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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
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Correct Answer: Option D
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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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts