If the maximum value of the term independent | vedclass

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Question

$\left( t ^{2} x ^{\frac{1}{5}}+\frac{(1- x )^{\frac{1}{10}}}{ t }\right)^{15}$

$T_{r+1}={ }^{15} C_{r}\left(t^{2} x^{\frac{1}{5}}\right)^{15-r} \c

Vedclass · Accepted Answer

$6006$

Vedclass · Answer

$\left( t ^{2} x ^{\frac{1}{5}}+\frac{(1- x )^{\frac{1}{10}}}{ t }\right)^{15}$

$T_{r+1}={ }^{15} C_{r}\left(t^{2} x^{\frac{1}{5}}\right)^{15-r} \cdot \frac{(1-x)^{\frac{r}{10}}}{t^{r}}$

For independent of $t$,

$30-2 r-r=0$

$r =10$

So, Maximum value of ${ }^{15} C _{10} x (1- x )$ will be at

$x=\frac{1}{2}$

i.e. $6006$

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

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