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

\(\left(\frac{(x+1)}{\left(x^{2 / 3}+1-x^{1 / 3}\right)}-\frac{(x+1)}{\left(x-x^{1 / 2}\right)}\right)^{10}, x\gt1\) ના વિસ્તરણમાં \(x\) થી સ્વતંત્ર પદ છે:

  1. A \(210\)
  2. B \(150\)
  3. C \(240\)
  4. D \(120\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(210\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \left(\frac{(x+1)}{\left(x^{\frac{2}{3}}+1-x^{\frac{1}{3}}\right)}-\frac{(x-1)}{\left(x-x^{\frac{1}{2}}\right)}\right)^{10} \\ & =\left(\left(x^{\frac{1}{3}}+1\right)-\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\right)^{10} \\ &…

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