1. Tardigrade
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  3. Mathematics
  4. Let f:(-1,1)→ B is defined as f(x)= tan -1(2x/1-x2). Function f is one-one and onto, then the interval B is

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Q. Let $ f:(-1,1)\to B $ is defined as $ f(x)={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}. $ Function $ f $ is one-one and onto, then the interval B is

Rajasthan PETRajasthan PET 2007

Solution:

$ x\in (-1,1)\Rightarrow {{\tan }^{-1}}x\in \left( -\frac{\pi }{4},\frac{\pi }{4} \right) $
$ \Rightarrow $ $ 2{{\tan }^{-1}}x\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right) $
and $ f(x)={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}=2{{\tan }^{-1}}x,({{x}^{2}}<1) $
$ \therefore $ $ f(x)\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right) $
Function is one-one onto. $ x\in (-1,1)\Rightarrow {{\tan }^{-1}}x\in \left( -\frac{\pi }{4},\frac{\pi }{4} \right) $
$ \Rightarrow $ $ 2{{\tan }^{-1}}x\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right) $
and $ f(x)={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}=2{{\tan }^{-1}}x,({{x}^{2}}<1) $
$ \therefore $ $ f(x)\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right) $
Function is one-one onto.