# For the function $f(x)=x+\large \frac{1}{x},\normalsize x\in[1,3]$,the value of c for mean value theorem is

$\begin{array}{1 1}(A)\;1 & (B)\;\sqrt 3\\(C)\;2 &(D)\;none\;of\;these\end{array}$

Toolbox:
• Let $f(x)$ be a function defined on $[a,b]$ such that
• $\quad(i)$ it is continuous on $[a,b]$.
• $\quad(ii)$ it is differentiable on $[a,b]$.Then there exists a real number $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$f(x)=x+\large\frac{1}{x}$$;[1,3]$
$f'(x)=1-\large\frac{1}{x^2}$
$f'(c)=\large\frac{f(b)-f(a)}{b-a}$
$f'(1)=1+1=2$
$f'(3)=3+\large\frac{1}{3}=\frac{10}{3}$
Therefore $f'(c)=1-\large\frac{1}{c^2}$
Step 2:
$1-\large\frac{1}{c^2}=\large\frac{\Large\frac{10}{3}-\normalsize 2}{3-1}$
$\large\frac{c^2-1}{c^2}=\frac{\Large\frac{4}{3}}{2}$
$\Rightarrow \large\frac{c^2-1}{c^2}=\frac{4}{6}$
$\Rightarrow 6c^2-6=4c^2$
$\Rightarrow 2c^2=6$
$c^2=3$
$c=\pm \sqrt 3$
Clearly $\sqrt 3$ lies in the given interval $[1,3]$
Hence $B$ is the correct option.