• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Messages
52
Reaction score
40
Points
28
Hey ! Amm is math compolsury in cie ? Lol i know it is :cry: but still if i can do some other subjects instead of this math :( mental math will really make me mental please tell me what can i do :oops:
 
Messages
174
Reaction score
371
Points
73
Hey ! Amm is math compolsury in cie ? Lol i know it is :cry: but still if i can do some other subjects instead of this math :( mental math will really make me mental please tell me what can i do :oops:
Why are you so afraid of mathematics?
It keeps our mind active and fresh.
Nothing is hard until you put your heart and all your grace in your work.
I would suggest you to take mathematics. No other suggestion from my side.
 
Messages
8,477
Reaction score
34,837
Points
698
Someone plz help me in this stupid confusing domain range chapter :/
If someone can help me understand ~Domain Range~ chapter and ~Vector Paper 3~ only, I would be really Thankful to him/her :/


Help needed!!!
I know that range of a function is the domain of the inverse function and vice versa but how to find domain in this question
part (iv) only
and PLZZ answer with deep knowledge ( I mean with more theory and working which is clearly understandable to solve rest of the questions):
Thanks and May Allah bless us all with knowledge and A/A*s in A Levels Ameen:)
View attachment 46644


kitkat <3 :P Angelina_25 shazmina fantastic girl Anyone :SSS
Plejj :'/
Here Rookaya did this. http://www.twiddla.com/1729925 Aly Emran
 
Last edited:
Messages
8,477
Reaction score
34,837
Points
698
Aly Emran
Part i) RANGE OF G INVERSE
Domain of g(x) = range of g inverse.
Domain of g(x) >= 4
Hence, Range of g inverse >=4
Part ii) DOMAIN OF G INVERSE
Range of g(x) = Domain of g inverse
First complete square :¬
- ( x - 4)^2 + 16 = g(x)
Now exchange g(x) with y.
y = - (x - 4)^2 +16
Now exchange x with y and vice-verse.
x = -(y - 4)^2 +16
Make y as subject.
y = 4 + root (16 - x)
Now exchange y with g inverse.
So g inverse is 4 + root (16 - x)
Now for g inverse, any value of x >= 16 will have undefined answer, yeah? so x <= 16
Another way is :¬
Due to the minus sign, it is the maximum point on the graph. So x <= 16 is range of g(x) and according to Range of g(x) = Domain of g inverse, domain of g inverse is x <= 16

Hope you got it.
 
Messages
52
Reaction score
40
Points
28
Why are you so afraid of mathematics?
It keeps our mind active and fresh.
Nothing is hard until you put your heart and all your grace in your work.
I would suggest you to take mathematics. No other suggestion from my side.
Actually i would love to learn something im not good at but the problem is that i study by myself so maths cant be done without tutor thats why i asked .
 
Messages
8,477
Reaction score
34,837
Points
698
Messages
8,477
Reaction score
34,837
Points
698
Oh Brillient (y)
got it :D
but cant we find the mid point and see that is the graph a min or max and then decide greater then and less than?
Ps: Is it necessary to do inverse of g(x) in part 1?? because part 2 demands inverse so the examiner might not accept the inverse answer of part 1 i guess :/
plzz help me on this issue too :)
Thanks in advance :)
No its not like that, till I know.
Go to this site :- http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/P1/module.php 2 chapter is function. :)
Everything in it. :)
All the best. ^_^
 
Messages
8,477
Reaction score
34,837
Points
698
Not much with it! Saved answer here :¬
complex-png.21999

As the question states that the argument of z is least, z must represent the lowest possible point in the shaded region. Draw a tangent from the center to the lowest point on the circle (this doesn't have to be accurate) as shown by the blue line. This line represents |z|. The angle between the radius and the tangent is always 90 so a right-angled triangle is formed. The opposite side, with is equal to the radius is 1 unit. The magnitude of complex u is the hypotenuse and the adjacent represents |z|. you can easily find |u|, √(2^2+2^) = √8.
So, |z| =√(√8)^2 -1^2 = √8-1 =√7
 
Top