JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 20)
If the term independent of $$x$$ in the expansion of $$\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}$$ is 105 , then $$\mathrm{a}^2$$ is equal to :
6
4
2
9
Explanation
$$\begin{aligned} & \left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10} \\ & T_{r+1}={ }^{10} C_r\left(\sqrt{a} x^2\right)^{10-r}\left(\frac{1}{2 x^3}\right)^r \end{aligned}$$
Independent of $$x \Rightarrow 20-2 r-3 r=0$$
$$r=4$$
Independent of $$x$$ is $${ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105$$
$$\begin{gathered} \frac{210}{2 \times 8} a^3=105 \\ \Rightarrow \quad a=2 \\ a^2=4 \end{gathered}$$
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