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

Answer 1

$$\begin{gathered} \left(c\right)3 \\ \mathrm{Given}:f\left(x\right)=\sum _{r=1}^5{\left(x-r\right)}^2 \\ \Rightarrow f\left(x\right)={\left(x-1\right)}^2+{\left(x-2\right)}^2+{\left(x-3\right)}^2+{\left(x-4\right)}^2+{\left(x-5\right)}^2 \\ \Rightarrow f\text{'}\left(x\right)=2\left(x-1+x-2+x-3+x-4+x-5\right) \\ \Rightarrow f\text{'}\left(x\right)=2\left(5x-15\right) \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{and}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 2\left(5x-15\right)=0 \\ \Rightarrow 5x-15=0 \\ \Rightarrow 5x=15 \\ \Rightarrow x=3 \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=10 \\ f\text{'}\text{'}\left(x\right)=10>0 \\ \mathrm{So},\mathit{}x=3\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{minima}. \end{gathered}$$