A) \[\left[ -1,-\frac{1}{\sqrt{2}} \right]\cup \left[ \frac{1}{\sqrt{2}},1 \right]\]
B) \[[-1,1]\]
C) \[\left( -\infty ,-\frac{1}{2} \right]\cup \left[ \frac{1}{\sqrt{2}},+\infty \right)\]
D) \[\left[ \frac{1}{\sqrt{2}},1 \right]\]
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, \[1-{{x}^{2}}\ge 0\,\,or\,\,{{x}^{2}}\le 1.\] |
Now. \[{{x}^{2}}\ge \frac{1}{2}\Rightarrow \left( x-\frac{1}{\sqrt{2}} \right)\left( x+\frac{1}{\sqrt{2}} \right)\ge 0\] |
\[\Rightarrow x\le -\frac{1}{\sqrt{2}}\] 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 \[{{x}^{2}}\le 1\Rightarrow x\in \left[ \frac{1}{\sqrt{2}},1 \right]\] |
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