Question

The minimum value of the expression $a+\frac{1}{a};a>0$ is:

• A. 1
• B. 2
• C. 4

Answer 1

The correct option is B 2

See the following graph of $y=x+\frac{1}{x}.$
Here, x = a
$y=x+\frac{1}{x}=a+\frac{1}{a}$

$\begin{array}{ccc}a& \frac{1}{a}& a+\frac{1}{a}\\ 1& 1& 2\\ 2& \frac{1}{2}& 2.5\\ 3& \frac{1}{3}& 3.33\\ 4& \frac{1}{4}& 4.25\\ 5& \frac{1}{5}& 5.20\\ \frac{1}{2}& 2& 2.5\\ \frac{1}{3}& 3& 3.33\\ \frac{1}{4}& 4& 4.25\end{array}$

Alternatively:
$y=a+\frac{1}{a}$ ...(i)
Diffrerentiating above eq. (i), we get
$\frac{dy}{da}=1-\frac{1}{a^2}$ ...(ii)
Equating eq. (ii) with zero, we get
$1-\frac{1}{a^2}=0\Rightarrow a^2=1\Rightarrow a=\pm 1$
But since a > 0, hence only a = 1 is acceptable. Now substituting a = 1 in the original eq. (i) we get the required minimim value of the expression.
Hence, minimum of
$y=1+\frac{1}{1}=2$