Mathematics: Post your doubts here!

[asexamskillme111]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf

question 7 part (iii)

i dont know how to go about solving the question. help pleaase

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf

question 7 part (iii)

i dont know how to go about solving the question. help pleaase

7iii)
As the question states that the argument of z is least, z must represent the lowest possible point in the shaded region. Draw a tangent from the center to the lowest point on the circle (this doesn't have to be accurate) as shown by the blue line. This line represents |z|. The angle between the radius and the tangent is always 90 so a right-angled triangle is formed. The opposite side, with is equal to the radius is 1 unit. The magnitude of complex u is the hypotenuse and the adjacent represents |z|. you can easily find |u|, √(2^2+2^) = √8.
So, |z| =√(√8)^2 -1^2 = √8-1 =√7

 

[Sarkerms]

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.

 

[minie23]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

Please help me for no. 5

 

[Ahmedraza73]

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

 

[Sarkerms]

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..

 

[Ahmedraza73]

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

iz this Question of AS lEVEL OR A LEVEL?

 

[Sarkerms]

iz this Question of AS lEVEL OR A LEVEL?

As Level

 

[Manobilly]

Plz someone write the fully working for this and explain me!
It is October November 10 Paper 11!

 

[littlecloud11]

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)

 

[cool Asviva]

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

What's the answer in marking scheme..? My answers are p =1/2 and unit vector is ( 1/3, 1/6, 1/6)

 

[Manobilly]

Plz someone write the fully working for this and explain me!
It is October November 10 Paper 11

 

[Manobilly]

This !

 

[Gémeaux]

This !

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

 

[Sarkerms]

What's the answer in marking scheme..? My answers are p =1/2 and unit vector is ( 1/3, 1/6, 1/6)

The answer is 2. Thanks anyways, I got the answer from littlecloud121

 

[minie23]

Can anyone help me for Maths 9709 June 2006 P3 no. 5 ? plz

 

[Dug]

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.

 

[minie23]

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

 

[Dug]

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

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)

 

[minie23]

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