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Q.
The term independent of $x$ in the expansion of $\left(\left[\left(t^{- 1} - 1\right) x + \left(t^{- 1} + 1\right)^{- 1} x^{- 1}\right]\right)^{8}$ is :
NTA AbhyasNTA Abhyas 2022
Solution:
$
\begin{array}{l}
{\left[\left(t^{-1}-1\right) x+\left(t^{-1}+1\right)^{-1} x^{-1}\right]^8} \\
=\left[\left(\frac{1}{t}-1\right) x+\left(\frac{1}{t}+1\right)^{-1} \frac{1}{x}\right]^8
\end{array}
$
Let $T _{r+1}$ be the term independent of $x$, then
$
\begin{array}{l}
T _{r+1}={ }^8 C_r\left(\frac{1}{t}-1\right)^{8-r} x^{8-r} \cdot\left(\frac{1}{t}+1\right)^{-r}\left(\frac{1}{x}\right)^r \\
={ }^8 C _r\left(\frac{1}{t}-1\right)^{8-r} \cdot\left(\frac{1}{t}+1\right)^{-r} \cdot x^{8-2 r} \\
\therefore 8-2 r=0 \Rightarrow r=4
\end{array}
$
$\therefore T _5$ is the term independent of $x$ and
$
\begin{array}{l}
T _5={ }^8 C_4\left(\frac{1}{t}-1\right)^4 \cdot\left(\frac{1}{t}+1\right)^{-4} \\
={ }^8 C _4\left(\frac{1-t}{t}\right)^4 \cdot\left(\frac{1+t}{t}\right)^{-4} \\
={ }^8 C _4\left(\frac{1-t}{1+t}\right)^4=\frac{8.7 \cdot 6.5}{4.3 \cdot 2.1}\left(\frac{1-t}{1+t}\right)^4 \\
=70 \cdot\left(\frac{1-t}{1+t}\right)^4
\end{array}
$