Inverse Trigonometric Functions Question 137
Question: The range of $ y=(co{t^{-1}}x)(co{t^{-1}}(-x)) $ is
Options:
A) $ (. 0,\frac{{{\pi }^{2}}}{4} ] $
B) $ (0,\pi ) $
C) $ (0,2\pi ] $
D) $ (0,1] $
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Answer:
Correct Answer: B
$ y=(co{t^{-1}}x)(co{t^{-1}}(-x)) $
$ ={{\cot }^{-1}}(x)(\pi -co{t^{-1}}(x)) $
Now $ {{\cot }^{-1}}(x) $ and $ (\pi -co{t^{-1}}(x))>0 $
Using A.M $ \ge G.M. $ we get $ \frac{{{\cot }^{-1}}x+(\pi -co{t^{-1}}x)}{2}\ge \sqrt{(co{t^{-1}}x)(\pi -co{t^{-1}}x)} $
$ \Rightarrow 0<{{\cot }^{-1}}(x)(\pi -co{t^{-1}}(x)) $
$ \le ( \frac{{{\cot }^{-1}}x+(\pi -co{t^{-1}}x)}{2} )=\frac{{{\pi }^{2}}}{4} $
$ \Rightarrow 0<y\le \frac{{{\pi }^{2}}}{4} $
BETA
