Here's my understanding:
$$y(t) = x(t)~ \cos(\Omega_0 t)$$
I take the Fourier transform of y(t) and I get this result:
$$Y(\Omega) = \frac{1}{2}X(\Omega - \Omega_0) + \frac{1}{2}X(\Omega + \Omega_0)$$
If the Highest frequency of $X(\Omega)$ is $\Omega_{N_X}$ ,shouldn't the highest frequency of $Y(\Omega)$ just be:
$$\Omega_{N_Y} = \Omega_0 + \Omega_{N_X}$$
instead of:
$$\Omega_{N_Y} = \Omega_{N_X} + 2 \Omega_0$$
as the textbook claims?
Update: I feel there is no need to bring concept of Bandpass sampling (as mentioned in comments below) because this is specifically regarding a simple misunderstanding that OP has. The text referenced asks about Nyquist Rate (which wrongly mentions it as Nyquist Frequency) but OP asks about highest Frequency in $y(t)$.
For the component $\frac{1}{2}X(\Omega-\Omega_0)$ the highest frequency content is at $\Omega_0+\Omega_N/2$. The lowest (starting frequency) is at $\Omega_0-\Omega_N/2$. Similary for the other component $\frac{1}{2}X(\Omega+\Omega_0)$, the starting frequency is at $-\Omega_0-\Omega_N/2$ and ending frequency is at $-\Omega_0+\Omega_N/2$.
So if you see the overall frequency content of $y(t)$, lowest = $-\Omega_0-\Omega_N/2$ and highest = $\Omega_0+\Omega_N/2$. This is your signal $y(t)$ whose highest frequency is $\Omega_0+\Omega_N/2$.
The Nyquist rate for this is $2\Omega_0+\Omega_N$ if you consider $y(t)$ as baseband signal in the picture below. But the minimum sampling rate will be much lesser than this if you consider it as Passband signal (concept of bandpass sampling as mentioned in another answer)
If the Highest frequency of $X(\Omega)$ is $\Omega_N$
No. The highest frequency content in this case is $\Omega_N/2$ which is half the Nyquist rate. Otherwise this would cause aliasing.
Firstly, your book uses the term Nyquist frqeuency as $\Omega_N + 2\Omega_o$, this is incorrect, this is the Nyquist rate (minimum sampling rate) if we consider the signal to be baseband. The maximum frequency content is then $\Omega_o + \frac{\Omega_N}{2}$, since you have defined maximum frequency of $x(t)$ as $\Omega_{N_x}$ this is nothing but $\frac{\Omega_{N}}{2}$ i.e. half of critical sampling rate or nyqusit rate. So it's essentially the same thing, Nyquist rate means sampling at the minimum rate which means the maximum frqeuency in the signal is half of it.
Secondly,
The signal $y(t)$ is a bandpass signal, if you have a Khz signal around a MHz band, we don't require to sample at Mhz.
Note: As soon as the term Nyquist frequency is aksed it is always in the context of sampling, because by definition Nyquist frqeuency is half the sampling rate.
This being a bandpass signal this has to be looked in context of bandpass sampling.
The book should be mentioning the maximum frequency or maximum bandwidth taking into account the signal is baseband. It is asking about Nyquist frqeuency which is half of sampling rate
Otherwise consider this example:
$\Omega_N=30$Khz and $\Omega_o = 1$ GHz, what is the minimum sampling rate required for this signal. Well if you define the Nyquist frequency by the term aksed in the questions this would be in GHz and minimum sampling rate (which is double of nyqusit frqeuency) would also be in GhZ, which is incorrect. minimum sampling rate for this example would be in Khz.
There is a good discussion available in this question by @Matt L.
Answer : What you are considering as $\Omega_{N_x}$ is equal to $\frac{\Omega_N}{2}$ according to question. So, what you are saying is same as what answer mentions given we are considering Baseband Samping of $y_a(t)$.
I think you are confused because of the terms Nyquist rate and Nyquist Frequency.
Nyquist rate and Nyquist frequency are two different things. Nyquist rate is sampling rate satisfying the Nyquist criterion.
But Nyquist frequency is maximum frequency in the sampled signal. Nyquist frequency is always half of the sampling rate. It doesnot matter whether sampling rate is Nyquist rate or not. Nyquist frequency is the property of sampling itself.
But all sampling rates are not Nyquist rates. Only those sampling rates are Nyquist rates which satisfy Nyquist sampling criterion. Note that Nyquist sampling criterion is satisfied by both baseband and bandpass sampling where $f_s \ge 2B$.
The Question says Nyquist Rate of the baseband signal $x_a(t)$ is $\Omega_N$, meaning that the highest frequency in baseband $x_a(t)$ before modulation by cosine is $\frac{\Omega_N}{2}$. And then the question asks for Nyquist Rate for the derived signal $y_a(t)$.
And the answer given mentions Nyquist Frequency but then it gives Nyquist Rate for Baseband Sampling of $y_a(t)$. The answer $2\Omega_o + \Omega_N$ is correct for Nyquist Rate with Baseband Sampling of $y_a(t)$. So, the only wrong thing in the answer is mentioning "Frequency" instead of "Rate".
