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JEE Mains · Maths · STD 11 - 7. binomial theoram

यदि \(\left(\sqrt{\mathrm{ax}}{ }^2+\frac{1}{2 \mathrm{x}^3}\right)^{10}\) के प्रसार में \(x\) से स्वतंत्र पद 105 है, तो \(\mathrm{a}^2\) = ...........

  1. A \(4\)
  2. B \(9\)
  3. C \(6\)
  4. D \(2\)
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Correct Answer

(A) \(4\)

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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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