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

यदि \(\left(t^2 x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t}\right)^{15}, x \geq 0\), के प्रसार में \(t\), से स्वतंत्र पद का अधिकतम मान \(K\) है, तो \(8 K\) बराबर है \(...........\)

  1. A \(6006\)
  2. B \(6005\)
  3. C \(6007\)
  4. D \(6008\)
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Answer & Solution

Correct Answer

(A) \(6006\)

Step-by-step Solution

Detailed explanation

\(\left( t ^{2} x ^{\frac{1}{5}}+\frac{(1- x )^{\frac{1}{10}}}{ t }\right)^{15}\) \(T_{r+1}={ }^{15} C_{r}\left(t^{2} x^{\frac{1}{5}}\right)^{15-r} \cdot \frac{(1-x)^{\frac{r}{10}}}{t^{r}}\) For independent of \(t\), \(30-2 r-r=0\) \(r =10\) So, Maximum value of…
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