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Mathematics: Post your doubts here!

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Hi guys.... Please help me with these questions... would be appreciated... Thanks :)
Q5,Q6ii, Q7iii, Q8i, Q9ii
Q 6 part two a and b.
a)A conjugate in simple words mean that the sign ( that is negative or positive) of the imaginary number (the coefficient of i ) would change.
z= √3 + i
z*= √3 - i
1....things to remember = i = √-1 and thus i^2= -1
2.... if there is a term in i in the denominator and we have to remove it... we multiply the whole term with its conjugate and divide the whole term with its conjugate too.
we need to find
a) 2z +z* = 2(√3 +i) + √3- i
2 √3 +2i + √3 -i
3 √3 + i ...... ans.

b)because the format of the fraction isnt good here i have made u a pic to understand from

http://i1275.photobucket.com/albums/y444/Rutzaba/xpc_zpsd8fbcce5.png
 
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can someone plz help me with this..I need the solution asap!plz
Maths expert
Need it asap.plz attach a scan if possible showing all the steps.ok
 

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can someone plz help me with this..I need the solution asap!plz
Maths expert
Need it asap.plz attach a scan if possible showing all the steps.ok
c u can do it like this :
separate the 4 and the 10s :-

4^4 x (10^-1 x 10^-2 x 10^-3 x 10^1)
= 256 x 10^-1-2-3+1
= 256 x 10^-5
= 0.00256
=0.002560

don't be confused because of the last zero because it is a trick. it does not actually matter how many zeros we put at the end the answer is still 0.002560

get it?? :)
 
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untitled-png.22605
 

Dug

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w04_qp_1.pdf
Hi guys!
How to do number 9 ii )???
Also, I solved 9 iv and Im getting square root of x + 9 then + 3, hoever the answer key has a square root under everything including +3..why is that??
Pls asap..Thank you!
ii)
f(x) = g(x)
2x - a = x^2 - 6x
x^2 - 8x + a = 0

Since one real solution, D = 0
D = b^2 - 4ac = 0
(-8)^2 - 4(1)(a) = 0
64 - 4a = 0
a = 16

iv)
let y = x^2 - 6x
y = x^2 - 6x + (-3)^2 - (-3)^2
y = (x - 3)^2 - 9
(x - 3)^2 = y + 9
x - 3 = √(y + 9)
x = √(y + 9) + 3

h-1(x) = √(x + 9) + 3

Domain: x ≥ -9
 
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I'm guessing this is for the P3 paper.

9i) y = e^(-x/2) √(1+2x)
dy/dx = -1/2 e^(-x/2) * √(1+2x) + 1/2* 2 * (1+2x)^(-1/2) * e^(-x/2)
dy/dx = [-e^(-x/2) * √(1+2x)]/2 + e^(-x/2)/ √(1+2x)

For M, dy/dx = 0
[-e^(-x/2) * √(1+2x)]/2 + e^(-x/2)/ √(1+2x) = 0
Take LCM
[-e^(-x/2) * √(1+2x)* √(1+2x) + e^(-x/2) *2] / 2√(1+2x) =0
cross multiply, so the denominator becomes 0
-e^(-x/2) * (1+2x) +2e^(-x/2) = 0
take -e^(-x/2) common-
-e^(-x/2) [(1+2x) -2] =0
so,
-e^(-x/2) = 0 (NA) or -(1+2x) -1 = 0
1+2x -2 = 0
2x= 1
x =1/2
Jazaki Allah khairan...Thanks alottt...!! Yes sorry it was for paper 3 ....THAnk u so much...May Allah reward you with the best of the two worlds. In Sha Allah your will get A*'s Ameen...THANK YOU for you help...May Allah have mercy on you and your family Ameen
 
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hey i need help in these P3 questions ...
Question 7, winter 2012 / 33 . please explain as much as u can especially the limits .. how to find them out ?
 
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did u understand the question that i solved for u?
im gud at maths on a paper but it takes ten thousand years for me to type this stuff.i will solve other qsas soon as possible
Urrmm yes i think i didd... my concepts needed some refreshment so watched a tutorial then saw ur solution.. I got it.
THank you :D
 
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