# The maximum value of the term independent

Question:

The maximum value of the term independent of $\mathrm{t}^{\prime}$ in the expansion of $\left(\mathrm{tx}^{\frac{1}{5}}+\frac{(1-\mathrm{x})^{\frac{1}{10}}}{\mathrm{t}}\right)^{10}$ where $x \in(0,1)$ is

1. $\frac{10 !}{\sqrt{3}(5 !)^{2}}$

2. $\frac{2.10 !}{3 \sqrt{3}(5 !)^{2}}$

3. $\frac{2.10 !}{3(5 !)^{2}}$

4. $\frac{10 !}{3(5 !)^{2}}$

Correct Option: , 2

Solution:

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

so $\mathrm{T}_{6}={ }^{10} \mathrm{C}_{5}\left(\mathrm{tx}^{2} 5\right)^{5}\left(\frac{(1-\mathrm{x})^{\frac{1}{10}}}{\mathrm{t}}\right)^{5}$

$\mathrm{T}_{6}=f(\mathrm{x})={ }^{10} \mathrm{C}_{5}(\mathrm{x} \sqrt{1-\mathrm{x}}) ;$ for maximum

$f^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=\frac{2}{3} \& f^{\prime \prime}\left(\frac{2}{3}\right)<0$

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