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Author Topic: Additional Math Help HERE ONLY...!  (Read 76127 times)

Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #300 on: June 07, 2010, 03:46:58 pm »
Quote from: slvri on June 07, 2010, 03:39:25 pm
thats because theyve given you the range for x but the angle whos sine is being taken in the question is x/2-1, not x so whatever angle you find after taking inverse sine will be for x/2-1, not x itself. thats why you have to convert the given range for x to a range for x/2-1
that means for the limits given (0 and 6pi) you divide by 2 to get x/2 and then subtract 1 to get x/2-1 which is the required range
and by the way i gave add math in june 2009 and i gave a level math in june 2010
so yeah just felt like helping others out for the add math exam tomorrow :)
oh and HI ASTAR....remember me?

:o OH, so every range they give you, you have to sub it in the ... thing like (x/2-1) to get the MAX value? does this apply to EVERY equation? thanks again :) :)
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Offline J.Darren

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Re: Additional Math Help HERE ONLY...!
« Reply #301 on: June 07, 2010, 03:51:31 pm »
Quote from: jellybeans on June 07, 2010, 02:11:31 pm
Helloooo, :D
ummm i need help with Q2 please? :) i can find the first two values but... how do you find the rest? :o and how do you know how many 'values' can fit in within the range?  :-\
THANKS IN ADVANCE ;D
http://img69.imageshack.us/img69/3933/trigonometry.png

This is a good way for those that are not accustomed to radian calculation in attempting these sort of questions ...
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #302 on: June 07, 2010, 04:06:27 pm »
Guys, can you please explain to me Q1 ON 2009 Paper 2? I don't know how to get the answers as shown in the mark schemes.
Thanxx :D
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Offline J.Darren

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Re: Additional Math Help HERE ONLY...!
« Reply #303 on: June 07, 2010, 04:13:48 pm »
Quote from: BlackBunny103 on June 07, 2010, 04:06:27 pm
Guys, can you please explain to me Q1 ON 2009 Paper 2? I don't know how to get the answers as shown in the mark schemes.
Thanxx :D
i ) Subsitute x = 0, the range would be greater than e^-1.

ii ) f(x) = e^(x+1)

y = e^(x+1)

ln y = (x+1) ln e

ln y = x + 1

ln x = y + 1

ln x - 1 = y

iii ) the domain of f-1(x) = the range of f(x), i.e. x > e^-1
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Offline Dibss

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Re: Additional Math Help HERE ONLY...!
« Reply #304 on: June 07, 2010, 04:18:11 pm »
Anyone who can upload or send me a link to the O/N 09 papers? I think only the mark schemes are available on FEP and Xtreme Papers...
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Offline J.Darren

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Re: Additional Math Help HERE ONLY...!
« Reply #305 on: June 07, 2010, 04:19:08 pm »
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Offline Dibss

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Re: Additional Math Help HERE ONLY...!
« Reply #306 on: June 07, 2010, 04:21:20 pm »
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #307 on: June 07, 2010, 04:24:12 pm »
Quote from: J.Darren on June 07, 2010, 04:13:48 pm
i ) Subsitute x = 0, the range would be greater than e^-1.

ii ) f(x) = e^(x+1)

y = e^(x+1)

ln y = (x+1) ln e

ln y = x + 1

ln x = y + 1

ln x - 1 = y

iii ) the domain of f-1(x) = the range of f(x), i.e. x > e^-1

Thanxx Darren ;). +rep
But I don't understand why from going   ln y = (x+1) ln e  to ln y = x + 1 why did you cancel lne but still keep ln e?
And why is the domain of f-1(x) = the range of f(x) ?
Thanxx again :D
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Offline J.Darren

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Re: Additional Math Help HERE ONLY...!
« Reply #308 on: June 07, 2010, 04:27:16 pm »
Quote from: BlackBunny103 on June 07, 2010, 04:24:12 pm
Thanxx Darren ;). +rep
But I don't understand why from going   ln y = (x+1) ln e  to ln y = x + 1 why did you cancel lne but still keep ln e?
And why is the domain of f-1(x) = the range of f(x) ?
Thanxx again :D
I pretained ln e there just so that you can comprehend what exactly has happened :D

Bear in mind that the inverse of every function is the reflection of the original function on the line y = x, as we know plugging any value of the range of the original function in a one-to-one equation into its inverse would get us back to where we strated from ...
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #309 on: June 07, 2010, 11:39:53 pm »
Guys would you please help me out with these permutation & combination questions?
Thanxx ;)

My exam is in 4 hours :-\ ...
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Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #310 on: June 07, 2010, 11:45:22 pm »
letter M is selcted so in fact you are only choosing 3 from 6C3
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Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #311 on: June 07, 2010, 11:47:34 pm »
Since she must select her favourite beatles cd she is only choosing 3 from 8 so 8C3
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #312 on: June 07, 2010, 11:54:58 pm »
Quote from: astarmathsandphysics on June 07, 2010, 11:45:22 pm
letter M is selcted so in fact you are only choosing 3 from 6C3

Oh right, this makes sense. Thanxx :D



Quote from: astarmathsandphysics on June 07, 2010, 11:47:34 pm
Since she must select her favourite beatles cd she is only choosing 3 from 8 so 8C3
I understand the concept, but I'm thinking since she's selecting 4 from 9 and there are 4 beatles cs in total, so what if all of her selections contain 4 Beatles CD? What I did was 9C4.
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Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #313 on: June 08, 2010, 12:01:57 am »
her FAVOURITE beatles cd so remove one from the selection sincve this is taken
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #314 on: June 08, 2010, 12:48:16 am »
I need help with this question, how can you find a?
Thanxx :D
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