Question

If $f(x)=\sqrt{3\left| x \right|-x-2} and g(x)=sinx$, then domain of definition of fog (x) is
A. $\left\{ 2n\pi +\dfrac{\pi }{2} \right\},n\in I$
B. $\left( 2n\pi +\dfrac{7\pi }{6},2n\pi +\dfrac{11\pi }{6} \right)$
C. $\left( 2n\pi +\dfrac{7\pi }{6} \right),n\in I$
D. $\{(4m+1)\dfrac{\pi }{2}:m\in I\}\underset{n\in I}{\mathop{\bigcup }}\,\left[ \left( 2n\pi +\dfrac{7\pi }{6},2n\pi +\dfrac{11\pi }{6} \right) \right]$

Answer 1

Hint:In this case, we will use the domain of composite function as we are provided with two functions $f(x)=\sqrt{3\left| x \right|-x-2}$ and $g(x)=sinx$. Fog cannot exist unless the domain of f is a subset of the range of g (x).


Complete step by step solution:We have given functions $f(x)=\sqrt{3\left| x \right|-x-2}$ and $g(x)=sinx$.
Fog cannot exist unless the domain of f is a subset of the range of g (x). Therefore for fog to exist in the $range of g \subseteq domain of f$.
So the domain of f,
$\Rightarrow 3\left| x \right|-x-2\ge 0$
$\Rightarrow 3\left| x \right|-x\ge 2$
When $x\ge 0$
$\Rightarrow x\ge 1$
When$ x<0$
$\Rightarrow x<-\dfrac{1}{2}$
$\therefore \,\,\,\,\sin x\ge 1\,\,and\,\,x<-\dfrac{1}{2}$
$\sin \,\,x=1\,\,and\,\,-1\,\le \,\sin \,x<-\dfrac{1}{2}$
$\therefore x=(4m+1)\dfrac{\pi }{2}$ and $2n\pi +\dfrac{7\pi }{6}\le x\le 2n\pi +\dfrac{11\pi }{6}$
which means
$\{(4m+1)\dfrac{\pi }{2}:m\in I\}\underset{n\in I}{\mathop{\bigcup }}\,\left[ \left( 2n\pi +\dfrac{7\pi }{6},2n\pi +\dfrac{11\pi }{6} \right) \right]$

Thus, Option (C) is correct.

Note:The fog cannot exist unless the range of g is a subset of the domain of f(x). A mathematical function with real values is one that has real numbers as its values.