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# Solution for The Function F(X) = 5 ∑ R = 1 (X − R)2 Assumes Minimum Value at X = (A) 5 (B) 5 2 (C) 3 (D) 2 - CBSE (Commerce) Class 12 - Mathematics

#### Question

The function f(x) = $\sum^5_{r = 1}$ (x $-$ r)2 assumes minimum value at x =

(a) 5
(b) $\frac{5}{2}$

(c) 3
(d) 2

#### Solution

$(c) 3$

$\text { Given:} f\left( x \right) = \sum^5_{r = 1} \left( x - r \right)^2$

$\Rightarrow f\left( x \right) = \left( x - 1 \right)^2 + \left( x - 2 \right)^2 + \left( x - 3 \right)^2 + \left( x - 4 \right)^2 + \left( x - 5 \right)^2$

$\Rightarrow f'\left( x \right) = 2\left( x - 1 + x - 2 + x - 3 + x - 4 + x - 5 \right)$

$\Rightarrow f'\left( x \right) = 2\left( 5x - 15 \right)$

$\text { For a local maxima and a local minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow 2\left( 5x - 15 \right) = 0$

$\Rightarrow 5x - 15 = 0$

$\Rightarrow 5x = 15$

$\Rightarrow x = 3$

$\text { Now,}$

$f''\left( x \right) = 10$

$f''\left( x \right) = 10 > 0$

$\text { So, x = 3 is a local minima }.$

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Solution for question: The Function F(X) = 5 ∑ R = 1 (X − R)2 Assumes Minimum Value at X = (A) 5 (B) 5 2 (C) 3 (D) 2 concept: Graph of Maxima and Minima. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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