# The maximum value of the term independent of 't' in the expansion of $\left ( tx^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t} \right )^{10}$ where $x\; \epsilon \; \left ( 0,1 \right )$ is :   Option: 1 $\frac{2.10!}{3\sqrt{3}(5!)^{2}}$ Option: 2 $\frac{10!}{3(5!)^{2}}$ Option: 3 $\frac{10!}{\sqrt{3}(5!)^{2}}$   Option: 4 $\frac{2.10!}{3(5!)^{2}}$

The term independent of t will be the middle term due to exact same magnitude but opposite sign powers of t in the binomial expression given

$\\\text { so } \mathrm{T}_{6}={ }^{10} \mathrm{C}_{5}\left(\mathrm{tx}^{2} 5\right)^{5}\left(\frac{(1-\mathrm{x})^{\frac{1}{10}}}{\mathrm{t}}\right)^{5}\\T_6=f(x)={ }^{10} \mathrm{C}_{5}(\mathrm{x} \sqrt{1-\mathrm{x}})\\f'(x)=^{10}C_5\left ( \sqrt{1-x}-\frac{x}{2\sqrt{1-x}} \right )=^{10}C_5\left ( -\frac{3 x-2}{2 \sqrt{1-x}} \right )$

for maximum ƒ'(x) = 0

$\Rightarrow x=\frac{2}{3}$

$\\f''(x)=\frac{2-3 x}{4(1-x)^{\frac{3}{2}}}-\frac{3}{2 \sqrt{1-x}}=\frac{3 x-4}{4(1-x)^{\frac{3}{2}}}\\f''\left (\frac{2}{3} \right )<0$

$\text { so } f(x)_{\max .}={ }^{10} C_{5}\left(\frac{2}{3}\right) \cdot \frac{1}{\sqrt{3}}$

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