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Mathematics: Post your doubts here!

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Is there anyone who has studied or opted S2 in Alevels ???? or maybe anyone studying or may have studied Advanced/high level Stats ???
 
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Can someone please solve the Q10 of W_14_33
It is the question of integration by substitution.
I integrated it but I am stuck at the limit and applying it.
 
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Have to give all four P1, P3, M1, S1 in May-June 2016. :p
You've got plenty of time. :p P1 is super easy. P3 is lengthy but you can do it. As far as M1 is concerned, you've studied physics so it'd be easy for you and S1, you should focus on permutations and combinations and probability. Rest all is pretty much formula-based and easy
You'd easily get an A*
 
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You've got plenty of time. :p P1 is super easy. P3 is lengthy but you can do it. As far as M1 is concerned, you've studied physics so it'd be easy for you and S1, you should focus on permutations and combinations and probability. Rest all is pretty much formula-based and easy
You'd easily get an A*

Yeah, I totally will. :p
Math is too easy. :D
 
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So do you know any book for S2 ??? or any solved book for its papers ???
I'm using this book :
http://www.amazon.com/Concise-Course-Advanced-Level-Statistics/dp/074875475X/ref=oosr
covers both S1 and S2
As for pastpapers, when I bought solved past papers, I had chosen S1 and M1 but later I changed my mind and found Statistics much more bearable than Mechanics, so I've solved papers for S1 but not for S2..But it doesn't matter, I'll print out unsolved ones and I can check them through mark schemes! What's your plan?
 
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Give me five minutes. Ill help you out :D
View attachment 57660
How to solve this question please??? I'm completely lost :cry:
Dm /dt = K(M) cos (0.02t)
dM / m = kcos0.02t dt
ʃ dM / m = ʃ kcos0.02t dt
(M ^(1/2)) / (1/2) = ksin0.02t/0.02 + c
Rearrange:
2M =(ksin0.02t)/0.02 + C
Now we have t = 0 and M = 100 so just substitute inorder to find the constant C
Therefore,
2100 = (ksin0.02(0))/0.02 + C
2(10) = 0 + C
C = 20
Now : substitute the value of C to find the relationship.
2M = (ksin0.02t)/0.02 +20
**1/0.02 = 50
So, it will become: 2M = 50ksin0.02t + 20
b) M = 196 and t = 50
Just plug in the values.
2196 = 50ksin0.02(50) +20
14*2 = 50ksin0.02(50) + 20
28 = 50ksin0.02(50) + 20
28 – 20 = 50ksin0.02(50)
8 = 50ksin(1)
8/50 = ksin(1)
K = 0.19 Ans.
c) 2M = 50ksin0.02t + 20
Make m the subject
You will get:
M = ((50ksin0.02t + 20)/2)^2
Plus in the values.
M =((50(0.19)sin0.02t + 20)/2)^2
Solve the square. And you will get around 27 or 28 as your answer
If you still have problem, Inbox me :)
I dont own a cellphone otherwise a picture wouldve cleared it out well. Sorry :p
 
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Dm /dt = K(M) cos (0.02t)
dM / m = kcos0.02t dt
ʃ dM / m = ʃ kcos0.02t dt
(M ^(1/2)) / (1/2) = ksin0.02t/0.02 + c
Rearrange:
2M =(ksin0.02t)/0.02 + C
Now we have t = 0 and M = 100 so just substitute inorder to find the constant C
Therefore,
2100 = (ksin0.02(0))/0.02 + C
2(10) = 0 + C
C = 20
Now : substitute the value of C to find the relationship.
2M = (ksin0.02t)/0.02 +20
**1/0.02 = 50
So, it will become: 2M = 50ksin0.02t + 20
b) M = 196 and t = 50
Just plug in the values.
2196 = 50ksin0.02(50) +20
14*2 = 50ksin0.02(50) + 20
28 = 50ksin0.02(50) + 20
28 – 20 = 50ksin0.02(50)
8 = 50ksin(1)
8/50 = ksin(1)
K = 0.19 Ans.
c) 2M = 50ksin0.02t + 20
Make m the subject
You will get:
M = ((50ksin0.02t + 20)/2)^2
Plus in the values.
M =((50(0.19)sin0.02t + 20)/2)^2
Solve the square. And you will get around 27 or 28 as your answer
If you still have problem, Inbox me :)
I dont own a cellphone otherwise a picture wouldve cleared it out well. Sorry :p
Thankyou soooooooo much!!
This is great! Your workings and steps are clear!!
Thanks a ton for your effort and time!!! (y):D
 
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Can someone please give me the final answer and working for this question.
 

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y = e^(-x) sinx

For stationary point of any curve put dy/dx equal to 0

For dy/dx

Let u = e^(-x)

v = sinx

u’ = -e^(-x)

v’ = cosx

dy/dx = vu’ + uv’

You have all the values now. Just plug in.

0 = (sinx)(-e^(-x)) + (e^(-x))(cosx)

0 = e^(-x) {-sinx + cosx}

e^(-x) {-sinx + cosx} = 0

-sinx +cosx = 0

-sinx=-cosx

sinx=cosx

sinx/cosx = 0

tanx = 0

x = 45® or π/4 Ans.



b) For nature determination, Differentiate again. That will give you d2y/dx2

e^(-x) {-sinx + cosx}

d2y/dx2 = vu’ + uv’

u = e^(-x)

v = {-sinx + cosx}

u’ = -e^(-x)

v’ = -cosx –sinx

d2y/dx2 = vu’ + uv’

= [{-sinx + cosx}*-e^(-x) ]+[( e^(-x))*( -cosx –sinx)]

e^(-x)[sinx -cosx – cosx –sinx] = 0

[sinx -cosx – cosx –sinx] = 0

-2cosx = 0 Put x = 45 or π/4

-2(π/4) = 0

-π/2 < 0 Therefore this point is maximum (y)
 
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