JEE Main 2023 — Binomial Theorem Question with Solution

Given,

The constant term in the binomial expansion of ${\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^l}\right)}^9$ is $-84$ and the coefficient of $x^{-3}$ is $2^{\alpha }\beta$

Now General Term of the binomial is given by, $T_{r+1}={}^{9}C_r{\left(\frac{x^{\frac{5}{2}}}{2}\right)}^{9-r}{\left(\frac{-4}{x^l}\right)}^r={\left(-1\right)}^r{C}_r{x}^{\frac{45}{2}-\frac{5r}{2}-lr}{2}^{3-9}$

Now for constant term $45-5r-2lr=0\Rightarrow r=\frac{45}{5+2l}$

So, coefficient of constant term will be,

${\left(-1\right)}^9{\times }^9{C}_r{2}^{3r-9}=-84\Rightarrow r=3$

Now putting the value of $r$ in $r=\frac{45}{5+2l}$ we get, $l=5$

Now coefficient of $x^{-3l}=x^{-15}\Rightarrow \frac{45}{2}-\frac{5r}{2}-lr=-15\Rightarrow r=5$ when $l=5$
Now coefficient of $x^{-15}={\left(-1\right)}^{59 }{C}_5{2}^6=-63\left(2^7\right)=\beta \left(2^{\alpha }\right)$

Hence, on comparing we get, $\alpha =7, \beta =-63 \& l=5$

So, the value of  $\left|\alpha l-\beta \right|=\left|7\times 5-\left(-63\right)\right|=98$