If x be real then the minimum value of $40 - 12 x + x^{2}$ is?

28

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4

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- 4

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0

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Solution

The correct option is B $4$
$40 - 12 x + x^{2} = x^{2} - 12 x + 40$
$= x^{2} - 2 \times 6 x + 36 + 4$
$= (x^{2} - 2 \times 6 x + 6^{2}) + 4$
$= (x - 6)^{2} \leq 0$
Now
$40 - 12 x + x^{2} = (x - 6)^{2} + 4 \leq 0 + 4 \leq 4$
So minimum value is $4$ .

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