QJEEAdv12Ma6546: JEE Advanced 2012 Mathematics - Class 11 JEEAdv12Ma6546 Solution

$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}...} } } } \right)$

Let $\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}} \sqrt {...} } = y$

$\therefore$ $y = \sqrt {4 - {1 \over {3\sqrt 2 }}y}$

$\Rightarrow {y^2} + {1 \over {3\sqrt 2 }}y - 4 = 0$

$\Rightarrow 3\sqrt 2 {y^2} + y - 12\sqrt 2 = 0$

$\therefore$ $y = {{ - 1 \pm 17} \over {6\sqrt 2 }}$ or $y = {8 \over {3\sqrt 2 }}$

Now,

$6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.y} \right) = 6 + {\log _{{3 \over 2}}}\left( {{1 \over {3\sqrt 2 }}.{8 \over {3\sqrt 2 }}} \right)$

$= 6 + {\log _{{3 \over 2}}}\left( {{4 \over 9}} \right) = 6 + {\log _{{3 \over 2}}}{\left( {{3 \over 2}} \right)^{ - 2}}$

$= 6 - 2.{\log _{{3 \over 2}}}\left( {{3 \over 2}} \right) = 4$