Question

If the Laplace transform of a function \( f(t) \) is given by \( \frac{2s + 1}{(s + 1)(s + 2)} \), then \( f(0) \) is equal to ...........

• A. 2
• B. 4
• C. -4
• D. $3e - 1$

Hint:

When evaluating the Laplace transform at $s = 0$, use the fact that the transform is often expressed in terms of a rational function where you can simply evaluate the function at $s = 0$ to find $f(0)$. This is crucial in solving differential equations.

Answer 1

The Correct Option is A

To find $f(0)$, we use the property of the Laplace transform that: \[ \mathcal{L}\{f(t)\} = F(s) \] where the Laplace transform of $f(t)$ is denoted as $F(s)$.
We are given that: \[ F(s) = \frac{2s + 1}{(s+1)(s+2)} \] To find $f(0)$, we need to compute the value of $F(s)$ at $s = \infty$. However, using the inverse of the Laplace transform property:
The formula for $f(0)$ from the inverse Laplace transform is: \[ f(0) = \lim_{s \to \infty} s \cdot F(s) \] Substituting for $F(s)$: \[ f(0) = \lim_{s \to \infty} s \cdot \frac{2s + 1}{(s+1)(s+2)} \] We now simplify:
As $s \to \infty$, the highest power of $s$ in both the numerator and denominator is $s^2$. So, the limit is: \[ f(0) = \frac{2s}{s^2} = 2 \] Thus, $f(0) = 2$.