Question

Find the domains of the following functions:

$f\left(x\right)=\frac{1}{\sqrt{4x^2-1}}+\ln x(x^2-1)$

• A. $(-1<x<-\frac{1}{2})\cup (x>1)$
• B. $(-1<x<-\frac{1}{2})$
• C. $(x>1)$
• D. none of these

Answer 1

The correct option is A $(-1<x<-\frac{1}{2})\cup (x>1)$

Given, $f\left(x\right)=\frac{1}{\sqrt{4x^2-1}}+ln\left(x\right(x^2-1\left)\right)$

If $\frac{1}{\sqrt{4x^2-1}}$ has to be defined quantity,

then $4x^2-1>0$

$⟹(-1<x<-\frac{1}{2})$

Similarly, if $ln\left(x\right(x^2-1\left)\right)$ has to be a defined quantity,

then $x(x^2-1)>0$

$⟹x>1$

So, the domain of the function is $(-1<x<-\frac{1}{2})\cup (x>1)$