JEE 2023 — Mathematics PYQ

Let $T_{r+1}$ be the constant term.

$T_{r+1}=7C_r(3x{)}^{2\gamma -r}{\left(\frac{-1}{2x^5}\right)}^r$

For constant term, power of $x$ should be zero.

i.e., $14-2r-5r=0$

$\Rightarrow 14=7r\Rightarrow r=2$

Now, constant term is $\alpha$.

$\Rightarrow 7C_2(3{)}^5{\left(\frac{-1}{2}\right)}^2=\alpha$

$\Rightarrow 21\times 243\times \frac{1}{4}=\alpha$

$\Rightarrow [\alpha ]=1275.75=1275$