Let f(x) = (x - a)^2+ (x - b)^2 + (x - c)^2. Then, f(x) has a min… - 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 $f'(x) = 0$
$2(x - a)^+ 2(x - b)^2 + 2(x - c) = 0$
$\Rightarrow \text{x}=\frac{\text{a}+\text{b}+\text{c}}{3}$
$f''(x) = 6 > 0$
function has minima at $\text{x}=\frac{\text{a}+\text{b}+\text{c}}{3}$.

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