Question

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 $x \in(0,1)$ is

• A. $\frac{10 !}{\sqrt{3}(5 !)^{2}}$
• B. $\frac{2.10 !}{3 \sqrt{3}(5 !)^{2}}$
• C. $\frac{2.10 !}{3(5 !)^{2}}$
• D. $\frac{10 !}{3(5 !)^{2}}$

Answer 1

Answer: (B)

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 $T _{6}={ }^{10} C _{5}\left( tx ^{2} 5\right)^{5}\left(\frac{(1- x )^{\frac{1}{10}}}{ t }\right)^{5}$
$T _{6}=f( x )={ }^{10} C _{5}( x \sqrt{1- x })$ ; for maximum
$f^{\prime}(x)=0 \Rightarrow x=\frac{2}{3} \& f^{\prime \prime}\left(\frac{2}{3}\right)<\,0$
so $f(x)_{\max }={ }^{10} C_{5}\left(\frac{2}{3}\right) \cdot \frac{1}{\sqrt{3}}$