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Question

The minimum value of f(x)=(x1)2+(x2)2++(x10)2 occurs at x=k. Then the value of [k] is
(where [.] represents the greatest integer function)

A
5.0
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B
5.00
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C
05
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D
5
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Solution

f(x)=2((x1)+(x2)++(x10))
f(x)=2(10x(1+2+10))
From f(x)=0
2(10x10×112)=0
10x=55x=112=5.5

Since, sign of f(x) changes from negative to positive as x crosses 112 from left to right, therefore x=112 is a point of local minima.
As there is only one point of local minima so, at this point function is minimum.
Hence, [k]=[112]=5

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