Question:
The function $f(x)=\sum_{r=1}^{5}(x-r)^{2}$ assumes minimum value at $x=$
(a) 5
(b) $\frac{5}{2}$
(c) 3
(d) 2
Solution:
(c) 3
Given : $f(x)=\sum_{r=1}^{5}(x-r)^{2}$
$\Rightarrow f(x)=(x-1)^{2}+(x-2)^{2}+(x-3)^{2}+(x-4)^{2}+(x-5)^{2}$
$\Rightarrow f^{\prime}(x)=2(x-1+x-2+x-3+x-4+x-5)$
$\Rightarrow f^{\prime}(x)=2(5 x-15)$
For a local maxima and a local minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow 2(5 x-15)=0$
$\Rightarrow 5 x-15=0$
$\Rightarrow 5 x=15$
$\Rightarrow x=3$
Now,
$f^{\prime \prime}(x)=10$
$f^{\prime \prime}(x)=10>0$
So, $x=3$ is a local minima.
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