Question

The minimum value of $f\left(x\right)=(x-1{)}^2+(x-2{)}^2+\cdots +(x-10{)}^2$ occurs at $x=k.$ Then the value of $\left[k\right]$ is
$($where $[.]$ represents the greatest integer function$)$

• A. 5.0
• B. 5.00
• C. 05
• D. 5

Answer 1

Answer: (A), (B), (C), (D)

$f^{\prime }\left(x\right)=2\left(\right(x-1)+(x-2)+\cdots +(x-10\left)\right)$
$\Rightarrow f^{\prime }\left(x\right)=2(10x-(1+2+\dots 10\left)\right)$
From $f^{\prime }\left(x\right)=0$
$\Rightarrow 2(10x-\frac{10\times 11}{2})=0$
$\Rightarrow 10x=55\Rightarrow x=\frac{11}{2}=5.5$

Since, sign of $f^{\prime }\left(x\right)$ changes from negative to positive as $x$ crosses $\frac{11}{2}$ from left to right, therefore $x=\frac{11}{2}$ is a point of local minima.
As there is only one point of local minima so, at this point function is minimum.
Hence, $\left[k\right]=\left[\frac{11}{2}\right]=5$