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

\(\left(\frac{ x +1}{ x ^{2 / 3}- x ^{1 / 3}+1}-\frac{ x -1}{ x - x ^{1 / 2}}\right)^{10}, x \neq 0,1\) के प्रसार में ' \(x\) ' से स्वतंत्र पद बराबर है

  1. A \(110\)
  2. B \(210\)
  3. C \(300\)
  4. D \(400\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(210\)

Step-by-step Solution

Detailed explanation

\(\left(\left(x^{1 / 3}+1\right)-\left(\frac{x^{1 / 2}+1}{x^{1 / 2}}\right)\right)^{10}\) \(=\left(x^{1 / 3} \frac{1}{x^{1 / 2}}\right)^{10}\) Now General Term…
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