Question

The expansion of ${\left(3x+2\right)}^{\frac{-1}{2}}$ is valid in ascending powers of $x$ if $x$ lies in the interval.

• A. $\left(0,\frac{2}{3}\right)$
• B. $\left(\frac{-3}{2},\frac{3}{2}\right)$
• C. $\left(\frac{-2}{3},\frac{2}{3}\right)$
• D. $\left(-\infty ,-3\right)$

Answer 1

Answer: (A)

$\left(0,\frac{2}{3}\right)$

$(3x+2{)}^{\frac{-1}{2}}=2{(1+\frac{3x}{2})}^{\frac{-1}{2}}$

Expansion is valid if and only if $-1<\frac{3x}{2}<1$

$\Rightarrow -\frac{2}{3}<x<\frac{2}{3}$

For ascending powers of x, $x\in (0,\frac{2}{3})$