Question

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 ______.

Answer 1

$\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 $
$ \Rightarrow r =10$
So, Maximum value of ${ }^{15} C _{10} x (1- x )$ will be at
$x =\frac{1}{2}$
i.e. $6006$