TS EAMCET · Maths · Binomial Theorem
If \(0 < y < 2^{1 / 3}\) and \(x\left(y^3-1\right)=1\), then \(\frac{2}{x}+\frac{2}{3 x^3}+\frac{2}{5 x^5}+\ldots\) is equal to:
- A \(\log \left(\frac{y^3}{2-y^3}\right)\)
- B \(\log \left(\frac{y^3}{1-y^3}\right)\)
- C \(\log \left(\frac{2 y^3}{1-y^3}\right)\)
- D \(\log \left(\frac{y^3}{1-2 y^3}\right)\)
Answer & Solution
Correct Answer
(A) \(\log \left(\frac{y^3}{2-y^3}\right)\)
Step-by-step Solution
Detailed explanation
We have, \(x\left(y^3-1\right)=1 \Rightarrow x=\frac{1}{y^3-1}=\frac{1}{k} \text { (say) }\) \(\Rightarrow \quad k=\frac{1}{x}\) Then, \(\frac{2}{x}+\frac{2}{3 x^3}+\frac{2}{5 x^5}+\ldots\) \(=2 k+\frac{2}{3} k^3+\frac{2}{5} k^5+\ldots\)…
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