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=\(\frac{10}{5}=2\)
Forces \(F_{1} = (8N, 030°)\) and \(F_{2} = (10N, 150°)\) act on a particle. Find the horizo...
Express \(\frac{4π}{2}\) radians in degrees....
Find the coefficient of \(x^{3}\) in the expansion of \([\frac{1}{3}(2 + x)]^{6}\)...
Find the variance of 1, 2, 0, -3, 5, -2, 4....
Find the equation of a circle with centre (-3, -8) and radius \(4\sqrt{6}\)...
How many ways can 6 students be seated around a circular table? ...