If frac{A}{x-a}+frac{B x+C}{x^2+b^2}=frac{1}{ | vedclass

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

Question

(C) Given the equation: $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)(x^2+b^2)}$Multiplying both sides by $(x-a)(x^2+b^2)$,we get:$A(x^2+b^2)+(B

Vedclass · Accepted Answer

$\frac{-a}{a^2+b^2}$

Vedclass · Answer

(C) Given the equation: $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)(x^2+b^2)}$Multiplying both sides by $(x-a)(x^2+b^2)$,we get:$A(x^2+b^2)+(B x+C)(x-a)=1$Expanding the terms:$(A+B)x^2+(C-a B)x+(A b^2-a C)=1$Comparing the coefficients of $x^2$,$x$,and the constant term on both sides:$A+B=0$  ....$(i)$$C-a B=0$  ....$(ii)$$A b^2-a C=1$  ....$(iii)$From $(i)$,$B=-A$. Substituting this into $(ii)$:$C-a(-A)=0 \Rightarrow C+a A=0 \Rightarrow A=\frac{-C}{a}$Substituting $A=\frac{-C}{a}$ into $(iii)$:$(\frac{-C}{a})b^2-a C=1$$-C(\frac{b^2+a^2}{a})=1$$C=\frac{-a}{a^2+b^2}$

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

Similar Questions

The coefficient of $x^{3}$ in the infinite series expansion of $\frac{2}{(1-x)(2-x)},$ for $|x| < 1,$ is

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

The partial fraction decomposition of $\frac{9x-7}{(x+3)(x^2+1)}$ is

If $\frac{2 x+7}{\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}=\frac{A x+1}{x^2+4}+\frac{B x+m}{x^2+9}+\frac{C x+n}{x^2+16}$,then $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module