Question

Let $f\left(x\right)=(x-p{)}^2+(x-q{)}^2+(x-r{)}^2$. Then $f\left(x\right)$ has a minimum at $x=\lambda$, where $\lambda$ is equal to

• A. $\frac{p+q+r}{3}$
• B. ${}^3\sqrt{pqr}$
• C. $\frac{3}{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}$
• D. none of these

Answer 1

Answer: (A)

$\frac{p+q+r}{3}$

$f\left(x\right)=(x-p{)}^2+(x-q{)}^2+(x-r{)}^2$
For minima of $f\left(x\right)$ at $x=\lambda$
$f^{\prime }\left(\lambda \right)=0=2(\lambda -p)+2(\lambda -q)+2(\lambda -r)=0$
$\Rightarrow \lambda =\frac{p+q+r}{3}$