JEE Math Practice Question
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 :
- A.6
- B.4
- C.2
- D.9
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