Explanation

(b)
Given, $\frac{3x + 2}{(x + 1)(2x^{2} + 3)} = \frac{A}{(x + 1)} + \frac{Bx + C}{(2x^{2} + 3)}$
$\Rightarrow \ \ 3x + 2 = A\left( 2x^{2} + 3 \right) + (Bx + C)(x + 1)$
On putting $x + 1 = 0\ ie,\ x = - 1$
We get $3( - 1) + 2 = A\left\lbrack 2( - 1)^{2} + 3 \right\rbrack$
$\Rightarrow \ \ \ A = - \frac{1}{5}$
Now, on comparing the coefficients of $x^{2}$ and $x$, we get
$0 = 2A + B$
$\Rightarrow \ \ \ \ B = \frac{2}{5}$
And $3 = B + C$
$\Rightarrow \ \ \ C = 3 - \frac{2}{5} = \frac{13}{5}$
$\therefore\ \ \ \ A + C - B = - \frac{1}{5} + \frac{13}{5} - \frac{2}{5} = \frac{10}{5} = 2$