(a) If \(f(x) = \frac{2x - 3}{(x^{2} - 1)(x + 2)}\) (i) find the values
(a) If \(f(x) = \frac{2x - 3}{(x^{2} - 1)(x + 2)}\)
(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.
Explanation
(a)(i) \(f(x) = \frac{2x - 3}{(x^{2} - 1)(x + 2)}\)
f(x) is undefined at \((x^{2} - 1)(x + 2) = 0\).
Either \((x^{2} - 1) = \text{0 or (}x + 2) = 0\)
\((x + 2) = 0 \implies x = -2\)
\((x^{2} - 1) = 0 \implies x^{2} = 1 \)
\(x = \pm 1\)
At the points x = -1, 1 or 2, f(x) is undefined.
(ii) \(\frac{2x - 3}{(x^{2} - 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{x + 2}\)
\(\frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{x + 2} = \frac{A(x - 1)(x + 2) + B(x + 1)(x + 2) + C(x - 1)(x + 1)}{(x^{2} - 1)(x + 2)}\)
When x = -1,
\(-2A = 2(-1) - 3 \implies -2A = -5\)
\(A = \frac{5}{2}\)
When x = 1,
\(6B = 2(1) - 3 = -1\)
\(B = \frac{-1}{6}\)
When x = -2,
\(3C = 2(-2) - 3 = -7\)
\(C = \frac{-7}{3}\)
\(\frac{2x - 3}{(x^{2} - 1)(x + 2)} = \frac{5}{2(x + 1)} - \frac{1}{6(x - 1)} - \frac{7}{3(x + 2)}\)
(c) Equation of circle : \((x + 3)^{2} + (y - 1)^{2} = r^{2}\)
Circle passes through (3,1) :
\(\therefore (3 + 3)^{2} + (1 - 1)^{2} = r^{2}\)
\(6^{2} = r^{2} \implies r = 6\)
Equation becomes :
\((x + 3)^{2} + (y - 1)^{2} = 6^{2}\)
\(x^{2} + 6x + 9 + y^{2} - 2y + 1 - 36 = 0\)
\(x^{2} + 6x + y^{2} - 2y - 26 = 0\)
\(x^{2} + y^{2} + 6x - 2y - 26 = 0\)