Mathematics: Post your doubts here!

[LMGD33]

So after we integrate this completely, we still have to multiply the -1 at the end, right?

Sorry if it's unclear, I just relaized!! I multiplied by negative one outside the integral, and the numerator is supposed to be -10 u
which would give us +10 u and the answer they are looking for , sorry again!

 

[SararaIH]

Sorry if it's unclear, I just relaized!! I multiplied by negative one outside the integral, and the numerator is supposed to be -10 u
which would give us +10 u and the answer they are looking for , sorry again!

It's okay! You don't have to apologize
But I'm confused :/
I mean, do we have to keep the -1 outside the integration sign?

 

[SararaIH]

You can get the free trial version here, its a pretty handy tool so its good to have it in your PC. =) http://www.rarlab.com/download.htm

Thank you!

 

[D0cEngi]

Plz help me out with ques 7 part (iii) of this paper: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

 

[Anika Raisa]

Plz help me out with ques 7 part (iii) of this paper: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

take a=1+2i n b=1-3i

nw u=a/b

theres this formula dat arg(a/b)=arg a - arg b

n u knw arg=tan(inverse) imaginary/real

so here
arg a= tan(inverse) 2/1

arg b= tan(inverse) -3/1


Hope it helped!! BOL!

 

[LMGD33]

It's okay! You don't have to apologize
But I'm confused :/
I mean, do we have to keep the -1 outside the integration sign?

Not really, I just show it as a step to acknowledge the flipping of the limits.. since the upper limit was smaller than the lower, that is why they asked us to flip them and as a general rule you multiple by negative one, doesn't matter where really

 

[Anika Raisa]

SOME HELP HERE PLEASE WITH P3!!!
No. 7 iv Please draw n upload the diagram ... any1? Thank u in advance!!

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_31.pdf


SararaIH
VampBhums
LMGD33
Esme
Alice123 .... or any1?

 

[AsadZaidi]

guys i m stuck on m/j 2004 q10 part c... the integration one.... plz help...

 

[Esme]

SOME HELP HERE PLEASE WITH P3!!!
No. 7 iv Please draw n upload the diagram ... any1? Thank u in advance!!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf


SararaIH
VampBhums
LMGD33
Esme
Alice123 .... or any1?

 

[Anika Raisa]

Esme Thank u so very much!!! U r a saviour! May Allah Bless u!!!

 

[Esme]

Esme Thank u so very much!!! U r a saviour! May Allah Bless u!!!

You're welcome And remember me in your prayers

 

[xhizors]

Q10 iii
working with little explanation plz, anyone!
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_32.pdf

 

[AlishaK]

Q10 iii
working with little explanation plz, anyone!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

Esme ??!

 

[applepie1996]

Q10 iii
working with little explanation plz, anyone!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

wait

 

[applepie1996]

applepie1996
We have to find the point p where the perpendicular distance to the two planes is same.

first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

I dint do every step in detail, (very long) if u dont get any step, just ask. [Pray for me please.]

xhizors
credits to PhyZac
Pray for him

 

[LMGD33]

Q10 iii
working with little explanation plz, anyone!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_32.pdf

Lamda will be denoted as t
You know that P is a point on L , so it has x,y,z coordinates as follows : (1 + 2t), (1 + t), (-1 + 2t) , and m = x + 2y -2z = 1 and n = 2x - 2y + z = 7
Using the formula they provided, you find the distance of P to m and of P to n keeping in mind that they should be equal, so distance of P to m
| (1 + 2t) + 2 (1 +t) -2(-1 + 2t) - 1 | / square root ( 1^2 + 2^2 + (-2)^2 ) to get |4|/3
distance of P to n using the same equation will get you |4t - 8|/3
These two answers are equal, their numerators are equal so you square both sides to get rid of the modulus, leaving you with an equation such as :
4^2 = (4t - 8)^2
16 = 16t^2 - 64t - 64, solve for t and obtain t = 3 or t =1
substitute those two value in the original coordinates of P to get the two position vectors that are (7i + 4j + 5k) and (3i +2j + k) and then find the distance between these two points using your skill from P1! Good luck!

OR

Chem champ having prob with maths???
Here u go.... This was done by Physac
We have to find the point p where the perpendicular distance to the two planes is same.


first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6
ask if u dont get

 

[Anika Raisa]

You're welcome And remember me in your prayers

Sure will!

 

[messi10]

Someone please help :'(

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
question no 10

Its an substitution integration question.

thanks in advance.

 

[biba]

guys i m stuck on m/j 2004 q10 part c... the integration one.... plz help...

y= lnx/x^2...
we have to use integration by parts ..
u = lnx so du/dx=(1/x)
dv/dx=(1/x^2) so, integral of v=(-1/x)
(i am using ! as the sign of integral)
the formula is: !u(dv/dx) = uv - !v(du/dx)
= lnx(-1/x) - ! (-1/x)(1/x)
= -lnx/x + ! (1/x^2)
obtain the second integral:
= (-lnx/x) - (1/x)
the limits are x=e and x = 1
now u can put in the limits and obtain the answer !

 

[iKhaled]

hey ppl..when a plane passes through a point lets say for example point A..what will be the relation between the plane and the point it has passed ?