Mathematics: Post your doubts here!

[Translucent231]

Co prime numbers are numbers which have no factor common ,

like 2 and 3 are coprime 16 and 25 are coprime but 2 and 4 are not co prime...
8 and 5 are coprime and so on..

 

[PhyZac]

Re-arrange the eq-
ln y = 2x+1/x
differentiate both sides
1/y dy/dx = [2x- 1(2x+1)]/x^2 {use u/v formula for RHS and ln y = 1/y dy/dx for LHS}
1/y dy/dx = 2x-2x -1/ x^2
1/y dy/dx = -1/x^2
dy/dx = (-1/x^2)/ (1/y)
dy/dx = -y/x^2

Thank You so much !!!!! JazakAllah khiaran !! May Allah grant you and your family with highest grades in this life and hereafter. Ameen !

Alhamdulilah i got it !!

 

[snowbrood]

i dont get the second part the angle one. answer is 119.9 or60.1

 

[daredevil]

i dont get the second part the angle one. answer is 119.9 or60.1

As OA --> 4i
OC --> 2j
OD --> 3k
and it is a cuboid so:

OG --> 2j+ 3k
OA --> 4i
AG = OG - OA = -4i + 2j + 3k

and OB = 4i + 2j


that said... now use the scalar product to find the angle between them
(ai)(bi) + (aj)(bj) + (ak)(bk) = lal . lbl . Cos (angle between a and b)

where a and b are AG and OB respectively and LaL --> magnitude of vector a

solving this:
put in all the values and u'll get it down to:
-12 = (20)^1/2 . (29)^1/2 . Cos (angle between them)
Cos (angle b/w them) = -0.49827...
angle b/w them = 119.8858...
= 119.9

 

[maham khan]

h

Assalamoalaikum Wr Wb!

Post your doubts here. Make sure you give the link to the question paper when posting your doubts.


May Allah give us all success in this world as well as the HereAfter...Aameen!!

And oh yeah, let me put up some links for A level Maths Notes here.

Check this out! - Nice website, with video tutorials for everything! MUST CHECK

My P1 Notes! - Only few chapters available at the moment!

Maths Notes by destined007

Some Notes for P1 and P3 - shared by hamidali391

A LEVEL MATHS TOPIC WISE NOTES

MATHS A LEVEL LECTURES

compiled pastpapers p3 and p4 - by haseebriaz

Permutations and Combinations (my explanation)- P6

Permutations and Combinations - P6

Vectors - P3

Complex No. max/min IzI and arg(z) - P3

Sketcing Argand Diagrams - P3 (click to download..shared by ffaadyy)

Range of a function. - P1

help needed with question 3 of a level s1 paper 62 may june 2012

 

[kronix6]

Hi can anyone please slove 61/o/n/2010 question 3 part 2. and m/j/61/2012 question 6 part 2. and can u explain plx why in 2012 paper we have taken p(-1<x<+1) and why we have only taken p(x<1.645) in octnov 2012 paper. i will be grateful.

 

[PhyZac]

Asslamu Alikum Wa Rahmatulahi Wa Barakatoho.......!!! littlecloud11

Can anyone please solve question 8 part (ii)

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

 

[Omar99]

question no. 9
oct/nov 2012 p32

 

[Omar99]

Assalamoalaikum Wr Wb!

Post your doubts here. Make sure you give the link to the question paper when posting your doubts.


May Allah give us all success in this world as well as the HereAfter...Aameen!!

And oh yeah, let me put up some links for A level Maths Notes here.

Check this out! - Nice website, with video tutorials for everything! MUST CHECK

My P1 Notes! - Only few chapters available at the moment!

