Mathematics: Post your doubts here!

[The Godfather]

please help with question 7 (ii)
I can put it into the form -In(100-x) = 0.02t-In(k)
but I can not get K to equal 95 like they appear to do in the mark scheme.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_3.pdf

Hey dude. Thought Blocker is currently not available for few days. He is outta town! And he asked me to help you and Jie, I am not much sharper. But I will try my best to explain

 

[The Godfather]

Thx for ur help!! Appreciate it very much^^
Hmmm...on/9709/2008/03 question 8 and mj/9709/2008/03 quest8 too..(

w08 is tough! Lemme do it. I will solve in one paper. Hope it is visible, coz so much to write, and I get tired.

 

[The Godfather]

Finally both solved.

Thx for ur help!! Appreciate it very much^^
Hmmm...on/9709/2008/03 question 8 and mj/9709/2008/03 quest8 too..(

Winter 2008 :
As I said, its too long calculation, here is the pic :
8a)

8b)

8c)

please help with question 7 (ii)
I can put it into the form -In(100-x) = 0.02t-In(k)
but I can not get K to equal 95 like they appear to do in the mark scheme.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_3.pdf

For part (i) we are told that the rate of formation of X is proportional to the mass of Y.

Therefore at any instant we can say dx/dt = ky where k is some constant.

But we are also told that x + y = 100

Thus y = 100 - x

So dx/dt = k(100 - x)

We are also told that when x = 5, dx/dt = 1.9

1.9 = k(100 - 5)

1.9 = k95

k = 1.9 / 95

k = 0.02

dx/dt = 0.02(100 - x)

For part (ii) integrate this.

dx/dt = 0.02(100 - x)

Separate variables

dx / (100 - x) = 0.02 dt

Integrating gives

- ln (100 - x) = 0.02t + c (Note the appearance of the first minus sign)

When t = 0, x = 5

- ln (95) = c

So we now have

- ln (100 - x) = 0.02t - ln 95

ln (100 - x) = ln 95 - 0.02t

ln (100 - x) - ln 95 = -0.02t

ln [(100 - x) / 95] = -0.02t

(100 - x) / 95 = e^(-0.02t)

100 - x = 95e^(-0.02t)

100 - 95e^(-0.02t) = x

x = 100 - 95e^(-0.02t)

For part (iii) when t becomes large e^(-0.02t) becomes very small.

Thus x approaches 100

 

[Thought blocker]

Finally both solved.

Winter 2008 :
As I said, its too long calculation, here is the pic :
8a)

8b)

8c)

For part (i) we are told that the rate of formation of X is proportional to the mass of Y.

Therefore at any instant we can say dx/dt = ky where k is some constant.

But we are also told that x + y = 100

Thus y = 100 - x

So dx/dt = k(100 - x)

We are also told that when x = 5, dx/dt = 1.9

1.9 = k(100 - 5)

1.9 = k95

k = 1.9 / 95

k = 0.02

dx/dt = 0.02(100 - x)

For part (ii) integrate this.

dx/dt = 0.02(100 - x)

Separate variables

dx / (100 - x) = 0.02 dt

Integrating gives

- ln (100 - x) = 0.02t + c (Note the appearance of the first minus sign)

When t = 0, x = 5

- ln (95) = c

So we now have

- ln (100 - x) = 0.02t - ln 95

ln (100 - x) = ln 95 - 0.02t

ln (100 - x) - ln 95 = -0.02t

ln [(100 - x) / 95] = -0.02t

(100 - x) / 95 = e^(-0.02t)

100 - x = 95e^(-0.02t)

100 - 95e^(-0.02t) = x

x = 100 - 95e^(-0.02t)

For part (iii) when t becomes large e^(-0.02t) becomes very small.

Thus x approaches 100

Yes, like that. KEEP IT UP.

 

[AnujaK]

Heeyy everyone! I've started a site that'll give you SOLVED past papers! Really helpful! Do check it out Also has some worksheets and revision notes!
https://sites.google.com/site/fromatoscom/

 

[Thought blocker]

Finally both solved.

