Question

Find the set of values of x for which the expansion of ${(9+5x)}^{\frac{-1}{2}}$ is valid in ascending powers of x.

• A. $(0,\frac{9}{5})$
• B. $(\frac{-9}{5},0)$
• C. $(\frac{-9}{5},\frac{9}{5})$
• D. None of these

Answer 1

The correct option is C

$(\frac{-9}{5},\frac{9}{5})$

${(9+5x)}^{-\frac{1}{2}}=9^{\frac{-1}{2}}{(1+\frac{5x}{9})}^{\frac{-1}{2}}$

${(1+x)}^{-n}=1-nx+\frac{n(n+1))}{2}$ $x^2$ + ..........if |x| < 1

${(1+\frac{5x}{9})}^{\frac{-1}{2}}$ can be expanded in ascending powers of x - if
$|\frac{5x}{9}|$ < 1

⇒ $-1<\frac{5x}{9}<1$

$-\frac{9}{5}<x<\frac{9}{5}$.