TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

If ‌

2x+1

(x−1)2(x2+1)

=‌

A

x−1

+‌

B

(x−1)2

+‌

Cx+D

x2+1

⇒‌‌2x+1=A(x−1)(x2+1)+B(x2+1)
+(Cx+D)(x−1)2
⇒‌‌2x+1=x3(A+C)+x2(−A+B−2C+D)
+x(A+C−2D)+(−A+B+D)
On comparing coefficients of x3,x2,x and constant terms both sides we get.
A+C=0 . . . (i)
−A+B−2C+D=0 . . . (ii)
A+C−2D=2 . . . (iii)
and ‌‌−A+B+D=1 . . . (iv)
On solving Eqs. (i), (ii), (iii) and (iv), we get
A=‌

−1

2

,B=‌

3

2

,C=‌

1

2

and D=−1
∴‌‌A+B+C+D=‌

−1

2

+‌

3

2

+‌

1

2

−1=‌

1

2