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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