Question

The maximum value of the term independent of ‘$t$’ in the expansion of ${\left[t{x}^{\frac{1}{5}}+\left\{\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{10}}}}{t}\right\}\right]}^{10}$ where $x\in \left(0,1\right)$ is :

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

Answer 1

The correct option is D

$\frac{2·10!}{3\sqrt{3}{\left(5!\right)}^2}$

Explanation for the correct option:

Step 1: Find the term independent of $t$.

A binomial expression ${\left[t{x}^{\frac{1}{5}}+\left\{\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{10}}}}{t}\right\}\right]}^{10}$ is given.

Find the general term $r+1$ as follows:

$$\begin{gathered} T_{r+1}{=}^{10}{C}_r{\left(t{x}^{\frac{1}{5}}\right)}^{10-r}{\left[\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{10}}}}{t}\right]}^r\Rightarrow T_{r+1}{=}^{10}{C}_r\left(t^{10-r}{x}^{\frac{10-r}{5}}\right)\left[\frac{{\left(1-x\right)}^{{\displaystyle \frac{r}{10}}}}{t^r}\right] \\ \Rightarrow T_{r+1}{=}^{10}{C}_r·t^{10-2r}·x^{\frac{10-r}{5}}·{\left(1-x\right)}^{\frac{r}{10}} \end{gathered}$$

Find the value of $r$ such that the term is independent of $t$ as follows:

$$\begin{gathered} 10-2r=0 \\ \Rightarrow r=5 \end{gathered}$$

Therefore, the sixth term is independent of $t$.

Compute the sixth term as follows:

$$\begin{gathered} T_6{=}^{10}{C}_5·t^{10-2·5}·x^{\frac{10-5}{5}}·{\left(1-x\right)}^{\frac{5}{10}} \\ \Rightarrow T_6{=}^{10}{C}_5·x\sqrt{1-x}...1 \end{gathered}$$

Step 2: Find the maximum term of the sixth term.

Assume that, $f\left(x\right)=T_6$.

Therefore, $f\left(x\right){=}^{10}{C}_5·x·\sqrt{1-x}$

Differentiate both sides with respect to $x$.

$$\begin{gathered} f\text{'}\left(x\right){=}^{10}{C}_5\left[\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}\right] \\ \Rightarrow f\text{'}\left(x\right){=}^{10}{C}_5\left[\frac{2-2x-x}{2\sqrt{1-x}}\right] \\ \Rightarrow f\text{'}\left(x\right){=}^{10}{C}_5\left[\frac{2-3x}{2\sqrt{1-x}}\right] \end{gathered}$$

Put $f\text{'}\left(x\right)=0$ to find the critical points.

$$\begin{gathered} {}^{10}{C}_5\left[\frac{2-3x}{2\sqrt{1-x}}\right]=0 \\ \Rightarrow 2-3x=0 \\ \Rightarrow x=\frac{2}{3} \end{gathered}$$

Therefore, the term independent of $t$ is maximum for $x=\frac{2}{3}$.

So, the maximum value of the term independent of $t$ is as follows:

$$\begin{gathered} max\left(T_6\right){=}^{10}{C}_5·\frac{2}{3}·\sqrt{1-\frac{2}{3}} \\ \Rightarrow max\left(T_6\right){=}^{10}{C}_5·\frac{2}{3}·\sqrt{\frac{1}{3}} \\ \Rightarrow max\left(T_6\right)=\frac{2·10!}{3\sqrt{3}{(5!)}^2} \end{gathered}$$

Therefore, The maximum value of the term independent of ‘$t$’ in the given expansion is $\frac{2·10!}{3\sqrt{3}{\left(5!\right)}^2}$.

Hence, option $D$ is the correct answer.