If the solve the problem


Let $x, y$ be two variables and $x>0, x y=1$, then minimum value of $x+y$ is

(a) 1

(b) 2

(c) $2 \frac{1}{2}$

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


(b) 2

Given : $x y=1$

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


$\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$

For a local maxima or a local minima, we must have


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

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

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm 1$

$\Rightarrow x=1$        (Given : $x>1$ )

$\Rightarrow y=1$


$f^{\prime \prime}(x)=\frac{2}{x^{3}}$

$\Rightarrow f^{\prime \prime}(1)=2>0$

So, $x=1$ is a local minima.

$\therefore$ Minimum value of $f(x)=f(1)=1+1=2$

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