5. Continuity and Differentiation
easy

For the function$x + {1 \over x},x \in [1,\,3]$, the value of $ c$  for the mean value theorem is

A

$1$

B

$\sqrt 3 $

C

$2$

D

None of these

Solution

(b) $f'(x) = 1 – \frac{1}{{{x^2}}} \Rightarrow f'(c) = 1 – \frac{1}{{{c^2}}}$

$\therefore 1 – \frac{1}{{{c^2}}} = \frac{{\frac{{10}}{3} – 2}}{2} $

$\Rightarrow 1 – \frac{1}{{{c^2}}} = \frac{2}{3} \Rightarrow {c^2} = 3$.

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.