Question

Resolve into partial fractions $\frac{23x-11x^2}{(2x-1)(9-x^2)}$

• A. $\frac{1}{(2x-1)}+\frac{-1}{(3-x)}+\frac{4}{(3+x)}$
• B. $\frac{-1}{(2x-1)}+\frac{-1}{(3-x)}+\frac{4}{(3+x)}$
• C. $\frac{4}{(2x-1)}+\frac{-1}{(3-x)}+\frac{1}{(3+x)}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{(2x-1)}+\frac{-1}{(3-x)}+\frac{4}{(3+x)}$

We know $\frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{23x-11x^2}{(2x-1)(3-x)(3+x)}$
Let $\frac{23x-11x^2}{(2x-1)(x-3)(x+3)}=\frac{A}{(2x-1)}+\frac{B}{(3-x)}+\frac{C}{(3+x)}$ .....(1)
$23x-11x^2=A(3-x)(3+x)+B(2x-1)(3+x)+C(2x-1)(3-x)23x-11x^2=A\left({9-x}^2\right)+B(2x^2+5x-3)+C(-2x^2+7x-3)$
On comparing coefficients we get
$-11=-A+2B-2C,23=5B+7C,0=9A-3B-3C$
Solving them we get
$A=1,B=-1$ and $C=4$

Putting this values in equation (1), we get

$\frac{23x-11x^2}{(2x-1)(x-3)(x+3)}=\frac{1}{(2x-1)}+\frac{-1}{(3-x)}+\frac{4}{(3+x)}$
Hence, option 'A' is correct.