Let f(x) = (x - a)2 + (x - b)2 + (x - c)2. Then, f(x) has a minimum at x = 1. a+b+c 3 2. [3] abc 3. 3 1 a +1 b +1 c 4. None of these.

1. a+b+c 3 Solution: f(x) = (x - a)2 + (x - b)2 + (x - c)2 ⇒ 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 x= a+b+c 3 f''(x) = 6 > 0 function has minima at x= a+b+c 3 .

Let f(x) = (x - a)2 + (x - b)2 + (x - c)2. Then, f(x) has a minim…

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Question

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

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

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Answer

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

Solution:

f(x) = (x - a)²+ (x - b)² + (x - c)²

⇒ 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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