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(B) The bandwidth $(BW)$ of an $AM$ signal is given by $BW = 2f_m$,where $f_m$ is the maximum modulating frequency.Given $BW = 10 \, kHz$,we have $2f_

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$95 \, kHz$ and $105 \, kHz$

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(B) The bandwidth $(BW)$ of an $AM$ signal is given by $BW = 2f_m$,where $f_m$ is the maximum modulating frequency.Given $BW = 10 \, kHz$,we have $2f_m = 10 \, kHz$,which implies $f_m = 5 \, kHz$.The carrier frequency $f_c = 100 \, kHz$.The Lower Sideband $(LSB)$ frequency is $f_{LSB} = f_c - f_m = 100 \, kHz - 5 \, kHz = 95 \, kHz$.The Upper Sideband $(USB)$ frequency is $f_{USB} = f_c + f_m = 100 \, kHz + 5 \, kHz = 105 \, kHz$.Therefore,the sideband frequencies are $95 \, kHz$ and $105 \, kHz$.

In $AM$ transmission,the bandwidth of transmission is $10 \, kHz$ and the carrier frequency is $100 \, kHz$. The sideband frequencies are given by:

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