Qualification > Math
Pure 3 help.
mousa:
--- Quote from: Requiem on October 02, 2010, 01:25:42 am ---PS:This is my teacher's work.
I have just started studying this chapter.
More will come tomorrow
Sorry i have exams
and the working was too much.
Cheers
--- End quote ---
Thanks Alott MAAN!!!
mousa:
--- Quote from: Deadly_king on October 02, 2010, 04:55:50 am ---Jun 09
3.
(i) Let pheta be x
Prove cosec 2x +cot 2x = cot x
cosec 2x + cot 2x = 1/(sin 2x) + (cos 2x)/(sin 2x)
Take sin 2x as common denominator to get (1 + cos 2x)/(sin 2x)
Use double angle formula sin 2x = 2sinx cosx and 1 + cos 2x = 2cos2x
You'll get (2cos2x)/(2sinx cosx)
Eliminate cosx from the numerator and denominator to obtain cosx / sinx = cot x
(ii) cosec 2x + cot 2x = 2
This also implies that cot x = 2 ----> 1/tanx = 2
therefore tanx = 1/2
Key angle x = tan-1(1/2) = 26.6o
tan is positive in first and third quadrant.
Hence x = pheta = 26.6o and 206.6o
8.
Let 100/x2(10-x) = A/x + B/x2 + C/(10-x)
Take x2(10-x) as common denominator on the right hand side :
A(x2(10-x)) + B((10-x) + C(x2) = 100
Now to find C : take x=10 in the equation above -----> C=1
Now to find B : take x=0 ----> B 10
Now to find A : take x=1 ----> 9A + 9B + C = 100
Replace values of B and C calculated above to obtain A=1
Hence 100/x2(10-x) = 1/x + 10/x2 + 1/(10-x)
(ii) Give dx/dt = 1/100*x2(10-x)
Send the terms in x to the left side and you'll obtain :
100/x2(10-x) dx = dt
Use the answer obtained in part (i) to replace 100/x2(10-x)
Integrate on both sides to get :
ln x - 10/x + ln(10-x) = t + c
Given when x=1, t=0 ----> replace in the equation to get the value of c
c = -10 - ln 9
Therefore ln x - 10/x - ln(10-x) = t -10 - ln 9
Make t subject of formula and use logarithm rules to obtain answer as :
t = ln (9x/(10-x)) - 10/x + 10
--- End quote ---
Thnaks amillion!! but can you plz help me in the other questions???! ???
Deadly_king:
--- Quote from: mousa on October 02, 2010, 05:50:25 am ---
Thnaks amillion!! but can you plz help me in the other questions???! ???
--- End quote ---
the questions are quite long. Thus i'll take some time to answer all of them. I've already done all of them but i don't have a scanner. So i'll have to post them 1 by 1.
If you could specify the parts that you don't understand we can work it out faster :)
mousa:
--- Quote from: Deadly_king on October 02, 2010, 05:57:56 am ---the questions are quite long. Thus i'll take some time to answer all of them. I've already done all of them but i don't have a scanner. So i'll have to post them 1 by 1.
If you could specify the parts that you don't understand we can work it out faster :)
--- End quote ---
Sorry man, but I dont know all of the parts, they are quite annoying.I have been studying the subject alone this summer and my exams are this NOV session!!
Please do you have any good resources to revise from or tips??
Thanks
Deadly_king:
--- Quote from: mousa on October 02, 2010, 06:00:51 am ---Sorry man, but I dont know all of the parts, they are quite annoying.I have been studying the subject alone this summer and my exams are this NOV session!!
Please do you have any good resources to revise from or tips??
Thanks
--- End quote ---
Well the CIE textbook can be quite helpful. But alone it is quite difficult to understand all that. A teacher would have been really helpful!
Anyway i'll be modifying my posts now and then until I answer all the questions.
However if you do not understand something.......do ask me and i'll try to elaborate more....ok??
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