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

यदि \(\left(\mathrm{x}^{\frac{2}{3}}+\frac{\alpha}{\mathrm{x}^3}\right)^{22}\) के प्रसार में \(\mathrm{x}\) से स्वतंत्र पद 7315 है, तो \(|\alpha|\) बराबर है______________.

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

Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

\(T _{ r +1}={ }^{22} C _{ r } \cdot\left( x ^{\frac{2}{3}}\right)^{22- r } \cdot(\alpha)^{ r }, x ^{-3 r }\) \(={ }^{22} C _{ r } \cdot x ^{\frac{44}{3}-\frac{2 r }{3}-3 r }(\alpha)^{ r }\) \(\frac{44}{3}=\frac{11 r }{3}\) \(r =4\) \({ }^{22} C _4 \cdot \alpha^4=7315\)…
From JEE Mains
Explore more questions on app