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Suppose that the modulating signal is $m(t)=2 \cos \left(2 \pi f_{m} t\right)$ and the carrier signal is $x_{c}(t)=A_{C} \cos \left(2 \pi f_{c}t\right)$. Which one of the following is a conventional $\text{AM}$ signal without over-modulation?

- $x(t)=A_{c} m(t) \cos \left(2 \pi f_{c} t\right)$
- $x(t)=A_{c}[1+m(t)] \cos \left(2 \pi f_{c} t\right)$
- $x(t)=A_{c} \cos \left(2 \pi f_{c} t\right)+\frac{A_{C}}{4} m(t) \cos \left(2 \pi f_{c} t\right)$
- $x(t)=A_{c} \cos \left(2 \pi f_{m} t\right) \cos \left(2 \pi f_{c} t\right)+A_{c} \sin \left(2 \pi f_{m} t\right) \sin \left(2 \pi f_{c} t\right)$