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Let $g(n) $ is a continuous differential function which takes positive value for $ x \leq 0$ and satisfy $\int \limits_0^2 g(t) dt = x \sqrt {g(x)} $ with $f(1) =V_2$. find the value of $f(\sqrt 2 +1)$

$(a)\;\frac{1}{4} \\ (b)\;4 \\ (c)\;\sqrt 2 +1 \\ (d)\;\sqrt 2 -1 $
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$\int \limits_0^x g(t) d(t) =x \sqrt {g(x)}$
$g(x)=\sqrt {g(x)}-\large\frac{x(n)}{2 \sqrt g (n)}$
$g(n) =y$
$(y\sqrt y)=-\large\frac{x}{2 \sqrt y }\frac{dy}{dx}$
$\int \large\frac{-2dx}{x} =\int \large\frac{dy}{\sqrt y(y-\sqrt y)}$
$-2 \log x =\int \large\frac {dx}{y^{3/2} (\frac{1}{\sqrt y}-1)}$
$-2 \log x =\int \large\frac{-2 dt}{t}$
$\log x =\log t + c$
$\log x/t =\log c'$
$x= c't$
$x= \large\frac{c'(1-\sqrt {9(n)})}{\sqrt 9 (n)}$
$g(1)= \large\frac{1}{2}$
$1= \large\frac{c'(1- 1/ \sqrt 2)}{1/ \sqrt 2}$
$\large\frac{1}{\sqrt 2}= c' \bigg(\large\frac{\sqrt 2 -1}{\sqrt 2}\bigg)$
$c'=\large\frac{1}{\sqrt 2 -1}$
$c'= \large\frac{1}{(\sqrt 2 -1)}$$(\sqrt 2 +1)$
$n=(\sqrt 2 +1) \bigg( \large\frac{1}{\sqrt {g(n)}}-1\bigg)$
$g(\sqrt 2+1)=>$
$\sqrt 2 +1 =\sqrt 2+1 \bigg( \large\frac{1}{\sqrt g(\sqrt 2 +1)}-1\bigg)$
$2= \large\frac{1}{\sqrt {g(\sqrt 2 +1)}}$
$g( \sqrt 2 +1)=\large\frac{1}{4}$
Hence a is the correct answer.
answered Feb 3, 2014 by meena.p
 

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