If the function f(x) = √(2+cos x) if x ≠ π and f(x) = k if x = π is continuous at x = π, then k equals:

For f(x) to be continuous at x = π, we must have limx→π f(x) = f(π).

We have f(π) = k.

Let's find the limit:

limx→π √(2 + cos x)

Since cos x is continuous, we can substitute x = π:

limx→π √(2 + cos x) = √(2 + cos π) = √(2 - 1) = √1 = 1

Therefore, we must have k = 1. However, 1 is not among the given options.

Let's reconsider the problem statement. It appears there might be a typographical error or a misunderstanding in the original question. The provided function is:

f(x) = { √(2+cos x) if x ≠ π
{ k if x = π

The limit as x approaches π is:

limx→π √(2 + cos x) = 1

For continuity at x = π, we require:

limx→π f(x) = f(π)

1 = k

Therefore, k = 1. Since 1 is not one of the options, there's likely an error in the question or options provided.