If the function g(x) = k√x + 1, 0 ≤ x ≤ 3 and g(x) = mx + 2, 3 < x ≤ 5 is differentiable, then the value of k + m is?

Since, the function is differentiable at x = 3, we need to use the fact that it is continuous as well as differentiable at x = 3.

For continuity at x = 3:
limx→3- g(x) = limx→3+ g(x) = g(3)
limx→3- (k√x + 1) = limx→3+ (mx + 2)
k√3 + 1 = 3m + 2

For differentiability at x = 3:
limx→3- g'(x) = limx→3+ g'(x)
limx→3- (k/(2√x)) = limx→3+ (m)
k/(2√3) = m

Now we have two equations:
k√3 + 1 = 3m + 2
k/(2√3) = m
Substituting the second equation into the first:
k√3 + 1 = 3(k/(2√3)) + 2
2(k√3) + 2 = 3k/(√3) + 4
2k√3 - k√3 = 2
k√3 = 2
k = 2/√3

Now substitute k back into the second equation to find m:
m = k/(2√3) = (2/√3)/(2√3) = 1/3

k + m = 2/√3 + 1/3 = (2√3 + 1)/3 ≈ 1.609

None of the options match this value, there might be a mistake in the question or options provided.