Question:
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) $3 \frac{1}{3}$
Solution:
(b) 2
Given : $x y=1$
$\Rightarrow y=\frac{1}{x}$
$f(x)=x+\frac{1}{x}$
$\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$
For a local maxima or a local minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow 1-\frac{1}{x^{2}}=0$
$\Rightarrow x^{2}-1=0$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm 1$
$\Rightarrow x=1$ (Given : $x>1$ )
$\Rightarrow y=1$
Now,
$f^{\prime \prime}(x)=\frac{2}{x^{3}}$
$\Rightarrow f^{\prime \prime}(1)=2>0$
So, $x=1$ is a local minima.
$\therefore$ Minimum value of $f(x)=f(1)=1+1=2$
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