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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:

Concept:

Use partial fractions and equate numerators, then substitute convenient values of

x

to find

P

and

Q

.

Explanation:

Since

x^2 - 4 = (x-2)(x+2)

, we have:

\frac{1}{(x-2)(x+2)} = \frac{P}{x+2} + \frac{Q}{x-2}

Multiply both sides by

(x-2)(x+2)

:

1 = P(x-2) + Q(x+2)

Substitute

x = 2

to eliminate

P

:

1 = Q(4)

Q = \frac{1}{4}

Substitute

x = -2

to eliminate

Q

:

1 = P(-4)

P = -\frac{1}{4}

Thus:

P + Q = -\frac{1}{4} + \frac{1}{4} = 0

Answer:

0

(Option D)

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