JEE 2023 — Mathematics PYQ

Solution

For ${\left(a{x}^3+\frac{1}{b{x}^{1/3}}\right)}^{15}$, general term $T_{r+1}=\left(\genfrac{}{}{0ex}{}{15}{r}\right)(a{x}^3{)}^{15-r}(b{x}^{-1/3}{)}^r$

Power of $x=3(15-r)-\frac{r}{3}=15⟹45-\frac{10r}{3}=15⟹r=9$

Coeff of $x^{15}=\left(\genfrac{}{}{0ex}{}{15}{9}\right)a^6{b}^{-9}$

For ${\left(a{x}^{1/3}-\frac{1}{b{x}^3}\right)}^{15}$, power of $x=\frac{15-r}{3}-3r=-15⟹15-10r=-45⟹r=6$

Coeff of $x^{-15}=\left(\genfrac{}{}{0ex}{}{15}{6}\right)a^9(-b{)}^{-6}=(\genfrac{}{}{0ex}{}{15}{6})a^9{b}^{-6}$

Given: $\left(\genfrac{}{}{0ex}{}{15}{9}\right)a^6{b}^{-9}=\left(\genfrac{}{}{0ex}{}{15}{6}\right)a^9{b}^{-6}$

$\frac{1}{b^3}=a^3⟹(ab{)}^3=1⟹ab=1$

Sahi Option: (2)