Mathematics: Post your doubts here!

[asexamskillme111]

man u serious ? u r an A level student and u dont know how to integrate (x^2-2x)(1/x)..

(x^2-2x)/x = x(x-2)/x cancel the x's!! then u will have x-2 and thats pretty easy to integrate eh..i guess u were a bit confused right ?

LOL. No idea what I was thinking. Must've had a dumb moment.

 

[CaptainDanger]

If y = tan^-1 x, show that dy/dx = (1/x^2 + 1)?

 

[Kandinsky]

If y = tan^-1 x, show that dy/dx = (1/x^2 + 1)?

Should use the identity (dy/dx) * (dx/dy) = 1.
In this case x = tan y. then dx/dy = sec^2 y

hence dy/dx = cos^2 y now substitute y = tan^(-1) x:

dy/dx = cos^2( tan^(-1) x).
denote the angle tan^(-1) x as B and find cos^2 B if tan B = x;
using identity sec^2(B) = 1 + tan^2 B, obtain

dy/dx = 1/(1+x^2)

 

[iKhaled]

how to solve question 9 ( ii) http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

 

[Kandinsky]

how to solve question 9 ( ii) http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s08_qp_3.pdf

(-1/2;0) is the point of intersection of the graph with x-axis hence the lower limit is -1/2
it's given by:

pi* integral from -1/2 to 0 of y^2

 

[iKhaled]

(-1/2;0) is the point of intersection of the graph with x-axis hence the lower limit is -1/2
it's given by:

pi* integral from -1/2 to 0 of y^2

duuuhhh -_- i know that but how can i integrate that ? i know we have to use integration by parts but it is not working for me !

 

[Kandinsky]

why not. take exponent as dv and the polynomial as u:

integral u dv = uv - integral v du

 

[iKhaled]

why not. take exponent as dv and the polynomial as u:

integral u dv = uv - integral v du

yeah i tried to but its failing..can u do it please ?

 

[Kandinsky]

so your integral is given by
I = integral(-0.5 ; 0) of e^(-x) * (1+2x) dx
u = 2x+1, dv = e^(-x) dx. hence du = 2dx and v = -e^(-x)

I = (2x+1)*(-e^(-x)) +2 integral (e^(-x) dx)
I = -e^(-x) * (2x+1+2) = -e^(-x)*(2x+3)
inserting limits, obtain:

I = -3 + e^(0.5)*2;
hence the volume is pi*(2 sqrt(e) - 3)

 

[iKhaled]

can someone please explain to me the first question in this paper ?
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

 

[Tharindu]

question 5) (ii) May June, paper 3 (M1) ....how do we find the tension of S2 ?

 

[whitecorp]

can someone please explain to me the first question in this paper ?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf

ln( 2+ e^-x )=2
2+e^-x = e^2
e^-x = e^2 - 2
-x = ln (e^2 - 2)
Hence, x = - ln (e^2 - 2)= -1.68 (shown)

Hope this helps. Peace.

 

[whitecorp]

question 5) (ii) May June, paper 3 (M1) ....how do we find the tension of S2 ?

Posting the question paper would be nice, you know?

 

[iKhaled]

ln( 2+ e^-x )=2
2+e^-x = e^2
e^-x = e^2 - 2
-x = ln (e^2 - 2)
Hence, x = - ln (e^2 - 2)= -1.68 (shown)

Hope this helps. Peace.

how did uk that 2+e^-x = e^2

 

[Yousif Mukkhtar]

Guys for this question in the attached file why cant we used this method:
S(x-x bar)^2
= Sx^2-x bar^2

Mean of x= Sx/n=645/150=4.3
x bar^2= 4.3^2=18.49
S(x-x bar)^2=8287.5/18.49
=8269.01

 

[whitecorp]

how did uk that 2+e^-x = e^2

You simply take the exponential base on both sides.

Hope this helps. Peace.

 

[iKhaled]

how this stupid question can be solved!!!

(2^x + 1 ) / (2^x - 1 ) = 5

first question in m/j 2010 32

 

[VelaneDeBeaute]

how this stupid question can be solved!!!

(2^x + 1 ) / (2^x - 1 ) = 5

first question in m/j 2010 32

Studied logs? Well, try it that way.

 

[iKhaled]

Studied logs? Well, try it that way.

yeah nevermind..i got it now i just didn't know how to multiply 5(2^x-1)

 

[Alice123]

how do i differentiate ln(xy) with respect to x???