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

\({\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{\frac{1}{3}}} + 1\;}}--\frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}\)ના વિસ્તરણમાં અચળ પદ મેળવો. 

  1. A \(4\)
  2. B \(120\)
  3. C \(210\)
  4. D \(310\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(210\)

Step-by-step Solution

Detailed explanation

\(\left(\left(x^{1 / 3}+1\right)-\left(\frac{\sqrt{x}+1}{\sqrt{x}}\right)\right)^{10}\) \(\left(x^{13}-x^{-1 / 2}\right)^{10}\) \({T_{r + 1}}{ = ^{10}}{C_r}{({x^{1/3}})^{10 - r}}{( - {x^{ - 1/2}})^r}\) \(\frac{10-r}{3}-\frac{r}{2}=0\) \(\Rightarrow \quad 20-2 r-3 r=0\)…
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