What is the value of x, when f(x)=6+(x−2)2 is at its minimum?
A
−6
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B
−4
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C
0
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D
2
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E
5
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Solution
The correct option is D2 Given, f(x)=6+(x−2)2 For f(x) to be minimum, 6+(x−2)2 must be minimum, which implies (x−2)2 should be minimum. The minimum value of any square term is 0. So, the minimum value of (x−2)2 is 0 and occurs at x=2.