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

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

Question

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

Vedclass · Accepted Answer

$B$

Vedclass · Answer

(D) Given the partial fraction decomposition: $\frac{x-2}{x^2(2x-3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{2x-3}$.Multiplying both sides by $x^2(2x-3)$, we get: $x-2 = Ax(2x-3) + B(2x-3) + Cx^2$.Setting $x = 0$: $-2 = B(-3) \Rightarrow B = \frac{2}{3}$.Setting $x = \frac{3}{2}$: $\frac{3}{2} - 2 = C(\frac{3}{2})^2$ $\Rightarrow -\frac{1}{2} = C(\frac{9}{4})$ $\Rightarrow C = -\frac{2}{9}$.Comparing the coefficients of $x^2$: $0 = 2A + C$ $\Rightarrow 2A = -C = \frac{2}{9}$ $\Rightarrow A = \frac{1}{9}$.Now, calculate $2(A-C) = 2(\frac{1}{9} - (-\frac{2}{9})) = 2(\frac{1+2}{9}) = 2(\frac{3}{9}) = 2(\frac{1}{3}) = \frac{2}{3}$.Since $B = \frac{2}{3}$, we have $2(A-C) = B$.

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

Similar Questions

If $\frac{(x - a)(x - b)}{(x - c)(x - d)} = \frac{A}{x - c} - \frac{B}{x - d} + C$,then $C =$

The fraction $\frac{x^2}{(x-a)(x-b)}$ is

If $\frac{x^3}{(2x - 1)(x + 2)(x - 3)} = p + \frac{q}{2x - 1} + \frac{r}{x + 2} + \frac{s}{x - 3}$,then:

If $\frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 2} - \frac{5}{2x + 3}$,then:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module