Question

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

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

Answer 1

The correct option is D $\frac{2\times 10!}{3\sqrt{3}(5!{)}^2}$

$T_{r+1}={}^{10}{C}_r(t{x}^{1/5}{)}^{10-r}\times {\left[\frac{(1-x{)}^{1/10}}{t}\right]}^r$
$={}^{10}{C}_r\cdot t^{(10-2r)}\times x^{(10-r)/5}\times (1-x{)}^{r/10}$
For a term to be independent of $t,$
$10-2r=0$
$\Rightarrow r=5$

$T_6={}^{10}{C}_{5}x\sqrt{1-x}$
$\frac{d{T}_6}{dx}={}^{10}{C}_5[\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}]=0$
$\Rightarrow 1-x=\frac{x}{2}$
$\Rightarrow x=\frac{2}{3}$
$\therefore max\left(T_6\right)=\frac{10!}{5!5!}\times \frac{2}{3\sqrt{3}}$