The term independent of \(x\) in the expansion of \(\left(x - \frac{1}{x}\right)^4 \left(x + \frac{1}{x}\right)^3\), is
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- \(-3\)
- 0
- 1
- 3
The Correct Option is B
Solution and Explanation
To find the term independent of \(x\) in the expansion of \(\left(x - \frac{1}{x}\right)^4 \left(x + \frac{1}{x}\right)^3\), we can follow these steps:
Firstly, consider the general term from the binomial expansion of \(\left(x - \frac{1}{x}\right)^4\):
\(T_{r+1} = \binom{4}{r} x^{4-r} \left(-\frac{1}{x}\right)^r = \binom{4}{r} (-1)^r x^{4-2r}\)
Similarly, the general term from the binomial expansion of \(\left(x + \frac{1}{x}\right)^3\) is:
\(T_{s+1} = \binom{3}{s} x^{3-s} \left(\frac{1}{x}\right)^s = \binom{3}{s} x^{3-2s}\)
We need to find the combined term from both expansions such that the power of \(x\) is zero (i.e., independent of \(x\)). This means:
\((4 - 2r) + (3 - 2s) = 0\)
Solving the equation:
\(7 - 2(r + s) = 0 \Rightarrow r + s = \frac{7}{2}\)
Since both \(r\) and \(s\) must be integers, and their sum being a non-integer indicates that there is no valid integer pair \((r, s)\) that satisfies \(r + s = \frac{7}{2}\).
Therefore, there is no term in the expansion that is independent of \(x\).
The correct answer is 0.
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