WAEC 2017 · Paper 2 · Q10

(a)
If $f(x) = \dfrac{2x - 3}{(x^2 - 1)(x + 2)}$ , (i) find the values of $x$ for which $f(x)$ is undefined; (ii) express $f(x)$ in partial fractions.
(b)
A circle with centre $(-3, 1)$ passes through the point $(3, 1)$ . Find its equation.
Show the answer
$x^2 + y^2 + 6x - 2y - 26 = 0$

Worked solution (try it first)

(a)(i)

Factorise: $(x^2 - 1)(x + 2) = (x - 1)(x + 1)(x + 2)$ .
$f(x)$ is undefined where this is zero: $x = 1$ , $-1$ or $-2$ .

(ii)

Write $\dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{x + 2}$ and multiply through: $2x - 3 = A(x + 1)(x + 2) + B(x - 1)(x + 2) + C(x - 1)(x + 1)$ .
Put $x = 1$ : $-1 = 6A$ , so $A = -\frac16$ .
Put $x = -1$ : $-5 = -2B$ , so $B = \frac52$ .
Put $x = -2$ : $-7 = 3C$ , so $C = -\frac73$ .
So $f(x) = -\dfrac{1}{6(x - 1)} + \dfrac{5}{2(x + 1)} - \dfrac{7}{3(x + 2)}$ .

(b)

The radius is the distance from $(-3, 1)$ to $(3, 1)$ : $r = 6$ .
The equation: $(x + 3)^2 + (y - 1)^2 = 36$ .
Expand: $x^2 + 6x + 9 + y^2 - 2y + 1 = 36$ , so $x^2 + y^2 + 6x - 2y - 26 = 0$ .

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