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

यदि \(\left(\alpha x^3+\frac{1}{\beta x}\right)^{11}\) के प्रसार में \(x^9\) का गुणांक एवं \(\left(\alpha \mathrm{x}-\frac{1}{\beta \mathrm{x}^3}\right)^{11}\) के प्रसार में \(\mathrm{x}^{-9}\) का गुणांक बराबर हैं तब \((\alpha \beta)^2\) बराबर है____________. 

  1. A \(2\)
  2. B \(4\)
  3. C \(1\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(1\)

Step-by-step Solution

Detailed explanation

Coefficient of \(x ^9\) in \(\left(\alpha x^3+\frac{1}{\beta x}\right)={ }^{11} C_6 \cdot \frac{\alpha^5}{\beta^6}\) \(\because\) Both are equal \(\therefore \frac{11}{C_6} \cdot \frac{\alpha^5}{\beta^6}=-\frac{11}{C_5} \cdot \frac{\alpha^6}{\beta^5}\)…
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