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Mathematics: Post your doubts here!

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okay
2 - 1/2 + 2ln2 - 3/2ln3
3/2 + 2ln2 - 3/2ln3
3/2 + 1/2 (4ln2 - 3ln3) [i took 1/2 as common factor]
3/2 +1/2 (ln 2^4 - ln3^3)
3/2 + 1/2 (ln 2^4/3^3)
3/2 + 1/2 (ln 16/27)
Im so sorry, I did such a dumb mistake and u had to write this all! :/
Jazak Allah khair! May Allah bless u ! :')
 
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Tried ten to the best of my abilities!! I hope you can do part two now that one is cleared up :D

tumblr_inline_mn1p77GXSj1qz4rgp.jpg
tumblr_inline_mn1p7cTUrE1qz4rgp.jpg
 
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you know that the center of the circle is 2+2i
form a right angled triangle for which the hypotenuse is from the origin to the center of the circle. to find the length of this hypotenuse its sqrt(2^2 +2^2)= sqrt(8)
2 is the radius of the circle. sqrt(8) + 2 should give you the hypotenuse of the larger triangle which you can use to find the ReZ. The adjacent of this triangle which I marked in the diagram is the ReZ.

Feel free to ask questions if you're still confused.
Perfect sense! can't believe I was such an idiot D: Thank you!!
 
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hey can any one prove this.....
1/(sinx cosx) ==> ( secx)whole sqre /tanx
pls some try this thnkx
 
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10i) iw^2= (2-2i)^2
iw^2 = 4 - 2*2*2i + (2i)^2
iw^2 = 4- 8i - 4
iw^2 = - 8i
w^2 = -8
w= sqrt(8)i and -sqrt(8)i

ii) http://imgur.com/gVGylCL

I drew the same diagram twice. The first is for the argument.
The green angle reprsents the smallest possible value and the blue represents the largest possible value.
First find angle 'a' (the black one)
Sin a = 2/√32
a= .362 rad
The green angle is therefore pi/4 - .361 =.424
and the blue angle is pi/4 +.361 = 1.15

In the second diagram, the blue line is the least modulus and the blue+ red line is the greatest possible modulus.
For the blue line, it's √32- 2 = 3.66
and for the red line it's √32 +2 +2 = 7.66

LMGD33 here.
 
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y= lnx/x^2...
we have to use integration by parts ..
u = lnx so du/dx=(1/x)
dv/dx=(1/x^2) so, integral of v=(-1/x)
(i am using ! as the sign of integral)
the formula is: !u(dv/dx) = uv - !v(du/dx)
= lnx(-1/x) - ! (-1/x)(1/x)
= -lnx/x + ! (1/x^2)
obtain the second integral:
= (-lnx/x) - (1/x)
the limits are x=e and x = 1
now u can put in the limits and obtain the answer (y) !
thanks ....u r a life saver...
 
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hey can any one prove this.....
1/(sinx cosx) ==> ( secx)whole sqre /tanx
pls some try this thnkx
i am taking R.H.S:
(secx)^2 = 1/(cosx)^2
so,
[1/(cosx)^2] / tanx..
[1/(cosx)^2] x (1/tanx)
[1/(cosx)^2] x (cosx/sinx)
it becomes : (cosx)/[(cosx)^2 sinx]
simplify u get : 1/cosx sinx
 
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