 4?

My Solution: I wrote out the relationship like: $$\frac{x*\sqrt{w}}{y*z^2}= K \\$$where k is a constant

and then I performed the procedures stated in the problem:

$$\frac{2*2}{(\frac{1}{2})^2}$$

After this I simplified and got 4/ (1/4) and I noticed that y needs to be 16 in order to balance out the equation to make it equal 1. Does this method always worth with these type of variations/relationship problems? Thanks.

5. Aug 11, 2007

### learningphysics

$$\frac{x*\sqrt{w}}{y*z^2}= K \\$$

but to make it slightly more rigorous and I think you'll be less likely to make a mistake... I'd do this... solve for y first since your question is asking what happens to y.

so $$y = \frac{x*\sqrt{w}}{K*z^2}$$

so this is a function where x,y,z,w can take on different values... let y1 be the value of y, when x=x1, z=z1 and w=w1

so $$y_1 = \frac{x_1*\sqrt{w_1}}{K*z_1^2}$$

now let y2 be the y value when x is doubled , z is halfed and w is multiplied by 4... so in the formula I put in 2*x1 insted of x1, (1/2)*z1 instead of z1 etc...

so $$y_2 = \frac{(2*x_1)*\sqrt{4*w_1}}{K*(\frac{1}{2}*z_1)^2}$$

if you simplify and bring the constant out to the left... you'll get:

$$y_2 = 16* \frac{x_1*\sqrt{w_1}}{K*z_1^2}$$

so $$y_2 = 16*y_1$$

so y gets multiplied by 16.

6. Aug 11, 2007

### lLovePhysics

Is there a good site that teaches you Probability really well? I'm really confused about probablity sometimes. Thanks again!

7. Aug 11, 2007

### lLovePhysics

Ah I see. That is a very good method also! I will try to practice that more

8. Aug 12, 2007

### lLovePhysics

I also have trouble on this problem:

If the probability that John will buy a certain product is 3/5, that Bill will buy that product is 2/3, and that Sue will buy that product is 1/4, what is the probability that at least one of them will buy the product?

My Solution:

I just multiplied all of their probablities that they will buy the product together because they are independent events so:

(3/5)*(2/3)*(1/4)= 1/10

However, my answer is wrong but I do not know why it differs from the book's solution. The book tells you to consider the probability that none of them buys the product, which turns out to be 1/10. Then you need to subtract that from 1 to get 9/10. Why doesn't my solution work and why does it turn out to be the number in which none of them buys the product?

9. Aug 12, 2007

### Curious3141

What you have here is the probability that John buys AND Bill buys AND Sue buys. What you need is the probability that ONE OR MORE of them buys.

The list of cases you need to consider to do that would go :

OR

OR

OR

OR ...
...

See how exhausting it is to do it the long way?

The simpler way is to realise that the complementary case to at least one of them buying (=one or more buying) is NONE of them buying. That's easy to work out as

(2/5)*(1/3)*(3/4) = 1/10

and since you want the complementary event to that, the required probability is one minus that = 9/10.

Coincidence.

10. Aug 13, 2007

### lLovePhysics

Oh ok, I see now. So do I use the same strategy when I encounter these type of "at least one" problems? Generally, I would find the complementary "None" and then subract that by 1 to find the "at least one"?

