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

\(\left(1-x^2+3 x^3\right)\left(\frac{5}{2} x^3-\frac{1}{5 x^2}\right)^{11}, x \neq 0\) के प्रसार में \(x\) से स्वतंत्र पद है

  1. A \(\frac{7}{40}\)
  2. B \(\frac{33}{200}\)
  3. C \(\frac{39}{200}\)
  4. D \(\frac{11}{50}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{33}{200}\)

Step-by-step Solution

Detailed explanation

\(\left(1-x^{2}+3 x^{3}\right)\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}\) General term of \(\left(\frac{5}{2} x ^{3}-\frac{1}{5 x ^{2}}\right)^{11}\) is \({ }^{11} C_{r}\left(\frac{5}{2} x^{3}\right)^{11-r}\left(-\frac{1}{5 x^{2}}\right)^{ r }\) General term is…
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