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

\(\left(2 x^3-\frac{1}{3 x^2}\right)^5\) के प्रसार में \(\mathrm{x}^5\) का गुणांक है

  1. A \(8\)
  2. B \(9\)
  3. C \(\frac{80}{9}\)
  4. D \(\frac{26}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{80}{9}\)

Step-by-step Solution

Detailed explanation

\(\left(2 x^3-\frac{1}{3 x^2}\right)^5\) \(T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r}\) \(\therefore 15-5 r =5\) \(\therefore r =2\) \(T_3=10\left(\frac{8}{9}\right) x^5\) So, coefficient is…
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