D

Step 1: Consider the linear density \( \lambda(x) = kx^n \) where \( k \) is a constant and \( n \) can be zero or any positive number.
Step 2: The center of mass \( x_{CM} \) for a rod with variable density is calculated using the formula:
\[ x_{CM} = \frac{1}{M} \int_0^L x \lambda(x) \, dx \]
where \( M \) is the total mass given by
\[ M = \int_0^L \lambda(x) \, dx = \int_0^L kx^n \, dx = \frac{kL^{n+1}}{n+1} \]
Step 3: Substituting \( \lambda(x) \) into the center of mass equation leads to:
\[ x_{CM} = \frac{1}{M} \int_0^L x kx^n \, dx = \frac{1}{M} \frac{kL^{n+2}}{n+2} = \frac{(n+1)}{(n+2)L}
Step 4: Analyze the behavior of \( x_{CM} \) as \( n \) increases:
- As \( n \) approaches 0, \( x_{CM} \) approaches \( \frac{L}{2} \) (uniform density).
- As \( n \) increases, \( x_{CM} \) approaches L, as the density becomes more concentrated towards the right end of the rod.
Step 5: The graph showing this trend is a smooth curve that approaches L as n increases, which matches Option D.