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QJEEM23Ma8c37: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8c37 Solution: If the term without in the expansion of is 7315 , then is equal …

If the term without xx in the expansion of (x23+αx3)22\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22} is 7315 , then ∣α∣|\alpha| is equal to ___________.

Step-by-Step Verified Solution

Given expansion (x23+αx3)22\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}

Tr+1=22Cr(x23)22−r(αx3)rT_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r

For constant term

amp;44−2r3−3r=0amp;⇒r=4\begin{aligned}& \frac{44-2 r}{3}-3 r=0 \\\\& \Rightarrow r=4\end{aligned}

Now 22C4α4=7315{ }^{22} \mathrm{C}_4 \alpha^4=7315

amp;22×21×20×194×3×2×1α4=7315amp;∴α4=1amp;∴∣α∣=1\begin{aligned}& \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315 \\\\& \therefore \alpha^4=1 \\\\& \therefore |\alpha|=1\end{aligned}

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