Question

HEEELP

Answer 1

Answer: (c)

Step-by-step explanation:

Given

$f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}$

Here, $\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)$

To get $f\left(g(x)\right)$, replace $x$ in $f(x)$ by $g(x)\ \text{i.e. by}\ \sqrt{x+5}$

$\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)$

Using $(i)$ and $(ii)$  it can be concluded that the domain of $f\left(g(x)\right)$ is all real numbers except 0.

Therefore, its domain is given by

$x\in [-5,4)\cup (4,\infty)$

Option (c) is correct.