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

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

Question

(B) 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 b

Vedclass · Accepted Answer

$-5$

Vedclass · Answer

(B) 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: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$.Expanding the terms: $3x-2 = A(x^2+4x+3) + B(x+3) + C(x^2+2x+1)$.Equating coefficients of $x^2$: $A + C = 0 \Rightarrow C = -A$.Equating coefficients of $x$: $4A + B + 2C = 3$.Equating constants: $3A + 3B + C = -2$.Substitute $C = -A$ into the other equations:$4A + B - 2A = 3$ $\Rightarrow 2A + B = 3$ $\Rightarrow B = 3 - 2A$.$3A + 3(3 - 2A) - A = -2$ $\Rightarrow 3A + 9 - 6A - A = -2$ $\Rightarrow -4A = -11$ $\Rightarrow A = \frac{11}{4}$.Then $C = -\frac{11}{4}$ and $B = 3 - 2(\frac{11}{4}) = 3 - \frac{11}{2} = -\frac{5}{2}$.We need to find $4A + 2B + 4C = 4(\frac{11}{4}) + 2(-\frac{5}{2}) + 4(-\frac{11}{4}) = 11 - 5 - 11 = -5$.

Given $\frac{3x-2}{(x+1)^2(x+3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+3}$,find the value of $4A + 2B + 4C$.

Similar Questions

If $\frac{ax+5}{(x^2+b)(x+3)}=\frac{x+21}{12(x^2+b)}+\frac{c}{12(x+3)}$,then $b^2=$

If $\frac{x+2}{x^2-3}$ is one of the partial fractions of $\frac{3x^3-x^2-2x+17}{x^4+x^2-12}$,then the other partial fraction is:

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

The partial fraction of $\frac{x^2}{x^2+3x-4}$ is

If $\frac{3x^2+x+1}{(x-1)^4} = \frac{a}{(x-1)} + \frac{b}{(x-1)^2} + \frac{c}{(x-1)^3} + \frac{d}{(x-1)^4}$,then $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module