Question

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

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

Answer 1

The correct option is D $(–\infty ,–1)\cup (1,4]$

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

Let $g\left(x\right)=\sqrt{(}4–x)$ and $h\left(x\right)=\frac{1}{\sqrt{x^2-1}}$

$\Rightarrow f\left(x\right)=g\left(x\right)+h\left(x\right)$

$\because rg\left(x\right)=\sqrt{(4–x)}$

$\Rightarrow 4-x\ge 0$

$[\because \surd f(x)$ is defined when $f\left(x\right)\ge 0]$

$\Rightarrow x\le 4\dots \left(i\right)$

$\because h\left(x\right)=\frac{1}{\sqrt{x^2-1}}$

$\Rightarrow x2–1>0$

$[\because \frac{1}{\sqrt{f\left(x\right)}}$ is defined when $f\left(x\right)>0]$

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

Plotting$x=–1,1$on number line


Now find intersection of (i) and (ii) we get $xϵ(–\infty ,–1)\cup (1,4]$

Therefore, Domain of$f=(-\infty ,–1)\cup (1,4]$

Hence, Option (A) is correct