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

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

Question

(A) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ Equating the numerators: $3x+2 = A(2x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ Equating the numerators: $3x+2 = A(2x^2+3) + (x+1)(Bx+C)$ $3x+2 = 2Ax^2 + 3A + Bx^2 + Cx + Bx + C$ $3x+2 = (2A+B)x^2 + (B+C)x + (3A+C)$ Comparing coefficients: $2A+B = 0$ $(i)$ $B+C = 3$ (ii) $3A+C = 2$ (iii) From $(i)$,$B = -2A$. Substitute into (ii): $-2A+C = 3$ (iv) Subtract (iv) from (iii): $(3A+C) - (-2A+C) = 2 - 3 \implies 5A = -1 \implies A = -\frac{1}{5}$ Then $B = -2(-\frac{1}{5}) = \frac{2}{5}$ And $C = 3 - B = 3 - \frac{2}{5} = \frac{13}{5}$ Finally,$A-B+C = -\frac{1}{5} - \frac{2}{5} + \frac{13}{5} = \frac{10}{5} = 2$

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

Similar Questions

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

Let $a, b$, and $c$ be such that $\frac{1}{(1-x)(1-2x)(1-3x)} = \frac{a}{1-x} + \frac{b}{1-2x} + \frac{c}{1-3x}$. Then $\frac{a}{1} + \frac{b}{3} + \frac{c}{5}$ is equal to:

If $\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=a+\frac{b}{x+p}+\frac{c}{q x+3}+\frac{d}{3 x+p}$ then $\frac{a+b}{p+q}=$

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

If $\frac{3x^3-7x+1}{(x-2)^5} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4} + \frac{E}{(x-2)^5}$,then $A(B+C+D+E) =$ ?

Vedclass Test Series

Exam Paper Generator

Online Exam Module