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General
Easy

Question

If open vertical bar 2 t a n to the power of negative 1 end exponent invisible function application x plus s i n to the power of negative 1 end exponent invisible function application fraction numerator 2 x over denominator 1 plus x to the power of 2 end exponent end fraction close vertical baris independent of x then

  1. x [1, )    
  2. x [-1, 1]    
  3. x(-, -1] [1, )    
  4. none of these    

The correct answer is: x(-, -1] [1, )


    |2tan-1 x + sin-1 fraction numerator 2 x over denominator 1 plus x to the power of 2 end exponent end fraction|
    tan-1x = theta element of open parentheses negative fraction numerator pi over denominator 2 end fraction comma fraction numerator pi over denominator 2 end fraction close parentheses
    sin-1 fraction numerator 2 x over denominator 1 plus x to the power of 2 end exponent end fraction= 2tan-1x x
    or  - 2tan-1x or - - 2 tan-1 x
    according as x [1, ) or [-1, 1]
    or (-, -1].

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