$(a)\;\frac{1}{2\sqrt{2}}\qquad(b)\;\frac{1}{2}-\frac{1}{\sqrt{2}}\qquad(c)\;\frac{1}{2}+\frac{1}{\sqrt{2}}\qquad(d)\;\frac{1}{2}$

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Answer : (c) $\;\frac{1}{2}+\frac{1}{\sqrt{2}}$

Explanation : $\;a+b+c=\frac{3}{2}$

$3b=\;\frac{3}{2}$

$b=\frac{1}{2}$

$a^2\;c^2=b^4=\frac{1}{16}\quad\;a+c=1$

$a^2\;(1-a)^2=\frac{1}{16}$

$a=\frac{1}{2}\pm\;\frac{1}{\sqrt{2}}\quad\;c=1-a=\frac{1}{2}\pm\frac{1}{\sqrt{2}}$

$a\;<\;b\;<\;c$

$c=\frac{1}{2}+\frac{1}{\sqrt{2}}.$

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