Step 1: Understanding the Problem:
This question asks us to analyze the periodicity and symmetry properties of a continuous-time signal consisting of the sum of two sinusoidal components.
Step 2: Key Formula or Approach:
1. For a sum of periodic continuous-time signals $x(t) = x_1(t) + x_2(t)$, the combined signal is periodic if the ratio of their fundamental periods $\frac{T_1}{T_2}$ is a rational number.
2. An even signal satisfies $x(-t) = x(t)$.
3. An odd signal satisfies $x(-t) = -x(t)$.
Step 3: Detailed Explanation:
• Let $x_1(t) = \cos(2\pi t)$. Its angular frequency is $\omega_1 = 2\pi$ rad/s.
• The period $T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{2\pi} = 1$ s.
• Let $x_2(t) = \sin(3\pi t)$. Its angular frequency is $\omega_2 = 3\pi$ rad/s.
• The period $T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{3\pi} = \frac{2}{3}$ s.
• Now, we check the ratio of the periods:
\[ \frac{T_1}{T_2} = \frac{1}{2/3} = \frac{3}{2} \]
• Since $\frac{3}{2}$ is a rational number (a ratio of two integers), the sum signal $x(t)$ is periodic.
• The fundamental period of $x(t)$ is $T_0 = 2 T_1 = 3 T_2 = 2$ seconds.
• Next, we evaluate the symmetry of $x(t)$:
\[ x(-t) = \cos(-2\pi t) + \sin(-3\pi t) \]
• Since $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$, we get:
\[ x(-t) = \cos(2\pi t) - \sin(3\pi t) \]
• This result is neither equal to $x(t)$ nor equal to $-x(t)$.
• Therefore, the signal is neither even nor odd. It is periodic only.
Step 4: Final Answer
Thus, the signal is periodic only, matching option (A).