QJEEM22Ma4d1a: JEE Main 2022 Mathematics - Class 11 JEEM22Ma4d1a Solution: The term independent of x in the expansion of is :

General term of Binomial expansion ${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$ is

${T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}$

$= {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left( { - {1 \over 5}} \right)^r}\,.\,{x^{33 - 5r}}$

In the term,

$\left( {1 - {x^2} + 3{x^3}} \right){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$

Term independent of x is when

(1) $33 - 5r = 0$

$\Rightarrow r = {{33} \over 5}\,\, \notin$ integer

(2) $33 - 5r = -2$

$\Rightarrow 5r = 35$

$\Rightarrow r = 7\,\, \in$ integer

(3) $33 - 5r = - 3$

$\Rightarrow 5r = 36$

$\Rightarrow r = {{36} \over 5}\,\, \notin$ integer

$\therefore$ Only for r = 7 independent of x term possible.

$\therefore$ Independent of x term

$= - \left( {{}^{11}{C_7}{{\left( {{5 \over 2}} \right)}^4}\,.\,{{\left( { - {1 \over 5}} \right)}^7}} \right)$

$= - \left( {{{11\,.\,10\,.\,9\,.\,8} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{{{5^4}} \over {{2^4}}}\,.\, - {1 \over {{5^7}}}} \right)$

$= {{11\,.\,10\,.\,3} \over {{2^4}\,.\,{5^3}}}$

$= {{11\,.\,3} \over {{2^3}\,.\,{5^2}}}$

$= {{33} \over {200}}$