inverse trig and couting probability question (1 Viewer)

[saa]

Hi,

I'm looking for answers to these questions if someone could help out, as I'm a bit rusty and am helping a friend..
(arcsin = sin^-1 doesn't it?)

Domain and Range of:
xcos^-1x (xarccosx)
xsin^-1x (xarcsinx)
mainly range I got stuck on.

Derivitive of:
tan^-1(1/x) + tan^-1x (arctan(1/x) + arctanx)

Probabilities involving the word ENTERTAINMENT:
a) 2 e's together and 1 apart
b) all e's apart
c) all e's together (I can get that one I think. 11!3!/13!)
d) e's at each end (I got this one too. 6*11!/13!)

cheers

 

[mojako]

domain of acos(x), or arccos(x) I think they're both valid.. arccos might be the fuller term:
-1 <= x <= 1 (just punch ur calculator to check)
the domain of asin(x) is the same
range of acos(x) is 0 <= acos(x) <= pi... again, punch ur calculator.. find acos(1) and acos(-1)
range of asin(x) is -pi/2 <= asin(x) <= pi/2.. also check with calculator
Also, as a side note, the graph of asin(x) doesnt look like x^3. near the origin the gradient of asin(x) is 1, and near the endpoints the tangent tends to vertical
in x^3 the curve around origin is flat because gradient = 0



tan^-1(1/x) + tan^-1x (arctan(1/x) + arctanx)
Is that an x (=ex) there?
If the expression is tan^-1(1/x) + tan^-1(x) [tan^-1(1/x) + tan^-1(x)]
then use the product rule
If its not an x and the expression is tan^-1(1/x) + tan<sup>-1</sup[tan^-1(1/x) + tan^-1(x)], I think you can still use the chain rule (treat the thing inside [..] as a pronumeral)
Anyway, the derivative of tan^-1(1/x) or tan^-1(x^-1) is
1 / [1^2 + (x^-1)^2] * derivative of x^-1

Probabilities (I think):
(a) favourable outcomes = 2 E's are together - 3 E's are together
12!*2! - 11!*3!
(b) fav outcomes = all possible outcomes - 2 E's are together (at least 2.. can be 3 as well)
13! - 12!*2!
Do I get it right??

 

[saa]

thanks mojako, but for the domain/range questions I was actually after D/R of:

x*acos(x)
x*asin(x)

(I probably should've made it clearer)


oh and with the tan one, I wrote the bit in brackets not meaning to times the two, but to write the same thing in a different way to fully show people what I was asking..

Anyway, the derivative of tan^-1(1/x) or tan^-1(x^-1) is
1 / [1^2 + (x^-1)^2] * derivative of x^-1

that's wher I was going wrong.. didn't multiply by the derivative of x^-1
why do you do that?

 

[mojako]

oh sorry... its just ive never seen anyone asking the domain of x*acos(x)
but anyway, the domain is the same as that of acos(x)
now, the range of acos(x) is between 0 and pi right? (0 when x = 1 and pi when x = -1)
so, if its x*acos(x), it'll be between 1*0 and -1*pi
its between -pi and 1.. WAIT.. its not!!
you need to use calculus to find the range... differentiate x*acos(x) and find the stationary point and its nature (max, min, horiz inflexion). you'll find its a max. suppose it occurs at x=x1
you can now find x1*acos(x1) and this will be the upper limit of the range.
now, the lower limit is -pi from what you did before. why is it definitely -pi? becoz theres only 1 turning point within the domain (-1 <=x <= 1) and the endpoints of the function are (-1, -pi) and (1, 0) and -pi is the lower one.

for the tan one,
its because... the inside of the tan(..) is not simply x.
its like.. derivative of atan(2x) is 1 / (1 + (2x)^2) * 2,
the 2 being the derivative of 2x.
u do it according to the chain rule.

another example: derivative of sin(2x) = cos(2x) * 2

 

[saa]

damn, have to differentiate.. never thought of that. not normally requried for domain/range questions!

thanks heaps for your help

 

[Xayma]

and yes arcsin=sin^-1
That is arcsin x=y
is the inverse function of y=sin x

 

[mojako]

any reason whi you wrote
arcsin x=y
y=sin x
as opposed to putting the y's on the left??

