Solution:

Given transfer function is G(s) = 2/s(s+3)

Step 1: Finding the differential equation of the system

The transfer function of the system is given by

G(s) = Y(s)/X(s)

where Y(s) is the Laplace transform of output signal y(t) and X(s) is the Laplace transform of input signal x(t).

The Laplace transform of the unit step function u(t) is given by

U(s) = 1/s

The Laplace transform of the output signal y(t) can be written as

Y(s) = G(s) X(s)

Substituting the given transfer function in the above equation, we get

Y(s) = (2/s(s+3)) X(s)

Taking inverse Laplace transform on both sides, we get

y(t) = 2(u(t) - e^(-3t)u(t))

y(t) = 2u(t) - 2e^(-3t)u(t)

y(t) = u(t) - 2e^(-3t)u(t) + u(t)

y(t) = 1 - 2e^(-3t) + u(t)

Step 2: Simplifying the expression

The unit step response of the system is given by y(t) = 1 - 2e^(-3t) + u(t)

Taking the derivative of y(t), we get

dy/dt = 6e^(-3t)

Taking the limit of y(t) as t tends to infinity, we get

lim y(t) = lim (1 - 2e^(-3t) + u(t))

t->infinity t->infinity

= 1 - 0 + 1

= 2

Therefore, the steady-state value of the unit step response is 2.

Step 3: Checking the options

Option (a) 1 + 2e^(-t) - e^(-2t)

Taking the limit of the above expression as t tends to infinity, we get

lim (1 + 2e^(-t) - e^(-2t))

t->infinity

= 1 + 0 - 0

= 1

Therefore, the steady-state value of option (a) is incorrect.

Option (b) 1 + e^(-t) - 2e^(-2t)

Taking the limit of the above expression as t tends to infinity, we get

lim (1 + e^(-t) - 2e^(-2t))

t->infinity

= 1 + 0 - 0

= 1

Therefore, the steady-state value of option (b) is incorrect.

Option (c) 1 - e^(-t) + 2e^(-2t)

Taking the limit of the above expression as t tends to infinity, we get

lim (1 - e^(-t) + 2e^(-2t))

t->infinity

= 1 - 0 + 0

= 1

Therefore, the steady-state value of option (c) is incorrect.

Option (d) 1 - 2e^(-t) + e^(2t)

Taking the limit of the above expression as t tends to infinity, we get

lim (1 - 2e^(-t