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 the differential equation for the Fourier transform $F\left(\omega \right)$:
$\frac{dF\left(\omega \right)}{d\omega }=-\omega F\left(\omega \right)$

We can solve this first-order ordinary differential equation by separating variables:
$\frac{dF\left(\omega \right)}{F\left(\omega \right)}=-\omega d\omega$

Integrating both sides:
$\int \frac{dF\left(\omega \right)}{F\left(\omega \right)}=-\int \omega d\omega$
$\ln \left(F\right(\omega \left)\right)=-\frac{{\omega }^2}{2}+C$
Exponentiating both sides gives:
$F\left(\omega \right)=A{e}^{-\frac{{\omega }^2}{2}}$
where $A=e^C$ is a constant.

Using the initial condition $F\left(0\right)=1$:
$F\left(0\right)=A{e}^0=A=1$
Thus, the Fourier transform is:
$F\left(\omega \right)=e^{-\frac{{\omega }^2}{2}}$

The inverse Fourier transform is defined as:
$f\left(t\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }F\left(\omega \right)e^{j\omega t}d\omega$

To find the value at $t=0$, we substitute $t=0$ into the inverse Fourier transform formula:
$f\left(0\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{-\frac{{\omega }^2}{2}}d\omega$

Using the standard Gaussian integral formula, ${\int }_{-\infty }^{\infty }{e}^{-a{x}^2}dx=\sqrt{\frac{\pi }{a}}$ with $a=\frac{1}{2}$:
${\int }_{-\infty }^{\infty }{e}^{-\frac{{\omega }^2}{2}}d\omega =\sqrt{\frac{\pi }{1/2}}=\sqrt{2\pi }$

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

Since $\pi \approx 3.14159$, we have $2\pi \approx 6.283$, which means $\sqrt{2\pi }>1$. Therefore:
$f\left(0\right)=\frac{1}{\sqrt{2\pi }}<1$