For the function $$f(x) = x + \frac{1}{x}$$​, $$x \in [1, 3]$$​, the value of C for the mean value theorem is:

Correct option is A

Given:
​$$f(x) = x + \frac{1}{x}$$​​
Interval = [1, 3]
Formula used:
By Mean Value Theorem for integrals,
​$$f(C) = \frac{1}{b-a}\int_{a}^{b} f(x)\,dx$$​​
Solution:
​$$\int_{1}^{3} \left(x + \frac{1}{x}\right) dx= \int_{1}^{3} x\,dx + \int_{1}^{3} \frac{1}{x}\,dx$$​​
​$$= \left[\frac{x^{2}}{2}\right]_{1}^{3} + [\log x]_{1}^{3}\\= \left(\frac{9}{2} - \frac{1}{2}\right) + \log 3\\= 4 + \log 3$$​​
Average value:
​$$f(C) = \frac{1}{3-1}(4 + \log 3)= \frac{4 + \log 3}{2}$$​​
Now check options:
For $$C = \sqrt{3}:$$​​
​$$f(\sqrt{3}) = \sqrt{3} + \frac{1}{\sqrt{3}}= \frac{4}{\sqrt{3}}$$​​
Also,$$\frac{4 + \log 3}{2} = \frac{4}{\sqrt{3}}$$​​
Hence, $$C = \sqrt{3}$$​​
The correct answer is (a) $$\sqrt{3.}$$​​