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

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

Question

(C) Given the partial fraction decomposition: $\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$ Multiplying both sides by $(x+

Vedclass · Accepted Answer

$-\frac{5}{2}$

Vedclass · Answer

(C) Given the partial fraction decomposition: $\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$ Multiplying both sides by $(x+1)^2(x+3)$,we get: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$ Setting $x = -1$: $3(-1)-2 = B(-1+3)$ $\Rightarrow -5 = 2B$ $\Rightarrow B = -\frac{5}{2}$ Setting $x = -3$: $3(-3)-2 = C(-3+1)^2$ $\Rightarrow -11 = 4C$ $\Rightarrow C = -\frac{11}{4}$ Comparing the coefficients of $x^2$ on both sides: $0 = A + C \Rightarrow A = -C = \frac{11}{4}$ Thus,$A+B+C = \frac{11}{4} - \frac{5}{2} - \frac{11}{4} = -\frac{5}{2}$

If $\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$,then $A+B+C=$

Similar Questions

If $\frac{x^4}{(x-1)(x-2)}=f(x)+\frac{A}{x-1}+\frac{B}{x-2}$,then $f(-2)+A+B=$

If $\frac{3x+1}{(x-1)(x^2+2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+2}$,then $5(A-B)=$

If $\frac{1}{x^4+x^2+1}=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{x^2-x+1}$,then $\cos^{-1}(A+B+C+D)=$

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$. Then $A_r$ is equal to:

If $|x| < 1$,then the coefficient of $x^5$ in the expansion of $\frac{3x}{(x-2)(x+1)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module