If 9x-7/(x+3)(x²+1) = A/x+3 + Bx+C/x²+1 where A, B, C R, then A+B+C =

Correct Answer is : −65-6/55−6

Correct Answer is : −65-6/55−6

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TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 19 of 160

Marks: +1, -0

\frac{9x-7}{(x+3)(x^2+1)}

\frac{A}{x+3} + \frac{Bx+C}{x^2+1}

A, B, C \in R

A+B+C

Solution:

\;\frac{9x-7}{(x+3)(x^{2}+1)} = \;\frac{A}{x+3} + \;\frac{Bx+C}{x^{2}+1}

\Rightarrow 9x-7 = A(x^{2}+1) + (Bx+C)(x+3)

\Rightarrow 9x-7 = Ax^{2}+A+Bx^{2}+3Bx+Cx+3C

Comparing coefficients of

x^{2}, x

and constant to both the sides, we have

\; A+B=0

. . . (i)

\; 3B+C=9

. . . (ii)

A+3C=-7 \;\; \dots

(iii)

From Eq. (i),

A = -B

\therefore

Eq. (ii) becomes,

-3A + C = 9

. . . (iv)

Solving Eqs. (iii) and (iv), we have

C = \;\frac{-6}{5}

Now,

A+B+C = 0 + \left(\;\frac{-6}{5}\right) = \;\frac{-6}{5}

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