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Solution for Let F(X) = (X − A)2 + (X − B)2 + (X − C)2. Then, F(X) Has a Minimum at X = (A) a + B + C 3 (B) 3 √ a B C (C) 3 1 a + 1 B + 1 C (D) None of These - CBSE (Science) Class 12 - Mathematics

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Question

Let f(x) = (x \[-\] a)2 + (x \[-\] b)2 + (x \[-\] c)2. Then, f(x) has a minimum at x =

(a) \[\frac{a + b + c}{3}\]

(b) \[\sqrt[3]{abc}\]

(c) \[\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}\]

(d) none of these

Solution

\[(a)  \frac{a + b + c}{3}\] 
\[\text { Given }:   f\left( x \right) =    \left( x - a \right)^2  +  \left( x - b \right)^2  +  \left( x - c \right)^2 \] 
\[ \Rightarrow f'\left( x \right) =   2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right)\] 
\[\text { For  a  local  maxima  or  a  local  minima, we  must  have }  \] 
\[f'\left( x \right) = 0\] 
\[ \Rightarrow 2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right) = 0\] 
\[ \Rightarrow 2x - 2a + 2x - 2b + 2x - 2c = 0\] 
\[ \Rightarrow 6x = 2\left( a + b + c \right)\] 
\[ \Rightarrow x = \frac{a + b + c}{3}\] 
\[\text { Now },   \] 
\[f''\left( x \right) = 2 + 2 + 2 = 6 > 0\] 
\[\text { So },   x = \frac{a + b + c}{3} \text {  is  a  localminima. }\] 
  Is there an error in this question or solution?

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Solution for question: Let F(X) = (X − A)2 + (X − B)2 + (X − C)2. Then, F(X) Has a Minimum at X = (A) a + B + C 3 (B) 3 √ a B C (C) 3 1 a + 1 B + 1 C (D) None of These concept: Graph of Maxima and Minima. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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