Functions Question 381
Question: The domain of the function $ f(x)=\sqrt{x-\sqrt{1-x^{2}}} $ is
Options:
A) $ [ -1,-\frac{1}{\sqrt{2}} ]\cup [ \frac{1}{\sqrt{2}},1 ] $
B) $ [-1,1] $
C) $ ( -\infty ,-\frac{1}{2} ]\cup [ \frac{1}{\sqrt{2}},+\infty ) $
D) $ [ \frac{1}{\sqrt{2}},1 ] $
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Answer:
Correct Answer: D
Solution:
[d] For f(x) to be defined, we must have $ x-\sqrt{1-x^{2}}\ge 0 $ or $ x\ge \sqrt{1-x^{2}}>0 $ or $ x^{2}\ge 1-x^{2} $ or $ x^{2}\ge \frac{1}{2}. $ Also, $ 1-x^{2}\ge 0orx^{2}\le 1. $ Now. $ x^{2}\ge \frac{1}{2}\Rightarrow ( x-\frac{1}{\sqrt{2}} )( x+\frac{1}{\sqrt{2}} )\ge 0 $
$ \Rightarrow x\le -\frac{1}{\sqrt{2}} $ or $ x\ge \frac{1}{\sqrt{2}} $ Also, $ x^{2}\le 1\Rightarrow (x-1)(x+1)\le 0\Rightarrow -1\le x\le 1 $ Thus, $ x>0,x^{2}\ge \frac{1}{2} $ and $ x^{2}\le 1\Rightarrow x\in [ \frac{1}{\sqrt{2}},1 ] $
BETA
