Option 2 states the two defining facts about FM — constant total power, and bandwidth that grows with modulation depth — and both are correct.

Why the power is constant. An FM wave is

\(s(t)=A_{c}\cos\left(\omega_{c}t+m_{f}\sin\omega_{m}t\right)\)

Its amplitude \(A_{c}\) never changes — only the argument of the cosine does — so the transmitted power is fixed at \(A_{c}^{2}/2R\) whatever the modulation index. Modulation redistributes power between carrier and sidebands but adds none, which is expressed by the Bessel identity

\(J_{0}^{2}(m_{f})+2\sum_{n=1}^{\infty}J_{n}^{2}(m_{f})=1\)

Why the bandwidth grows. A larger index puts significant amplitude into more sideband pairs, and Carson's rule captures it:

\(BW\approx2\left(\Delta f+f_{m}\right)=2f_{m}\left(m_{f}+1\right)\)

So FM buys its noise advantage by spending bandwidth rather than power — exactly the trade the statement describes.

Why option 1 fails. Because the message enters inside the cosine, FM is a non-linear modulation and its spectrum contains components at \(f_{c}\pm nf_{m}\) for every integer n. The number is infinite; only the number of significant pairs is finite, roughly \(m_{f}+1\), and that is what Carson's rule counts.

Why option 4 is exactly backwards. The carrier amplitude is \(A_{c}J_{0}(m_{f})\), and \(J_{0}\) has zeros at \(m_{f}=2.405,\ 5.520,\ 8.654\) and so on. At those "eigen values" the carrier disappears completely rather than appearing completely — a fact used in the laboratory to calibrate deviation, by raising the index until the carrier nulls on a spectrum analyser.

Why option 3 is rejected. The carrier component of an AM wave does have a fixed amplitude, but the statement is offered as a distinguishing truth and is misleading in that role: AM's total power rises with modulation as \(P_{c}\left(1+m^{2}/2\right)\), which is the very contrast option 2 draws with FM.

Hence, option 2 is the true statement.