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

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