Question

If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____

Answer 1

Correct Answer: 98

The correct answer is 98.
In, ${\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{\ell }}\right)}^9$
$T_{r+1}={}^9{C}_r\frac{{\left(x^{5/2}\right)}^{9-r}}{2^{9-r}}{\left(\frac{-4}{x^{\ell }}\right)}^r$
$=(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r{x}^{\frac{45}{2}-\frac{5r}{2}-r}$
$=45-5r-2lr=0$
$r=\frac{45}{5+21}$ ....(1)
Now, according to the question, $(-1{)}^r\frac{{}^9{C}_r}{2^{9-r}}{4}^r=-84$
$=(-1{)}^r{}^9{C}_r{2}^{3r-9}=21\times 4$
Only natural value of $r$ possible if $3r-9=0$
$r=3$ and ${}^9{C}_3=84$
$\therefore 1=5$ from equation (1)
Now, coefficient of $x^{-31}=x^{\frac{45}{2}-\frac{5r}{2}-\mathrm{lr}}$ at $1=5$, gives
$r=5$
$\therefore {}^9{c}_5(-1)\frac{4^5}{2^4}=2^{\alpha }\times \beta$
$=-63\times 2^7$
$\Rightarrow \alpha =7,\beta =-63$
$\therefore \text{value of}|\alpha \ell -\beta |=98$