QJEEM22Mae4de: JEE Main 2022 Mathematics - Class 11 JEEM22Mae4de Solution: If the maximum value of the term independent of in the…

General term of ${\left( {{t^2}{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{15}}$ is

${T_{r + 1}} = {}^{15}{C_r}\,.\,{\left( {{t^2}{x^{{1 \over 5}}}} \right)^{15 - r}}\,.\,{\left( {{{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^r}$

$= {}^{15}{C_r}\,.\,{t^{30 - 2r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}\,.\,{t^{ - r}}$

$= {}^{15}{C_r}\,.\,{t^{30 - 3r}}\,.\,{x^{{{15 - r} \over 5}}}\,.\,{\left( {1 - x} \right)^{{r \over {10}}}}$

Term will be independent of $\mathrm{t}$ when $30 - 3r = 0 \Rightarrow r = 10$

$\therefore$ ${T_{10 + 1}} = {T_{11}}$ will be independent of $\mathrm{t}$

$\therefore$ ${T_{11}} = {}^{15}{C_{10}}\,.\,{x^{{{15 - 10} \over 5}}}\,.\,{\left( {1 - x} \right)^{{{10} \over {10}}}}$

$= {}^{15}{C_{10}}\,.\,{x^1}\,.\,{\left( {1 - x} \right)^1}$

$\mathrm{T_{11}}$ will be maximum when $x(1 - x)$ is maximum.

Let $f(x) = x(1 - x) = x - {x^2}$

$f(x)$ is maximum or minimum when $f'(x) = 0$

$\therefore$ $f'(x) = 1 - 2x$

For maximum/minimum $f'(x) = 0$

$\therefore$ $1 - 2x = 0$

$\Rightarrow x = {1 \over 2}$

Now, f''(x) = - 2 < 0

$\therefore$ At $x = {1 \over 2}$ , $f(x)$ maximum

$\therefore$ Maximum value of $\mathrm{T_{11}}$ is

$= {}^{15}{C_{10}}\,.\,{1 \over 2}\left( {1 - {1 \over 2}} \right)$

$= {}^{15}{C_{10}}\,.\,{1 \over 4}$

Given $K = {}^{15}{C_{10}}\,.\,{1 \over 4}$

Now, $8K = 2\left( {{}^{15}{C_{10}}} \right)$

$= 6006$