A) \[\left[ 0,1 \right]\]
B) \[\left[ 0,\frac{1}{2} \right]\]
C) \[\left[ \frac{1}{2},1 \right]\]
D) \[\left\{ -1,0,1 \right\}\]
Correct Answer: C
Solution :
| [c] Case 1: |
| If \[0\le x\le \frac{1}{2}\], then \[{{\cos }^{-1}}\left( \frac{x}{2}+\frac{1}{2}\sqrt{3-3{{x}^{2}}} \right)\] |
| \[{{\cos }^{-1}}\left( x\times \frac{1}{2}+\sqrt{1-{{x}^{2}}}\frac{\sqrt{3}}{2} \right)\] |
| \[={{\cos }^{-1}}x-{{\cos }^{-1}}\frac{1}{2}\] |
| Therefore, the equation is |
| \[{{\cos }^{-1}}x+{{\cos }^{-1}}x-{{\cos }^{-1}}\frac{1}{2}=\frac{\pi }{3}\Rightarrow x=\frac{1}{2}.\] |
| Case 2: if \[\frac{1}{2}\le x\le 1,\]then |
| \[{{\cos }^{-1}}\left( \frac{x}{2}+\frac{1}{2}\sqrt{3-3{{x}^{2}}} \right)={{\cos }^{-1}}\frac{1}{2}-{{\cos }^{-1}}x\] |
| Therefore, the equation is |
| \[{{\cos }^{-1}}x+{{\cos }^{-1}}\frac{1}{2}-{{\cos }^{-1}}x=\frac{\pi }{3},\] |
| which is an identity. |
| Hence, the identity holds good for \[x\in \left[ \frac{1}{2},1 \right]\]. |
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