Option : (C)
\(\displaystyle\sum_{r=1}^{5} (x-r)^2\)
f(x) = (x - 1)2 + (x - 2)2 + (x - 3)2 + (x - 4)2 + (x - 5)2
f’(x) = 2 [5x - 15]
f’(x) = 0 ; x = 3
Hence by second derivative test
f’’(x) > 0 so it’s a point of minimum.
f”(x) = 1 > 0 so At x = 3 minimum value.