Let f(t) be an even function i.e. f(-t) = f(t) for all t. Let the Fourier transform of f(t) be defined as $F(\omega )={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)e^{-j\omega t}dt}$. Suppose ${\displaystyle \frac{dF(\omega )}{d\omega }}=-\omega F(\omega )$ for all ω, and F(0) = 1. Then

The correct option is f(0) < 1.

We are given a differential equation for the Fourier transform $F\left(\omega \right)$ of an even function $f\left(t\right)$:
$$\frac{dF\left(\omega \right)}{d\omega }=-\omega F\left(\omega \right)$$
subject to the initial condition $F\left(0\right)=1$.

Let us solve this first-order ordinary differential equation using the separation of variables method:
$$\frac{dF}{F}=-\omega d\omega$$
Integrating both sides, we get:
$$\ln \left(F\right(\omega \left)\right)=-\frac{{\omega }^2}{2}+C$$
Taking the exponential of both sides:
$$F\left(\omega \right)=A{e}^{-\frac{{\omega }^2}{2}}$$
where $A=e^C$ is a constant. Using the initial condition $F\left(0\right)=1$, we find:
$$F\left(0\right)=A{e}^0=A=1$$
Thus, the Fourier transform of the function is:
$$F\left(\omega \right)=e^{-\frac{{\omega }^2}{2}}$$

Now, we find $f\left(t\right)$ using the inverse Fourier transform formula:
$$f\left(t\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }F\left(\omega \right)e^{j\omega t}d\omega$$
Substituting $F\left(\omega \right)=e^{-\frac{{\omega }^2}{2}}$ and setting $t=0$ to evaluate $f\left(0\right)$, we have:
$$f\left(0\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{-\frac{{\omega }^2}{2}}d\omega$$

Recall the standard Gaussian integral formula:
$${\int }_{-\infty }^{\infty }{e}^{-a{x}^2}dx=\sqrt{\frac{\pi }{a}}$$
For our integral, $a=\frac{1}{2}$. Therefore, the integral evaluates to:
$${\int }_{-\infty }^{\infty }{e}^{-\frac{{\omega }^2}{2}}d\omega =\sqrt{\frac{\pi }{1/2}}=\sqrt{2\pi }$$

Substituting this value back into the expression for $f\left(0\right)$:
$$f\left(0\right)=\frac{1}{2\pi }\cdot \sqrt{2\pi }=\frac{\sqrt{2\pi }}{2\pi }=\frac{1}{\sqrt{2\pi }}$$

Since $2\pi \approx 6.28>1$, it follows that:
$$\sqrt{2\pi }>1$$
Taking the reciprocal:
$$f\left(0\right)=\frac{1}{\sqrt{2\pi }}<1$$
Thus, the value of $f\left(0\right)$ is strictly less than 1.