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

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the partial fraction decomposition: $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$.Multiplying both sides by $(x+1)(2x^2+3)$,we get: $3x+2 = A(2x^2+3) + (x+1)(Bx+C)$.Setting $x = -1$: $3(-1)+2 = A(2(-1)^2+3) + 0$ $\Rightarrow -1 = 5A$ $\Rightarrow A = -\frac{1}{5}$.Comparing the coefficients of $x^2$: $0 = 2A + B \Rightarrow B = -2A = -2(-\frac{1}{5}) = \frac{2}{5}$.Comparing the constant terms: $2 = 3A + C \Rightarrow C = 2 - 3A = 2 - 3(-\frac{1}{5}) = 2 + \frac{3}{5} = \frac{13}{5}$.Now,calculate $A-B+C = -\frac{1}{5} - \frac{2}{5} + \frac{13}{5} = \frac{-1-2+13}{5} = \frac{10}{5} = 2$.

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 two sides of a triangle are represented by $3x^2-5xy+2y^2=0$ and its orthocentre is $(2,1)$,then the equation of the third side is

If $f(x) = (p - x^n)^{1/n}$,$p > 0$ and $n$ is a positive integer,then $f[f(x)]$ is equal to

If a gas is compressed isothermally,then the r.m.s. velocity of the molecules

Let $\alpha, \beta$ be the roots of the equation $x^2-px+r=0$ and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2-qx+r=0$. Then the value of $r$ is

If a line makes angles $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with the $X$-axis and $Y$-axis respectively, then the angle made by the line with the $Z$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module