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  • If the maximum value of the term independent of t in the expansion of  <img alt="\mathrm{\left(\mathrm{t}^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0

If the maximum value of the term independent of t in the expansion of  \mathrm{\left(\mathrm{t}^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0} is \mathrm{K}, then \mathrm{8K} is equal to __________.

Option: 1

6006


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\left(t^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t}\right)^{15}}

\mathrm{T_{r+1}={ }^{1 5_{r}}\left(t^{2} x^{\frac{1}{5}}\right)^{15-r} \cdot \frac{(1-x)^{\frac{r}{10}}}{t^r}}

\text{for independent of t},

\mathrm{30-2 r-r=0} \\

\mathrm{\Rightarrow r=10}

So maximum value of \mathrm{{ }^{15} c_{10} x(1-x)} will be at \mathrm{x=\frac{1}{2}}

i.e. 6006

Hence answer is 6006

 

Posted by

manish

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