The Laplace transform of a continuous-time signal x(t)

The Laplace transform of a continuous-time signal x(t) is,
X(s) = 5 - s
s² - s - 2

If the Fourier transform of this signal exists, then x(t) is

1. e^2t u(t) – 2e^{– t} u(t)
2. – e^2t u(– t) + 2e^{– t} u(t)
3. – e^2t u(– t) – 2e^{– t}u(t)
4. none of these

X(s) =	5 - s	=	A	+	B
	(s + 1)(s - 2)		(s + 1)		(s - 2)

A =	5 - s			= - 2; and B	5 - s			= 1
	s - 1		_s=-1		s + 1		_s=2

then, X(t) = L^{– 1} X(s) = L^{– 1}		- 2	+	1
		s + 1		s - 2

= [e^2t – 2e^{– t}]u(t)

X(s) =	5 - s
	s² - s - 2