Maths Notes by destined007

Some Notes for P1 and P3 - shared by hamidali391

A LEVEL MATHS TOPIC WISE NOTES

MATHS A LEVEL LECTURES

compiled pastpapers p3 and p4 - by haseebriaz

Permutations and Combinations (my explanation)- P6

Permutations and Combinations - P6

Vectors - P3

Complex No. max/min IzI and arg(z) - P3

Sketcing Argand Diagrams - P3 (click to download..shared by ffaadyy)

Range of a function. - P1

oct/nov 2012
p32

 

[littlecloud11]

Asslamu Alikum Wa Rahmatulahi Wa Barakatoho.......!!! littlecloud11

Can anyone please solve question 8 part (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

dx/dt = x^2 (10−x)/100
Cross multiply
∫100/ (x^2 10−x) dx = ∫ dt
use your answer to part (i) for the LHS
∫1/x + 10/x^2 + 1/(10-x) dx = ∫ dt
lnx - 10/x - ln|10-x| = t + c -------1
when x=1, t=0
ln1 - 10/1 - ln| 10-1| = 0 + c
-10 - ln9 =c
substitute the value of c in eq 1
lnx - 10/x -ln|10-x| = t - 10 - ln9
t = -10/x + 10 +ln9 +lnx - ln|10-x|
t= -10/x + 10 + ln | 9*x/ (10-x)|

 

[PhyZac]

dx/dt = x^2 (10−x)/100
Cross multiply
∫100/ (x^2 10−x) dx = ∫ dt
use your answer to part (i) for the LHS
∫1/x + 10/x^2 + 1/(10-x) dx = ∫ dt
lnx - 10/x - ln|10-x| = t + c -------1
when x=1, t=0
ln1 - 10/1 - ln| 10-1| = 0 + c
-10 - ln9 =c
substitute the value of c in eq 1
lnx - 10/x -ln|10-x| = t - 10 - ln9
t = -10/x + 10 +ln9 +lnx - ln|10-x|
t= -10/x + 10 + ln | 9*x/ (10-x)|

Thank you thank you thank you soooooo much !!!!!!!!!!! Jazki Allah khairan !!! THANKS ALOT for your help, i am sorry for tagging all time !! May Allah reward you with best grades , Ameen. May Allah have mercy on you and your family, Ameen. May Allah bless you and your family with happiness and success, Ameen. May Allah grant you with Jannah tul firdous Ameeen!!!!

 

[littlecloud11]

Thank you thank you thank you soooooo much !!!!!!!!!!! Jazki Allah khairan !!! THANKS ALOT for your help, i am sorry for tagging all time !! May Allah reward you with best grades , Ameen. May Allah have mercy on you and your family, Ameen. May Allah bless you and your family with happiness and success, Ameen. May Allah grant you with Jannah tul firdous Ameeen!!!!

You tagging me is worth it just for the good wishes I get afterwards.
So tag away!
And thank YOU.

 

[PhyZac]

Assalamu Alikum littlecloud11

Can anyone solve question 10 part (ii)

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

 

[littlecloud11]

Assalamu Alikum littlecloud11

Can anyone solve question 10 part (ii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

Sin^2(θ) = [1 - Cos 2(θ)]/2
so, sin^2 (2θ) = [1 - Cos 2(2θ)]/2
sin^2 (2θ) = [1 - Cos 4θ]/2

Now substitute sin^2 (2θ) as [1 - Cos 4θ]/2 for the integration:
1/4∫ sin^2 2θ =1/4 ∫[1 - Cos 4θ]/2
= 1/4 * 1/2 ∫1 - Cos 4θ
= 1/8 [θ - Sin 4θ/4]
put the limits
=1/8 [(π/2 -0) - (0 -0)]
=π/16
so A = π/16

 

[PhyZac]

Sin^2(θ) = [1 - Cos 2(θ)]/2
so, sin^2 (2θ) = [1 - Cos 2(2θ)]/2
sin^2 (2θ) = [1 - Cos 4θ]/2

Now substitute sin^2 (2θ) as [1 - Cos 4θ]/2 for the integration:
1/4∫ sin^2 2θ =1/4 ∫[1 - Cos 4θ]/2
= 1/4 * 1/2 ∫1 - Cos 4θ
= 1/8 [θ - Sin 4θ/4]
put the limits
=1/8 [(π/2 -0) - (0 -0)]
=π/16
so A = π/16

Thankss Alot!!! JAzaki Allah khairan !! May Allah grant you A*'s in all the subjects you are giving, Ameen And In Sha Allah you get into a good university and In Shaa Allah you get a scholarship too ! Ameeennn!! THAnks thanks thanks !!

sister, thou i have this question as doubt too, but this is part (iii) and i needed for (ii) as well...!!BUT THANKS because i was stuck here too...thank you..!!!

