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QJEEM23Ma73a0: JEE Main 2023 Mathematics - Class 11 JEEM23Ma73a0 Solution: If the constant term in the binomial expansion of is and the…

If the constant term in the binomial expansion of (x522−4xl)9\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9} is −84-84 and the coefficient of x−3lx^{-3 l} is 2αβ2^{\alpha} \beta , where \beta<0 is an odd number, then ∣αl−β∣|\alpha l-\beta| is equal to ________.

Step-by-Step Verified Solution

Given binomial expansion of (x522−4xl)9\left(\frac{x^{\frac{5}{2}}}{2}-\frac{4}{x^{l}}\right)^{9}

Tr+1amp;=9Cr(x522)9−r(−4xl)ramp;=9Crx45−5r2−lr⋅2r−9⋅4r⋅(−1)r\begin{aligned}T_{r+1} & ={ }^{9} C_{r}\left(\frac{x^{\frac{5}{2}}}{2}\right)^{9-r}\left(\frac{-4}{x^{l}}\right)^{r} \\& ={ }^{9} C_{r} x^{\frac{45-5 r}{2}-lr} \cdot 2^{r-9} \cdot 4^{r} \cdot(-1)^{r}\end{aligned}

For constant term, power of x is zero.

So, 45−5r2=lr⇒2lr+5r=45\frac{45-5 r}{2}=l r \Rightarrow 2 l r+5 r=45

Now constant term =−84=-84

and 9Cr⋅23r−9(−1)r=−84{ }^{9} C_{r} \cdot 2^{3 r-9}(-1)^{r}=-84

So, r=3r=3 and l=5l=5

Now for x−15,45−5r2−5r=−15x^{-15}, \frac{45-5 r}{2}-5 r=-15

⇒\Rightarrow 45−15r=−3045-15 r=-30

⇒\Rightarrow r=5r=5

∴\therefore Coefficient =−9C526=−63.27=-{ }^{9} C_{5} 2^{6}=-63.2^{7}

∴α=7,β=−63\therefore \alpha=7, \beta=-63

and ∣αl−β∣=∣7×5+63∣=98|\alpha l-\beta|=|7 \times 5+63|=98

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