JEE 2023 — Mathematics PYQ

Solution:

General term $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$ ka use karte hue:

$T_{r+1}=\left(\genfrac{}{}{0ex}{}{5}{r}\right)(2x^3{)}^{5-r}{\left(-\frac{1}{3x^2}\right)}^r$

$T_{r+1}=\left(\genfrac{}{}{0ex}{}{5}{r}\right)2^{5-r}{x}^{15-3r}{\left(-\frac{1}{3}\right)}^r{x}^{-2r}$

$T_{r+1}=\left(\genfrac{}{}{0ex}{}{5}{r}\right)2^{5-r}{\left(-\frac{1}{3}\right)}^r{x}^{15-5r}$

Humein $x^5$ ka coefficient chahiye, isliye $15-5r=5\Rightarrow 5r=10\Rightarrow r=2$.

Coefficient $=\left(\genfrac{}{}{0ex}{}{5}{2}\right)2^{5-2}{\left(-\frac{1}{3}\right)}^2=10\cdot 2^3\cdot \frac{1}{9}=\frac{80}{9}$

Correct Option: (1) 80/9