A modulated signal $$C_m(t)$$ has the form $$C_m(t) = 30\sin 300\pi t + 10(\cos 200\pi t - \cos 400\pi t)$$. The carrier frequency $$f'_c$$, the modulating frequency (message frequency) $$f_\omega$$ and the modulation index $$\mu$$ are respectively given by:

We start by recalling the mathematical model of a conventional amplitude-modulated (AM) wave. In its most compact form it is written as

$$s(t)=A_c\,[1+\mu\cos\omega_m t]\sin\omega_c t,$$

where $$A_c$$ is the unmodulated carrier amplitude, $$\omega_c=2\pi f_c$$ is the carrier angular frequency, $$\omega_m=2\pi f_m$$ is the message (or modulating) angular frequency and $$\mu$$ is the modulation index. Using the trigonometric identity

$$\cos\alpha\,\sin\beta=\tfrac12\,[\sin(\beta+\alpha)+\sin(\beta-\alpha)],$$

the same signal can be rewritten as

$$s(t)=A_c\sin\omega_c t+\frac{\mu A_c}{2}\Bigl[\sin(\omega_c+\omega_m)t+\sin(\omega_c-\omega_m)t\Bigr].$$

Thus an AM wave is recognised by one term at the carrier frequency $$\omega_c$$ and two equal-amplitude sideband terms at $$\omega_c+\omega_m$$ and $$\omega_c-\omega_m$$. Each sideband amplitude equals $$\dfrac{\mu A_c}{2}$$.

Now examine the given modulated signal

$$C_m(t)=30\sin 300\pi t+10\bigl(\cos 200\pi t-\cos 400\pi t\bigr).$$

Although the sidebands appear with cosine instead of sine, a phase shift of $$\tfrac{\pi}{2}$$ does not change their frequencies or amplitudes, so frequency identification is unaffected. We therefore match every term with the standard AM form.

First, identify the carrier term. The only pure sine term is

$$30\sin 300\pi t.$$

Hence

$$A_c=30,\qquad \omega_c=300\pi\;\text{rad s}^{-1}.$$

Converting the carrier angular frequency to ordinary frequency, we have

$$f_c=\frac{\omega_c}{2\pi}=\frac{300\pi}{2\pi}=150\;\text{Hz}.$$

Next, look at the two cosine terms which form the sidebands:

$$10\cos 200\pi t\quad\text{and}\quad -10\cos 400\pi t.$$

Their angular frequencies are

$$\omega_1=200\pi\;\text{rad s}^{-1},\qquad\omega_2=400\pi\;\text{rad s}^{-1}.$$

Dividing by $$2\pi$$ gives the corresponding ordinary frequencies

$$f_1=\frac{200\pi}{2\pi}=100\;\text{Hz},\qquad f_2=\frac{400\pi}{2\pi}=200\;\text{Hz}.$$

In a standard AM spectrum these frequencies should be $$f_c-f_m$$ and $$f_c+f_m$$ respectively. Setting them equal we obtain

$$f_c-f_m=100,\qquad f_c+f_m=200.$$

Substituting $$f_c=150$$ Hz (found earlier) into either equation gives

$$150-f_m=100 \;\;\Longrightarrow\;\; f_m=50\;\text{Hz},$$

and this also satisfies $$150+f_m=200$$. Thus the message (modulating) frequency is

$$f_m=50\;\text{Hz}.$$

Finally, determine the modulation index $$\mu$$. The amplitude of each sideband term is 10. According to the AM formula each sideband amplitude equals $$\dfrac{\mu A_c}{2}$$, so we write

$$\frac{\mu A_c}{2}=10.$$

Substituting $$A_c=30$$ gives

$$\frac{\mu(30)}{2}=10 \;\;\Longrightarrow\;\; 15\mu=10 \;\;\Longrightarrow\;\; \mu=\frac{10}{15}=\frac{2}{3}.$$

We have now obtained

$$f_c=150\;\text{Hz},\qquad f_m=50\;\text{Hz},\qquad \mu=\frac{2}{3}.$$

Comparing these results with the given options, we see that they match Option B.

Hence, the correct answer is Option B.