Given that \(\frac{1}{x^2 - 4} = \frac{p}{(x + 2)} + \frac{Q}{(x - 2})\)
x \(\neq \pm 2\)
Find the value of (P + Q)
\(\frac{3}{2}\)
1
\(\frac{1}{2}\)
0
Correct answer is D
\(\frac{1}{x^2 - 4} = \frac{P}{(x + 2)} + \frac{Q}{(x - 2)}\)
I = p(x - 2) + Q(x + 2)
Let x = 2
I = P(2 - 2) + Q(2+ 2)
I = -4Q
Q = \(\frac{1}{4}\)
Let x = -2
I = P(-2 - 2) + Q(-2 + 2)
I = -4p
P = \(\frac{1}{-4}\)
PQQ = - \(\frac{1}{4} + \frac{1}{4}\)
= 0