MCQ
If $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}, \text{x} > 0$, then its greatest value is :
  • A
    $-2$
  • B
    $0$
  • C
    $3$
  • None of these.

Answer

Correct option: D.
None of these.
Given, $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$
$\Rightarrow \text{f}'(\text{x})=1-\frac{1}{\text{x}^{2}}$
For a local maxima or a local minima, we must have $f\ '(x) = 0$
$\Rightarrow 1-\frac{1}{\text{x}^{2}}=0$
$\Rightarrow \text{x}^{2}-1=0$
$\Rightarrow \text{x}^{2}=1$
$\Rightarrow \text{x}=\pm1$
$\Rightarrow \text{x}=1$
Now, $\text{f}\ ''(\text{x})=\frac{2}{\text{x}^{3}}$
$\text{f}\ ''(1)=2 > 0$
So, $x = 1$ is a local minima.

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