JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 24)
Let $$[t]$$ denote the greatest integer $$\leq t$$. If the constant term in the expansion of $$\left(3 x^{2}-\frac{1}{2 x^{5}}\right)^{7}$$ is $$\alpha$$, then $$[\alpha]$$ is equal to ___________.
Answer
1275
Explanation
Let $\mathrm{T}_{r+1}$ be the constant term.
$$ \mathrm{T}_{r+1}={ }^7 \mathrm{C}_r\left(3 x^2\right)^{7-r}\left(\frac{-1}{2 x^5}\right)^r $$
For constant term, power of $x$ should be zero.
$$ \begin{aligned} & \text { i.e., } 14-2 r-5 r=0 \\\\ & \Rightarrow 14=7 r \Rightarrow r=2 \end{aligned} $$
Now, constant term $=\alpha$
$$ \begin{aligned} & \Rightarrow{ }^7 C_2(3)^5\left(\frac{-1}{2}\right)^2=\alpha \\\\ & \Rightarrow 21 \times 243 \times \frac{1}{4}=\alpha \\\\ & \Rightarrow[\alpha]=[1275.75]=1275 \end{aligned} $$
$$ \mathrm{T}_{r+1}={ }^7 \mathrm{C}_r\left(3 x^2\right)^{7-r}\left(\frac{-1}{2 x^5}\right)^r $$
For constant term, power of $x$ should be zero.
$$ \begin{aligned} & \text { i.e., } 14-2 r-5 r=0 \\\\ & \Rightarrow 14=7 r \Rightarrow r=2 \end{aligned} $$
Now, constant term $=\alpha$
$$ \begin{aligned} & \Rightarrow{ }^7 C_2(3)^5\left(\frac{-1}{2}\right)^2=\alpha \\\\ & \Rightarrow 21 \times 243 \times \frac{1}{4}=\alpha \\\\ & \Rightarrow[\alpha]=[1275.75]=1275 \end{aligned} $$
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