If frac{9x-7}{(x+3)(x^2+1)} = frac{A}{x+3} + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given,$\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$ Equating the numerators: $9x-7 = A(x^2+1) + (Bx+C)(x+3)$ $9x-7 = Ax^2 + A +

Vedclass · Accepted Answer

$\frac{-6}{5}$

Vedclass · Answer

(B) Given,$\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$ Equating the numerators: $9x-7 = A(x^2+1) + (Bx+C)(x+3)$ $9x-7 = Ax^2 + A + Bx^2 + 3Bx + Cx + 3C$ $9x-7 = (A+B)x^2 + (3B+C)x + (A+3C)$ Comparing coefficients on both sides: $A+B = 0$ $3B+C = 9$ $A+3C = -7$ From $A+B=0$,we get $B = -A$. Substituting $B = -A$ into $3B+C=9$,we get $-3A+C=9$,so $C = 9+3A$. Substituting $C = 9+3A$ into $A+3C=-7$: $A + 3(9+3A) = -7$ $A + 27 + 9A = -7$ $10A = -34 \implies A = -\frac{34}{10} = -\frac{17}{5}$ Then $B = -A = \frac{17}{5}$ And $C = 9 + 3(-\frac{17}{5}) = 9 - \frac{51}{5} = \frac{45-51}{5} = -\frac{6}{5}$ Finally,$A+B+C = -\frac{17}{5} + \frac{17}{5} - \frac{6}{5} = -\frac{6}{5}$

If $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$,where $A, B, C \in \mathbb{R}$,then $A+B+C = $

Similar Questions

What is the quotient when $x^3-5x^2+2x+7$ is divided by $(x-1)$?

If $\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$ and $a>b$ then $B+D=$

The partial fraction of $\frac{3x - 1}{(1 - x + x^2)(2 + x)}$ is:

If $\frac{ax - 1}{(1 - x + x^2)(2 + x)} = \frac{x}{1 - x + x^2} - \frac{1}{2 + x}$,then $a = $

If $\frac{x^2+x+1}{x^2+2x+1}=A+\frac{B}{x+1}+\frac{C}{(x+1)^2}$,then $A-B$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module