JEE Advanced 2023 — Paper 1 — Question 12

General term of $\left(a x^{2}+\frac{70}{27 b x}\right)^{4}$ is ${ }^{4} C_{r}\left(a x^{2}\right)^{4-r}\left(\frac{70}{27 b x}\right)^{r}={ }^{4} C_{r} a^{4-r} \frac{(70)^{r}}{(27 b)^{r}} x^{8-3 r}$ for coefficient of $x^{5}$ we put $8-3 r=5 \Rightarrow r=1$ $\therefore$ coeff. of $\mathrm{x}^{5}$ is ${ }^{4} \mathrm{C}_{1} \frac{\mathrm{a}^{3} \times 70}{27 \mathrm{~b}}=\frac{280}{27} \frac{\mathrm{a}^{3}}{\mathrm{~b}}$ General term of $\left(a x-\frac{1}{b x^{2}}\right)^{7}$ is ${ }^{7} C_{r} \frac{a^{7-r}(-1)^{r}}{b^{r}} x^{7-3 r}$ for coeff. of $\mathrm{x}^{-5}$ we put $7-3 \mathrm{r}=-5 \Rightarrow \mathrm{r}=4$ $\therefore$ coeff. of $\mathrm{x}^{-5}$ is $^{7} \mathrm{C}_{4} \frac{\mathrm{a}^{3}}{\mathrm{~b}^{4}}=\frac{35 \mathrm{a}^{3}}{\mathrm{~b}^{4}}$ Given $\frac{280}{27} \frac{\mathrm{a}^{3}}{\mathrm{~b}}=\frac{35 \mathrm{a}^{3}}{\mathrm{~b}^{4}} \Rightarrow \mathrm{~b}^{3}=\frac{27}{8} \Rightarrow \mathrm{~b}=\frac{3}{2}$ $\Rightarrow 2 \mathrm{~b}=3$