Step 1 — match the signal to the standard FM form.

\(s(t)=A_c\sin\left(\omega_c t+\beta\cos\omega_m t\right)\)

Comparing with \(s(t)=10\sin\left(4\pi\times10^{6}t+9\cos\left(2\pi\times10^{3}t\right)\right)\):

Carrier: \(\omega_c=4\pi\times10^{6} \Rightarrow f_c = 2\ \text{MHz}\). Modulating signal: \(\omega_m=2\pi\times10^{3} \Rightarrow f_m = 1\ \text{kHz}\). Modulation index (the coefficient of the phase-modulating term): \(\beta = 9\).

Step 2 — frequency deviation. By definition the modulation index of an FM wave is the ratio of peak deviation to modulating frequency:

\(\beta = \dfrac{\Delta f}{f_m} \Rightarrow \Delta f = \beta f_m\)

\(\Delta f = 9 \times 1\ \text{kHz} = 9\ \text{kHz}\)

Physically, the instantaneous frequency swings 9 kHz above and 9 kHz below the 2 MHz carrier. (You can also get this by differentiating the phase: \(f_i = f_c - \beta f_m \sin\omega_m t\), whose peak excursion is βf_m.)

Step 3 — bandwidth by Carson's rule.

\(BW = 2\left(\Delta f + f_m\right) = 2\left(\beta+1\right)f_m\)

\(BW = 2(9+1)\ \text{kHz} = 20\ \text{kHz}\)

Why Carson's rule and not "infinite bandwidth". An FM spectrum strictly contains infinitely many sidebands spaced f_m apart, with amplitudes given by Bessel functions J_n(β). Carson's rule keeps the (β + 1) significant pairs that carry about 98 % of the power, which is the accepted practical bandwidth.

Watch the distractors. 10 kHz would come from wrongly reading the deviation as (β + 1)f_m; 30 kHz would come from \(2(\Delta f + 2f_m)\) or from using β = 14. Note also that β = 9 ≫ 1 means this is wideband FM, so Carson's rule is the correct tool.

Hence, the frequency deviation is 9 kHz and the bandwidth is 20 kHz.