(got me thinking for a few secs)

 

[saa]

oh and with the counting problems, part of the reason I was getting it wrong was because I forgot to compensate for the 3 T's and 3 N's...

so should've divided by 3!3! too

 

[mojako]

oh and with the counting problems, part of the reason I was getting it wrong was because I forgot to compensate for the 3 T's and 3 N's...

so should've divided by 3!3! too

I dont think you divide by 3!3! for probabilities..
(you do for counting [only])
can somebody explain?

 

[dr nick]

you divide by 3! to cover the number of identical arrangements you get by arranging 3 N's.

 

[Xayma]

any reason whi you wrote
arcsin x=y
y=sin x
as opposed to putting the y's on the left??

(got me thinking for a few secs)

Nope. None at all.

 

[mojako]

you divide by 3! to cover the number of identical arrangements you get by arranging 3 N's.

and you still do that for probability questions??

 

[dr nick]

(a) favourable outcomes = 2 E's are together - 3 E's are together
12!*2! - 11!*3!

12!/(3!3!) - 11!/(3!3!)

(b) fav outcomes = all possible outcomes - 2 E's are together (at least 2.. can be 3 as well)
13! - 12!*2!
Do I get it right??

11!/(3!3!)

 

[dr nick]

and you still do that for probability questions??

how many ways to arrange EEE?

one way. 3! ways if they were all different, but divide by # of identical letters.

So yes.

Probability just = count them up and divide by total number of possible arrangements.

 

[dr nick]

so, for my answers, you will have to divide by 13!/(3!3!3!) to get the probabilities

 

[CrashOveride]

Probabilities:

a) Draw up 13 little _ and put two E's together. Notice when the "block" is at either end, the other E can occupy any 10 places (not being adjacent) and others can go in 10!/(3!.3!).
Now thats the case when the block is at either end, otherwise its a similar situation only this time when its someplace in the middle (non-end etc) the extra E cant be adjacent to the double E block so it leaves only 9 places for the extra E. This can happen 10 ways so 10 [ 9. 10! / (3!.3!) ] . Now you add them up, divided by 13!/(3!)^3 and you should get 5/13.

c) Group em all together , you know the drill. P(3E) = 1/26

b) P(e apart) = 1 - (1/26 + 5/13) = 15/26

d) Put E at either end. Remaining letters in 11!/36 . Now divide that by 28828800 and 1/26.

 

[mojako]

In probabilities, you can regard identical elements as being un-identical, as "saa" has demonstrated in his/her answer in post #1, to parts (c) and (d).

CrashOveride method can also be done by treating all letters as different (all E's are "different, all N and T's are different) when counting favourable outcomes AND when counting all possible outcomes.

As for my method here:
---
Probabilities (I think):
(a) favourable outcomes = 2 E's are together - 3 E's are together
12!*2! - 11!*3!
(b) fav outcomes = all possible outcomes - 2 E's are together (at least 2.. can be 3 as well)
13! - 12!*2!
---
It was wrong, but not because we can't treat identical elements as not identical. Below is the correct one (although CrashOveride's way is much much better because it ensures that you get the picture right).
(a)
number of favourable outcomes: 12!*3! - 11!*3! - 11!*3!
the 12! is for choosing 12 groups (1 of them contains 2 E letters)
the 3! (in 12!*3! and in the other two terms as well) is for choosing which E is to be put in specific places
the first 11!*3! represents the number of outcomes where 3 E's are grouped together and the third E is to the right of the corresponding 2-E group in the 12!*3! part
the second 11!*3! represents the number of outcomes where the 3-E group is to the left of the corresponding 2-E group
number of possible outcomes would be 13!

(b)
You can substract 1 by the probabilities of (a) and (c).
Alternatively,
no of favourable outcomes = (13! - 12!*3!) - 11*10!*3!
the (13! - 12!*3!) represents the outcomes when two particular E's are not together
the 11*10!*3! represents those where the third E is next to any of the two E's; the 11 is how we can arrange the third E (which is anywhere except in the places of the other two E's), the 10! is how we can arrange the other letters

It's very hard to see without the little _ (and I had to work backwards from the little _)
So my method is not really a working method.