A real and non-negative signal x(t) has Fourier transform

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A real and non-negative signal x(t) has Fourier transform X(jω). The following facts are given 1. F^{– 1}{(1 + jω) X(jω)} = Ae^{– 2t}u(t), where A is constant
2.
The signal x(t) may be

1. √12(e^-2t - e^-t)u(t)
2. √12(e^-t - e^-2t)u(t)
3. √3(e^-t - e^-2t)u(t)
4. √3(e^-t - e^-2t)u(t)

Correct Option: B

Y(jω) =	1	-	1	=	1
	3 + jω		4 + jω		(3 + jω)(4 + jω)

X(jω) =	Y(jω)	=	1
	H(jω)		4 + jω

⇒ x(t) = e^{– 4t}u(t)