Functions Question 735
Question: If $ f(x)=\sqrt{3| x |-x-2} $ and $ g(x)=sinx, $ then domain of definition of fog (x) is
Options:
A) $ { 2n\pi +\frac{\pi }{2} },n\in I $
B) $ \underset{n\in I}{\mathop{\bigcup }}, $ $ ( 2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6} ) $
C) $ ( 2n\pi +\frac{7\pi }{6} ),n\in I $
D) $ {(4m+1)\frac{\pi }{2}:m\in I}\underset{n\in I}{\mathop{\bigcup }},[ ( 2n\pi +\frac{7\pi }{6},2n\pi +\frac{11\pi }{6} ) ] $
Show Answer
Answer:
Correct Answer: D
Solution:
For $ (fog)(x) $ to exist, the range of g $ \subseteq $ domain of f.
$ \therefore $ Domain of $ f\Rightarrow 3|x| -x -2\ge 0 $
$ \Rightarrow 3| x |-x\ge 2 $ When $ x\ge 0\Rightarrow x\ge 1 $ when $ x<0\Rightarrow x<-\frac{1}{2} $
$ \therefore \sin x\ge 1andx<-\frac{1}{2} $ for $ f{g(x)} $ to exists. i.e. $ \sin x=1and-1,\le ,\sin ,x\le -\frac{1}{2} $
$ \therefore x=(4m+1)\frac{\pi }{2} $ and $ 2n\pi +\frac{7\pi }{6}\le x\le 2n\pi +\frac{11\pi }{6} $ $ ie, $ $ { (4m+1)\frac{\pi }{2}:m\in \mathbb{Z} }\cup_{n\in \mathbb{Z}},{ 2n\pi +\frac{7\pi }{6}\le x\le 2n\pi +\frac{11\pi }{6} } $
BETA
