A) \[x=0\]
B) \[x=1\]
C) \[x=A\]
D) \[x=A+1\]
Correct Answer: C
Solution :
Let, \[f\left( x \right)=\sum\limits_{i=1}^{n}{{{\left( x-{{a}_{i}} \right)}^{2}}}\] \[\Rightarrow \]\[f'(x)=\sum\limits_{i=1}^{n}{2(x-{{a}_{i}})}\] |
\[=2nx-({{a}_{1}}+{{a}_{2}}.....+{{a}_{n}})\] |
\[=2nx-2nA\]where \[A=\frac{{{a}_{1}}+{{a}_{2}}+....+{{a}_{n}}}{n}\] |
\[f'\left( x \right)=0\Rightarrow x=\operatorname{A}.\,Again\,f''\left( x \right)=2n>0\] |
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