Let f(x)=(x-a)^2+(x-b)^2+(x-c)^2. Then, f(x) has a minimum at x= - Vidyadip

MCQ

Let $f(x)=(x-a)^2+(x-b)^2+(x-c)^2.$ Then, $f(x)$ has a minimum at $x=$

✓
$\frac{\text{a}+\text{b}+\text{c}}{3}$
B
$\sqrt[3]{\text{a}\text{b}\text{c}}$
C
$\frac{3}{\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}}$
D
None of these.

✓

Answer

Correct option: A.

$\frac{\text{a}+\text{b}+\text{c}}{3}$

$f(x)=(x-a)^2+(x-b)^2+(x-c)^2$
$\Rightarrow 2(x-a)+2(x-b)+2(x-c)$
to find minima or maxima $f ^{\prime}(x)=0$
$2(x-a)+2(x-b)^2+2(x-c)=0$
$\Rightarrow x=\frac{a+b+c}{3}$
$f^{\prime \prime}(x)=6>0$
function has minima at $\text{x}=\frac{\text{a}+\text{b}+\text{c}}{3}$.

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