JEE Mains · Maths · STD 11 - 7. binomial theoram
If the term independent of \(x\) in the expansion of \(\left(\sqrt{\mathrm{ax}}{ }^2+\frac{1}{2 \mathrm{x}^3}\right)^{10}\) is 105 , then \(\mathrm{a}^2\) is equal to :
- A \(4\)
- B \(9\)
- C \(6\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\( \left(\sqrt{\mathrm{a}} \mathrm{x}^2+\frac{1}{2 \mathrm{x}^3}\right)^{10} \) \( \text { General term }={ }^{10} \mathrm{C}_{\mathrm{r}}(\sqrt{\mathrm{ax}})^{10-\mathrm{r}}\left(\frac{1}{2 \mathrm{x}^3}\right)^{\mathrm{r}} \) \( 20-2 \mathrm{r}-3 \mathrm{r}=0 \)…
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