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 :
| For \[f(x)\] to be defined, we must have \[x-\sqrt{1-{{x}^{2}}}\ge 0\] or \[x\ge \sqrt{1-{{x}^{2}}}>0\] |
| \[\therefore \,\,{{x}^{2}}\ge 1-{{x}^{2}}\] 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 \[\Rightarrow \,\,x\ge \frac{1}{\sqrt{2}}\] |
| 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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