Question

If $\frac{(2x+3)(3x-4)}{(x-1)(4x+5)}=f\left(x\right)-\frac{5}{9(x-1)}+\frac{31}{18(4x+5)}$ then $f\left(x\right)=k/2$. Find the value of $k$.

Answer 1

$\frac{(2x+3)(3x-4)}{(x-1)(4x+5)}=f\left(x\right)-\frac{5}{9(x-1)}+\frac{31}{18(4x+5)}$
$\Rightarrow \frac{(2x+3)(3x-4)}{(x-1)(4x+5)}=\frac{k}{2}-\frac{5}{9(x-1)}+\frac{31}{18(4x+5)}$
$\Rightarrow \frac{(2x+3)(3x-4)}{(x-1)(4x+5)}=\frac{9k(4x+5)(x-1)-10(4x+5)+31(x-1)}{18(x-1)(4x+5)}4$
$\Rightarrow 18(2x+3)(3x-4)=9k(4x+5)(x-1)-10(4x+5)+31(x-1)\Rightarrow k=2$