Question

The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to

• A. $-2$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

The correct option is C

$1$

Compute the required value:

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

Differentiate with respect to $x$,

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

Put $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$

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

$\Rightarrow$ $x=\pm 1$

Differentiate $\mathrm{f}\text{'}\left(\mathrm{x}\right)$ with respect to $x$,

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

Substitute the value of $x$,

$$\begin{gathered} \mathrm{f}\text{'}\text{'}\left(1\right)=2 \\ \mathrm{f}\text{'}\text{'}(-1)=-2 \end{gathered}$$

Since, value of $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)$ is positive for $\mathrm{x}=1$,

Therefore, for $\mathrm{x}=1$, $\mathrm{f}\left(\mathrm{x}\right)$ have minimum value.

Hence, option (C) is the correct option.