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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) a+b+c3

(b) abc3

(c) 31a+1b+1c

(d) none of these

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Solution

(a) a+b+c3Given: fx= x-a2+x-b2+x-c2f'x= 2x-a+2x-b+2x-cFor a local maxima or a local minima, we must have f'x=02x-a+2x-b+2x-c=02x-2a+2x-2b+2x-2c=06x=2a+b+cx=a+b+c3Now, f''x=2+2+2=6>0So, x=a+b+c3 is a local minima.

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