1 ( x^2 + 1 x ) ^ 4 3 can be expanded by binomial theorem, if - Vidyadip

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 ${x^{ - 8/3}}{\left( {1 + \frac{1}{{{x^3}}}} \right)^{ - 4/3}}$

Hence $\left| {\frac{1}{{{x^3}}}} \right|\, < 1 \Rightarrow |x|\, > 1$

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