JEE Main 202228 Jun, Evening shiftMathematicsMediumBinomial Theorem and Its Simple Applications131× since 2002The term independent of x in the expansion of (1−x2+3x3)(52x3−15x2)11, x≠0(1 - {x^2} + 3{x^3}){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}},\,x \ne 0(1−x2+3x3)(25x3−5x21)11,x=0 is :AOption A. 740{7 \over {40}}407BOption B. 33200{33 \over {200}}20033COption C. 39200{39 \over {200}}20039DOption D. 1150{11 \over {50}}5011CheckSkip · show solution
