[7][7]
how bhaiya???
SHOULD
\frac{x^{3}}{(x-1)^{3}(x-2)}
BE WRITTEN AS
\frac{A}{(x-1)^{3}}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)}+\frac{D}{(x-2)}
IF NOT WHAT IS THE CORRECT WAY TO SPLIT IT?
nahi ani i now think u r rite
wat i m saying is true whaen we hav a cubic or quadratic expression
i think[7][7]
ani is right but tat's a long way:::: i dunno whether such ques are asked in JEE
use bhool ja ani main sab kuch mila ke solve karta huin these r my fattas
\frac{x^{3}}{(x-1)^{3}(x-2)} = \frac{A}{(x-1)^{3}}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)}+\frac{D}{(x-2)}
multiply by (x-1)^3, and put x=1
A=1/-1 = -1
mulitply by x-2 and put x=2
8=D
now take x=0,
0=-1/-1 + B/1-c-4
thus, B-C=3
also x=3,
27/8=-1/8+B/4+C/2+8
-9=B/2+C
thus, -9=3b/2-3
-6=3b/2
b=-4
c=-7
see by putting x=4 if this satisfies!
64/27x2 = -1/27-4/9-7/3+8/2
64/27x2 = -1/27-12/27-63/27+8/2
32/27 = -76/27+8/2
108/27=4
which is true..
hence this above holds..
can someone give a good justification?
\frac{x^{2}}{(x-1)^{2}(x-2)} = \frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x-2}
can also be done.. there is no mistake at all..
multiply by (x-1)^2
put x=1
B=-1
now multiply by x-2 and put x=2
4=C
put x=0
0=A/-1 + B/1 + c/-2
0=-A-1-2
A=-3
now substitute any value for x and verify :)
You all are confused becasue it could have been written as
\frac{Ax^2+Bx+C}{(x-1)^3}+\frac{D}{(x-2)}
as well.
exactly bhaiya.....
and i still don't understand why
\frac{x^{2}}{(x-2)^{2}(x-1)}
can't be written as
\frac{Ax+B}{(x-2)^2}+\frac{C}{x-1}
A/(x-2)+B/(x-2)2+C/(x-1) ..(i)
=A(x-2)+B/(x-2)2+C/(x-1)
Ax-2A+B/(x-2)2+C/(x-1)
Let B'=B-2A
Ax+B'/(x-2)2+C/(x-1) ...(ii)
So U can write both i and ii.. botha are correct.. but (i) is preferred since its integration is easy.. :)
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u can if u want to ani.. or ..... but even im having a feeling that it will be wrong.. y not ask prophet sir?
cube ke uper Ax3+Bx2+Cx+D likh
aur square term ke upar
Ax2+Bx+C likh
Samajh me nahin aaya, yaar...
If u don't mind, can u pls represent it? (using latex)
\frac{Ax^{3}+Bx^{2}+Cx+D}{(x-1)^{3}}+\frac{Ex^{2}+Fx+G}{(x-1)^{2}}+\frac{H}{(x-1)}+\frac{I}{(x-2)}