Inverse Trigonometric Functions Question 49
Question: The domain of the function $ f(x)=si{n^{-1}}{ {\log_2}( \frac{1}{2}x^{2} ) } $ is
Options:
A) $ [-2,-1)\cup [1,2] $
B) $ (-2,-1]\cup [1,2] $
C) $ [-2,-1]\cup [1,2] $
D) $ (-2,-1)\cup (1,2) $
Show Answer
Answer:
Correct Answer: C
For $ f(x) $ to be defined we must have $ -1\le {\log_2}( \frac{1}{2}x^{2} )\le 1\Rightarrow {2^{-1}}\le \frac{1}{2}x^{2}\le 2^{1} $
$ \Rightarrow 1\le x^{2}\le 4 $ -.(1)
Now, $ 1\le x^{2}\Rightarrow x^{2}-1\ge 0 $ i.e. $ (x-1)(x+1)\ge 0 $
$ \Rightarrow x\le -1orx\ge 1 $ -.(2)
Also $ x^{2}\le 4\Rightarrow x^{2}-4\le 0 $ i.e. $ (x-2)(x+2)\le 0 $
$ \Rightarrow -2\le x\le 2 $ -.(3)
Form (2) and (3), we get the domain of $ f={(-\infty ,-1]\cup [1,\infty )}\cap [-2,2] $
$ =[-2,-1]\cup [1,2] $
BETA
