Question
Let $$f\left( x \right) = {\left( {x - p} \right)^2} + {\left( {x - q} \right)^2} + {\left( {x - r} \right)^2}.$$ Then $$f\left( x \right)$$ has a minimum at $$x = \lambda ,$$ where $$\lambda $$ is equal to :
A.
$$\frac{{p + q + r}}{3}$$
B.
$$\root 3 \of {pqr} $$
C.
$$\frac{3}{{\frac{1}{p} + \frac{1}{q} + \frac{1}{r}}}$$
D.
none of these
Answer :
$$\frac{{p + q + r}}{3}$$
Solution :
$$\eqalign{
& f\left( x \right) = 3{x^2} - 2\left( {p + q + r} \right)x + {p^2} + {q^2} + {r^2} \cr
& \therefore \,f'\left( x \right) = 6x - 2\left( {p + q + r} \right) \cr
& \therefore \,f'\left( x \right) = 0 \cr
& \Rightarrow x = \frac{{p + q + r}}{3} = {\text{AM of }}p,\,q,\,r \cr
& {\text{Also, }}f''\left( x \right) = 6 > 0 \cr} $$