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Question : 19 of 160
Marks:
+1,
-0
Solution:
=+ ⇒ =| (Ax+B)(x2+2x+2)+(Cx+D)(x2−2x+2) |
| (x2+2)−2x2 |
⇒x2+1=(A+C)x2+(2A+B−2C+D)x2 +(2A+2B+2C−2D)x+(2B+2D) On Comparing both sides:
A+C=0⇒A=−C,2B+2D=1 and
2A+B−2C+D=1 ⇒2(A−C)+B+D=1 ⇒4A+=1 ⇒4A=1−= ⇒A= and
C=− Now,
2A+2B+2C−2D=0 ⇒2(A+C)+2(B−D)=0 ⇒B=D and
2B+2D=1 ⇒4B=1⇒B=D= Now,
3A+2B+3C=2B ∵3A+3C=0 =2D
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