Question

The least value of the sum of any positive real number and its reciprocal is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$2$

…Step 1: Assume a variable and form a function

Let, $x$ be any positive real number.

Then, the function representing $x$ and its reciprocal is,

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

$\Rightarrow$ $f\left(x\right)=x+x^{-1}$

On differentiating the above with respect to $x$,

$f\text{'}\left(x\right)=1-x^{-2}$ …$\left[\because \frac{d}{dx}{x}^n=n{x}^{n-1}\right]$

Again, differentiating the above with respect to $x$,

$f\text{'}\text{'}\left(x\right)=0+2x^{-3}$

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

Step 2: Apply the conditions of maxima

For the function $f\left(x\right)$ to be maximum or minimum, it is necessary that

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

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

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

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

$\Rightarrow$ $x^2=1$

$\Rightarrow$ $x=\pm 1$ … [taking square root on both sides]

Now, since $x$ is a positive real number.

So, $x=1$

Step 3: Apply the conditions of minima

Now, as we know, for a function $f\left(x\right)$ to be minimum, the necessary condition is $f\text{'}\text{'}\left(x\right)>0$.

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

$\Rightarrow$ $f\text{'}\text{'}\left(x\right)=\frac{2}{1^3}$ …$\left[\because x=1\right]$

$\Rightarrow$ $f\text{'}\text{'}\left(x\right)=2>0$

Thus, the function $f\left(x\right)$ is minimum at $x=1$.

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

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

$\Rightarrow$ $f\left(x\right)=1+1$

$\Rightarrow$ $f\left(x\right)=2$

that is, the least value of the sum of any positive real number and its reciprocal is $2$.

Hence, option (B) is the correct option.