Suppose signal y(t) is obtained by the time-reversal of signal x(t) . i.e., y(t) = x(-t), -∞ < t < ∞. Which one of the following options is always true for the convolution of x(t) and y(t) ?

The correct option is It is an even signal.

Let the convolution of the signals $x\left(t\right)$ and $y\left(t\right)$ be denoted as $z\left(t\right)$.
By definition, the convolution of two continuous-time signals is given by:

$z\left(t\right)=x\left(t\right)*y\left(t\right)={\int }_{-\infty }^{\infty }x(\tau )y(t-\tau )d\tau$

We are given that $y\left(t\right)=x(-t)$. Substituting this relationship into the convolution integral, we get:

$z\left(t\right)={\int }_{-\infty }^{\infty }x(\tau )x(-(t-\tau \left)\right)d\tau ={\int }_{-\infty }^{\infty }x(\tau )x(\tau -t)d\tau$

To determine whether the resulting signal $z\left(t\right)$ is even or odd, we evaluate $z(-t)$ by substituting $-t$ for $t$:

$z(-t)={\int }_{-\infty }^{\infty }x(\tau )x(\tau -(-t\left)\right)d\tau ={\int }_{-\infty }^{\infty }x(\tau )x(\tau +t)d\tau$

Let us perform a change of variables by setting $u=\tau +t$, which implies $\tau =u-t$ and $d\tau =du$.
The limits of integration remain from $-\infty$ to $\infty$. Substituting these into the integral:

$z(-t)={\int }_{-\infty }^{\infty }x(u-t)x\left(u\right)du$

Since $u$ is a dummy variable of integration, we can rename it back to $\tau$:

$z(-t)={\int }_{-\infty }^{\infty }x(\tau -t)x(\tau )d\tau$

Rearranging the terms inside the integrand:

$z(-t)={\int }_{-\infty }^{\infty }x(\tau )x(\tau -t)d\tau =z\left(t\right)$

Since $z(-t)=z\left(t\right)$ is satisfied for all $t$, the convolution of $x\left(t\right)$ and its time-reversed counterpart $y\left(t\right)$ is always an even signal.