#Mark the correct alternative in each of the following
Let f(x) = (x – a)2 + (x – b)2 + (x – c)2. Then, f(x) has a minimum at x =
f(x)=(x-a)2 +(x-b)2 +(x - c)2
f'(x)=2[x – a + x – b + x - c]
f'(x)=2[3x-a-b-c]
f’(x)=0
2[3x-a-b-c] =0
3x-a-b-c=0
3x=a + b + c
Hence by second derivative test
f’’(x)>0 so it’s a point of minimum.
f” (x)=2(3(1))
=6>0
so point of minimum.
Option(A).