Question

${(3+x)}^{p/q}$ can be expanded in ascending powers of $x$ if

• A. $-1<x<1$
• B. $x>3$
• C. $-3<x<3$
• D. $x<-3$

Answer 1

The correct option is C $-3<x<3$

$(x+3{)}^{p/q}$
$=3^{p/q}(\frac{x}{3}+1{)}^{p/q}$
This can be expanded in the powers of x only when
$|\frac{x}{3}|<1$
$-1<\frac{x}{3}<1$
$-3<x<3$