Explanation:
The initial value theorem states that the value of the function at time t = 0 is equal to the limit of the Laplace transform of the function as s approaches infinity.

Let's consider each option and see if the initial value theorem holds good for it or not.

a) Ramp function:
The ramp function is defined as f(t) = t for t ≥ 0.
Using the Laplace transform, we get F(s) = 1/s^2.
Now, applying the initial value theorem, we get f(0) = lim (s→∞) sF(s) = lim (s→∞) s/s^2 = 0.
Therefore, the initial value theorem holds good for the ramp function.

b) Step function:
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Using the Laplace transform, we get F(s) = 1/s.
Now, applying the initial value theorem, we get f(0) = lim (s→∞) sF(s) = lim (s→∞) 1 = 1.
Therefore, the initial value theorem does not hold good for the step function.

c) Delta function:
The delta function is defined as δ(t) = 0 for t ≠ 0 and ∫δ(t)dt = 1.
Using the Laplace transform, we get F(s) = 1.
Now, applying the initial value theorem, we get δ(0) = lim (s→∞) sF(s) = lim (s→∞) s = ∞.
Therefore, the initial value theorem does not hold good for the delta function.

d) Hyperbolic function:
Hyperbolic functions include sinh, cosh, and tanh.
Using the Laplace transform, we get the Laplace transforms of these functions.
Now, applying the initial value theorem, we can find the value of the function at t=0.
Therefore, the initial value theorem holds good for hyperbolic functions.

Conclusion:
The initial value theorem holds good for the ramp function and hyperbolic functions, but it does not hold good for the step function and delta function.