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WAEC Further Mathematics 2019 Paper

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Question : 11

Total: 39

Given that

\frac{1}{x^2-4} = \frac{p}{x+2} + \frac{Q}{x-2}

$\frac{3}{2}$
1
$\frac{1}{2}$
0

Solution:

$\frac{1}{x^2-4} = \frac{P}{x+2} + \frac{Q}{x-2}$

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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