JEE Main 2022 (Online) 28th June Evening ShiftQuestion 1
The term independent of $x$ in the expansion of $\left(1-x^{2}+3x^{3}\right)\left(\frac{5}{2}x^{3}-\frac{1}{5x^{2}}\right)^{11}$, $x\neq0$ is :
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. The binomial gives only the powers $33-5r$; check which term of $1-x^2+3x^3$ can be matched. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - General term. $$T_{r+1}={}^{11}C_r\left(\frac52\right)^{11-r}\left(-\frac15\right)^r x^{33-5r}.$$ Step 2 - Which pairs give $x^0$? With $1$: $33-5r=0$, no integer $r$. With $-x^2$: $33-5r=-2$, $r=7$. With $3x^3$: $33-5r=-3$, no integer $r$. Step 3 - The term. $$-{}^{11}C_7\left(\frac52\right)^4\left(-\frac15\right)^7=330\cdot\frac{625}{16}\cdot\frac{1}{78125}=\frac{330}{2000}=\frac{33}{200}.$$ Answer: (B)