The function $f(x)=\sum\limits_{k=1}^5(x-k)^2$ assumes minimum value for x given by
Answer & explanation
Correct answer: option 2
$f(x)=(x-1)^2+(x-2)^2+(x-3)^2+(x-4)^2+(x-5)^2$
$\Rightarrow f'(x)=2[x-1+x-2+x-3+x-4+x-5] \Rightarrow f'(x)$
$=2[x-1+x-2+x-3+x-4+x-5]$
$=2(5 x-15) \Rightarrow f''(x)=10>0$ for each x
∴ f'(x) = 0 ⇒ x = 3
∴ f (x) is min. for x = 3.