2
-2
1
-1
Correct answer is A
\(\frac{3x + 4}{(x - 2)(x + 3)}≡\frac{P}{x + 3}+\frac{Q}{x - 2}\)
\(\frac{3x + 4}{(x - 2)(x + 3)}=\frac{P(x-2)+Q(x+3)}{(x-2)(x+3)}\)
=\(3x+4=P(x-2)+Q(x+3)\)
=\(3x+4=Px-2P+Qx+3Q\)
=\(3x+4=Px+Qx-2P+3Q\)
=\(3x+4=(P+Q)x-2P+3Q\)
Equating x and the constants
P+Q=3------(i)
2P+3Q=4-------(ii)
From equation (i)
P=3-Q------(iii)
Substitute (3-Q) for P inequation (ii)
=-2(3-Q)+3Q=4
=-6+2Q+3Q=4
=-6+5Q=4
=5Q=4+6
=5Q=10
∴Q=\(\frac{10}{5}=2\)