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# Solution for Let X, Y Be Two Variables and X>0, Xy=1, Then Minimum Value of X+Y is (A) 1 (B) 2 (C) 2 1 2 (D) 3 1 3 - CBSE (Commerce) Class 12 - Mathematics

#### Question

Let x, y be two variables and x>0, xy=1, then minimum value of x+y is
(a) 1
(b) 2
(c) $2\frac{1}{2}$

(d) $3\frac{1}{3}$

#### Solution

(b) 2

$\text { Given }: xy = 1$

$\Rightarrow y = \frac{1}{x}$

$f\left( x \right) = x + \frac{1}{x}$

$\Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}$

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

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

$\Rightarrow 1 - \frac{1}{x^2} = 0$

$\Rightarrow x^2 - 1 = 0$

$\Rightarrow x^2 = 1$

$\Rightarrow x = \pm 1$

$\Rightarrow x = 1 \left( \text { Given }: x > 1 \right)$

$\Rightarro \[\text { Now,}$

$f''\left( x \right) = \frac{2}{x^3}$

$\Rightarrow f''\left( 1 \right) = 2 > 0$

$\text { So, x = 1 is a local minima } .$

$\therefore \text { Minimum value of } f\left( x \right) = f\left( 1 \right) = 1 + 1 = 2$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution for question: Let X, Y Be Two Variables and X>0, Xy=1, Then Minimum Value of X+Y is (A) 1 (B) 2 (C) 2 1 2 (D) 3 1 3 concept: Graph of Maxima and Minima. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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