Given that \(\frac{1}{x^2 - 4} = \frac{p}{(x + 2)} + \frac{Q}{(x - 2})\) x \(\neq
FURTHER MATHEMATICS
WAEC 2019
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)
- A. \(\frac{3}{2}\)
- B. 1
- C. \(\frac{1}{2}\)
- D. 0
Correct Answer: D. 0
Explanation
\(\frac{1}{x^2 - 4} = \frac{P}{(x + 2)} + \frac{Q}{(x - 2)}\)
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
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