Question

To expand ${(4-\frac{3}{x^2})}^{\frac{-1}{2}}$ as the sum of infinite series, then $|x|>$ must be greater than:

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

$(4-\frac{3}{x^2}{)}^{\frac{-1}{2}}$
Taking 4 out of the bracket we get
$\frac{(1-\frac{3}{4x^2}{)}^{\frac{-1}{2}}}{2}$
Hence $|\frac{3}{4x^2}|<1$
Therefore
$|\frac{4x^2}{3}|>1$
$x^2<\frac{-3}{4}$ ...(i) and $x^2>\frac{3}{4}$ ...(ii)
Hence,$|x|>\frac{\sqrt{3}}{2}$