JEE Main 2023 — Mathematics Algebra

Given,

The constant term in the binomial expansion of ${(\frac{{x}^{\frac{5}{2}}}{2}-\frac{4}{{x}^{l}})}^{9}$(2x25​​−xl4​)9 is $-84$−84 and the coefficient of ${x}^{-3}$x−3 is ${2}^{\alpha }\beta$2αβ

Now General Term of the binomial is given by, ${T}_{r+1}=Cr9{(\frac{{x}^{\frac{5}{2}}}{2})}^{9-r}{(\frac{-4}{{x}^{l}})}^{r}={(-1)}^{r}{C}_{r}{x}^{\frac{45}{2}-\frac{5r}{2}-lr}{2}^{3-9}$Tr+1​=Cr9(2x25​​)9−r(xl−4​)r=(−1)rCr​x245​−25r​−lr23−9

Now for constant term $45-5r-2lr=0\Rightarrow r=\frac{45}{5+2l}$45−5r−2lr=0⇒r=5+2l45​

So, coefficient of constant term will be,

$${(-1)}^{9}{\times }^{9}{C}_{r}{2}^{3r-9}=-84\Rightarrow r=3$$(−1)9×9Cr​23r−9=−84⇒r=3

Now putting the value of $r$r in $r=\frac{45}{5+2l}$r=5+2l45​ we get, $l=5$l=5

Now coefficient of ${x}^{-3l}={x}^{-15}\Rightarrow \frac{45}{2}-\frac{5r}{2}-lr=-15\Rightarrow r=5$x−3l=x−15⇒245​−25r​−lr=−15⇒r=5 when $l=5$l=5

Now coefficient of ${x}^{-15}={(-1)}^{59}{C}_{5}{2}^{6}=-63({2}^{7})=\beta ({2}^{\alpha })$x−15=(−1)59C5​26=−63(27)=β(2α)

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

So, the value of $|\alpha l-\beta |=|7\times 5-(-63)|=98$∣αl−β∣=∣7×5−(−63)∣=98