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# Let F(X) = (X − A)2 + (X − B)2 + (X − C)2. Then, F(X) Has a Minimum at X = (A) a + B + C 3 (B) 3 √ a B C (C) 3 1 a + 1 B + 1 C (D) None of These - CBSE (Science) Class 12 - Mathematics

#### Question

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

• $\frac{a + b + c}{3}$

• $\sqrt[3]{abc}$

• $\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

• none of these

#### Solution

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

$\text { Given }: f\left( x \right) = \left( x - a \right)^2 + \left( x - b \right)^2 + \left( x - c \right)^2$
$\Rightarrow f'\left( x \right) = 2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right)$
$\text { For a local maxima or a local minima, we must have }$
$f'\left( x \right) = 0$
$\Rightarrow 2\left( x - a \right) + 2\left( x - b \right) + 2\left( x - c \right) = 0$
$\Rightarrow 2x - 2a + 2x - 2b + 2x - 2c = 0$
$\Rightarrow 6x = 2\left( a + b + c \right)$
$\Rightarrow x = \frac{a + b + c}{3}$
$\text { Now },$
$f''\left( x \right) = 2 + 2 + 2 = 6 > 0$
$\text { So }, x = \frac{a + b + c}{3} \text { is a local minima. }$
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Solution Let F(X) = (X − A)2 + (X − B)2 + (X − C)2. Then, F(X) Has a Minimum at X = (A) a + B + C 3 (B) 3 √ a B C (C) 3 1 a + 1 B + 1 C (D) None of These Concept: Graph of Maxima and Minima.
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