QJEEM22Macef0: JEE Main 2022 Mathematics - Class 11 JEEM22Macef0 Solution: Let be a function defined by . If the function , then the…

Given,

$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)\left( {2 + {x^{25}}} \right)} \right)^{{1 \over {50}}}}$

and $g(x) = f\left( {f\left( {f\left( x \right)} \right)} \right) + f\left( {f\left( x \right)} \right)$

$\therefore$ $g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$

Now, $f(1) = {\left( {2\left( {1 - {{{1^{25}}} \over 2}} \right)\left( {2 + {1^{25}}} \right)} \right)^{{1 \over {50}}}}$

$= {\left( {2\left( {1 - {1 \over 2}} \right)\left( {2 + 1} \right)} \right)^{{1 \over {50}}}}$

$= {\left( 3 \right)^{{1 \over {50}}}}$

$\therefore$ $f\left( {f\left( 1 \right)} \right) = f\left( {{3^{{1 \over {50}}}}} \right)$

$= {\left( {2\left( {1 - {{{{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \over 2}} \right)\left( {2 + {{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \right)} \right)^{{1 \over {50}}}}$

$= {\left( {2\left( {1 - {{{3^{{1 \over 2}}}} \over 2}} \right)\left( {2 + {3^{{1 \over 2}}}} \right)} \right)^{{1 \over {50}}}}$

$= {\left( {2 \times \left( {{{2 - \sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)} \right)^{{1 \over {50}}}}$

$= {\left[ {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} \right]^{{1 \over {50}}}}$

$= {\left( {4 - 3} \right)^{{1 \over {50}}}}$

$= {1^{{1 \over {50}}}} = 1$

Now, $f\left( {f\left( {f\left( 1 \right)} \right)} \right) = f(1) = {3^{{1 \over {50}}}}$

$\therefore$ $g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$

$= {3^{{1 \over {50}}}} + 1$

Now, greatest integer less than or equal to $g(1)$

$= \left[ {g(1)} \right]$

$= \left[ {{3^{{1 \over {50}}}} + 1} \right]$

$= \left[ {{3^{{1 \over {50}}}}} \right] + \left[ 1 \right]$

$= [1.02] + 1$

$= 1 + 1 = 2$