The function f(x)= _ r=1 ^5(x-r)^ 2 assume minimum value at x = - Vidyadip

MCQ

The function $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$ assume minimum value at $x =$

A
$5$
B
$\frac{5}{2}$
✓
$3$
D
$2$

✓

Answer

Correct option: C.

$3$

Given, $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$
lmplies that $f(x) = (x - 1)^2 + (x - 2)^2+ (x - 3)^2 + (x - 4)^2 + x - 5^2$
lmplies that $f'(x) = 2(x - 1 + x - 2 + x - 3 + x - 4 + x - 5)$
lmplies that $f'(x) = 2(5x - 15)$
For a local maxima and a local minima, we must have $f'(x) = 0$
limplies that $2(5x - 15) = 0$
limplies that $5x - 15 = 0$
limplies that $x = 3$
Now, $f''(x) = 10$
f$''(x) = 10 > 0$
Therefore, $x = 3$ is a local minima.

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