If the Laplace transform of a function 𝒇(𝒕) is given by $\frac{s+3}{(s+1)(s+2)}$ , then 𝒇(𝟎) is

The correct option is 1.

To find the value of $f(0)$ from its Laplace transform $F(s)=L\{f(t)\}$, we can use the Initial Value Theorem of Laplace transforms.

The Initial Value Theorem states that if a function $f(t)$ and its derivative are Laplace transformable, then:
$$f(0)=\underset{t\to 0}{lim}f(t)=\underset{s\to \infty }{lim}sF(s)$$

Here, the Laplace transform $F(s)$ is given by:
$$F(s)=\frac{s+3}{(s+1)(s+2)}$$

Now, let's construct the expression $sF(s)$:
$$sF(s)=s\cdot \frac{s+3}{(s+1)(s+2)}=\frac{s^2+3s}{s^2+3s+2}$$

Next, we evaluate the limit of $sF(s)$ as $s$ approaches infinity. To simplify this limit, divide the numerator and the denominator by the highest power of $s$, which is $s^2$:
$$\underset{s\to \infty }{lim}sF(s)=\underset{s\to \infty }{lim}\frac{1+\frac{3}{s}}{1+\frac{3}{s}+\frac{2}{s^2}}$$

As $s\to \infty$, the terms $\frac{3}{s}$ and $\frac{2}{s^2}$ approach $0$:
$$f(0)=\frac{1+0}{1+0+0}=1$$

Thus, the value of $f(0)$ is $1$.