Q8 of 137 Page 19

#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).

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