It is kind of hard to visualize or comprehend how subtracting the probability of that none would do something by 1 would give a probablity that *at least one* would do something. It seems more like.. ALL would do something. They just don't seem that complementary to me. =/

~~~~~~~~~~~~~~~~~~~~

Also, can I graph this function: $$y=3 \sin 2 (x+\frac{2\pi}{3})$$ right onto my HP 50g in this mode?: $$y=3 \sin (2(x+\frac{2\pi}{3}))$$ I've tried it and it works. I thought the 2 isn't part of the parentheses and can never be. Why does it work when I insert it on my graphing calculator?

You surely can't change $$y=3 \sin 2 (x+\frac{2\pi}{3})$$ into $$y=3 \sin (2(x+\frac{2\pi}{3}))$$ when you try to solve it by hand right? Don't you need to use the Double-Angle formula? or.. can you do it that way?

11. Aug 13, 2007

### lLovePhysics

Another problem:

If $$(x-4)^{2}+4(y-3)^{2}=16$$ is graphed, the sum of the distances, form any fixed point on the curve to the two foci is.

Books solution: It tells me to find a (the distance from the center to one of the vertices) and then multiply that by 2 to find the distance. Is this correct? Is it a specific forumla (i.e. how can I find it?)? Thanks.

12. Aug 13, 2007

### Curious3141

Yes, although you should verify the wording of the problem to make sure you've read it right.

13. Aug 13, 2007

### lLovePhysics

I mean like: 1- "None"= "At Least one" sounds kind of weird.

When I first encountered this type of problem I thought that: 1- "None" = "the total probability" or something, not just "at least one."

I can't find a way to comprehend it fully... =/

14. Aug 13, 2007

### Curious3141

Venn diagram. Makes it clearer.

15. Aug 13, 2007

### learningphysics

A way to think about the complementary probability thing... If an event A happens 2/3 of the time... then how often does it not happen? The rest of the time... which is 1-2/3 = 1/3 of the time...

If it is not the case that nobody is buying... that's exactly the same as saying that at least one person is buying... at least one means greater than or equal to 1... so everything except 0...

A way to categorize the the possibiliites is:

All events in your question fit into these 4 categories... so The probability that 0 people buy + the probability that 1 person buys + prob that 2 people buy + prob that 3 people buy= 1

Do you see that the probability that at least one person buys = probability 1 person buys + probability 2 people buy + probability 3 people buy ?

ie: everything except the possibilty that 0 people buy.

I agree with Curious that a venn diagram can make it clearer.

16. Aug 13, 2007

### learningphysics

Not sure I understood you fully with regards to the graphing... I agree that it is ambiguous without all the parentheses... $$y=3 \sin 2 (x+\frac{2\pi}{3})$$ I think is usually interpreted as:

$$y=3 \sin (2 (x+\frac{2\pi}{3}))$$

or I think more rarely it could be interpreted as:

$$y=3 (\sin 2) (x+\frac{2\pi}{3})$$

17. Aug 13, 2007

### learningphysics

yes, that works, but you need the vertex on the major axis... ie it's the length of the major axis you're looking for.

If you rewrite the equation of the ellipse in the right form, it will make it easier to see what you need...

18. Aug 13, 2007

### lLovePhysics

Thanks for all of your help guys! I really appreciate it!

I'm kind of confused about inverse fucntions. My text says: "The inverse of any function f can always be made a function by limiting the domain of f"

So if you limit the domain of f, you are basically limiting the range of $$f^{-1}$$ right? The inverse of any function always has the same values of the function's domain and range except that they are switched?

Also, is the universal procedure for finding inverses swapping the x and y in the function and then solving for y? (except trig functions I think) and do you always keep the magnitudes of the domain and range the same but swap them?

Added: My text also says, "A function must pass the horiz. line test to possess a valid inverse." Does it mean inverse function because even if a function doesn't pass the horiz. line test it could still have an "inverse" right?

The horizontal line test is basically the one-to-one test? I just want to clear this up: If a function passes the horiz. line test it has an inverse function because the test-every y, only has one x *turns into* every x value only has one y value because the x and y are switched? So basically inverses have all properties of x and y switched, since you switched the x and y in the first place to find the inverse right? Anything involving the independent variable x and dependent variable y are swtiched, the domain, range, points, etc... Is this 100% correct?

Last edited: Aug 13, 2007
19. Aug 13, 2007

### lLovePhysics

So basically 2 x the major axis = the distance from the two foci to one point on the ellipse? Is there any proof or way of verifying this visually/analytically? Or should I just memorize the relationship?

20. Aug 13, 2007

### learningphysics

Hmm... that's not exactly right. Does your text distinguish between codomain and range?

If you presume that the codomain is the range of the function, then what your text says is right... because that automatically makes the function onto...

Otherwise, f may not be an onto function. In which case you'll have to restrict the codomain also to make $$f^{-1}$$ a function.

Yes, exactly...

do you see why you need to restrict the domain of f? suppose f is a many to one function... there's no inverse function for f because in the inverse relation one element maps to many... and that's not a function...

draw a mapping to see this...

yes, that's the procedure as long as the function is invertible... what do you mean by magnitudes of the domain and range?

Last edited: Aug 13, 2007