Discussion
If that message frequency doubles at constant amplitude, the PM bandwidth becomes
Answer: C
The index is unchanged at 6 because amplitude is unchanged, so
\[ B = 2 \times 4(1+6) = 56 \ \mathrm{kHz} \]
which is double the previous figure. That strict proportionality to message frequency is peculiar to PM: in FM the same change would lower the index and widen the bandwidth only modestly. Keeping the bandwidth unchanged ignores the frequency term. 34 kHz and 14 kHz follow from no correct substitution. The contrast is worth computing once for each scheme side by side.
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