A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:

In amplitude modulation, the carrier signal and the modulating (audio) signal combine according to the standard formula

$$s(t)=\bigl[A_c + A_m\cos(2\pi f_m t)\bigr]\cos(2\pi f_c t),$$

where $$f_c$$ is the carrier frequency and $$f_m$$ is the modulating frequency. We now expand this product. Using the trigonometric identity

$$\cos\alpha \cos\beta=\frac{1}{2}\bigl[\cos(\alpha+\beta)+\cos(\alpha-\beta)\bigr],$$

we write

$$$ \begin{aligned} s(t) & = A_c\cos(2\pi f_c t)+A_m\cos(2\pi f_m t)\cos(2\pi f_c t)\\[4pt] & = A_c\cos(2\pi f_c t)+A_m\;\frac{1}{2}\Bigl[\cos\!\bigl(2\pi(f_c+f_m)t\bigr)+\cos\!\bigl(2\pi(f_c-f_m)t\bigr)\Bigr]. \end{aligned} $$$

This expression clearly contains three cosine terms whose frequencies are

$$f_c,\;f_c+f_m,\;f_c-f_m.$$

Now substitute the numerical values. The carrier frequency is given as $$f_c = 2\ \text{MHz}.$$ Converting to kilohertz,

$$2\ \text{MHz}=2\times1000\ \text{kHz}=2000\ \text{kHz}.$$

The modulating signal frequency is $$f_m = 5\ \text{kHz}.$$ Therefore

$$f_c+f_m = 2000\ \text{kHz}+5\ \text{kHz}=2005\ \text{kHz},$$

and

$$f_c-f_m = 2000\ \text{kHz}-5\ \text{kHz}=1995\ \text{kHz}.$$

Hence the amplitude-modulated wave consists of three distinct frequency components:

$$2005\ \text{kHz},\;2000\ \text{kHz},\;1995\ \text{kHz}.$$

All three appear together in the spectrum of the AM signal.

Hence, the correct answer is Option D.