Inverse Trigonometric Functions Question 75
Question: The set of values of x for which the identity $ {{\cos }^{-1}}x+{{\cos }^{-1}}( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} )=\frac{\pi }{3} $ holds good is
Options:
A) $ [ 0,1 ] $
B) $ [ 0,\frac{1}{2} ] $
C) $ [ \frac{1}{2},1 ] $
D) $ { -1,0,1 } $
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Answer:
Correct Answer: C
Case 1: If $ 0\le x\le \frac{1}{2} $ , then $ {{\cos }^{-1}}( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} ) $
$ {{\cos }^{-1}}( x\times \frac{1}{2}+\sqrt{1-x^{2}}\frac{\sqrt{3}}{2} ) $
$ ={{\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}}( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} )={{\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 [ \frac{1}{2},1 ] $ .
BETA
