Mathematics: Post your doubts here!

[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!

 

[ASstudent_96]

Could anyone show me how to solve question 11 oct/02 please ?! I'll be gratefull
Thanks all

 

[zunair]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Assalamoalaikum!!

Sure..anytime!

Im having problem in W12/13 Q6a ..... my answer for that part is 5pi/3 where as marking scheme says 4pi/3.
kindly help and inbox me the procedure. Thnak u

 

[zunair]

Im having problem in W12/13 Q6a ..... my answer for that part is 5pi/3 where as marking scheme says 4pi/3.
kindly help and inbox me the procedure. Thnak u

 

[mominzahid]

Hi guys.... Please help me with these questions... would be appreciated... Thanks
Q5,Q6ii, Q7iii, Q8i, Q9ii

 

[syed1995]

Im having problem in W12/13 Q6a ..... my answer for that part is 5pi/3 where as marking scheme says 4pi/3.
kindly help and inbox me the procedure. Thnak u

It's neither of those two. The answer for the question would be -π/3 ...

f(x)=x/2 + π/6
g(x)=Cos(x)
g(f(x)) = Cos((x/2+π/6))
Cos((x/2+π/6)) = 1
x/2+π/6= Cos^-1(1) (Cos Inverse of 1 = 0)
x/2+π/6=0
x/2=-π/6
x=-2π/6
x=-π/3

The working done here is perfect. There is an error in the marking scheme. I always check my answers against the Examiner's Report.

Cheers

 

[parthrocks]

dude u post a question hre once and once ok? we will find u k?

Arey my net was down and so got repeated many times.m really very sorry

 

[Rutzaba]

Arey my net was down and so got repeated many times.m really very sorry

xD dun wrry mere sath b hota hy

 

[Rutzaba]

Hi guys.... Please help me with these questions... would be appreciated... Thanks
Q5,Q6ii, Q7iii, Q8i, Q9ii

bKI bhi kar deti hun ... 3 4 tou reh gye hyn
i thinka whole day wud be required to explain all these Qs

 

[HubbaBubba]

Can somebody explain 2 (ii)?

 

[Hassan Ali Abid]

Can somebody explain 2 (ii)?

let sigma(x-100)=sigma y =72

y bar =sigmay /n : y bar= 72/n

y bar= x bar -100
72/n= 104.8-100

solve it and n =15

 

[Rutzaba]

Hi guys.... Please help me with these questions... would be appreciated... Thanks
Q5,Q6ii, Q7iii, Q8i, Q9ii

Q.5(a) let x be 2sina

where x= 2sin a

dx/da = 2 cos a

dx= 2 cos a da
(2sin a)^2 /√ (4- (2sin a) ^2) x dx
4sin^2 a / √((4)(1-sin^2 a) x dx
4sin^2 a / √(4)(cos ^2 a) x dx
4 sin ^2 a / 2cos a x 2cos a da
4 sin^2 a = proved
now for the limits... the function is x= 2sin a
if x=1 then sina = 1/2 and a = sin inverse of 0.5 which is π/6
if x=0 then 2sina =0 and sin inverse of 0 is 0.

sin
Q.5 (b
∫4sin ^2 a
here we wud convert sin ^2 a into the formula= (1-cos2a)/2
∫4/2 (1-cos2a)
∫2(1-cos2a)
∫ 2 -2cos 2a

2a- sin 2a apply limitsnow

 

[mominzahid]

bKI bhi kar deti hun ... 3 4 tou reh gye hyn
i thinka whole day wud be required to explain all these Qs

Hahah Im sorryy
The thing is i wasted my whole year and i literally studied p3 and m1 in like 20 days.. khud andaza laga lo agay kaise atay hngay..
That is why i face alot of problems in papers initially. I hope it gets better :/

 

[Rutzaba]

Hahah Im sorryy
The thing is i wasted my whole year and i literally studied p3 and m1 in like 20 days.. khud andaza laga lo agay kaise atay hngay..
That is why i face alot of problems in papers initially. I hope it gets better :/

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

 

[syed1995]

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

haha so true.. same with me.. doing maths on paper is way easier than typing the solution out...