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# The values of $a\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}b$ if $\frac{4+3\sqrt{2}}{4\sqrt{48}-\sqrt{128}+\sqrt{200}-8\sqrt{12}+5\sqrt{8}}$ is

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a
$a=\frac{1}{4},b=\frac{1}{6}$
b
$a=\frac{1}{4},b=-\frac{1}{6}$
c
$a=-\frac{1}{4},b=-\frac{1}{6}$
d
$a=-\frac{1}{4},b=\frac{1}{6}$
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detailed solution

Correct option is A

Given $\frac{4+3\sqrt{2}}{4\sqrt{48}-\sqrt{128}+\sqrt{200}-8\sqrt{12}+5\sqrt{8}}$

$=\frac{4+3\sqrt{2}}{12\sqrt{2}}$

$=\frac{4+3\sqrt{2}}{16\sqrt{3}-8\sqrt{2}+10\sqrt{2}-16\sqrt{3}+10\sqrt{2}}$

$=\frac{4\sqrt{2}+6}{24}$

$=\frac{4\sqrt{2}}{24}+\frac{6}{24}$

$=\frac{1}{4}+\frac{\sqrt{2}}{6}$

$=a+b\sqrt{2}$

$a=\frac{1}{4},b=\frac{1}{6}$

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