Question

The real number $x$ when added to its inverse gives the minimum value of the sum at $x=$
$(x\in +R)$

• A. $1$
• B. $3$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $1$

Real number$=x$

Inverse of this number$=\frac{1}{x}$

Now, we have to find value of x at which $(x+\frac{1}{x})$ is minimum

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

$f^{\prime }\left(x\right)\Rightarrow 1-\frac{1}{x^2}=0$

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

$x^2=1$

$x=\pm 1$

As $xϵ+R$

$f\left(x\right)$ is minimum at $x=+1$