Question

The domain of the function $f\left(x\right)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$ is

• A. $x\in (-\infty ,-1]\cup (1,4]$
• B. $x\in (-\infty ,-1)\cup (1,4]$
• C. $x\in (-\infty ,-1)\cup [1,4]$
• D. $x\in (-\infty ,-1]\cup [1,4]$

Answer 1

The correct option is B $x\in (-\infty ,-1)\cup (1,4]$

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

For domain of the given function

both $\left(4-x\right)\ge 0$

and

$(x^2-1)>0$

$(4-x)⩾0$

$\Rightarrow x⩽4$

and

$x^2-1>0$

$\Rightarrow \left(x-1\right)\left(x+1\right)>0$

$x<-1$ or $x>1$

on combining both conditions and finding the combined other regions, we get

$x\in \left(-\infty ,-1\right)\cup (1,4]$