Winter 2008 :
As I said, its too long calculation, here is the pic :
8a)

8b)

8c)

For part (i) we are told that the rate of formation of X is proportional to the mass of Y.

Therefore at any instant we can say dx/dt = ky where k is some constant.

But we are also told that x + y = 100

Thus y = 100 - x

So dx/dt = k(100 - x)

We are also told that when x = 5, dx/dt = 1.9

1.9 = k(100 - 5)

1.9 = k95

k = 1.9 / 95

k = 0.02

dx/dt = 0.02(100 - x)

For part (ii) integrate this.

dx/dt = 0.02(100 - x)

Separate variables

dx / (100 - x) = 0.02 dt

Integrating gives

- ln (100 - x) = 0.02t + c (Note the appearance of the first minus sign)

When t = 0, x = 5

- ln (95) = c

So we now have

- ln (100 - x) = 0.02t - ln 95

ln (100 - x) = ln 95 - 0.02t

ln (100 - x) - ln 95 = -0.02t

ln [(100 - x) / 95] = -0.02t

(100 - x) / 95 = e^(-0.02t)

100 - x = 95e^(-0.02t)

100 - 95e^(-0.02t) = x

x = 100 - 95e^(-0.02t)

For part (iii) when t becomes large e^(-0.02t) becomes very small.

Thus x approaches 100

Hey question 8 is blundered. Here you go :¬

iii) part is easy do it yourself.

 

[Jie Xi]

thx thx very much!!!!!but i cant see the image ehh..dunnoe why...can u post again??i can see it few days ago but now they are corrupted....=( so sorrry...

 

[Edison Weeho]

can you guys solve for me maths P3 O/N 07 Q7 iii)

 

[Thought blocker]

Hey question 8 is blundered. Here you go :¬

iii) part is easy do it yourself.

 

[Thought blocker]

can you guys solve for me maths P3 O/N 07 Q7 iii)

If possible post the link, my iPhone is working damn slow.

 

[Sajidul Siraji]

Could someone please solve his for me:
The number of defect lamps found in every house has a Poisson distribution with mean 3.
a) Find the probability that for 7 particular houses, no defect lamps were found in 2 houses.
b) Find the probability that for 120 particular houses, at least 2 defect lamps were found in less than 70 houses

 

[Edison Weeho]

If possible post the link, my iPhone is working damn slow.

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf

Well whats big deal here?
Just replace the sin function with -1 that's all.
"Just replace sin 2t by -1"
Thats all in it.

Got it? Or want a work solution of whole 7th question?

 

[Harsh Poddar]

Can someone help me with the Q7 - http://gopapers.net/alevel.html?_9709_s13_qm_12
Thanks

 

[Thought blocker]

Can someone help me with the Q7 - http://gopapers.net/alevel.html?_9709_s13_qm_12
Thanks

Hey this is the only way to do it, like its just inverse of a method we use to do "midpoint thingy"
Tell me if you dont get it.

 

[Thought blocker]

Could someone please solve his for me:
The number of defect lamps found in every house has a Poisson distribution with mean 3.
a) Find the probability that for 7 particular houses, no defect lamps were found in 2 houses.
b) Find the probability that for 120 particular houses, at least 2 defect lamps were found in less than 70 houses

I haven't took Statistics. Sorry.

 

[Harsh Poddar]

Hey this is the only way to do it, like its just inverse of a method we use to do "midpoint thingy"
Tell me if you dont get it.

Thanks bro

 

[Thought blocker]

robinhoodmustafa M in hurry. gtg.
Here:¬

 

[jungly krinsky]

Thanks a lot.

 

[Farheen1201]

Salam,
I need help with mechanics 1....Can somebody help?...
Chapter 4...
Question 7
A particle P of mass 4m kg is at rest on a horizontal table. A force of magnitude 50m Newtons acting upwards at an acute angle α to the horizontal, is applied to the particle. Given that tan α = (3/4) and that there is a resistance to the motion of magnitude 20m Newtons, find the acceleration with which P moves. Find in terms of m the magnitude of the normal contact force of the table on P.
Please help asap....