The function f(x) = \(\displaystyle\sum_{r=1}^{5} (x-r)^2\) assumes m

1.	The function f(x) = \(\displaystyle\sum_{r=1}^{5} (x-r)^2\) assumes minimum value at x = A. 5 B. \(\frac{5}{2}\)C. 3 D. 2
Answer» 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.

The function f(x) = \(\displaystyle\sum_{r=1}^{5} (x-r)^2\) assumes minimum value at x = A. 5 B. \(\frac{5}{2}\)C. 3 D. 2

Answer»

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.