x_R and x_A are, respectively, the rms and average values of x(t) = x(t - T), and similarly, yR and yA are, respectively, the rms and average values of y(t) = kx(t), k, T are independent of t. Which of the following is true?

The correct option is y_A = kx_A; y_R ≠ kx_R.

To understand why this is the case, let us analyze the average value and the root-mean-square (rms) value of the functions step-by-step.

1. Relationship between the average values (y_A and x_A):
The average value of a periodic signal x(t) with period T is defined as:
$$x_A=\frac{1}{T}{\int }_0^{T}x(t)dt$$
Similarly, the average value of y(t) = kx(t) is defined as:
$$y_A=\frac{1}{T}{\int }_0^{T}y(t)dt=\frac{1}{T}{\int }_0^{T}kx(t)dt$$
Since k is independent of t, it can be factored out of the integral:
$$y_A=k(\frac{1}{T}{\int }_0^{T}x(t)dt)=k{x}_A$$
Thus, the relationship y_A = kx_A always holds true.

2. Relationship between the rms values (y_R and x_R):
The rms value of x(t) is defined as:
$$x_R=\sqrt{\frac{1}{T}{\int }_0^T{x}^2(t)dt}$$
Similarly, the rms value of y(t) = kx(t) is:
$$y_R=\sqrt{\frac{1}{T}{\int }_0^T{y}^2(t)dt}=\sqrt{\frac{1}{T}{\int }_0^T{[kx(t)]}^{2}dt}$$
Simplifying the term inside the integral:
$$y_R=\sqrt{k^2(\frac{1}{T}{\int }_0^T{x}^2(t)dt)}=\sqrt{k^2}{x}_R$$
Since the square root of k² is the absolute value |k|, we have:
$$y_R=\left|k\right|x_R$$
Because the constant k can be negative (k < 0), the absolute value |k| is not generally equal to k. Therefore, y_R is not always equal to kx_R (for example, if k = -2, y_R = 2x_R, which is not equal to -2x_R). Thus, y_R ≠ kx_R.

Consequently, the correct relation is y_A = kx_A and y_R ≠ kx_R.