MCQ
$\frac{1}{{{{\left( {{x^2} + \frac{1}{x}} \right)}^{\frac{4}{3}}}}}$ can be expanded by binomial theorem, if
- A$x < 1$
- B$|x| < 1$
- C$x > 1$
- ✓$|x| > 1$
✓
Answer
Correct option: D.
$|x| > 1$
d
(d) The expression is ${x^{ - 8/3}}{\left( {1 + \frac{1}{{{x^3}}}} \right)^{ - 4/3}}$
(d) The expression is ${x^{ - 8/3}}{\left( {1 + \frac{1}{{{x^3}}}} \right)^{ - 4/3}}$
Hence $\left| {\frac{1}{{{x^3}}}} \right|\, < 1 \Rightarrow |x|\, > 1$
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