Question

The domain of definition of function $f\left(x\right)=\frac{1+2(x+4{)}^{-0.5}}{2-(x+4{)}^{0.5}}+(x+4{)}^{0.5}+4(x+4{)}^{0.5}$ is

• A. $R$
• B. $(-4,4)$
• C. $R^{+}$
• D. $(-4,0)\cup (0,\infty )$

Answer 1

Answer: (D)

$(-4,0)\cup (0,\infty )$

We have,

$f\left(x\right)=\frac{1+2(x+4{)}^{-0.5}}{2-(x+4{)}^{0.5}}+(x+4{)}^{0.5}+4(x+4{)}^{0.5}$

$\Rightarrow f\left(x\right)=\frac{1+\frac{2}{\sqrt{x+4}}}{2-\sqrt{x+4}}+\sqrt{x+4}+4\sqrt{x+4}$
Clearly, $f\left(x\right)$ is defined for $x+4>0$ and $x\ne 0$. So, domain of $f\left(x\right)$ is $(-4,0)\cup (0,\infty )$.