Question

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

• A. $\frac{-1}{13}$
• B. $\frac{-2}{13}$
• C. $\frac{6}{13}$
• D. $\frac{1}{13}$

Answer 1

The correct option is B $\frac{-2}{13}$

Consider the given equation.

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

$x=A(1+x^2)+(Bx+C)(3-2x)$

$x=A+A{x}^2+3Bx-2B{x}^2+3C-2Cx$ …….. (1)

On comparing coefficient of $x^2$, we get

$0=A-2B$

$A=2B$ ……. (2)

On comparing coefficient of $x$, we get

$1=3B-2C$ …….. (3)

On comparing constant term, we get

$0=A+3C$ ……. (4)

From equation (2) and (4), we have

$0=2B+3C$

$B=-\frac{3}{2}C$

On putting $B$ in equation (3), we get

$1=3\left(\frac{-3}{2}C\right)-2C$

$1=\left(\frac{-9}{2}C\right)-2C$

$1=\left(\frac{-13}{2}C\right)$

$C=-\frac{2}{13}$

Hence, the value of $C$ is $-\frac{2}{13}$.