JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 22)
If the term without $$x$$ in the expansion of $$\left(x^{\frac{2}{3}}+\frac{\alpha}{x^{3}}\right)^{22}$$ is 7315 , then $$|\alpha|$$ is equal to ___________.
Answer
1
Explanation
Given expansion $\left(x^{\frac{2}{3}}+\frac{\alpha}{x^3}\right)^{22}$
$$ T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r $$
For constant term
$$ \begin{aligned} & \frac{44-2 r}{3}-3 r=0 \\\\ & \Rightarrow r=4 \end{aligned} $$
Now ${ }^{22} \mathrm{C}_4 \alpha^4=7315$
$$ \begin{aligned} & \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315 \\\\ & \therefore \alpha^4=1 \\\\ & \therefore |\alpha|=1 \end{aligned} $$
$$ T_{r+1}={ }^{22} C_r\left(x^{\frac{2}{3}}\right)^{22-r}\left(\frac{\alpha}{x^3}\right)^r $$
For constant term
$$ \begin{aligned} & \frac{44-2 r}{3}-3 r=0 \\\\ & \Rightarrow r=4 \end{aligned} $$
Now ${ }^{22} \mathrm{C}_4 \alpha^4=7315$
$$ \begin{aligned} & \frac{22 \times 21 \times 20 \times 19}{4 \times 3 \times 2 \times 1} \alpha^4=7315 \\\\ & \therefore \alpha^4=1 \\\\ & \therefore |\alpha|=1 \end{aligned} $$
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