 

[littlecloud11]

Thankss Alot!!! JAzaki Allah khairan !! May Allah grant you A*'s in all the subjects you are giving, Ameen And In Sha Allah you get into a good university and In Shaa Allah you get a scholarship too ! Ameeennn!! THAnks thanks thanks !!

sister, thou i have this question as doubt too, but this is part (iii) and i needed for (ii) as well...!!BUT THANKS because i was stuck here too...thank you..!!!

Oh, I'm sorry, I misread.
Here you go-
ii) y = x^2 √(1-x^2)
when y = 0
x^2 √(1-x^2) = 0
x^2 =o or √(1-x^2) = o
x= 0 or x=1

Now for area, ​

x= Sinθ​

dx = Cosθ dθ

∫ x^2 √(1-x^2) dx ​

x= sinθ​

∫ Sin^2θ √(1-Sin^2θ) * Cosθ dθ​

∫Sin^2θ √(Cos^2θ) *Cosθ dθ
∫Sin^2θ Cos^2θ dθ

Sin2θ = 2Sin(θ) Cos(θ)
4Sin^2 (2θ) Cos^2(2θ) = Sin^2 (2θ)

so,
1/4∫4Sin^2θ Cos^2θ dθ = ∫Sin^2 (2θ)

For the limits, x=1 to x=0-
x= Sinθ
when x= 0, θ=0
when x-=1, θ=π/2

 

[PhyZac]

Oh, I'm sorry, I misread.
Here you go-
ii) y = x^2 √(1-x^2)
when y = 0
x^2 √(1-x^2) = 0
x^2 =o or √(1-x^2) = o
x= 0 or x=1

Now for area,
x= Sinθ
dx = Cosθ dθ
∫ x^2 √(1-x^2) dx
x= sinθ
∫ Sin^2θ √(1-Sin^2θ) * Cosθ dθ
∫Sin^2θ √(Cos^2θ) *Cosθ dθ
∫Sin^2θ Cos^2θ dθ

Sin2θ = 2Sin(θ) Cos(θ) STEP 1
Sin^2 (2θ) Cos^2(2θ) = Sin^2 (2θ) STEP 2

so,
∫Sin^2θ Cos^2θ dθ = ∫Sin^2 (2θ)

For the limits, x=1 to x=0-
x= Sinθ
when x= 0, θ=0
when x-=1, θ=π/2

Jazaki Allah khairan !! Thanks Alot Alot Alot !!!!!!!! May Allah grant you the best of the two worlds !!!! And In Shaa Allah you always be happy in your life, May Allah bless you and your parents ....!! That really really helped ALOt..!!!

I have only a doubt how to go to 2 from 1 !! I cant figure tht out ! But i got the rest, Thank youuuuu!

 

[littlecloud11]

Jazaki Allah khairan !! Thanks Alot Alot Alot !!!!!!!! May Allah grant you the best of the two worlds !!!! And In Shaa Allah you always be happy in your life, May Allah bless you and your parents ....!! That really really helped ALOt..!!!

I have only a doubt how to go to 2 from 1 !! I cant figure tht out ! But i got the rest, Thank youuuuu!

For step 2 I missed a 4 before.
2sinθCosθ = Sin2θ
(2sinθCosθ)^2 = (Sin2θ)^2
so
4 sin^2θ Cos^2θ = Sin^2(2θ)

 

[PhyZac]

For step 2 I missed a 4 before.
2sinθCosθ = Sin2θ
(2sinθCosθ)^2 = (Sin2θ)^2
so
4 sin^2θ Cos^2θ = Sin^2(2θ)

Okay i got it now ! Thanks alottt !! Thank you for all the help! Thanks alootttt!!!! May Allah reward withmore and more deeds for all the help you gave me and other students !! Jazaki Allah khairan !!! May Allah keep you happy all the time, Ameen

 

[Sheikh Nahiyan]

P1 - Oct-Nov'10 Qp-11 no. 7 (iii)

Please help!