Diploma 1st year
question
RESOLVE INTO PARTIAL FRACRION?
x+23x
'(x+3) (x+1)*
Answers
Answered by
2
Answer:
Since x
2
+2 can not be factored
So,
(x−1)(x
2
+2)
2x
2
+3x+4
=
x−1
A
+
x
2
+2
Bx+C
Now cross multiply, 2x
2
+3x+4=A(x
2
+2)+(Bx+C)(x−1)
⇒2x
2
+3x+4=A(x
2
+2)+Bx
2
−Bx+Cx−C
⇒2x
2
+3x+4=(A+B)x
2
+(C−B)x+(2A+C)
Comparing the coefficients, A+B=2
(1),C−B=3(2) and 2A−C=4 (3)
Adding the three equations, we get
3A=9⇒A=3
From (1), A+B=2⇒3+B=2⇒B=−1
From (3), 2A−C=4⇒6−C=4⇒C=2
A=3,B=−1,C=2
Hence,
(x−1)(x +2)
2x 2 +3x=
x−1+ x+2−x+2
Attachments:
Answered by
0
Answer:
Resolve into partial fractions
A
B
C
D
Medium
Solution
Verified by Toppr
Correct option is
C
Let
On comparing coefficients we get
Hence
Hence, option 'C' is correct.
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