If the constant term in the binomial expansio | vedclass

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(D) The general term $T_{r+1}$ in the expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^{\ell}}\right)^9$ is given by:$T_{r+1} = \binom{9}{r} \left(\

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$98$

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(D) The general term $T_{r+1}$ in the expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^{\ell}}ight)^9$ is given by:$T_{r+1} = \binom{9}{r} \left(\frac{x^{5/2}}{2}ight)^{9-r} \left(-\frac{4}{x^{\ell}}ight)^r$$T_{r+1} = \binom{9}{r} \frac{1}{2^{9-r}} (-4)^r x^{\frac{5(9-r)}{2} - r\ell}$For the constant term,the exponent of $x$ must be $0$:$\frac{45-5r}{2} - r\ell = 0 \implies 45 - 5r - 2r\ell = 0 \implies r(5 + 2\ell) = 45$Given the constant term is $-84$:$(-1)^r \binom{9}{r} \frac{4^r}{2^{9-r}} = -84$$(-1)^r \binom{9}{r} 2^{2r - 9 + r} = -84 \implies (-1)^r \binom{9}{r} 2^{3r-9} = -84$For $r=3$,$\binom{9}{3} = 84$. Thus,$(-1)^3 (84) 2^{3(3)-9} = -84 	imes 2^0 = -84$. This matches.Substituting $r=3$ into $r(5+2\ell) = 45$:$3(5+2\ell) = 45 \implies 5+2\ell = 15 \implies 2\ell = 10 \implies \ell = 5$.Now,find the coefficient of $x^{-3\ell} = x^{-15}$:$\frac{45-5r}{2} - 5r = -15 \implies 45 - 5r - 10r = -30 \implies 15r = 75 \implies r = 5$.The coefficient is $(-1)^5 \binom{9}{5} \frac{4^5}{2^{9-5}} = -126 	imes \frac{2^{10}}{2^4} = -126 	imes 2^6 = -63 	imes 2^7$.Given $2^{\alpha}\beta = -63 	imes 2^7$,we have $\alpha = 7$ and $\beta = -63$.Finally,$|\alpha\ell - \beta| = |7(5) - (-63)| = |35 + 63| = 98$.

If the constant term in the binomial expansion of $\left(\frac{x^{5/2}}{2} - \frac{4}{x^{\ell}}\right)^9$ is $-84$ and the coefficient of $x^{-3\ell}$ is $2^{\alpha}\beta$,where $\beta < 0$ is an odd number,then $|\alpha\ell - \beta|$ is equal to

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