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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

  1. A \(\left(\frac{3}{4 x}\right)^{4 / 5}\left(1-\frac{4}{3 x}-\frac{7}{5 x^2}\right)\)
  2. B \(\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}+\frac{13}{5 x^2}\right)\)
  3. C \(\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}-\frac{13}{5 x^2}\right)\)
  4. D \(\left(\frac{3}{4 x}\right)^{4 / 5}\left(1-\frac{4}{3 x}+\frac{7}{5 x^2}\right)\)
Verified Solution

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

\(\left(\frac{3 x-5}{4 x^2+3}\right)^{-4 / 5} = \left(\frac{3x(1 - 5/(3x))}{4x^2(1 + 3/(4x^2))}\right)^{-4 / 5}\) \(= \left(\frac{3}{4x}\right)^{-4/5} \left(1 - \frac{5}{3x}\right)^{-4/5} \left(1 + \frac{3}{4x^2}\right)^{-(-4/5)}\)…