(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)