AP EAMCET · Maths · Binomial Theorem
If \(x\) is so large that terms containing \(x^{-3}, x^{-4}, x^{-5}, \ldots\) can be neglected, then the approximate value of \(\left(\frac{3 x-5}{4 x^2+3}\right)^{-4 / 5}\) is
- A \(\left(\frac{3}{4 x}\right)^{4 / 5}\left(1-\frac{4}{3 x}-\frac{7}{5 x^2}\right)\)
- B \(\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}+\frac{13}{5 x^2}\right)\)
- C \(\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}-\frac{13}{5 x^2}\right)\)
- D \(\left(\frac{3}{4 x}\right)^{4 / 5}\left(1-\frac{4}{3 x}+\frac{7}{5 x^2}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}+\frac{13}{5 x^2}\right)\)
Step-by-step Solution
Detailed explanation
\(\left(\frac{3 x-5}{4 x^2+3}\right)^{-4 / 5} = \left(\frac{3x(1 - 5/(3x))}{4x^2(1 + 3/(4x^2))}\right)^{-4 / 5}\) \(= \left(\frac{3}{4x}\right)^{-4/5} \left(1 - \frac{5}{3x}\right)^{-4/5} \left(1 + \frac{3}{4x^2}\right)^{-(-4/5)}\)…
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