First slide

Mean value theorems

Question

The value of c In Lagrange’s theorem for the following for the function $f (x) = \log \sin x$ in the interval $[π / 6, 5 π / 6]$

Moderate

A

$π / 4$
B

$π / 2$
C

$2 π / 3$
D

None of these

Solution

$f (\frac{5 π}{6}) = \log \sin (\frac{5 π}{6}) = \log \sin \frac{π}{6}$
$= \log \frac{1}{2} = \log 2,$

$f (\frac{π}{6}) = \log \sin \frac{π}{6} = - \log 2.$

$f^{'} (c) = \frac{1}{\sin x} \cos x = \cot x$

By Lagrange’s mean value theorem,

$\frac{f (5 π / 6) - f (π / 6)}{(5 π / 6) (π / 6)} = \cot c$

$\Rightarrow \cot c = 0 \Rightarrow c = \frac{π}{2}$

Thus, $c = \frac{π}{2} \in (π / 6, 5 π / 6) .$

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