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The value of$'c'$ of Lagranges Mean Value Theorem for $f(x)=\sqrt{x}$ when $a=1$ and $b=4$ is

\[\begin{array} (1)\frac{9}{4}&(2)\frac{3}{2}\\(3)\frac{1}{2}&(4)\frac{1}{4} \end{array}\]

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$f(x)= \sqrt x$
$f'(x) =\large\frac{1}{2 \sqrt x}$
$f(1) =\sqrt 1$
$f(4)=\sqrt 4=2$
$f'(c)= \large\frac{f(b)-f(a)}{b-a}$
$\qquad= \large\frac{f(4)-f(1) }{4-1}$
$\qquad= \large\frac{2-1}{4-1} $
$\qquad= \large\frac{1}{3}$-----(1)
$f'(x) =\large\frac{1}{2 \sqrt x}$
$f'(c) =\large\frac{1}{2 \sqrt c}$
(1) => $ \large\frac{1}{2\sqrt c}=\large\frac{1}{3}$
$=> \large\frac{1}{\sqrt c}=\frac{2}{3}$
$=> c= \large\frac{9}{4} \in (1,4)$
Hence 1 is the correct answer.
answered May 19, 2014 by meena.p
 
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