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प्रश्न
Let f(x) = (x \[-\] a)2 + (x \[-\] b)2 + (x \[-\] c)2. Then, f(x) has a minimum at x = _____________ .
पर्याय
\[\frac{a + b + c}{3}\]
\[\sqrt[3]{abc}\]
\[\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\]
none of these
MCQ
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उत्तर
\[\frac{a + b + c}{3}\]
\[\text { Given }: f\left( x \right) = \left( x - a \right)^2 + \left( x - b \right)^2 + \left( x - c \right)^2 \]
\[ \Rightarrow f'\left( x \right) = 2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right)\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow 2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right) = 0\]
\[ \Rightarrow 2x - 2a + 2x - 2b + 2x - 2c = 0\]
\[ \Rightarrow 6x = 2\left( a + b + c \right)\]
\[ \Rightarrow x = \frac{a + b + c}{3}\]
\[\text { Now }, \]
\[f''\left( x \right) = 2 + 2 + 2 = 6 > 0\]
\[\text { So }, x = \frac{a + b + c}{3} \text { is a local minima. }\]
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