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Mathematics: Post your doubts here!

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How to find MAX OR MIN point or Stationary points in differntiation of AS level?
Please help me:(

By finding the dy/dx of an equation and then solving dy/dx=0 you will get the min or max point (stationary point) or both depending on the equation of the curve. After that if you differentiate dy/dx you will get the equation of d^2y/dx^2 and when you put the x values you got from the earlier dy/dx=0 in the d^2y/dx^2 you will get an answer. If this is +ve then it is a minimum point while if it is -ve then it is the maximum point..
 
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Plz someone write the fully working for this and explain me!
It is October November 10 Paper 11!
 
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Ok i am seriously stuck in this question and some help the marking scheme turned out to be. Can anyone help me solve the following question.

View attachment 22000

As OAB is a straight line the point O, A and B are co-linear. You can write it as OA= sOB, where s is a constant because OA and OB have the same direction vector.
so,
(p 1 1) = s (4 2 p)
You can form 2 equation from this.
p =4s -----1
1= 2s ----2
from eq 2
s = .5 now substitute this in eq 1 and you get p as 2

So, OA= (2 1 1), unit vector in the direction OA = {1/√(4+1+1)} * (2 1 1) = 1/√6 * (2 1 1)
 
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Plz someone write the fully working for this and explain me!
It is October November 10 Paper 11
 
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Take the integral:
(x+1/x)^2 dx
For (x+1/x)^2, do long division:
= (x^2 + 1/x^2 + 2) dx
= 1/x^2 dx + x^2 dx+2 1 dx
The integral of 1 is x:
= 1/x^2 dx + x^2 dx + 2 x
The integral of 1/x^2 is -1/x:
= x^2 dx + 2 x - 1/x
The integral of x^2 is x^3/3:
Answer: x^3/3 + 2x - 1/x + constant
 

Dug

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Can anyone help me for Maths 9709 June 2006 P3 no. 5 ? plz :(
x = 2Θ + sin2Θ
y = 1 - cos2Θ

dx/dΘ = 2 + 2cos2Θ
dy/dΘ = 2sin2Θ

dy/dx = dy/dΘ . dΘ/dx

dΘ/dx = 1/(2 + 2cos2Θ)

dy/dx = 2sin2Θ . 1/(2 + 2cos2Θ)

Prove that,
2sin2Θ . 1/(2 + 2sinΘcosΘ) = tanΘ

L.H.S = 2(2sinΘcosΘ)/[2(1 + cos2Θ)]
= 2sinΘcosΘ/(1 + cos2Θ)

We know that cos2Θ = 2cos^2(Θ) - 1

2sinΘcosΘ/(1+2cos^2(Θ) - 1)

Simplify it and you get tanΘ.
 
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x = 2Θ + sin2Θ
y = 1 - cos2Θ

dx/dΘ = 2 + 2cos2Θ
dy/dΘ = 2sin2Θ

dy/dx = dy/dΘ . dΘ/dx

dΘ/dx = 1/(2 + 2cos2Θ)

dy/dx = 2sin2Θ . 1/(2 + 2cos2Θ)

Prove that,
2sin2Θ . 1/(2 + 2sinΘcosΘ) = tanΘ

L.H.S = 2(2sinΘcosΘ)/[2(1 + cos2Θ)]
= 2sinΘcosΘ/(1 + cos2Θ)

We know that cos2Θ = 2cos^2(Θ) - 1

2sinΘcosΘ/(1+2cos^2(Θ) - 1)

Simplify it and you get tanΘ.

Thanks, but I think you missed the question. Its no. 5 about differential equation :p
 

Dug

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Thanks, but I think you missed the question. Its no. 5 about differential equation :p
Sorry... :(
Q5)

i)
dx/dt = kx - 25
75 = 1000k - 25
k = 0.1

dx/dt = 0.1x - 25 = 0.1(x - 250)

ii)
Separating variables:
1/(x - 250) dx = 0.1 dt
⌡1/(x - 250) dx = ⌡0.1 dt

ln|x - 250| = 0.05t + c

When t = 0, x = 1000.
ln|1000 - 250| = c
c = ln|750|

ln|x - 250| = 0.05t + ln|750|
ln|x - 250| - ln|750| = 0.05t
ln|(x-250)/750| = 0.05t
e^(0.05t) = (x-250)/750
750e^(0.05t) + 250 = x
x = 250(3e^0.05t + 1)
 
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Sorry... :(
Q5)

i)
dx/dt = kx - 25
75 = 1000k - 25
k = 0.1

dx/dt = 0.1x - 25 = 0.1(x - 250)

ii)
Separating variables:
1/(x - 250) dx = 0.1 dt
⌡1/(x - 250) dx = ⌡0.1 dt

ln|x - 250| = 0.05t + c

When t = 0, x = 1000.
ln|1000 - 250| = c
c = ln|750|

ln|x - 250| = 0.05t + ln|750|
ln|x - 250| - ln|750| = 0.05t
ln|(x-250)/750| = 0.05t
e^(0.05t) = (x-250)/750
750e^(0.05t) + 250 = x
x = 250(3e^0.05t + 1)

Thank you loads genius :)
 
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Can anyone help me out with the b part of this Q. I have my exams tomorrow and I cant figure this out :(Untitled.jpg
 
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Can anyone help me out with the b part of this Q. I have my exams tomorrow and I cant figure this out :(View attachment 22026

(ii) Any unit vector has a magnitude of 1. If AB is a unit vector |AB| = 1
AB = (k -k 2k) - (1 0 2) = (k-1, -k, 2k-2)
|AB| = √{(k-1)^2 + (-k)^2 + (2k-2)^2}
1 = √(k^2 -2k + 1 + k^2 + 4k^2 -8k +4)
1^2 = 6k^2 - 10k +5
1 = 6k^2 - 10k + 5
6k^2 -10k +4 =0
Solve and you get k as 1 and 2/3
 
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