Q10 of 137 Page 19

#Mark the correct alternative in each of the following

The function assumes minimum value at x =



f(x)=(x-1)2+(x-2)2+(x-3)2+(x-4)2+(x-5)2


f’(x)=2[5x-15]


f’(x)=0;x=3


Hence by second derivative test


f’’(x)>0 so it’s a point of minimum.


f”(x)=1>0 so At x=3 minimum value.Option(C)

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