SATHEE: Application of Derivatives 4 Question 58

Application of Derivatives 4 Question 58

####61. If $x$ and $y$ be two real variables such that $x > 0$ and $x y = 1$ . Then, find the minimum value of $x + y$ .

(1981, 2M)

Integer Answer Type Questions

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Answer:

Correct Answer: 61. (7)

Solution:

Let $f (x) = x + y$ , where $x y = 1$

$\begin{array}{ll} \Rightarrow & f (x) = x + \frac{1}{x} \\ \Rightarrow & f^{'} (x) = 1 - \frac{1}{x^{2}} = \frac{x^{2} - 1}{x^{2}} \\ Also, & f^{''} (x) = 2 / x^{3} \end{array}$

On putting $f^{'} (x) = 0$ , we get

$x = \pm 1, but x > 0 [neglecting x = - 1]$

$(x) > 0, for x = 1$

Hence, $f (x)$ attains minimum at $x = 1, y = 1$

$\Rightarrow (x + y)$ has minimum value 2 .

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Application of Derivatives 4 Question 58

Integer Answer Type Questions

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