Mathematics: Post your doubts here!

[Fatima18]

http://papers.xtremepapers.com/CIE/...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!

 

[Dug]

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

 

[PhyZac]

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

 

[talalz94]

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 ?

 

[mominzahid]

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

 

[Muhammad Bin Anis]

Can any one of you explain Q8 part (ii) and Q9 part (ii)??
(this is from nov 2012 / p32)

 

[minie23]

Can anyone please help me for no. 5, 7, 9 please ? Thank you

 

[syed1995]

Can you link me the paper?

 

[D0cEngi]

Plz help me in solving q.9 part 2,q.10 of June 2007 paper 3....Plz reply asap..I have a mock tmrw.

 

[Rutzaba]

hey can anyone help me solve these Qs of integration... i kno they are solved by that tan formula but i dunno how to apply that...


1)integral of ((x^2 -2x + 3))/ (x-1)( x^2 +2x +2))

 

[Fatima18]

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

Thanx alot!! ....

 

[applepie1996]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_3.pdf
question 4 part 2 and question 5 please
http://papers.xtremepapers.com/CIE/...hematics (9709)/9709_s03_ms_1+2+3+4+5+6+7.pdf

 

[Fatima18]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
Hi guys..How to do number 11 i)?

 

[mutilated_grass]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_13.pdf
Hey guys. need help in question 8 part i) and ii). I think they got the answer wrong in the MS and ER for part i because it's only differentiated once and it's supposed to be differentiated twice.

 

[scouserlfc]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
Hi guys..How to do number 11 i)?

sub both equation in f(x)=g(x) like u do normally and bring all the terms on one side so that it becomes ..... = 0 now use discriminant b(square) - 4ac =0 and like u normally use it get values of k !
we use =0 because it says their are two equal real roots ! !

 

[scouserlfc]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_3.pdf
question 4 part 2 and question 5 please
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_ms_1 2 3 4 5 6 7.pdf

in first one
after finding a put it in the equation and find its derivative,put the derivative as 0 so that u find the stationary values of x,now do double derivative and put these values of x and see if points are a minimum if they are then put in the following step if one is then put that and forget the other ,put the values in f(x) and see if u get negative answer which u shudnt as f(x) stays positive !
this proves that the graph never went beyond x axis and was always +,u use stationary values to find the values of x for minimum points because these values have f(x) at the lowest value so if they dont give u minus nothing will
U will get a cubic equation so u will need to solve it (hahaha thats why i didnt solve it its really boring )

For the other one try it urself it really is straightforward,u need to use those different forms of expressing complex no. to get w then its quite easy ! Ill INSHAALLAH solve it tomorrow or soon as for now im really out of time sorry

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf
Hi guys..How to do number 11 i)?

f(x)=g(x)
So
k-x=9/(x+2)
(k-x)(x+2)=9
kx+2k-x^2-2x=9
-x^2+(k-2)x+2k-9=0
or
x^2+(2-k)x+9-2k=0

For Two Equal and real roots > b^2-4ac=0

(k-2)^2-4(-1)(2k-9)=0
k^2-4k+4+8k-36=0
k^2+4k-32=0

k = 4 or -8

now solve substituting values of k

for k=4
x^2+(2-k)x+9-2k=0
x^2+(2-4)x+9-2(4)=0
x^2-2x+1=0

root = 1

for k=-8
x^2+(2-k)x+9-2k=0
x^2+(2-(-8))x+9-2(-8)=0
x^2+10x+25=0

root = -5

so the values of k are 4 or -8. and the roots for these values are 1 and -5.

PS. Well i might have made some sign mix-ups while copying the solution from my copy ..

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_13.pdf
Hey guys. need help in question 8 part i) and ii). I think they got the answer wrong in the MS and ER for part i because it's only differentiated once and it's supposed to be differentiated twice.

That's because the equation given is dy/dx itself .. so to make it d2y/dx2 we just need to diffrentiate it once!

8 i)

dy/dx=2(3x + 4)^3/2 -6x -8
d2y/dx2= 9(3x+4)^1/2-6

8 ii)

put x=-1 in dy/dx=2(3x + 4)^3/2 -6x -8 and see if it equals zero.
dy/dx=2(3(-1) + 4)^3/2 -6(-1) -8
2(-3+4)^3/2+6-8
2+6-8=0

hence there is a stationary point at x=-1

Nature:
d2y/dx2= 9(3(-1)+4)^1/2-6
9(1)-6
d2y/dx2=3
d2y/dx2>0 so minimum.

 

[mutilated_grass]

thanks! Lol I must read the question properly

 

[Rutzaba]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_3.pdf
question 4 part 2 and question 5 please
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_ms_1 2 3 4 5 6 7.pdf

part 4 (ii)

We know by part (i) that the factors are (x^2 +2x +2) (x^2 -4x +4)
to factorize (x^2 -4x +4)
this can be done as (x-2) ^2
then we d factorize (x^2 +2x +2)
we wud get ((x+1) ^2) +1)
now if we multiply ((x+1)^2) +1)) (x-2) ^2 all the negative terms wud be squared to give positive results thus
((x+1)^2) +1)) (x-2) ^2 > 0