WAEC 2009 · Paper 2 · Q9

(a)
Express $\dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)}$ in partial fractions.
(b)
If $y = \sqrt{x + 1}$ , determine the value of $x$ for which $\dfrac{dy}{dx} = y$ .

Worked solution (try it first)

(a)

The factor $x^2 + 1$ doesn't factorise, so write $\dfrac{4x^2 - x + 3}{(x^2 + 1)(x - 1)} = \dfrac{P}{x - 1} + \dfrac{Qx + R}{x^2 + 1}$ .
Multiply through by the denominator: $4x^2 - x + 3 = P(x^2 + 1) + (Qx + R)(x - 1)$ .
Put $x = 1$ : $4 - 1 + 3 = 2P$ , so $P = 3$ .
Compare $x^2$ terms: $4 = P + Q$ , so $Q = 1$ .
Compare constants: $3 = P - R$ , so $R = 0$ .
So the partial fractions are $\dfrac{3}{x - 1} + \dfrac{x}{x^2 + 1}$ .

(b)

$y = (x + 1)^{\frac12}$ , so $\dfrac{dy}{dx} = \frac12(x + 1)^{-\frac12}$
$= \dfrac{1}{2\sqrt{x + 1}}$ .
Set $\dfrac{1}{2\sqrt{x + 1}} = \sqrt{x + 1}$ and multiply by $2\sqrt{x + 1}$ : $1 = 2(x + 1)$ .
So $x + 1 = \frac12$ and $x = -\frac12$ .

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