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

ધારોકે \(\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30}\)ના વિસ્તરણમાં \(x^{-\alpha}\) વાળો પદ હોય તેવો \(\alpha > 0\) નાનામાં નાની સંખ્યા \(\beta x^{-\alpha}, \beta \in N\) છે. તો \(\alpha\) ની  કિમંત મેળવો.

  1. A \(2\)
  2. B \(4\)
  3. C \(6\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

\(T _{ r +1}={ }^{30} C _{ r }\left( x ^{2 / 3}\right)^{30- r }\left(\frac{2}{ x ^3}\right)^{ r }\) \(={ }^{30} C _{ r } \cdot 2^{ r } \cdot x ^{\frac{60-11 r }{3}}\) \(\frac{60-11 r }{3} < 0 \Rightarrow 11 r > 60 \Rightarrow r >\frac{60}{11} \Rightarrow r =6\)…
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