resolve into partial fraction X + 1/ (x+2) (x+3)
Answers
Given faction is a proper fraction.
Let
(x−1)(x−2)(x−3)
x
2
=
x−1
A
+
x−2
B
+
x−3
C
....(i)
Then
(x−1)(x−2)(x−3)
x
2
=
(x−1)(x−2)(x−3)
A(x−2)(x−3)+B(x−3)(x−1)+C(x−1)(x−2)
⟹x
2
=A(x−2)(x−3)+B(x−3)(x−1)+C(x−1)(x−2) ....(ii)
Substituting x−1=0orx=1 in (ii), we get
1
2
=a(1−2)(1−3)+0+0
⟹A=
2
1
Substituting x−2=0 or x=2 in (ii), we get
2
2
=0+B(2−3)(2−1)+0
∴B=−4
Substituting x−3=0 or x=3 in (ii), we get
3
2
=0+0+C(3−1)(3−2)
∴C=
2
9
Substituting the value of A, B, and C in (i) then
(x−1)(x−2)(x−3)
x
2
=
2(x−1)
1
−
(x−2)
4
+
2(x−3)
9
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Answer:
Correct option is B)
Given faction is a proper fraction.
Let
(x−1)(x−2)(x−3)
x
2
=
x−1
A
+
x−2
B
+
x−3
C
....(i)
Then
(x−1)(x−2)(x−3)
x
2
=
(x−1)(x−2)(x−3)
A(x−2)(x−3)+B(x−3)(x−1)+C(x−1)(x−2)
⟹x
2
=A(x−2)(x−3)+B(x−3)(x−1)+C(x−1)(x−2) ....(ii)
Substituting x−1=0orx=1 in (ii), we get
1
2
=a(1−2)(1−3)+0+0
⟹A=
2
1
Substituting x−2=0 or x=2 in (ii), we get
2
2
=0+B(2−3)(2−1)+0
∴B=−4
Substituting x−3=0 or x=3 in (ii), we get
3
2
=0+0+C(3−1)(3−2)
∴C=
2
9
Substituting the value of A, B, and C in (i) then
(x−1)(x−2)(x−3)
x
2
=
2(x−1)
1
−
(x−2)
4
+
2(x−3)
9
Hence, option 'B' is correct.