Mathematics: Post your doubts here!

[MustafaMotani]

CAn Not dothefirst Part Have Done Therest I Know Its Silly But Can U Plz Show Da Wrking For Part 1 Iam Trying For So Long THANX

(3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k

comparing coefficients of k
(1+λ) = (2-µ)

λ = 1 - µ

comparing coefficients of i

(3-λ) = (4 +µa)
3 -1 + µ = 4 + µa
µ = 2/(1-a)

comparing coefficients of j

(-2+2λ) = (4 +µb)
-2 + 2 - 2µ = 4 + µb
0 = 4 + µ(2+b)
0 = 4 + (4+2b)/(1-a)

2a-b = 4

 

[Igcse stuff]

hey guys, can someone please share a link or book(s) with good AS and A2 math notes

 

[monoho]

can someone solve q2 http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf

 

[Rutzaba]

i had some ambiguity in below quest:
1) P(Z>a)=0.8888 , a is negativ,
1-fi(-a)=o.8888
1-[1-fi(a) ]=o.8888
the other solution of quest i found:
2)p(z<t)=0.25, t is negativ
fi(-t)=0.25
1-fi(t)=0.25..
can anyone tell me this difference and concept how to solve such type of quest plzzz

did anyone help you?

 

[MustafaMotani]

Alright... ii) part please ?

its 2 + 2^(1/2)?

 

[studen12345]

(3-λ)i + (-2+2λ)j + (1+λ)k = (4 +µa)i + (4 +µb)j + (2-µ)k

comparing coefficients of k
(1+λ) = (2-µ)

λ = 1 - µ

comparing coefficients of i

(3-λ) = (4 +µa)
3 -1 + µ = 4 + µa
µ = 2/(1-a)

comparing coefficients of j

(-2+2λ) = (4 +µb)
-2 + 2 - 2µ = 4 + µb
0 = 4 + µ(2+b)
0 = 4 + (4+2b)/(1-a)

2a-b = 4

THANX MAN you Are A Big Help In This Crucial Tym INSHALLAH ALLAH will Reward U

 

[Esme]

its 2 + 2^(1/2)?

Yep

 

[MustafaMotani]

THANX MAN you Are A Big Help In This Crucial Tym INSHALLAH ALLAH will Reward U

thanx for the prayer..

Yep

wait ..

 

[MustafaMotani]

Alright... ii) part please ?

Sorrry for the delay .. am not good at paint ..

 

[D0cEngi]

These might help, but do as many past paper questions on these topics as you can.
And most importantly pray to Allah, insha'Allah you will get it.

Thank you so much.

 

[Esme]

Sorrry for the delay .. am not good at paint ..

It's ok. Chill !
This really helped! Got it now! Thanks a lot

 

[D0cEngi]

Esme Can you plz help me with q.4 of this paper.: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_32.pdf

 

[MustafaMotani]

It's ok. Chill !
This really helped! Got it now! Thanks a lot

no need to say thanx again and again or I wil have to say "anytime" again and again ..

 

[Esme]

no need to say thanx again and again or I wil have to say "anytime" again and again ..

Lol okay... I guess it does get annoying after a while

 

[MustafaMotani]

Esme Can you plz help me with q.4 of this paper.: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_32.pdf

as far as i can comprehend you would be having difficult in integrating
cos(2x)/sin(2x) dx
u see differential of sin(2x) is 2cos(2x) ...
to get 2 cos(2x) in numerator just multiply the integral with 2/2
u shall get 1/2 S 2 cos(2x)/sin(2x) ( S denotes integral sign)
now u can integrate using integration by recognition.. it will become
1/2 ln (sin(2x))

 

[iKhaled]

no need to say thanx again and again or I wil have to say "anytime" again and again ..

u dont have to keep sayin anytime everytime...just like the post dude liking it is like sayin u welcome

 

[MustafaMotani]

u dont have to keep sayin anytime everytime...just like the post dude liking it is like sayin u welcome

really ?? .. ohkay..
where can i get "made up rules" of the forum ..

 

[monoho]

as far as i can comprehend you would be having difficult in integrating
cos(2x)/sin(2x) dx
u see differential of sin(2x) is 2cos(2x) ...
to get 2 cos(2x) in numerator just multiply the integral with 2/2
u shall get 1/2 S 2 cos(2x)/sin(2x) ( S denotes integral sign)
now u can integrate using integration by recognition.. it will become
1/2 ln (sin(2x))

can someone solve q2 http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf

 

[MustafaMotani]

can someone solve q2 http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_32.pdf

its logarithm question
substitute 5^x by a
yu shall get
a/5 = a - 5 (if yu dont get this step .. tell me)
a = 5a - 25
a =25/4

now
5^x = 25 /4 (apply logs to both sides)

x log (5) = log (25/4)
x = 1.14

 

[MustafaMotani]

I am retiring for today ... its 1 am here and my brain has given up ..