Mathematics: Post your doubts here!

[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...

 

[PhyZac]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. littlecloud11
May June 2008
and 9 (i)
Sorry for not posting link ....but xtremepapers isnt working good.

 

[Dug]

Can somebody explain 2 (ii)?

n = 15
Σx = 1572

Σ(x - 104.8)^2 = Σx^2 - (2)(104.8)Σx + Σ(104.8)^2 = Σx^2 - (2)(104.8)(1572) + (15)(104.8^2) --- i

Σ(x - 100)^2 = 499.2
Expanding LHS:
Σx^2 - 2(100)Σx + Σ(100^2) = 499.2
Σx^2 - (2)(100)(1572) + 150000 = 499.2
Σx^2 = 164899.2 --- ii

Put (ii) in (i):
Σ(x - 104.8)^2 = 153.6

 

[littlecloud11]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.. littlecloud11
May June 2008
and 9 (i)
Sorry for not posting link ....but xtremepapers isnt working good.

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

 

[Syed Abdul Haseeb]

Salam to one and all, i need some help in P1(Hugh Neill and Douglas Quadling) Ch:13 Vectors Miscellaneous 13 Q 9, 13, 14 and 15 thankz in advance

 

[zunair]

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

Thank you very much but there's this thinngy confusing me that inverse of cos 1 is 0 as well as 2pi so when put equal to 2 pi the answer turns out to be 5pi/3 as well

 

[zunair]

Thank you very much but there's this thinngy confusing me that inverse of cos 1 is 0 as well as 2pi so when put equal to 2 pi the answer turns out to be 5pi/3 as well

where can i get examiner reports