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TS EAMCET · Maths · Binomial Theorem

The set of all real values of \(x\) for which the expansion of \(\left(125 x^2-\frac{27}{x}\right)^{-2 / 3}\) is valid, is

  1. A \(\left(-\frac{3}{5}, \frac{3}{5}\right)\)
  2. B \(\left(-\infty,-\frac{3}{5}\right) \cup\left(\frac{3}{5}, \infty\right)\)
  3. C \(\left(-\frac{5}{3}, \frac{5}{3}\right)\)
  4. D \(\left(-\infty,-\frac{1}{3}\right) \cup\left(\frac{1}{3}, \infty\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(-\infty,-\frac{3}{5}\right) \cup\left(\frac{3}{5}, \infty\right)\)

Step-by-step Solution

Detailed explanation

Given : \(\left(125 x^2-\frac{27}{x}\right)^{\frac{-2}{3}}=\frac{x^{\frac{2}{3}}}{\sqrt[3]{\left(125 x^3-27\right)^2}}\) For expansion to be valid, \(x \neq 0\) and \(125 x^3-27 \neq 0\) \(\Rightarrow x^3 \neq \frac{27}{125} \Rightarrow x \neq \frac{3}{5}\) \(\therefore\) From…