TS EAMCET 4-Aug-2021 Shift 2 Question Paper

‌

9x−7

(x+3)(x2+1)

=‌

A

x+3

+‌

Bx+C

x2+1

⇒9x−7=A(x2+1)+(Bx+C)(x+3)
⇒9x−7=Ax2+A+Bx2+3Bx+Cx+3C
Comparing coefficients of x2,x and constant to both the sides, we have
‌A+B=0 . . . (i)
‌3B+C=9 . . . (ii)
A+3C=−7‌‌... (iii)
From Eq. (i),
A=−B
∴ Eq. (ii) becomes, −3A+C=9 . . . (iv)
Solving Eqs. (iii) and (iv), we have
C=‌

−6

5

Now, A+B+C=0+(‌

−6

5

)=‌

−6

5